| $x$-bin | ran0 | ran1 | ran2 | ran3 |
|---|---|---|---|---|
| 0.0-0.1 | 1013 | 991 | 938 | 1047 |
| 0.1-0.2 | 1002 | 1009 | 1040 | 1030 |
| 0.2-0.3 | 989 | 999 | 1030 | 993 |
| 0.3-0.4 | 939 | 960 | 1023 | 937 |
| 0.4-0.5 | 1038 | 1001 | 1002 | 992 |
| 0.5-0.6 | 1037 | 1047 | 1009 | 1009 |
| 0.6-0.7 | 1005 | 989 | 1003 | 989 |
| 0.7-0.8 | 986 | 962 | 985 | 954 |
| 0.8-0.9 | 1000 | 1027 | 1009 | 1023 |
| 0.9-1.0 | 991 | 1015 | 961 | 1026 |
| $\\mu$ | 0.4997 | 0.5018 | 0.4992 | 0.4990 |
| $\\sigma$ | 0.2882 | 0.2892 | 0.2861 | 0.2915 |
| Relations | Name | matrix elements |
|---|---|---|
| $A = A^{T}$ | symmetric | $a_{ij} = a_{ji}$ |
| $A = \\left (A^{T} \\right )^{-1}$ | real orthogonal | $\\sum_k a_{ik} a_{jk} = \\sum_k a_{ki} a_{kj} = \\delta_{ij}$ |
| $A = A^{ * }$ | real matrix | $a_{ij} = a_{ij}^{ * }$ |
| $A = A^{\\dagger}$ | hermitian | $a_{ij} = a_{ji}^{ * }$ |
| $A = \\left (A^{\\dagger} \\right )^{-1}$ | unitary | $\\sum_k a_{ik} a_{jk}^{ * } = \\sum_k a_{ki}^{ * } a_{kj} = \\delta_{ij}$ |
![](\"fig/pandas.jpg\")
\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "Another useful Python package is\n",
+ "[pandas](https://pandas.pydata.org/), which is an open source library\n",
+ "providing high-performance, easy-to-use data structures and data\n",
+ "analysis tools for Python. **pandas** stands for panel data, a term borrowed from econometrics and is an efficient library for data analysis with an emphasis on tabular data.\n",
+ "**pandas** has two major classes, the **DataFrame** class with two-dimensional data objects and tabular data organized in columns and the class **Series** with a focus on one-dimensional data objects. Both classes allow you to index data easily as we will see in the examples below. \n",
+ "**pandas** allows you also to perform mathematical operations on the data, spanning from simple reshapings of vectors and matrices to statistical operations. \n",
+ "\n",
+ "The following simple example shows how we can, in an easy way make tables of our data. Here we define a data set which includes names, place of birth and date of birth, and displays the data in an easy to read way. We will see repeated use of **pandas**, in particular in connection with classification of data."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "import pandas as pd\n",
+ "from IPython.display import display\n",
+ "data = {'First Name': [\"Frodo\", \"Bilbo\", \"Aragorn II\", \"Samwise\"],\n",
+ " 'Last Name': [\"Baggins\", \"Baggins\",\"Elessar\",\"Gamgee\"],\n",
+ " 'Place of birth': [\"Shire\", \"Shire\", \"Eriador\", \"Shire\"],\n",
+ " 'Date of Birth T.A.': [2968, 2890, 2931, 2980]\n",
+ " }\n",
+ "data_pandas = pd.DataFrame(data)\n",
+ "display(data_pandas)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "In the above we have imported **pandas** with the shorthand **pd**, the latter has become the standard way we import **pandas**. We make then a list of various variables\n",
+ "and reorganize the aboves lists into a **DataFrame** and then print out a neat table with specific column labels as *Name*, *place of birth* and *date of birth*.\n",
+ "Displaying these results, we see that the indices are given by the default numbers from zero to three.\n",
+ "**pandas** is extremely flexible and we can easily change the above indices by defining a new type of indexing as"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 18,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "data_pandas = pd.DataFrame(data,index=['Frodo','Bilbo','Aragorn','Sam'])\n",
+ "display(data_pandas)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Thereafter we display the content of the row which begins with the index **Aragorn**"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 19,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "display(data_pandas.loc['Aragorn'])"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We can easily append data to this, for example"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 20,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "new_hobbit = {'First Name': [\"Peregrin\"],\n",
+ " 'Last Name': [\"Took\"],\n",
+ " 'Place of birth': [\"Shire\"],\n",
+ " 'Date of Birth T.A.': [2990]\n",
+ " }\n",
+ "data_pandas=data_pandas.append(pd.DataFrame(new_hobbit, index=['Pippin']))\n",
+ "display(data_pandas)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Here are other examples where we use the **DataFrame** functionality to handle arrays, now with more interesting features for us, namely numbers. We set up a matrix \n",
+ "of dimensionality $10\\times 5$ and compute the mean value and standard deviation of each column. Similarly, we can perform mathematial operations like squaring the matrix elements and many other operations."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 21,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "import numpy as np\n",
+ "import pandas as pd\n",
+ "from IPython.display import display\n",
+ "np.random.seed(100)\n",
+ "# setting up a 10 x 5 matrix\n",
+ "rows = 10\n",
+ "cols = 5\n",
+ "a = np.random.randn(rows,cols)\n",
+ "df = pd.DataFrame(a)\n",
+ "display(df)\n",
+ "print(df.mean())\n",
+ "print(df.std())\n",
+ "display(df**2)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Thereafter we can select specific columns only and plot final results"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 22,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "df.columns = ['First', 'Second', 'Third', 'Fourth', 'Fifth']\n",
+ "df.index = np.arange(10)\n",
+ "\n",
+ "display(df)\n",
+ "print(df['Second'].mean() )\n",
+ "print(df.info())\n",
+ "print(df.describe())\n",
+ "\n",
+ "from pylab import plt, mpl\n",
+ "plt.style.use('seaborn')\n",
+ "mpl.rcParams['font.family'] = 'serif'\n",
+ "\n",
+ "df.cumsum().plot(lw=2.0, figsize=(10,6))\n",
+ "plt.show()\n",
+ "\n",
+ "\n",
+ "df.plot.bar(figsize=(10,6), rot=15)\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We can produce a $4\\times 4$ matrix"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 23,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "b = np.arange(16).reshape((4,4))\n",
+ "print(b)\n",
+ "df1 = pd.DataFrame(b)\n",
+ "print(df1)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and many other operations. \n",
+ "\n",
+ "The **Series** class is another important class included in\n",
+ "**pandas**. You can view it as a specialization of **DataFrame** but where\n",
+ "we have just a single column of data. It shares many of the same features as _DataFrame. As with **DataFrame**,\n",
+ "most operations are vectorized, achieving thereby a high performance when dealing with computations of arrays, in particular labeled arrays.\n",
+ "As we will see below it leads also to a very concice code close to the mathematical operations we may be interested in.\n",
+ "For multidimensional arrays, we recommend strongly [xarray](http://xarray.pydata.org/en/stable/). **xarray** has much of the same flexibility as **pandas**, but allows for the extension to higher dimensions than two. We will see examples later of the usage of both **pandas** and **xarray**. \n",
+ "\n",
+ "\n",
+ "\n",
+ "## Reading Data and fitting\n",
+ "\n",
+ "In order to study various Machine Learning algorithms, we need to\n",
+ "access data. Acccessing data is an essential step in all machine\n",
+ "learning algorithms. In particular, setting up the so-called **design\n",
+ "matrix** (to be defined below) is often the first element we need in\n",
+ "order to perform our calculations. To set up the design matrix means\n",
+ "reading (and later, when the calculations are done, writing) data\n",
+ "in various formats, The formats span from reading files from disk,\n",
+ "loading data from databases and interacting with online sources\n",
+ "like web application programming interfaces (APIs).\n",
+ "\n",
+ "In handling various input formats, as discussed above, we will mainly stay with **pandas**,\n",
+ "a Python package which allows us, in a seamless and painless way, to\n",
+ "deal with a multitude of formats, from standard **csv** (comma separated\n",
+ "values) files, via **excel**, **html** to **hdf5** formats. With **pandas**\n",
+ "and the **DataFrame** and **Series** functionalities we are able to convert text data\n",
+ "into the calculational formats we need for a specific algorithm. And our code is going to be \n",
+ "pretty close the basic mathematical expressions.\n",
+ "\n",
+ "Our first data set is going to be a classic from nuclear physics, namely all\n",
+ "available data on binding energies. Don't be intimidated if you are not familiar with nuclear physics. It serves simply as an example here of a data set. \n",
+ "\n",
+ "We will show some of the\n",
+ "strengths of packages like **Scikit-Learn** in fitting nuclear binding energies to\n",
+ "specific functions using linear regression first. Then, as a teaser, we will show you how \n",
+ "you can easily implement other algorithms like decision trees and random forests and neural networks.\n",
+ "\n",
+ "But before we really start with nuclear physics data, let's just look at some simpler polynomial fitting cases, such as,\n",
+ "(don't be offended) fitting straight lines!\n",
+ "\n",
+ "\n",
+ "### Simple linear regression model using **scikit-learn**\n",
+ "\n",
+ "We start with perhaps our simplest possible example, using **Scikit-Learn** to perform linear regression analysis on a data set produced by us. \n",
+ "\n",
+ "What follows is a simple Python code where we have defined a function\n",
+ "$y$ in terms of the variable $x$. Both are defined as vectors with $100$ entries. \n",
+ "The numbers in the vector $\\hat{x}$ are given\n",
+ "by random numbers generated with a uniform distribution with entries\n",
+ "$x_i \\in [0,1]$ (more about probability distribution functions\n",
+ "later). These values are then used to define a function $y(x)$\n",
+ "(tabulated again as a vector) with a linear dependence on $x$ plus a\n",
+ "random noise added via the normal distribution.\n",
+ "\n",
+ "\n",
+ "The Numpy functions are imported used the **import numpy as np**\n",
+ "statement and the random number generator for the uniform distribution\n",
+ "is called using the function **np.random.rand()**, where we specificy\n",
+ "that we want $100$ random variables. Using Numpy we define\n",
+ "automatically an array with the specified number of elements, $100$ in\n",
+ "our case. With the Numpy function **randn()** we can compute random\n",
+ "numbers with the normal distribution (mean value $\\mu$ equal to zero and\n",
+ "variance $\\sigma^2$ set to one) and produce the values of $y$ assuming a linear\n",
+ "dependence as function of $x$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "y = 2x+N(0,1),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $N(0,1)$ represents random numbers generated by the normal\n",
+ "distribution. From **Scikit-Learn** we import then the\n",
+ "**LinearRegression** functionality and make a prediction $\\tilde{y} =\n",
+ "\\alpha + \\beta x$ using the function **fit(x,y)**. We call the set of\n",
+ "data $(\\hat{x},\\hat{y})$ for our training data. The Python package\n",
+ "**scikit-learn** has also a functionality which extracts the above\n",
+ "fitting parameters $\\alpha$ and $\\beta$ (see below). Later we will\n",
+ "distinguish between training data and test data.\n",
+ "\n",
+ "For plotting we use the Python package\n",
+ "[matplotlib](https://matplotlib.org/) which produces publication\n",
+ "quality figures. Feel free to explore the extensive\n",
+ "[gallery](https://matplotlib.org/gallery/index.html) of examples. In\n",
+ "this example we plot our original values of $x$ and $y$ as well as the\n",
+ "prediction **ypredict** ($\\tilde{y}$), which attempts at fitting our\n",
+ "data with a straight line.\n",
+ "\n",
+ "The Python code follows here."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 24,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# Importing various packages\n",
+ "import numpy as np\n",
+ "import matplotlib.pyplot as plt\n",
+ "from sklearn.linear_model import LinearRegression\n",
+ "\n",
+ "x = np.random.rand(100,1)\n",
+ "y = 2*x+np.random.randn(100,1)\n",
+ "linreg = LinearRegression()\n",
+ "linreg.fit(x,y)\n",
+ "xnew = np.array([[0],[1]])\n",
+ "ypredict = linreg.predict(xnew)\n",
+ "\n",
+ "plt.plot(xnew, ypredict, \"r-\")\n",
+ "plt.plot(x, y ,'ro')\n",
+ "plt.axis([0,1.0,0, 5.0])\n",
+ "plt.xlabel(r'$x$')\n",
+ "plt.ylabel(r'$y$')\n",
+ "plt.title(r'Simple Linear Regression')\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "This example serves several aims. It allows us to demonstrate several\n",
+ "aspects of data analysis and later machine learning algorithms. The\n",
+ "immediate visualization shows that our linear fit is not\n",
+ "impressive. It goes through the data points, but there are many\n",
+ "outliers which are not reproduced by our linear regression. We could\n",
+ "now play around with this small program and change for example the\n",
+ "factor in front of $x$ and the normal distribution. Try to change the\n",
+ "function $y$ to"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "y = 10x+0.01 \\times N(0,1),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $x$ is defined as before. Does the fit look better? Indeed, by\n",
+ "reducing the role of the noise given by the normal distribution we see immediately that\n",
+ "our linear prediction seemingly reproduces better the training\n",
+ "set. However, this testing 'by the eye' is obviouly not satisfactory in the\n",
+ "long run. Here we have only defined the training data and our model, and \n",
+ "have not discussed a more rigorous approach to the **cost** function.\n",
+ "\n",
+ "We need more rigorous criteria in defining whether we have succeeded or\n",
+ "not in modeling our training data. You will be surprised to see that\n",
+ "many scientists seldomly venture beyond this 'by the eye' approach. A\n",
+ "standard approach for the *cost* function is the so-called $\\chi^2$\n",
+ "function (a variant of the mean-squared error (MSE))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\chi^2 = \\frac{1}{n}\n",
+ "\\sum_{i=0}^{n-1}\\frac{(y_i-\\tilde{y}_i)^2}{\\sigma_i^2},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $\\sigma_i^2$ is the variance (to be defined later) of the entry\n",
+ "$y_i$. We may not know the explicit value of $\\sigma_i^2$, it serves\n",
+ "however the aim of scaling the equations and make the cost function\n",
+ "dimensionless. \n",
+ "\n",
+ "Minimizing the cost function is a central aspect of\n",
+ "our discussions to come. Finding its minima as function of the model\n",
+ "parameters ($\\alpha$ and $\\beta$ in our case) will be a recurring\n",
+ "theme in these series of lectures. Essentially all machine learning\n",
+ "algorithms we will discuss center around the minimization of the\n",
+ "chosen cost function. This depends in turn on our specific\n",
+ "model for describing the data, a typical situation in supervised\n",
+ "learning. Automatizing the search for the minima of the cost function is a\n",
+ "central ingredient in all algorithms. Typical methods which are\n",
+ "employed are various variants of **gradient** methods. These will be\n",
+ "discussed in more detail later. Again, you'll be surprised to hear that\n",
+ "many practitioners minimize the above function ''by the eye', popularly dubbed as \n",
+ "'chi by the eye'. That is, change a parameter and see (visually and numerically) that \n",
+ "the $\\chi^2$ function becomes smaller. \n",
+ "\n",
+ "There are many ways to define the cost function. A simpler approach is to look at the relative difference between the training data and the predicted data, that is we define \n",
+ "the relative error (why would we prefer the MSE instead of the relative error?) as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\epsilon_{\\mathrm{relative}}= \\frac{\\vert \\hat{y} -\\hat{\\tilde{y}}\\vert}{\\vert \\hat{y}\\vert}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The squared cost function results in an arithmetic mean-unbiased\n",
+ "estimator, and the absolute-value cost function results in a\n",
+ "median-unbiased estimator (in the one-dimensional case, and a\n",
+ "geometric median-unbiased estimator for the multi-dimensional\n",
+ "case). The squared cost function has the disadvantage that it has the tendency\n",
+ "to be dominated by outliers.\n",
+ "\n",
+ "We can modify easily the above Python code and plot the relative error instead"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 25,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "import numpy as np\n",
+ "import matplotlib.pyplot as plt\n",
+ "from sklearn.linear_model import LinearRegression\n",
+ "\n",
+ "x = np.random.rand(100,1)\n",
+ "y = 5*x+0.01*np.random.randn(100,1)\n",
+ "linreg = LinearRegression()\n",
+ "linreg.fit(x,y)\n",
+ "ypredict = linreg.predict(x)\n",
+ "\n",
+ "plt.plot(x, np.abs(ypredict-y)/abs(y), \"ro\")\n",
+ "plt.axis([0,1.0,0.0, 0.5])\n",
+ "plt.xlabel(r'$x$')\n",
+ "plt.ylabel(r'$\\epsilon_{\\mathrm{relative}}$')\n",
+ "plt.title(r'Relative error')\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Depending on the parameter in front of the normal distribution, we may\n",
+ "have a small or larger relative error. Try to play around with\n",
+ "different training data sets and study (graphically) the value of the\n",
+ "relative error.\n",
+ "\n",
+ "As mentioned above, **Scikit-Learn** has an impressive functionality.\n",
+ "We can for example extract the values of $\\alpha$ and $\\beta$ and\n",
+ "their error estimates, or the variance and standard deviation and many\n",
+ "other properties from the statistical data analysis. \n",
+ "\n",
+ "Here we show an\n",
+ "example of the functionality of **Scikit-Learn**."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 26,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "import numpy as np \n",
+ "import matplotlib.pyplot as plt \n",
+ "from sklearn.linear_model import LinearRegression \n",
+ "from sklearn.metrics import mean_squared_error, r2_score, mean_squared_log_error, mean_absolute_error\n",
+ "\n",
+ "x = np.random.rand(100,1)\n",
+ "y = 2.0+ 5*x+0.5*np.random.randn(100,1)\n",
+ "linreg = LinearRegression()\n",
+ "linreg.fit(x,y)\n",
+ "ypredict = linreg.predict(x)\n",
+ "print('The intercept alpha: \\n', linreg.intercept_)\n",
+ "print('Coefficient beta : \\n', linreg.coef_)\n",
+ "# The mean squared error \n",
+ "print(\"Mean squared error: %.2f\" % mean_squared_error(y, ypredict))\n",
+ "# Explained variance score: 1 is perfect prediction \n",
+ "print('Variance score: %.2f' % r2_score(y, ypredict))\n",
+ "# Mean squared log error \n",
+ "print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) )\n",
+ "# Mean absolute error \n",
+ "print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict))\n",
+ "plt.plot(x, ypredict, \"r-\")\n",
+ "plt.plot(x, y ,'ro')\n",
+ "plt.axis([0.0,1.0,1.5, 7.0])\n",
+ "plt.xlabel(r'$x$')\n",
+ "plt.ylabel(r'$y$')\n",
+ "plt.title(r'Linear Regression fit ')\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The function **coef** gives us the parameter $\\beta$ of our fit while **intercept** yields \n",
+ "$\\alpha$. Depending on the constant in front of the normal distribution, we get values near or far from $alpha =2$ and $\\beta =5$. Try to play around with different parameters in front of the normal distribution. The function **meansquarederror** gives us the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error or loss defined as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "MSE(\\hat{y},\\hat{\\tilde{y}}) = \\frac{1}{n}\n",
+ "\\sum_{i=0}^{n-1}(y_i-\\tilde{y}_i)^2,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The smaller the value, the better the fit. Ideally we would like to\n",
+ "have an MSE equal zero. The attentive reader has probably recognized\n",
+ "this function as being similar to the $\\chi^2$ function defined above.\n",
+ "\n",
+ "The **r2score** function computes $R^2$, the coefficient of\n",
+ "determination. It provides a measure of how well future samples are\n",
+ "likely to be predicted by the model. Best possible score is 1.0 and it\n",
+ "can be negative (because the model can be arbitrarily worse). A\n",
+ "constant model that always predicts the expected value of $\\hat{y}$,\n",
+ "disregarding the input features, would get a $R^2$ score of $0.0$.\n",
+ "\n",
+ "If $\\tilde{\\hat{y}}_i$ is the predicted value of the $i-th$ sample and $y_i$ is the corresponding true value, then the score $R^2$ is defined as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "R^2(\\hat{y}, \\tilde{\\hat{y}}) = 1 - \\frac{\\sum_{i=0}^{n - 1} (y_i - \\tilde{y}_i)^2}{\\sum_{i=0}^{n - 1} (y_i - \\bar{y})^2},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where we have defined the mean value of $\\hat{y}$ as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\bar{y} = \\frac{1}{n} \\sum_{i=0}^{n - 1} y_i.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Another quantity taht we will meet again in our discussions of regression analysis is \n",
+ " the mean absolute error (MAE), a risk metric corresponding to the expected value of the absolute error loss or what we call the $l1$-norm loss. In our discussion above we presented the relative error.\n",
+ "The MAE is defined as follows"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\text{MAE}(\\hat{y}, \\hat{\\tilde{y}}) = \\frac{1}{n} \\sum_{i=0}^{n-1} \\left| y_i - \\tilde{y}_i \\right|.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We present the \n",
+ "squared logarithmic (quadratic) error"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\text{MSLE}(\\hat{y}, \\hat{\\tilde{y}}) = \\frac{1}{n} \\sum_{i=0}^{n - 1} (\\log_e (1 + y_i) - \\log_e (1 + \\tilde{y}_i) )^2,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $\\log_e (x)$ stands for the natural logarithm of $x$. This error\n",
+ "estimate is best to use when targets having exponential growth, such\n",
+ "as population counts, average sales of a commodity over a span of\n",
+ "years etc. \n",
+ "\n",
+ "\n",
+ "Finally, another cost function is the Huber cost function used in robust regression.\n",
+ "\n",
+ "The rationale behind this possible cost function is its reduced\n",
+ "sensitivity to outliers in the data set. In our discussions on\n",
+ "dimensionality reduction and normalization of data we will meet other\n",
+ "ways of dealing with outliers.\n",
+ "\n",
+ "The Huber cost function is defined as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "H_{\\delta}(a)={\\begin{cases}{\\frac {1}{2}}{a^{2}}&{\\text{for }}|a|\\leq \\delta ,\\\\\\delta (|a|-{\\frac {1}{2}}\\delta ),&{\\text{otherwise.}}\\end{cases}}}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Here $a=\\boldsymbol{y} - \\boldsymbol{\\tilde{y}}$.\n",
+ "We will discuss in more\n",
+ "detail these and other functions in the various lectures. We conclude this part with another example. Instead of \n",
+ "a linear $x$-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 27,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "import matplotlib.pyplot as plt\n",
+ "import numpy as np\n",
+ "import random\n",
+ "from sklearn.linear_model import Ridge\n",
+ "from sklearn.preprocessing import PolynomialFeatures\n",
+ "from sklearn.pipeline import make_pipeline\n",
+ "from sklearn.linear_model import LinearRegression\n",
+ "\n",
+ "x=np.linspace(0.02,0.98,200)\n",
+ "noise = np.asarray(random.sample((range(200)),200))\n",
+ "y=x**3*noise\n",
+ "yn=x**3*100\n",
+ "poly3 = PolynomialFeatures(degree=3)\n",
+ "X = poly3.fit_transform(x[:,np.newaxis])\n",
+ "clf3 = LinearRegression()\n",
+ "clf3.fit(X,y)\n",
+ "\n",
+ "Xplot=poly3.fit_transform(x[:,np.newaxis])\n",
+ "poly3_plot=plt.plot(x, clf3.predict(Xplot), label='Cubic Fit')\n",
+ "plt.plot(x,yn, color='red', label=\"True Cubic\")\n",
+ "plt.scatter(x, y, label='Data', color='orange', s=15)\n",
+ "plt.legend()\n",
+ "plt.show()\n",
+ "\n",
+ "def error(a):\n",
+ " for i in y:\n",
+ " err=(y-yn)/yn\n",
+ " return abs(np.sum(err))/len(err)\n",
+ "\n",
+ "print (error(y))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "### To our real data: nuclear binding energies. Brief reminder on masses and binding energies\n",
+ "\n",
+ "Let us now dive into nuclear physics and remind ourselves briefly about some basic features about binding\n",
+ "energies. A basic quantity which can be measured for the ground\n",
+ "states of nuclei is the atomic mass $M(N, Z)$ of the neutral atom with\n",
+ "atomic mass number $A$ and charge $Z$. The number of neutrons is $N$. There are indeed several sophisticated experiments worldwide which allow us to measure this quantity to high precision (parts per million even). \n",
+ "\n",
+ "Atomic masses are usually tabulated in terms of the mass excess defined by"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\Delta M(N, Z) = M(N, Z) - uA,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $u$ is the Atomic Mass Unit"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "u = M(^{12}\\mathrm{C})/12 = 931.4940954(57) \\hspace{0.1cm} \\mathrm{MeV}/c^2.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The nucleon masses are"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "m_p = 1.00727646693(9)u,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "m_n = 939.56536(8)\\hspace{0.1cm} \\mathrm{MeV}/c^2 = 1.0086649156(6)u.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "In the [2016 mass evaluation of by W.J.Huang, G.Audi, M.Wang, F.G.Kondev, S.Naimi and X.Xu](http://nuclearmasses.org/resources_folder/Wang_2017_Chinese_Phys_C_41_030003.pdf)\n",
+ "there are data on masses and decays of 3437 nuclei.\n",
+ "\n",
+ "The nuclear binding energy is defined as the energy required to break\n",
+ "up a given nucleus into its constituent parts of $N$ neutrons and $Z$\n",
+ "protons. In terms of the atomic masses $M(N, Z)$ the binding energy is\n",
+ "defined by"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "BE(N, Z) = ZM_H c^2 + Nm_n c^2 - M(N, Z)c^2 ,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $M_H$ is the mass of the hydrogen atom and $m_n$ is the mass of the neutron.\n",
+ "In terms of the mass excess the binding energy is given by"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "BE(N, Z) = Z\\Delta_H c^2 + N\\Delta_n c^2 -\\Delta(N, Z)c^2 ,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $\\Delta_H c^2 = 7.2890$ MeV and $\\Delta_n c^2 = 8.0713$ MeV.\n",
+ "\n",
+ "\n",
+ "A popular and physically intuitive model which can be used to parametrize \n",
+ "the experimental binding energies as function of $A$, is the so-called \n",
+ "**liquid drop model**. The ansatz is based on the following expression"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "BE(N,Z) = a_1A-a_2A^{2/3}-a_3\\frac{Z^2}{A^{1/3}}-a_4\\frac{(N-Z)^2}{A},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $A$ stands for the number of nucleons and the $a_i$s are parameters which are determined by a fit \n",
+ "to the experimental data. \n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "To arrive at the above expression we have assumed that we can make the following assumptions:\n",
+ "\n",
+ " * There is a volume term $a_1A$ proportional with the number of nucleons (the energy is also an extensive quantity). When an assembly of nucleons of the same size is packed together into the smallest volume, each interior nucleon has a certain number of other nucleons in contact with it. This contribution is proportional to the volume.\n",
+ "\n",
+ " * There is a surface energy term $a_2A^{2/3}$. The assumption here is that a nucleon at the surface of a nucleus interacts with fewer other nucleons than one in the interior of the nucleus and hence its binding energy is less. This surface energy term takes that into account and is therefore negative and is proportional to the surface area.\n",
+ "\n",
+ " * There is a Coulomb energy term $a_3\\frac{Z^2}{A^{1/3}}$. The electric repulsion between each pair of protons in a nucleus yields less binding. \n",
+ "\n",
+ " * There is an asymmetry term $a_4\\frac{(N-Z)^2}{A}$. This term is associated with the Pauli exclusion principle and reflects the fact that the proton-neutron interaction is more attractive on the average than the neutron-neutron and proton-proton interactions.\n",
+ "\n",
+ "We could also add a so-called pairing term, which is a correction term that\n",
+ "arises from the tendency of proton pairs and neutron pairs to\n",
+ "occur. An even number of particles is more stable than an odd number. \n",
+ "\n",
+ "\n",
+ "### Organizing our data\n",
+ "\n",
+ "Let us start with reading and organizing our data. \n",
+ "We start with the compilation of masses and binding energies from 2016.\n",
+ "After having downloaded this file to our own computer, we are now ready to read the file and start structuring our data.\n",
+ "\n",
+ "\n",
+ "We start with preparing folders for storing our calculations and the data file over masses and binding energies. We import also various modules that we will find useful in order to present various Machine Learning methods. Here we focus mainly on the functionality of **scikit-learn**."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 28,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# Common imports\n",
+ "import numpy as np\n",
+ "import pandas as pd\n",
+ "import matplotlib.pyplot as plt\n",
+ "import sklearn.linear_model as skl\n",
+ "from sklearn.model_selection import train_test_split\n",
+ "from sklearn.metrics import mean_squared_error, r2_score, mean_absolute_error\n",
+ "import os\n",
+ "\n",
+ "# Where to save the figures and data files\n",
+ "PROJECT_ROOT_DIR = \"Results\"\n",
+ "FIGURE_ID = \"Results/FigureFiles\"\n",
+ "DATA_ID = \"DataFiles/\"\n",
+ "\n",
+ "if not os.path.exists(PROJECT_ROOT_DIR):\n",
+ " os.mkdir(PROJECT_ROOT_DIR)\n",
+ "\n",
+ "if not os.path.exists(FIGURE_ID):\n",
+ " os.makedirs(FIGURE_ID)\n",
+ "\n",
+ "if not os.path.exists(DATA_ID):\n",
+ " os.makedirs(DATA_ID)\n",
+ "\n",
+ "def image_path(fig_id):\n",
+ " return os.path.join(FIGURE_ID, fig_id)\n",
+ "\n",
+ "def data_path(dat_id):\n",
+ " return os.path.join(DATA_ID, dat_id)\n",
+ "\n",
+ "def save_fig(fig_id):\n",
+ " plt.savefig(image_path(fig_id) + \".png\", format='png')\n",
+ "\n",
+ "infile = open(data_path(\"MassEval2016.dat\"),'r')"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Before we proceed, we define also a function for making our plots. You can obviously avoid this and simply set up various **matplotlib** commands every time you need them. You may however find it convenient to collect all such commands in one function and simply call this function."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 29,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "from pylab import plt, mpl\n",
+ "plt.style.use('seaborn')\n",
+ "mpl.rcParams['font.family'] = 'serif'\n",
+ "\n",
+ "def MakePlot(x,y, styles, labels, axlabels):\n",
+ " plt.figure(figsize=(10,6))\n",
+ " for i in range(len(x)):\n",
+ " plt.plot(x[i], y[i], styles[i], label = labels[i])\n",
+ " plt.xlabel(axlabels[0])\n",
+ " plt.ylabel(axlabels[1])\n",
+ " plt.legend(loc=0)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Our next step is to read the data on experimental binding energies and\n",
+ "reorganize them as functions of the mass number $A$, the number of\n",
+ "protons $Z$ and neutrons $N$ using **pandas**. Before we do this it is\n",
+ "always useful (unless you have a binary file or other types of compressed\n",
+ "data) to actually open the file and simply take a look at it!\n",
+ "\n",
+ "\n",
+ "In particular, the program that outputs the final nuclear masses is written in Fortran with a specific format. It means that we need to figure out the format and which columns contain the data we are interested in. Pandas comes with a function that reads formatted output. After having admired the file, we are now ready to start massaging it with **pandas**. The file begins with some basic format information."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 30,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "\"\"\" \n",
+ "This is taken from the data file of the mass 2016 evaluation. \n",
+ "All files are 3436 lines long with 124 character per line. \n",
+ " Headers are 39 lines long. \n",
+ " col 1 : Fortran character control: 1 = page feed 0 = line feed \n",
+ " format : a1,i3,i5,i5,i5,1x,a3,a4,1x,f13.5,f11.5,f11.3,f9.3,1x,a2,f11.3,f9.3,1x,i3,1x,f12.5,f11.5 \n",
+ " These formats are reflected in the pandas widths variable below, see the statement \n",
+ " widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1), \n",
+ " Pandas has also a variable header, with length 39 in this case. \n",
+ "\"\"\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The data we are interested in are in columns 2, 3, 4 and 11, giving us\n",
+ "the number of neutrons, protons, mass numbers and binding energies,\n",
+ "respectively. We add also for the sake of completeness the element name. The data are in fixed-width formatted lines and we will\n",
+ "covert them into the **pandas** DataFrame structure."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 31,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# Read the experimental data with Pandas\n",
+ "Masses = pd.read_fwf(infile, usecols=(2,3,4,6,11),\n",
+ " names=('N', 'Z', 'A', 'Element', 'Ebinding'),\n",
+ " widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),\n",
+ " header=39,\n",
+ " index_col=False)\n",
+ "\n",
+ "# Extrapolated values are indicated by '#' in place of the decimal place, so\n",
+ "# the Ebinding column won't be numeric. Coerce to float and drop these entries.\n",
+ "Masses['Ebinding'] = pd.to_numeric(Masses['Ebinding'], errors='coerce')\n",
+ "Masses = Masses.dropna()\n",
+ "# Convert from keV to MeV.\n",
+ "Masses['Ebinding'] /= 1000\n",
+ "\n",
+ "# Group the DataFrame by nucleon number, A.\n",
+ "Masses = Masses.groupby('A')\n",
+ "# Find the rows of the grouped DataFrame with the maximum binding energy.\n",
+ "Masses = Masses.apply(lambda t: t[t.Ebinding==t.Ebinding.max()])"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We have now read in the data, grouped them according to the variables we are interested in. \n",
+ "We see how easy it is to reorganize the data using **pandas**. If we\n",
+ "were to do these operations in C/C++ or Fortran, we would have had to\n",
+ "write various functions/subroutines which perform the above\n",
+ "reorganizations for us. Having reorganized the data, we can now start\n",
+ "to make some simple fits using both the functionalities in **numpy** and\n",
+ "**Scikit-Learn** afterwards. \n",
+ "\n",
+ "Now we define five variables which contain\n",
+ "the number of nucleons $A$, the number of protons $Z$ and the number of neutrons $N$, the element name and finally the energies themselves."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 32,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "A = Masses['A']\n",
+ "Z = Masses['Z']\n",
+ "N = Masses['N']\n",
+ "Element = Masses['Element']\n",
+ "Energies = Masses['Ebinding']\n",
+ "print(Masses)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The next step, and we will define this mathematically later, is to set up the so-called **design matrix**. We will throughout call this matrix $\\boldsymbol{X}$.\n",
+ "It has dimensionality $p\\times n$, where $n$ is the number of data points and $p$ are the so-called predictors. In our case here they are given by the number of polynomials in $A$ we wish to include in the fit."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 33,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# Now we set up the design matrix X\n",
+ "X = np.zeros((len(A),5))\n",
+ "X[:,0] = 1\n",
+ "X[:,1] = A\n",
+ "X[:,2] = A**(2.0/3.0)\n",
+ "X[:,3] = A**(-1.0/3.0)\n",
+ "X[:,4] = A**(-1.0)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "With **scikitlearn** we are now ready to use linear regression and fit our data."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 34,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "clf = skl.LinearRegression().fit(X, Energies)\n",
+ "fity = clf.predict(X)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Pretty simple! \n",
+ "Now we can print measures of how our fit is doing, the coefficients from the fits and plot the final fit together with our data."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 35,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# The mean squared error \n",
+ "print(\"Mean squared error: %.2f\" % mean_squared_error(Energies, fity))\n",
+ "# Explained variance score: 1 is perfect prediction \n",
+ "print('Variance score: %.2f' % r2_score(Energies, fity))\n",
+ "# Mean absolute error \n",
+ "print('Mean absolute error: %.2f' % mean_absolute_error(Energies, fity))\n",
+ "print(clf.coef_, clf.intercept_)\n",
+ "\n",
+ "Masses['Eapprox'] = fity\n",
+ "# Generate a plot comparing the experimental with the fitted values values.\n",
+ "fig, ax = plt.subplots()\n",
+ "ax.set_xlabel(r'$A = N + Z$')\n",
+ "ax.set_ylabel(r'$E_\\mathrm{bind}\\,/\\mathrm{MeV}$')\n",
+ "ax.plot(Masses['A'], Masses['Ebinding'], alpha=0.7, lw=2,\n",
+ " label='Ame2016')\n",
+ "ax.plot(Masses['A'], Masses['Eapprox'], alpha=0.7, lw=2, c='m',\n",
+ " label='Fit')\n",
+ "ax.legend()\n",
+ "save_fig(\"Masses2016\")\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "### Seeing the wood for the trees\n",
+ "\n",
+ "As a teaser, let us now see how we can do this with decision trees using **scikit-learn**. Later we will switch to so-called **random forests**!"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 36,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "\n",
+ "#Decision Tree Regression\n",
+ "from sklearn.tree import DecisionTreeRegressor\n",
+ "regr_1=DecisionTreeRegressor(max_depth=5)\n",
+ "regr_2=DecisionTreeRegressor(max_depth=7)\n",
+ "regr_3=DecisionTreeRegressor(max_depth=9)\n",
+ "regr_1.fit(X, Energies)\n",
+ "regr_2.fit(X, Energies)\n",
+ "regr_3.fit(X, Energies)\n",
+ "\n",
+ "\n",
+ "y_1 = regr_1.predict(X)\n",
+ "y_2 = regr_2.predict(X)\n",
+ "y_3=regr_3.predict(X)\n",
+ "Masses['Eapprox'] = y_3\n",
+ "# Plot the results\n",
+ "plt.figure()\n",
+ "plt.plot(A, Energies, color=\"blue\", label=\"Data\", linewidth=2)\n",
+ "plt.plot(A, y_1, color=\"red\", label=\"max_depth=5\", linewidth=2)\n",
+ "plt.plot(A, y_2, color=\"green\", label=\"max_depth=7\", linewidth=2)\n",
+ "plt.plot(A, y_3, color=\"m\", label=\"max_depth=9\", linewidth=2)\n",
+ "\n",
+ "plt.xlabel(\"$A$\")\n",
+ "plt.ylabel(\"$E$[MeV]\")\n",
+ "plt.title(\"Decision Tree Regression\")\n",
+ "plt.legend()\n",
+ "save_fig(\"Masses2016Trees\")\n",
+ "plt.show()\n",
+ "print(Masses)\n",
+ "print(np.mean( (Energies-y_1)**2))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "### And what about using neural networks?\n",
+ "\n",
+ "The **seaborn** package allows us to visualize data in an efficient way. Note that we use **scikit-learn**'s multi-layer perceptron (or feed forward neural network) \n",
+ "functionality."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 37,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "from sklearn.neural_network import MLPRegressor\n",
+ "from sklearn.metrics import accuracy_score\n",
+ "import seaborn as sns\n",
+ "\n",
+ "X_train = X\n",
+ "Y_train = Energies\n",
+ "n_hidden_neurons = 100\n",
+ "epochs = 100\n",
+ "# store models for later use\n",
+ "eta_vals = np.logspace(-5, 1, 7)\n",
+ "lmbd_vals = np.logspace(-5, 1, 7)\n",
+ "# store the models for later use\n",
+ "DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)\n",
+ "train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n",
+ "sns.set()\n",
+ "for i, eta in enumerate(eta_vals):\n",
+ " for j, lmbd in enumerate(lmbd_vals):\n",
+ " dnn = MLPRegressor(hidden_layer_sizes=(n_hidden_neurons), activation='logistic',\n",
+ " alpha=lmbd, learning_rate_init=eta, max_iter=epochs)\n",
+ " dnn.fit(X_train, Y_train)\n",
+ " DNN_scikit[i][j] = dnn\n",
+ " train_accuracy[i][j] = dnn.score(X_train, Y_train)\n",
+ "\n",
+ "fig, ax = plt.subplots(figsize = (10, 10))\n",
+ "sns.heatmap(train_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n",
+ "ax.set_title(\"Training Accuracy\")\n",
+ "ax.set_ylabel(\"$\\eta$\")\n",
+ "ax.set_xlabel(\"$\\lambda$\")\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## A first summary\n",
+ "\n",
+ "The aim behind these introductory words was to present to you various\n",
+ "Python libraries and their functionalities, in particular libraries like\n",
+ "**numpy**, **pandas**, **xarray** and **matplotlib** and other that make our life much easier\n",
+ "in handling various data sets and visualizing data. \n",
+ "\n",
+ "Furthermore,\n",
+ "**Scikit-Learn** allows us with few lines of code to implement popular\n",
+ "Machine Learning algorithms for supervised learning. Later we will meet **Tensorflow**, a powerful library for deep learning. \n",
+ "Now it is time to dive more into the details of various methods. We will start with linear regression and try to take a deeper look at what it entails."
+ ]
+ }
+ ],
+ "metadata": {},
+ "nbformat": 4,
+ "nbformat_minor": 4
+}
diff --git a/doc/src/LectureNotes/chapter4.ipynb b/doc/src/LectureNotes/chapter4.ipynb
new file mode 100644
index 000000000..a8f98b7ba
--- /dev/null
+++ b/doc/src/LectureNotes/chapter4.ipynb
@@ -0,0 +1,5753 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Linear Regression and more Advanced Regression Analysis\n",
+ "\n",
+ "\n",
+ "## Why Linear Regression (aka Ordinary Least Squares and family)\n",
+ "\n",
+ "Fitting a continuous function with linear parameterization in terms of the parameters $\\boldsymbol{\\beta}$.\n",
+ "* Method of choice for fitting a continuous function!\n",
+ "\n",
+ "* Gives an excellent introduction to central Machine Learning features with **understandable pedagogical** links to other methods like **Neural Networks**, **Support Vector Machines** etc\n",
+ "\n",
+ "* Analytical expression for the fitting parameters $\\boldsymbol{\\beta}$\n",
+ "\n",
+ "* Analytical expressions for statistical propertiers like mean values, variances, confidence intervals and more\n",
+ "\n",
+ "* Analytical relation with probabilistic interpretations \n",
+ "\n",
+ "* Easy to introduce basic concepts like bias-variance tradeoff, cross-validation, resampling and regularization techniques and many other ML topics\n",
+ "\n",
+ "* Easy to code! And links well with classification problems and logistic regression and neural networks\n",
+ "\n",
+ "* Allows for **easy** hands-on understanding of gradient descent methods\n",
+ "\n",
+ "* and many more features\n",
+ "\n",
+ "For more discussions of Ridge and Lasso regression, [Wessel van Wieringen's](https://arxiv.org/abs/1509.09169) article is highly recommended.\n",
+ "Similarly, [Mehta et al's article](https://arxiv.org/abs/1803.08823) is also recommended.\n",
+ "\n",
+ "\n",
+ "## Regression analysis, overarching aims\n",
+ "\n",
+ "Regression modeling deals with the description of the sampling distribution of a given random variable $y$ and how it varies as function of another variable or a set of such variables $\\boldsymbol{x} =[x_0, x_1,\\dots, x_{n-1}]^T$. \n",
+ "The first variable is called the **dependent**, the **outcome** or the **response** variable while the set of variables $\\boldsymbol{x}$ is called the independent variable, or the predictor variable or the explanatory variable. \n",
+ "\n",
+ "A regression model aims at finding a likelihood function $p(\\boldsymbol{y}\\vert \\boldsymbol{x})$, that is the conditional distribution for $\\boldsymbol{y}$ with a given $\\boldsymbol{x}$. The estimation of $p(\\boldsymbol{y}\\vert \\boldsymbol{x})$ is made using a data set with \n",
+ "* $n$ cases $i = 0, 1, 2, \\dots, n-1$ \n",
+ "\n",
+ "* Response (target, dependent or outcome) variable $y_i$ with $i = 0, 1, 2, \\dots, n-1$ \n",
+ "\n",
+ "* $p$ so-called explanatory (independent or predictor) variables $\\boldsymbol{x}_i=[x_{i0}, x_{i1}, \\dots, x_{ip-1}]$ with $i = 0, 1, 2, \\dots, n-1$ and explanatory variables running from $0$ to $p-1$. See below for more explicit examples. \n",
+ "\n",
+ " The goal of the regression analysis is to extract/exploit relationship between $\\boldsymbol{y}$ and $\\boldsymbol{x}$ in or to infer causal dependencies, approximations to the likelihood functions, functional relationships and to make predictions, making fits and many other things.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Regression analysis, overarching aims II\n",
+ "\n",
+ "\n",
+ "Consider an experiment in which $p$ characteristics of $n$ samples are\n",
+ "measured. The data from this experiment, for various explanatory variables $p$ are normally represented by a matrix \n",
+ "$\\mathbf{X}$.\n",
+ "\n",
+ "The matrix $\\mathbf{X}$ is called the *design\n",
+ "matrix*. Additional information of the samples is available in the\n",
+ "form of $\\boldsymbol{y}$ (also as above). The variable $\\boldsymbol{y}$ is\n",
+ "generally referred to as the *response variable*. The aim of\n",
+ "regression analysis is to explain $\\boldsymbol{y}$ in terms of\n",
+ "$\\boldsymbol{X}$ through a functional relationship like $y_i =\n",
+ "f(\\mathbf{X}_{i,\\ast})$. When no prior knowledge on the form of\n",
+ "$f(\\cdot)$ is available, it is common to assume a linear relationship\n",
+ "between $\\boldsymbol{X}$ and $\\boldsymbol{y}$. This assumption gives rise to\n",
+ "the *linear regression model* where $\\boldsymbol{\\beta} = [\\beta_0, \\ldots,\n",
+ "\\beta_{p-1}]^{T}$ are the *regression parameters*. \n",
+ "\n",
+ "Linear regression gives us a set of analytical equations for the parameters $\\beta_j$.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Examples\n",
+ "In order to understand the relation among the predictors $p$, the set of data $n$ and the target (outcome, output etc) $\\boldsymbol{y}$,\n",
+ "consider the model we discussed for describing nuclear binding energies. \n",
+ "\n",
+ "There we assumed that we could parametrize the data using a polynomial approximation based on the liquid drop model.\n",
+ "Assuming"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "BE(A) = a_0+a_1A+a_2A^{2/3}+a_3A^{-1/3}+a_4A^{-1},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "we have five predictors, that is the intercept, the $A$ dependent term, the $A^{2/3}$ term and the $A^{-1/3}$ and $A^{-1}$ terms.\n",
+ "This gives $p=0,1,2,3,4$. Furthermore we have $n$ entries for each predictor. It means that our design matrix is a \n",
+ "$p\\times n$ matrix $\\boldsymbol{X}$.\n",
+ "\n",
+ "Here the predictors are based on a model we have made. A popular data set which is widely encountered in ML applications is the\n",
+ "so-called [credit card default data from Taiwan](https://www.sciencedirect.com/science/article/pii/S0957417407006719?via%3Dihub). The data set contains data on $n=30000$ credit card holders with predictors like gender, marital status, age, profession, education, etc. In total there are $24$ such predictors or attributes leading to a design matrix of dimensionality $24 \\times 30000$. This is however a classification problem and we will come back to it when we discuss Logistic Regression.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## General linear models\n",
+ "Before we proceed let us study a case from linear algebra where we aim at fitting a set of data $\\boldsymbol{y}=[y_0,y_1,\\dots,y_{n-1}]$. We could think of these data as a result of an experiment or a complicated numerical experiment. These data are functions of a series of variables $\\boldsymbol{x}=[x_0,x_1,\\dots,x_{n-1}]$, that is $y_i = y(x_i)$ with $i=0,1,2,\\dots,n-1$. The variables $x_i$ could represent physical quantities like time, temperature, position etc. We assume that $y(x)$ is a smooth function. \n",
+ "\n",
+ "Since obtaining these data points may not be trivial, we want to use these data to fit a function which can allow us to make predictions for values of $y$ which are not in the present set. The perhaps simplest approach is to assume we can parametrize our function in terms of a polynomial of degree $n-1$ with $n$ points, that is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "y=y(x) \\rightarrow y(x_i)=\\tilde{y}_i+\\epsilon_i=\\sum_{j=0}^{n-1} \\beta_j x_i^j+\\epsilon_i,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $\\epsilon_i$ is the error in our approximation.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Rewriting the fitting procedure as a linear algebra problem\n",
+ "For every set of values $y_i,x_i$ we have thus the corresponding set of equations"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\begin{align*}\n",
+ "y_0&=\\beta_0+\\beta_1x_0^1+\\beta_2x_0^2+\\dots+\\beta_{n-1}x_0^{n-1}+\\epsilon_0\\\\\n",
+ "y_1&=\\beta_0+\\beta_1x_1^1+\\beta_2x_1^2+\\dots+\\beta_{n-1}x_1^{n-1}+\\epsilon_1\\\\\n",
+ "y_2&=\\beta_0+\\beta_1x_2^1+\\beta_2x_2^2+\\dots+\\beta_{n-1}x_2^{n-1}+\\epsilon_2\\\\\n",
+ "\\dots & \\dots \\\\\n",
+ "y_{n-1}&=\\beta_0+\\beta_1x_{n-1}^1+\\beta_2x_{n-1}^2+\\dots+\\beta_{n-1}x_{n-1}^{n-1}+\\epsilon_{n-1}.\\\\\n",
+ "\\end{align*}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Rewriting the fitting procedure as a linear algebra problem, more details\n",
+ "Defining the vectors"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{y} = [y_0,y_1, y_2,\\dots, y_{n-1}]^T,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{\\beta} = [\\beta_0,\\beta_1, \\beta_2,\\dots, \\beta_{n-1}]^T,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{\\epsilon} = [\\epsilon_0,\\epsilon_1, \\epsilon_2,\\dots, \\epsilon_{n-1}]^T,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and the design matrix"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X}=\n",
+ "\\begin{bmatrix} \n",
+ "1& x_{0}^1 &x_{0}^2& \\dots & \\dots &x_{0}^{n-1}\\\\\n",
+ "1& x_{1}^1 &x_{1}^2& \\dots & \\dots &x_{1}^{n-1}\\\\\n",
+ "1& x_{2}^1 &x_{2}^2& \\dots & \\dots &x_{2}^{n-1}\\\\ \n",
+ "\\dots& \\dots &\\dots& \\dots & \\dots &\\dots\\\\\n",
+ "1& x_{n-1}^1 &x_{n-1}^2& \\dots & \\dots &x_{n-1}^{n-1}\\\\\n",
+ "\\end{bmatrix}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "we can rewrite our equations as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{y} = \\boldsymbol{X}\\boldsymbol{\\beta}+\\boldsymbol{\\epsilon}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The above design matrix is called a [Vandermonde matrix](https://en.wikipedia.org/wiki/Vandermonde_matrix).\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Generalizing the fitting procedure as a linear algebra problem\n",
+ "\n",
+ "We are obviously not limited to the above polynomial expansions. We\n",
+ "could replace the various powers of $x$ with elements of Fourier\n",
+ "series or instead of $x_i^j$ we could have $\\cos{(j x_i)}$ or $\\sin{(j\n",
+ "x_i)}$, or time series or other orthogonal functions. For every set\n",
+ "of values $y_i,x_i$ we can then generalize the equations to"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\begin{align*}\n",
+ "y_0&=\\beta_0x_{00}+\\beta_1x_{01}+\\beta_2x_{02}+\\dots+\\beta_{n-1}x_{0n-1}+\\epsilon_0\\\\\n",
+ "y_1&=\\beta_0x_{10}+\\beta_1x_{11}+\\beta_2x_{12}+\\dots+\\beta_{n-1}x_{1n-1}+\\epsilon_1\\\\\n",
+ "y_2&=\\beta_0x_{20}+\\beta_1x_{21}+\\beta_2x_{22}+\\dots+\\beta_{n-1}x_{2n-1}+\\epsilon_2\\\\\n",
+ "\\dots & \\dots \\\\\n",
+ "y_{i}&=\\beta_0x_{i0}+\\beta_1x_{i1}+\\beta_2x_{i2}+\\dots+\\beta_{n-1}x_{in-1}+\\epsilon_i\\\\\n",
+ "\\dots & \\dots \\\\\n",
+ "y_{n-1}&=\\beta_0x_{n-1,0}+\\beta_1x_{n-1,2}+\\beta_2x_{n-1,2}+\\dots+\\beta_{n-1}x_{n-1,n-1}+\\epsilon_{n-1}.\\\\\n",
+ "\\end{align*}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "**Note that we have $p=n$ here. The matrix is symmetric. This is generally not the case!**\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Generalizing the fitting procedure as a linear algebra problem\n",
+ "We redefine in turn the matrix $\\boldsymbol{X}$ as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X}=\n",
+ "\\begin{bmatrix} \n",
+ "x_{00}& x_{01} &x_{02}& \\dots & \\dots &x_{0,n-1}\\\\\n",
+ "x_{10}& x_{11} &x_{12}& \\dots & \\dots &x_{1,n-1}\\\\\n",
+ "x_{20}& x_{21} &x_{22}& \\dots & \\dots &x_{2,n-1}\\\\ \n",
+ "\\dots& \\dots &\\dots& \\dots & \\dots &\\dots\\\\\n",
+ "x_{n-1,0}& x_{n-1,1} &x_{n-1,2}& \\dots & \\dots &x_{n-1,n-1}\\\\\n",
+ "\\end{bmatrix}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and without loss of generality we rewrite again our equations as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{y} = \\boldsymbol{X}\\boldsymbol{\\beta}+\\boldsymbol{\\epsilon}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The left-hand side of this equation is kwown. Our error vector $\\boldsymbol{\\epsilon}$ and the parameter vector $\\boldsymbol{\\beta}$ are our unknow quantities. How can we obtain the optimal set of $\\beta_i$ values?\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Optimizing our parameters\n",
+ "We have defined the matrix $\\boldsymbol{X}$ via the equations"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\begin{align*}\n",
+ "y_0&=\\beta_0x_{00}+\\beta_1x_{01}+\\beta_2x_{02}+\\dots+\\beta_{n-1}x_{0n-1}+\\epsilon_0\\\\\n",
+ "y_1&=\\beta_0x_{10}+\\beta_1x_{11}+\\beta_2x_{12}+\\dots+\\beta_{n-1}x_{1n-1}+\\epsilon_1\\\\\n",
+ "y_2&=\\beta_0x_{20}+\\beta_1x_{21}+\\beta_2x_{22}+\\dots+\\beta_{n-1}x_{2n-1}+\\epsilon_1\\\\\n",
+ "\\dots & \\dots \\\\\n",
+ "y_{i}&=\\beta_0x_{i0}+\\beta_1x_{i1}+\\beta_2x_{i2}+\\dots+\\beta_{n-1}x_{in-1}+\\epsilon_1\\\\\n",
+ "\\dots & \\dots \\\\\n",
+ "y_{n-1}&=\\beta_0x_{n-1,0}+\\beta_1x_{n-1,2}+\\beta_2x_{n-1,2}+\\dots+\\beta_{n-1}x_{n-1,n-1}+\\epsilon_{n-1}.\\\\\n",
+ "\\end{align*}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "As we noted above, we stayed with a system with the design matrix \n",
+ " $\\boldsymbol{X}\\in {\\mathbb{R}}^{n\\times n}$, that is we have $p=n$. For reasons to come later (algorithmic arguments) we will hereafter define \n",
+ "our matrix as $\\boldsymbol{X}\\in {\\mathbb{R}}^{n\\times p}$, with the predictors refering to the column numbers and the entries $n$ being the row elements.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Our model for the nuclear binding energies\n",
+ "\n",
+ "In our [introductory notes](https://compphysics.github.io/MachineLearning/doc/pub/How2ReadData/html/How2ReadData.html) we looked at the so-called [liquid drop model](https://en.wikipedia.org/wiki/Semi-empirical_mass_formula). Let us remind ourselves about what we did by looking at the code.\n",
+ "\n",
+ "We restate the parts of the code we are most interested in."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "%matplotlib inline\n",
+ "\n",
+ "# Common imports\n",
+ "import numpy as np\n",
+ "import pandas as pd\n",
+ "import matplotlib.pyplot as plt\n",
+ "from IPython.display import display\n",
+ "import os\n",
+ "\n",
+ "# Where to save the figures and data files\n",
+ "PROJECT_ROOT_DIR = \"Results\"\n",
+ "FIGURE_ID = \"Results/FigureFiles\"\n",
+ "DATA_ID = \"DataFiles/\"\n",
+ "\n",
+ "if not os.path.exists(PROJECT_ROOT_DIR):\n",
+ " os.mkdir(PROJECT_ROOT_DIR)\n",
+ "\n",
+ "if not os.path.exists(FIGURE_ID):\n",
+ " os.makedirs(FIGURE_ID)\n",
+ "\n",
+ "if not os.path.exists(DATA_ID):\n",
+ " os.makedirs(DATA_ID)\n",
+ "\n",
+ "def image_path(fig_id):\n",
+ " return os.path.join(FIGURE_ID, fig_id)\n",
+ "\n",
+ "def data_path(dat_id):\n",
+ " return os.path.join(DATA_ID, dat_id)\n",
+ "\n",
+ "def save_fig(fig_id):\n",
+ " plt.savefig(image_path(fig_id) + \".png\", format='png')\n",
+ "\n",
+ "infile = open(data_path(\"MassEval2016.dat\"),'r')\n",
+ "\n",
+ "\n",
+ "# Read the experimental data with Pandas\n",
+ "Masses = pd.read_fwf(infile, usecols=(2,3,4,6,11),\n",
+ " names=('N', 'Z', 'A', 'Element', 'Ebinding'),\n",
+ " widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),\n",
+ " header=39,\n",
+ " index_col=False)\n",
+ "\n",
+ "# Extrapolated values are indicated by '#' in place of the decimal place, so\n",
+ "# the Ebinding column won't be numeric. Coerce to float and drop these entries.\n",
+ "Masses['Ebinding'] = pd.to_numeric(Masses['Ebinding'], errors='coerce')\n",
+ "Masses = Masses.dropna()\n",
+ "# Convert from keV to MeV.\n",
+ "Masses['Ebinding'] /= 1000\n",
+ "\n",
+ "# Group the DataFrame by nucleon number, A.\n",
+ "Masses = Masses.groupby('A')\n",
+ "# Find the rows of the grouped DataFrame with the maximum binding energy.\n",
+ "Masses = Masses.apply(lambda t: t[t.Ebinding==t.Ebinding.max()])\n",
+ "A = Masses['A']\n",
+ "Z = Masses['Z']\n",
+ "N = Masses['N']\n",
+ "Element = Masses['Element']\n",
+ "Energies = Masses['Ebinding']\n",
+ "\n",
+ "# Now we set up the design matrix X\n",
+ "X = np.zeros((len(A),5))\n",
+ "X[:,0] = 1\n",
+ "X[:,1] = A\n",
+ "X[:,2] = A**(2.0/3.0)\n",
+ "X[:,3] = A**(-1.0/3.0)\n",
+ "X[:,4] = A**(-1.0)\n",
+ "# Then nice printout using pandas\n",
+ "DesignMatrix = pd.DataFrame(X)\n",
+ "DesignMatrix.index = A\n",
+ "DesignMatrix.columns = ['1', 'A', 'A^(2/3)', 'A^(-1/3)', '1/A']\n",
+ "display(DesignMatrix)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "With $\\boldsymbol{\\beta}\\in {\\mathbb{R}}^{p\\times 1}$, it means that we will hereafter write our equations for the approximation as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{\\tilde{y}}= \\boldsymbol{X}\\boldsymbol{\\beta},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "throughout these lectures. \n",
+ "\n",
+ "\n",
+ "## Optimizing our parameters, more details\n",
+ "With the above we use the design matrix to define the approximation $\\boldsymbol{\\tilde{y}}$ via the unknown quantity $\\boldsymbol{\\beta}$ as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{\\tilde{y}}= \\boldsymbol{X}\\boldsymbol{\\beta},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and in order to find the optimal parameters $\\beta_i$ instead of solving the above linear algebra problem, we define a function which gives a measure of the spread between the values $y_i$ (which represent hopefully the exact values) and the parameterized values $\\tilde{y}_i$, namely"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "C(\\boldsymbol{\\beta})=\\frac{1}{n}\\sum_{i=0}^{n-1}\\left(y_i-\\tilde{y}_i\\right)^2=\\frac{1}{n}\\left\\{\\left(\\boldsymbol{y}-\\boldsymbol{\\tilde{y}}\\right)^T\\left(\\boldsymbol{y}-\\boldsymbol{\\tilde{y}}\\right)\\right\\},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "or using the matrix $\\boldsymbol{X}$ and in a more compact matrix-vector notation as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "C(\\boldsymbol{\\beta})=\\frac{1}{n}\\left\\{\\left(\\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta}\\right)^T\\left(\\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta}\\right)\\right\\}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "This function is one possible way to define the so-called cost function.\n",
+ "\n",
+ "\n",
+ "\n",
+ "It is also common to define\n",
+ "the function $C$ as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "C(\\boldsymbol{\\beta})=\\frac{1}{2n}\\sum_{i=0}^{n-1}\\left(y_i-\\tilde{y}_i\\right)^2,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "since when taking the first derivative with respect to the unknown parameters $\\beta$, the factor of $2$ cancels out.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Interpretations and optimizing our parameters\n",
+ "\n",
+ "The function"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "C(\\boldsymbol{\\beta})=\\frac{1}{n}\\left\\{\\left(\\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta}\\right)^T\\left(\\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta}\\right)\\right\\},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "can be linked to the variance of the quantity $y_i$ if we interpret the latter as the mean value. \n",
+ "When linking (see the discussion below) with the maximum likelihood approach below, we will indeed interpret $y_i$ as a mean value"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "y_{i}=\\langle y_i \\rangle = \\beta_0x_{i,0}+\\beta_1x_{i,1}+\\beta_2x_{i,2}+\\dots+\\beta_{n-1}x_{i,n-1}+\\epsilon_i,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $\\langle y_i \\rangle$ is the mean value. Keep in mind also that\n",
+ "till now we have treated $y_i$ as the exact value. Normally, the\n",
+ "response (dependent or outcome) variable $y_i$ the outcome of a\n",
+ "numerical experiment or another type of experiment and is thus only an\n",
+ "approximation to the true value. It is then always accompanied by an\n",
+ "error estimate, often limited to a statistical error estimate given by\n",
+ "the standard deviation discussed earlier. In the discussion here we\n",
+ "will treat $y_i$ as our exact value for the response variable.\n",
+ "\n",
+ "In order to find the parameters $\\beta_i$ we will then minimize the spread of $C(\\boldsymbol{\\beta})$, that is we are going to solve the problem"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "{\\displaystyle \\min_{\\boldsymbol{\\beta}\\in\n",
+ "{\\mathbb{R}}^{p}}}\\frac{1}{n}\\left\\{\\left(\\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta}\\right)^T\\left(\\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta}\\right)\\right\\}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "In practical terms it means we will require"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial C(\\boldsymbol{\\beta})}{\\partial \\beta_j} = \\frac{\\partial }{\\partial \\beta_j}\\left[ \\frac{1}{n}\\sum_{i=0}^{n-1}\\left(y_i-\\beta_0x_{i,0}-\\beta_1x_{i,1}-\\beta_2x_{i,2}-\\dots-\\beta_{n-1}x_{i,n-1}\\right)^2\\right]=0,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which results in"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial C(\\boldsymbol{\\beta})}{\\partial \\beta_j} = -\\frac{2}{n}\\left[ \\sum_{i=0}^{n-1}x_{ij}\\left(y_i-\\beta_0x_{i,0}-\\beta_1x_{i,1}-\\beta_2x_{i,2}-\\dots-\\beta_{n-1}x_{i,n-1}\\right)\\right]=0,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "or in a matrix-vector form as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial C(\\boldsymbol{\\beta})}{\\partial \\boldsymbol{\\beta}} = 0 = \\boldsymbol{X}^T\\left( \\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta}\\right).\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Interpretations and optimizing our parameters\n",
+ "We can rewrite"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial C(\\boldsymbol{\\beta})}{\\partial \\boldsymbol{\\beta}} = 0 = \\boldsymbol{X}^T\\left( \\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta}\\right),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X}^T\\boldsymbol{y} = \\boldsymbol{X}^T\\boldsymbol{X}\\boldsymbol{\\beta},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and if the matrix $\\boldsymbol{X}^T\\boldsymbol{X}$ is invertible we have the solution"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{\\beta} =\\left(\\boldsymbol{X}^T\\boldsymbol{X}\\right)^{-1}\\boldsymbol{X}^T\\boldsymbol{y}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We note also that since our design matrix is defined as $\\boldsymbol{X}\\in\n",
+ "{\\mathbb{R}}^{n\\times p}$, the product $\\boldsymbol{X}^T\\boldsymbol{X} \\in\n",
+ "{\\mathbb{R}}^{p\\times p}$. In the above case we have that $p \\ll n$,\n",
+ "in our case $p=5$ meaning that we end up with inverting a small\n",
+ "$5\\times 5$ matrix. This is a rather common situation, in many cases we end up with low-dimensional\n",
+ "matrices to invert. The methods discussed here and for many other\n",
+ "supervised learning algorithms like classification with logistic\n",
+ "regression or support vector machines, exhibit dimensionalities which\n",
+ "allow for the usage of direct linear algebra methods such as **LU** decomposition or **Singular Value Decomposition** (SVD) for finding the inverse of the matrix\n",
+ "$\\boldsymbol{X}^T\\boldsymbol{X}$.\n",
+ "\n",
+ "\n",
+ "\n",
+ "**Small question**: Do you think the example we have at hand here (the nuclear binding energies) can lead to problems in inverting the matrix $\\boldsymbol{X}^T\\boldsymbol{X}$? What kind of problems can we expect?\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Some useful matrix and vector expressions\n",
+ "\n",
+ "The following matrix and vector relation will be useful here and for the rest of the course. Vectors are always written as boldfaced lower case letters and \n",
+ "matrices as upper case boldfaced letters."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "2\n",
+ "6\n",
+ " \n",
+ "<\n",
+ "<\n",
+ "<\n",
+ "!\n",
+ "!\n",
+ "M\n",
+ "A\n",
+ "T\n",
+ "H\n",
+ "_\n",
+ "B\n",
+ "L\n",
+ "O\n",
+ "C\n",
+ "K"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "2\n",
+ "7\n",
+ " \n",
+ "<\n",
+ "<\n",
+ "<\n",
+ "!\n",
+ "!\n",
+ "M\n",
+ "A\n",
+ "T\n",
+ "H\n",
+ "_\n",
+ "B\n",
+ "L\n",
+ "O\n",
+ "C\n",
+ "K"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "2\n",
+ "8\n",
+ " \n",
+ "<\n",
+ "<\n",
+ "<\n",
+ "!\n",
+ "!\n",
+ "M\n",
+ "A\n",
+ "T\n",
+ "H\n",
+ "_\n",
+ "B\n",
+ "L\n",
+ "O\n",
+ "C\n",
+ "K"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial \\log{\\vert\\boldsymbol{A}\\vert}}{\\partial \\boldsymbol{A}} = (\\boldsymbol{A}^{-1})^T.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Interpretations and optimizing our parameters\n",
+ "The residuals $\\boldsymbol{\\epsilon}$ are in turn given by"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{\\epsilon} = \\boldsymbol{y}-\\boldsymbol{\\tilde{y}} = \\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and with"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X}^T\\left( \\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta}\\right)= 0,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "we have"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X}^T\\boldsymbol{\\epsilon}=\\boldsymbol{X}^T\\left( \\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta}\\right)= 0,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "meaning that the solution for $\\boldsymbol{\\beta}$ is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "Let us now return to our nuclear binding energies and simply code the above equations. \n",
+ "\n",
+ "## Own code for Ordinary Least Squares\n",
+ "\n",
+ "It is rather straightforward to implement the matrix inversion and obtain the parameters $\\boldsymbol{\\beta}$. After having defined the matrix $\\boldsymbol{X}$ we simply need to \n",
+ "write"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# matrix inversion to find beta\n",
+ "beta = np.linalg.inv(X.T.dot(X)).dot(X.T).dot(Energies)\n",
+ "# and then make the prediction\n",
+ "ytilde = X @ beta"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Alternatively, you can use the least squares functionality in **Numpy** as"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "fit = np.linalg.lstsq(X, Energies, rcond =None)[0]\n",
+ "ytildenp = np.dot(fit,X.T)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "And finally we plot our fit with and compare with data"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "Masses['Eapprox'] = ytilde\n",
+ "# Generate a plot comparing the experimental with the fitted values values.\n",
+ "fig, ax = plt.subplots()\n",
+ "ax.set_xlabel(r'$A = N + Z$')\n",
+ "ax.set_ylabel(r'$E_\\mathrm{bind}\\,/\\mathrm{MeV}$')\n",
+ "ax.plot(Masses['A'], Masses['Ebinding'], alpha=0.7, lw=2,\n",
+ " label='Ame2016')\n",
+ "ax.plot(Masses['A'], Masses['Eapprox'], alpha=0.7, lw=2, c='m',\n",
+ " label='Fit')\n",
+ "ax.legend()\n",
+ "save_fig(\"Masses2016OLS\")\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Adding error analysis and training set up\n",
+ "\n",
+ "We can easily test our fit by computing the $R2$ score that we discussed in connection with the functionality of **Scikit-Learn** in the introductory slides.\n",
+ "Since we are not using **Scikit-Learn** here we can define our own $R2$ function as"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "def R2(y_data, y_model):\n",
+ " return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and we would be using it as"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "print(R2(Energies,ytilde))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We can easily add our **MSE** score as"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "def MSE(y_data,y_model):\n",
+ " n = np.size(y_model)\n",
+ " return np.sum((y_data-y_model)**2)/n\n",
+ "\n",
+ "print(MSE(Energies,ytilde))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and finally the relative error as"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "def RelativeError(y_data,y_model):\n",
+ " return abs((y_data-y_model)/y_data)\n",
+ "print(RelativeError(Energies, ytilde))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## The $\\chi^2$ function\n",
+ "\n",
+ "Normally, the response (dependent or outcome) variable $y_i$ is the\n",
+ "outcome of a numerical experiment or another type of experiment and is\n",
+ "thus only an approximation to the true value. It is then always\n",
+ "accompanied by an error estimate, often limited to a statistical error\n",
+ "estimate given by the standard deviation discussed earlier. In the\n",
+ "discussion here we will treat $y_i$ as our exact value for the\n",
+ "response variable.\n",
+ "\n",
+ "Introducing the standard deviation $\\sigma_i$ for each measurement\n",
+ "$y_i$, we define now the $\\chi^2$ function (omitting the $1/n$ term)\n",
+ "as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\chi^2(\\boldsymbol{\\beta})=\\frac{1}{n}\\sum_{i=0}^{n-1}\\frac{\\left(y_i-\\tilde{y}_i\\right)^2}{\\sigma_i^2}=\\frac{1}{n}\\left\\{\\left(\\boldsymbol{y}-\\boldsymbol{\\tilde{y}}\\right)^T\\frac{1}{\\boldsymbol{\\Sigma^2}}\\left(\\boldsymbol{y}-\\boldsymbol{\\tilde{y}}\\right)\\right\\},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where the matrix $\\boldsymbol{\\Sigma}$ is a diagonal matrix with $\\sigma_i$ as matrix elements.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## The $\\chi^2$ function\n",
+ "\n",
+ "In order to find the parameters $\\beta_i$ we will then minimize the spread of $\\chi^2(\\boldsymbol{\\beta})$ by requiring"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial \\chi^2(\\boldsymbol{\\beta})}{\\partial \\beta_j} = \\frac{\\partial }{\\partial \\beta_j}\\left[ \\frac{1}{n}\\sum_{i=0}^{n-1}\\left(\\frac{y_i-\\beta_0x_{i,0}-\\beta_1x_{i,1}-\\beta_2x_{i,2}-\\dots-\\beta_{n-1}x_{i,n-1}}{\\sigma_i}\\right)^2\\right]=0,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which results in"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial \\chi^2(\\boldsymbol{\\beta})}{\\partial \\beta_j} = -\\frac{2}{n}\\left[ \\sum_{i=0}^{n-1}\\frac{x_{ij}}{\\sigma_i}\\left(\\frac{y_i-\\beta_0x_{i,0}-\\beta_1x_{i,1}-\\beta_2x_{i,2}-\\dots-\\beta_{n-1}x_{i,n-1}}{\\sigma_i}\\right)\\right]=0,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "or in a matrix-vector form as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial \\chi^2(\\boldsymbol{\\beta})}{\\partial \\boldsymbol{\\beta}} = 0 = \\boldsymbol{A}^T\\left( \\boldsymbol{b}-\\boldsymbol{A}\\boldsymbol{\\beta}\\right).\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where we have defined the matrix $\\boldsymbol{A} =\\boldsymbol{X}/\\boldsymbol{\\Sigma}$ with matrix elements $a_{ij} = x_{ij}/\\sigma_i$ and the vector $\\boldsymbol{b}$ with elements $b_i = y_i/\\sigma_i$.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## The $\\chi^2$ function\n",
+ "\n",
+ "We can rewrite"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial \\chi^2(\\boldsymbol{\\beta})}{\\partial \\boldsymbol{\\beta}} = 0 = \\boldsymbol{A}^T\\left( \\boldsymbol{b}-\\boldsymbol{A}\\boldsymbol{\\beta}\\right),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{A}^T\\boldsymbol{b} = \\boldsymbol{A}^T\\boldsymbol{A}\\boldsymbol{\\beta},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and if the matrix $\\boldsymbol{A}^T\\boldsymbol{A}$ is invertible we have the solution"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{\\beta} =\\left(\\boldsymbol{A}^T\\boldsymbol{A}\\right)^{-1}\\boldsymbol{A}^T\\boldsymbol{b}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## The $\\chi^2$ function\n",
+ "\n",
+ "If we then introduce the matrix"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{H} = \\left(\\boldsymbol{A}^T\\boldsymbol{A}\\right)^{-1},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "we have then the following expression for the parameters $\\beta_j$ (the matrix elements of $\\boldsymbol{H}$ are $h_{ij}$)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\beta_j = \\sum_{k=0}^{p-1}h_{jk}\\sum_{i=0}^{n-1}\\frac{y_i}{\\sigma_i}\\frac{x_{ik}}{\\sigma_i} = \\sum_{k=0}^{p-1}h_{jk}\\sum_{i=0}^{n-1}b_ia_{ik}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We state without proof the expression for the uncertainty in the parameters $\\beta_j$ as (we leave this as an exercise)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\sigma^2(\\beta_j) = \\sum_{i=0}^{n-1}\\sigma_i^2\\left( \\frac{\\partial \\beta_j}{\\partial y_i}\\right)^2,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "resulting in"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\sigma^2(\\beta_j) = \\left(\\sum_{k=0}^{p-1}h_{jk}\\sum_{i=0}^{n-1}a_{ik}\\right)\\left(\\sum_{l=0}^{p-1}h_{jl}\\sum_{m=0}^{n-1}a_{ml}\\right) = h_{jj}!\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## The $\\chi^2$ function\n",
+ "The first step here is to approximate the function $y$ with a first-order polynomial, that is we write"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "y=y(x) \\rightarrow y(x_i) \\approx \\beta_0+\\beta_1 x_i.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "By computing the derivatives of $\\chi^2$ with respect to $\\beta_0$ and $\\beta_1$ show that these are given by"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial \\chi^2(\\boldsymbol{\\beta})}{\\partial \\beta_0} = -2\\left[ \\frac{1}{n}\\sum_{i=0}^{n-1}\\left(\\frac{y_i-\\beta_0-\\beta_1x_{i}}{\\sigma_i^2}\\right)\\right]=0,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial \\chi^2(\\boldsymbol{\\beta})}{\\partial \\beta_1} = -\\frac{2}{n}\\left[ \\sum_{i=0}^{n-1}x_i\\left(\\frac{y_i-\\beta_0-\\beta_1x_{i}}{\\sigma_i^2}\\right)\\right]=0.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## The $\\chi^2$ function\n",
+ "\n",
+ "For a linear fit (a first-order polynomial) we don't need to invert a matrix!! \n",
+ "Defining"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\gamma = \\sum_{i=0}^{n-1}\\frac{1}{\\sigma_i^2},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\gamma_x = \\sum_{i=0}^{n-1}\\frac{x_{i}}{\\sigma_i^2},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\gamma_y = \\sum_{i=0}^{n-1}\\left(\\frac{y_i}{\\sigma_i^2}\\right),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\gamma_{xx} = \\sum_{i=0}^{n-1}\\frac{x_ix_{i}}{\\sigma_i^2},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\gamma_{xy} = \\sum_{i=0}^{n-1}\\frac{y_ix_{i}}{\\sigma_i^2},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "we obtain"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\beta_0 = \\frac{\\gamma_{xx}\\gamma_y-\\gamma_x\\gamma_y}{\\gamma\\gamma_{xx}-\\gamma_x^2},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\beta_1 = \\frac{\\gamma_{xy}\\gamma-\\gamma_x\\gamma_y}{\\gamma\\gamma_{xx}-\\gamma_x^2}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "This approach (different linear and non-linear regression) suffers\n",
+ "often from both being underdetermined and overdetermined in the\n",
+ "unknown coefficients $\\beta_i$. A better approach is to use the\n",
+ "Singular Value Decomposition (SVD) method discussed below. Or using\n",
+ "Lasso and Ridge regression. See below.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Fitting an Equation of State for Dense Nuclear Matter\n",
+ "\n",
+ "Before we continue, let us introduce yet another example. We are going to fit the\n",
+ "nuclear equation of state using results from many-body calculations.\n",
+ "The equation of state we have made available here, as function of\n",
+ "density, has been derived using modern nucleon-nucleon potentials with\n",
+ "[the addition of three-body\n",
+ "forces](https://www.sciencedirect.com/science/article/pii/S0370157399001106). This\n",
+ "time the file is presented as a standard **csv** file.\n",
+ "\n",
+ "The beginning of the Python code here is similar to what you have seen\n",
+ "before, with the same initializations and declarations. We use also\n",
+ "**pandas** again, rather extensively in order to organize our data.\n",
+ "\n",
+ "The difference now is that we use **Scikit-Learn's** regression tools\n",
+ "instead of our own matrix inversion implementation. Furthermore, we\n",
+ "sneak in **Ridge** regression (to be discussed below) which includes a\n",
+ "hyperparameter $\\lambda$, also to be explained below.\n",
+ "\n",
+ "## The code"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# Common imports\n",
+ "import os\n",
+ "import numpy as np\n",
+ "import pandas as pd\n",
+ "import matplotlib.pyplot as plt\n",
+ "import matplotlib.pyplot as plt\n",
+ "import sklearn.linear_model as skl\n",
+ "from sklearn.metrics import mean_squared_error, r2_score, mean_absolute_error\n",
+ "\n",
+ "# Where to save the figures and data files\n",
+ "PROJECT_ROOT_DIR = \"Results\"\n",
+ "FIGURE_ID = \"Results/FigureFiles\"\n",
+ "DATA_ID = \"DataFiles/\"\n",
+ "\n",
+ "if not os.path.exists(PROJECT_ROOT_DIR):\n",
+ " os.mkdir(PROJECT_ROOT_DIR)\n",
+ "\n",
+ "if not os.path.exists(FIGURE_ID):\n",
+ " os.makedirs(FIGURE_ID)\n",
+ "\n",
+ "if not os.path.exists(DATA_ID):\n",
+ " os.makedirs(DATA_ID)\n",
+ "\n",
+ "def image_path(fig_id):\n",
+ " return os.path.join(FIGURE_ID, fig_id)\n",
+ "\n",
+ "def data_path(dat_id):\n",
+ " return os.path.join(DATA_ID, dat_id)\n",
+ "\n",
+ "def save_fig(fig_id):\n",
+ " plt.savefig(image_path(fig_id) + \".png\", format='png')\n",
+ "\n",
+ "infile = open(data_path(\"EoS.csv\"),'r')\n",
+ "\n",
+ "# Read the EoS data as csv file and organize the data into two arrays with density and energies\n",
+ "EoS = pd.read_csv(infile, names=('Density', 'Energy'))\n",
+ "EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce')\n",
+ "EoS = EoS.dropna()\n",
+ "Energies = EoS['Energy']\n",
+ "Density = EoS['Density']\n",
+ "# The design matrix now as function of various polytrops\n",
+ "X = np.zeros((len(Density),4))\n",
+ "X[:,3] = Density**(4.0/3.0)\n",
+ "X[:,2] = Density\n",
+ "X[:,1] = Density**(2.0/3.0)\n",
+ "X[:,0] = 1\n",
+ "\n",
+ "# We use now Scikit-Learn's linear regressor and ridge regressor\n",
+ "# OLS part\n",
+ "clf = skl.LinearRegression().fit(X, Energies)\n",
+ "ytilde = clf.predict(X)\n",
+ "EoS['Eols'] = ytilde\n",
+ "# The mean squared error \n",
+ "print(\"Mean squared error: %.2f\" % mean_squared_error(Energies, ytilde))\n",
+ "# Explained variance score: 1 is perfect prediction \n",
+ "print('Variance score: %.2f' % r2_score(Energies, ytilde))\n",
+ "# Mean absolute error \n",
+ "print('Mean absolute error: %.2f' % mean_absolute_error(Energies, ytilde))\n",
+ "print(clf.coef_, clf.intercept_)\n",
+ "\n",
+ "# The Ridge regression with a hyperparameter lambda = 0.1\n",
+ "_lambda = 0.1\n",
+ "clf_ridge = skl.Ridge(alpha=_lambda).fit(X, Energies)\n",
+ "yridge = clf_ridge.predict(X)\n",
+ "EoS['Eridge'] = yridge\n",
+ "# The mean squared error \n",
+ "print(\"Mean squared error: %.2f\" % mean_squared_error(Energies, yridge))\n",
+ "# Explained variance score: 1 is perfect prediction \n",
+ "print('Variance score: %.2f' % r2_score(Energies, yridge))\n",
+ "# Mean absolute error \n",
+ "print('Mean absolute error: %.2f' % mean_absolute_error(Energies, yridge))\n",
+ "print(clf_ridge.coef_, clf_ridge.intercept_)\n",
+ "\n",
+ "fig, ax = plt.subplots()\n",
+ "ax.set_xlabel(r'$\\rho[\\mathrm{fm}^{-3}]$')\n",
+ "ax.set_ylabel(r'Energy per particle')\n",
+ "ax.plot(EoS['Density'], EoS['Energy'], alpha=0.7, lw=2,\n",
+ " label='Theoretical data')\n",
+ "ax.plot(EoS['Density'], EoS['Eols'], alpha=0.7, lw=2, c='m',\n",
+ " label='OLS')\n",
+ "ax.plot(EoS['Density'], EoS['Eridge'], alpha=0.7, lw=2, c='g',\n",
+ " label='Ridge $\\lambda = 0.1$')\n",
+ "ax.legend()\n",
+ "save_fig(\"EoSfitting\")\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The above simple polynomial in density $\\rho$ gives an excellent fit\n",
+ "to the data. \n",
+ "\n",
+ "We note also that there is a small deviation between the\n",
+ "standard OLS and the Ridge regression at higher densities. We discuss this in more detail\n",
+ "below.\n",
+ "\n",
+ "\n",
+ "## Splitting our Data in Training and Test data\n",
+ "\n",
+ "It is normal in essentially all Machine Learning studies to split the\n",
+ "data in a training set and a test set (sometimes also an additional\n",
+ "validation set). **Scikit-Learn** has an own function for this. There\n",
+ "is no explicit recipe for how much data should be included as training\n",
+ "data and say test data. An accepted rule of thumb is to use\n",
+ "approximately $2/3$ to $4/5$ of the data as training data. We will\n",
+ "postpone a discussion of this splitting to the end of these notes and\n",
+ "our discussion of the so-called **bias-variance** tradeoff. Here we\n",
+ "limit ourselves to repeat the above equation of state fitting example\n",
+ "but now splitting the data into a training set and a test set."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "import os\n",
+ "import numpy as np\n",
+ "import pandas as pd\n",
+ "import matplotlib.pyplot as plt\n",
+ "from sklearn.model_selection import train_test_split\n",
+ "# Where to save the figures and data files\n",
+ "PROJECT_ROOT_DIR = \"Results\"\n",
+ "FIGURE_ID = \"Results/FigureFiles\"\n",
+ "DATA_ID = \"DataFiles/\"\n",
+ "\n",
+ "if not os.path.exists(PROJECT_ROOT_DIR):\n",
+ " os.mkdir(PROJECT_ROOT_DIR)\n",
+ "\n",
+ "if not os.path.exists(FIGURE_ID):\n",
+ " os.makedirs(FIGURE_ID)\n",
+ "\n",
+ "if not os.path.exists(DATA_ID):\n",
+ " os.makedirs(DATA_ID)\n",
+ "\n",
+ "def image_path(fig_id):\n",
+ " return os.path.join(FIGURE_ID, fig_id)\n",
+ "\n",
+ "def data_path(dat_id):\n",
+ " return os.path.join(DATA_ID, dat_id)\n",
+ "\n",
+ "def save_fig(fig_id):\n",
+ " plt.savefig(image_path(fig_id) + \".png\", format='png')\n",
+ "\n",
+ "def R2(y_data, y_model):\n",
+ " return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)\n",
+ "def MSE(y_data,y_model):\n",
+ " n = np.size(y_model)\n",
+ " return np.sum((y_data-y_model)**2)/n\n",
+ "\n",
+ "infile = open(data_path(\"EoS.csv\"),'r')\n",
+ "\n",
+ "# Read the EoS data as csv file and organized into two arrays with density and energies\n",
+ "EoS = pd.read_csv(infile, names=('Density', 'Energy'))\n",
+ "EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce')\n",
+ "EoS = EoS.dropna()\n",
+ "Energies = EoS['Energy']\n",
+ "Density = EoS['Density']\n",
+ "# The design matrix now as function of various polytrops\n",
+ "X = np.zeros((len(Density),5))\n",
+ "X[:,0] = 1\n",
+ "X[:,1] = Density**(2.0/3.0)\n",
+ "X[:,2] = Density\n",
+ "X[:,3] = Density**(4.0/3.0)\n",
+ "X[:,4] = Density**(5.0/3.0)\n",
+ "# We split the data in test and training data\n",
+ "X_train, X_test, y_train, y_test = train_test_split(X, Energies, test_size=0.2)\n",
+ "# matrix inversion to find beta\n",
+ "beta = np.linalg.inv(X_train.T.dot(X_train)).dot(X_train.T).dot(y_train)\n",
+ "# and then make the prediction\n",
+ "ytilde = X_train @ beta\n",
+ "print(\"Training R2\")\n",
+ "print(R2(y_train,ytilde))\n",
+ "print(\"Training MSE\")\n",
+ "print(MSE(y_train,ytilde))\n",
+ "ypredict = X_test @ beta\n",
+ "print(\"Test R2\")\n",
+ "print(R2(y_test,ypredict))\n",
+ "print(\"Test MSE\")\n",
+ "print(MSE(y_test,ypredict))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "## The Boston housing data example\n",
+ "\n",
+ "The Boston housing \n",
+ "data set was originally a part of UCI Machine Learning Repository\n",
+ "and has been removed now. The data set is now included in **Scikit-Learn**'s \n",
+ "library. There are 506 samples and 13 feature (predictor) variables\n",
+ "in this data set. The objective is to predict the value of prices of\n",
+ "the house using the features (predictors) listed here.\n",
+ "\n",
+ "The features/predictors are\n",
+ "1. CRIM: Per capita crime rate by town\n",
+ "\n",
+ "2. ZN: Proportion of residential land zoned for lots over 25000 square feet\n",
+ "\n",
+ "3. INDUS: Proportion of non-retail business acres per town\n",
+ "\n",
+ "4. CHAS: Charles River dummy variable (= 1 if tract bounds river; 0 otherwise)\n",
+ "\n",
+ "5. NOX: Nitric oxide concentration (parts per 10 million)\n",
+ "\n",
+ "6. RM: Average number of rooms per dwelling\n",
+ "\n",
+ "7. AGE: Proportion of owner-occupied units built prior to 1940\n",
+ "\n",
+ "8. DIS: Weighted distances to five Boston employment centers\n",
+ "\n",
+ "9. RAD: Index of accessibility to radial highways\n",
+ "\n",
+ "10. TAX: Full-value property tax rate per USD10000\n",
+ "\n",
+ "11. B: $1000(Bk - 0.63)^2$, where $Bk$ is the proportion of [people of African American descent] by town\n",
+ "\n",
+ "12. LSTAT: Percentage of lower status of the population\n",
+ "\n",
+ "13. MEDV: Median value of owner-occupied homes in USD 1000s\n",
+ "\n",
+ "## Housing data, the code\n",
+ "We start by importing the libraries"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "import numpy as np\n",
+ "import matplotlib.pyplot as plt \n",
+ "\n",
+ "import pandas as pd \n",
+ "import seaborn as sns"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and load the Boston Housing DataSet from **Scikit-Learn**"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "from sklearn.datasets import load_boston\n",
+ "\n",
+ "boston_dataset = load_boston()\n",
+ "\n",
+ "# boston_dataset is a dictionary\n",
+ "# let's check what it contains\n",
+ "boston_dataset.keys()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Then we invoke Pandas"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "boston = pd.DataFrame(boston_dataset.data, columns=boston_dataset.feature_names)\n",
+ "boston.head()\n",
+ "boston['MEDV'] = boston_dataset.target"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and preprocess the data"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# check for missing values in all the columns\n",
+ "boston.isnull().sum()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We can then visualize the data"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# set the size of the figure\n",
+ "sns.set(rc={'figure.figsize':(11.7,8.27)})\n",
+ "\n",
+ "# plot a histogram showing the distribution of the target values\n",
+ "sns.distplot(boston['MEDV'], bins=30)\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "It is now useful to look at the correlation matrix"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# compute the pair wise correlation for all columns \n",
+ "correlation_matrix = boston.corr().round(2)\n",
+ "# use the heatmap function from seaborn to plot the correlation matrix\n",
+ "# annot = True to print the values inside the square\n",
+ "sns.heatmap(data=correlation_matrix, annot=True)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "From the above coorelation plot we can see that **MEDV** is strongly correlated to **LSTAT** and **RM**. We see also that **RAD** and **TAX** are stronly correlated, but we don't include this in our features together to avoid multi-colinearity"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "plt.figure(figsize=(20, 5))\n",
+ "\n",
+ "features = ['LSTAT', 'RM']\n",
+ "target = boston['MEDV']\n",
+ "\n",
+ "for i, col in enumerate(features):\n",
+ " plt.subplot(1, len(features) , i+1)\n",
+ " x = boston[col]\n",
+ " y = target\n",
+ " plt.scatter(x, y, marker='o')\n",
+ " plt.title(col)\n",
+ " plt.xlabel(col)\n",
+ " plt.ylabel('MEDV')"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Now we start training our model"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 18,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "X = pd.DataFrame(np.c_[boston['LSTAT'], boston['RM']], columns = ['LSTAT','RM'])\n",
+ "Y = boston['MEDV']"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We split the data into training and test sets"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 19,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "from sklearn.model_selection import train_test_split\n",
+ "\n",
+ "# splits the training and test data set in 80% : 20%\n",
+ "# assign random_state to any value.This ensures consistency.\n",
+ "X_train, X_test, Y_train, Y_test = train_test_split(X, Y, test_size = 0.2, random_state=5)\n",
+ "print(X_train.shape)\n",
+ "print(X_test.shape)\n",
+ "print(Y_train.shape)\n",
+ "print(Y_test.shape)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Then we use the linear regression functionality from **Scikit-Learn**"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 20,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "from sklearn.linear_model import LinearRegression\n",
+ "from sklearn.metrics import mean_squared_error, r2_score\n",
+ "\n",
+ "lin_model = LinearRegression()\n",
+ "lin_model.fit(X_train, Y_train)\n",
+ "\n",
+ "# model evaluation for training set\n",
+ "\n",
+ "y_train_predict = lin_model.predict(X_train)\n",
+ "rmse = (np.sqrt(mean_squared_error(Y_train, y_train_predict)))\n",
+ "r2 = r2_score(Y_train, y_train_predict)\n",
+ "\n",
+ "print(\"The model performance for training set\")\n",
+ "print(\"--------------------------------------\")\n",
+ "print('RMSE is {}'.format(rmse))\n",
+ "print('R2 score is {}'.format(r2))\n",
+ "print(\"\\n\")\n",
+ "\n",
+ "# model evaluation for testing set\n",
+ "\n",
+ "y_test_predict = lin_model.predict(X_test)\n",
+ "# root mean square error of the model\n",
+ "rmse = (np.sqrt(mean_squared_error(Y_test, y_test_predict)))\n",
+ "\n",
+ "# r-squared score of the model\n",
+ "r2 = r2_score(Y_test, y_test_predict)\n",
+ "\n",
+ "print(\"The model performance for testing set\")\n",
+ "print(\"--------------------------------------\")\n",
+ "print('RMSE is {}'.format(rmse))\n",
+ "print('R2 score is {}'.format(r2))"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 21,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# plotting the y_test vs y_pred\n",
+ "# ideally should have been a straight line\n",
+ "plt.scatter(Y_test, y_test_predict)\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Reducing the number of degrees of freedom, overarching view\n",
+ "\n",
+ "Many Machine Learning problems involve thousands or even millions of\n",
+ "features for each training instance. Not only does this make training\n",
+ "extremely slow, it can also make it much harder to find a good\n",
+ "solution, as we will see. This problem is often referred to as the\n",
+ "curse of dimensionality. Fortunately, in real-world problems, it is\n",
+ "often possible to reduce the number of features considerably, turning\n",
+ "an intractable problem into a tractable one.\n",
+ "\n",
+ "Later we will discuss some of the most popular dimensionality reduction\n",
+ "techniques: the principal component analysis (PCA), Kernel PCA, and\n",
+ "Locally Linear Embedding (LLE). \n",
+ "\n",
+ "\n",
+ "Principal component analysis and its various variants deal with the\n",
+ "problem of fitting a low-dimensional [affine\n",
+ "subspace](https://en.wikipedia.org/wiki/Affine_space) to a set of of\n",
+ "data points in a high-dimensional space. With its family of methods it\n",
+ "is one of the most used tools in data modeling, compression and\n",
+ "visualization.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Preprocessing our data\n",
+ "\n",
+ "Before we proceed however, we will discuss how to preprocess our\n",
+ "data. Till now and in connection with our previous examples we have\n",
+ "not met so many cases where we are too sensitive to the scaling of our\n",
+ "data. Normally the data may need a rescaling and/or may be sensitive\n",
+ "to extreme values. Scaling the data renders our inputs much more\n",
+ "suitable for the algorithms we want to employ.\n",
+ "\n",
+ "**Scikit-Learn** has several functions which allow us to rescale the\n",
+ "data, normally resulting in much better results in terms of various\n",
+ "accuracy scores. The **StandardScaler** function in **Scikit-Learn**\n",
+ "ensures that for each feature/predictor we study the mean value is\n",
+ "zero and the variance is one (every column in the design/feature\n",
+ "matrix). This scaling has the drawback that it does not ensure that\n",
+ "we have a particular maximum or minimum in our data set. Another\n",
+ "function included in **Scikit-Learn** is the **MinMaxScaler** which\n",
+ "ensures that all features are exactly between $0$ and $1$. The\n",
+ "\n",
+ "## More preprocessing\n",
+ "\n",
+ "\n",
+ "The **Normalizer** scales each data\n",
+ "point such that the feature vector has a euclidean length of one. In other words, it\n",
+ "projects a data point on the circle (or sphere in the case of higher dimensions) with a\n",
+ "radius of 1. This means every data point is scaled by a different number (by the\n",
+ "inverse of it’s length).\n",
+ "This normalization is often used when only the direction (or angle) of the data matters,\n",
+ "not the length of the feature vector.\n",
+ "\n",
+ "The **RobustScaler** works similarly to the StandardScaler in that it\n",
+ "ensures statistical properties for each feature that guarantee that\n",
+ "they are on the same scale. However, the RobustScaler uses the median\n",
+ "and quartiles, instead of mean and variance. This makes the\n",
+ "RobustScaler ignore data points that are very different from the rest\n",
+ "(like measurement errors). These odd data points are also called\n",
+ "outliers, and might often lead to trouble for other scaling\n",
+ "techniques.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Simple preprocessing examples, Franke function and regression"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 22,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# Common imports\n",
+ "import os\n",
+ "import numpy as np\n",
+ "import pandas as pd\n",
+ "import matplotlib.pyplot as plt\n",
+ "import sklearn.linear_model as skl\n",
+ "from sklearn.metrics import mean_squared_error\n",
+ "from sklearn.model_selection import train_test_split\n",
+ "from sklearn.preprocessing import MinMaxScaler, StandardScaler, Normalizer\n",
+ "\n",
+ "# Where to save the figures and data files\n",
+ "PROJECT_ROOT_DIR = \"Results\"\n",
+ "FIGURE_ID = \"Results/FigureFiles\"\n",
+ "DATA_ID = \"DataFiles/\"\n",
+ "\n",
+ "if not os.path.exists(PROJECT_ROOT_DIR):\n",
+ " os.mkdir(PROJECT_ROOT_DIR)\n",
+ "\n",
+ "if not os.path.exists(FIGURE_ID):\n",
+ " os.makedirs(FIGURE_ID)\n",
+ "\n",
+ "if not os.path.exists(DATA_ID):\n",
+ " os.makedirs(DATA_ID)\n",
+ "\n",
+ "def image_path(fig_id):\n",
+ " return os.path.join(FIGURE_ID, fig_id)\n",
+ "\n",
+ "def data_path(dat_id):\n",
+ " return os.path.join(DATA_ID, dat_id)\n",
+ "\n",
+ "def save_fig(fig_id):\n",
+ " plt.savefig(image_path(fig_id) + \".png\", format='png')\n",
+ "\n",
+ "\n",
+ "def FrankeFunction(x,y):\n",
+ "\tterm1 = 0.75*np.exp(-(0.25*(9*x-2)**2) - 0.25*((9*y-2)**2))\n",
+ "\tterm2 = 0.75*np.exp(-((9*x+1)**2)/49.0 - 0.1*(9*y+1))\n",
+ "\tterm3 = 0.5*np.exp(-(9*x-7)**2/4.0 - 0.25*((9*y-3)**2))\n",
+ "\tterm4 = -0.2*np.exp(-(9*x-4)**2 - (9*y-7)**2)\n",
+ "\treturn term1 + term2 + term3 + term4\n",
+ "\n",
+ "\n",
+ "def create_X(x, y, n ):\n",
+ "\tif len(x.shape) > 1:\n",
+ "\t\tx = np.ravel(x)\n",
+ "\t\ty = np.ravel(y)\n",
+ "\n",
+ "\tN = len(x)\n",
+ "\tl = int((n+1)*(n+2)/2)\t\t# Number of elements in beta\n",
+ "\tX = np.ones((N,l))\n",
+ "\n",
+ "\tfor i in range(1,n+1):\n",
+ "\t\tq = int((i)*(i+1)/2)\n",
+ "\t\tfor k in range(i+1):\n",
+ "\t\t\tX[:,q+k] = (x**(i-k))*(y**k)\n",
+ "\n",
+ "\treturn X\n",
+ "\n",
+ "\n",
+ "# Making meshgrid of datapoints and compute Franke's function\n",
+ "n = 5\n",
+ "N = 1000\n",
+ "x = np.sort(np.random.uniform(0, 1, N))\n",
+ "y = np.sort(np.random.uniform(0, 1, N))\n",
+ "z = FrankeFunction(x, y)\n",
+ "X = create_X(x, y, n=n) \n",
+ "# split in training and test data\n",
+ "X_train, X_test, y_train, y_test = train_test_split(X,z,test_size=0.2)\n",
+ "\n",
+ "\n",
+ "clf = skl.LinearRegression().fit(X_train, y_train)\n",
+ "\n",
+ "# The mean squared error and R2 score\n",
+ "print(\"MSE before scaling: {:.2f}\".format(mean_squared_error(clf.predict(X_test), y_test)))\n",
+ "print(\"R2 score before scaling {:.2f}\".format(clf.score(X_test,y_test)))\n",
+ "\n",
+ "scaler = StandardScaler()\n",
+ "scaler.fit(X_train)\n",
+ "X_train_scaled = scaler.transform(X_train)\n",
+ "X_test_scaled = scaler.transform(X_test)\n",
+ "\n",
+ "print(\"Feature min values before scaling:\\n {}\".format(X_train.min(axis=0)))\n",
+ "print(\"Feature max values before scaling:\\n {}\".format(X_train.max(axis=0)))\n",
+ "\n",
+ "print(\"Feature min values after scaling:\\n {}\".format(X_train_scaled.min(axis=0)))\n",
+ "print(\"Feature max values after scaling:\\n {}\".format(X_train_scaled.max(axis=0)))\n",
+ "\n",
+ "clf = skl.LinearRegression().fit(X_train_scaled, y_train)\n",
+ "\n",
+ "\n",
+ "print(\"MSE after scaling: {:.2f}\".format(mean_squared_error(clf.predict(X_test_scaled), y_test)))\n",
+ "print(\"R2 score for scaled data: {:.2f}\".format(clf.score(X_test_scaled,y_test)))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## The singular value decomposition\n",
+ "\n",
+ "\n",
+ "The examples we have looked at so far are cases where we normally can\n",
+ "invert the matrix $\\boldsymbol{X}^T\\boldsymbol{X}$. Using a polynomial expansion as we\n",
+ "did both for the masses and the fitting of the equation of state,\n",
+ "leads to row vectors of the design matrix which are essentially\n",
+ "orthogonal due to the polynomial character of our model. Obtaining the inverse of the design matrix is then often done via a so-called LU, QR or Cholesky decomposition. \n",
+ "\n",
+ "\n",
+ "\n",
+ "This may\n",
+ "however not the be case in general and a standard matrix inversion\n",
+ "algorithm based on say LU, QR or Cholesky decomposition may lead to singularities. We will see examples of this below.\n",
+ "\n",
+ "There is however a way to partially circumvent this problem and also gain some insights about the ordinary least squares approach, and later shrinkage methods like Ridge and Lasso regressions. \n",
+ "\n",
+ "This is given by the **Singular Value Decomposition** algorithm, perhaps\n",
+ "the most powerful linear algebra algorithm. Let us look at a\n",
+ "different example where we may have problems with the standard matrix\n",
+ "inversion algorithm. Thereafter we dive into the math of the SVD.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Linear Regression Problems\n",
+ "\n",
+ "One of the typical problems we encounter with linear regression, in particular \n",
+ "when the matrix $\\boldsymbol{X}$ (our so-called design matrix) is high-dimensional, \n",
+ "are problems with near singular or singular matrices. The column vectors of $\\boldsymbol{X}$ \n",
+ "may be linearly dependent, normally referred to as super-collinearity. \n",
+ "This means that the matrix may be rank deficient and it is basically impossible to \n",
+ "to model the data using linear regression. As an example, consider the matrix"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\begin{align*}\n",
+ "\\mathbf{X} & = \\left[\n",
+ "\\begin{array}{rrr}\n",
+ "1 & -1 & 2\n",
+ "\\\\\n",
+ "1 & 0 & 1\n",
+ "\\\\\n",
+ "1 & 2 & -1\n",
+ "\\\\\n",
+ "1 & 1 & 0\n",
+ "\\end{array} \\right]\n",
+ "\\end{align*}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The columns of $\\boldsymbol{X}$ are linearly dependent. We see this easily since the \n",
+ "the first column is the row-wise sum of the other two columns. The rank (more correct,\n",
+ "the column rank) of a matrix is the dimension of the space spanned by the\n",
+ "column vectors. Hence, the rank of $\\mathbf{X}$ is equal to the number\n",
+ "of linearly independent columns. In this particular case the matrix has rank 2.\n",
+ "\n",
+ "Super-collinearity of an $(n \\times p)$-dimensional design matrix $\\mathbf{X}$ implies\n",
+ "that the inverse of the matrix $\\boldsymbol{X}^T\\boldsymbol{X}$ (the matrix we need to invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\begin{align*}\n",
+ "\\boldsymbol{X} & = \\left[\n",
+ "\\begin{array}{rr}\n",
+ "1 & -1\n",
+ "\\\\\n",
+ "1 & -1\n",
+ "\\end{array} \\right].\n",
+ "\\end{align*}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We see easily that $\\mbox{det}(\\boldsymbol{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \\times (-1) - 1 \\times (-1) = 0$. Hence, $\\mathbf{X}$ is singular and its inverse is undefined.\n",
+ "This is equivalent to saying that the matrix $\\boldsymbol{X}$ has at least an eigenvalue which is zero.\n",
+ "\n",
+ "\n",
+ "## Fixing the singularity\n",
+ "\n",
+ "If our design matrix $\\boldsymbol{X}$ which enters the linear regression problem"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ "\\boldsymbol{\\beta} = (\\boldsymbol{X}^{T} \\boldsymbol{X})^{-1} \\boldsymbol{X}^{T} \\boldsymbol{y},\n",
+ "\\label{_auto1} \\tag{1}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "has linearly dependent column vectors, we will not be able to compute the inverse\n",
+ "of $\\boldsymbol{X}^T\\boldsymbol{X}$ and we cannot find the parameters (estimators) $\\beta_i$. \n",
+ "The estimators are only well-defined if $(\\boldsymbol{X}^{T}\\boldsymbol{X})^{-1}$ exits. \n",
+ "This is more likely to happen when the matrix $\\boldsymbol{X}$ is high-dimensional. In this case it is likely to encounter a situation where \n",
+ "the regression parameters $\\beta_i$ cannot be estimated.\n",
+ "\n",
+ "A cheap *ad hoc* approach is simply to add a small diagonal component to the matrix to invert, that is we change"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X}^{T} \\boldsymbol{X} \\rightarrow \\boldsymbol{X}^{T} \\boldsymbol{X}+\\lambda \\boldsymbol{I},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $\\boldsymbol{I}$ is the identity matrix. When we discuss **Ridge** regression this is actually what we end up evaluating. The parameter $\\lambda$ is called a hyperparameter. More about this later. \n",
+ "\n",
+ "\n",
+ "\n",
+ "## Basic math of the SVD\n",
+ "\n",
+ "\n",
+ "From standard linear algebra we know that a square matrix $\\boldsymbol{X}$ can be diagonalized if and only it is \n",
+ "a so-called [normal matrix](https://en.wikipedia.org/wiki/Normal_matrix), that is if $\\boldsymbol{X}\\in {\\mathbb{R}}^{n\\times n}$\n",
+ "we have $\\boldsymbol{X}\\boldsymbol{X}^T=\\boldsymbol{X}^T\\boldsymbol{X}$ or if $\\boldsymbol{X}\\in {\\mathbb{C}}^{n\\times n}$ we have $\\boldsymbol{X}\\boldsymbol{X}^{\\dagger}=\\boldsymbol{X}^{\\dagger}\\boldsymbol{X}$.\n",
+ "The matrix has then a set of eigenpairs"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "(\\lambda_1,\\boldsymbol{u}_1),\\dots, (\\lambda_n,\\boldsymbol{u}_n),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and the eigenvalues are given by the diagonal matrix"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{\\Sigma}=\\mathrm{Diag}(\\lambda_1, \\dots,\\lambda_n).\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The matrix $\\boldsymbol{X}$ can be written in terms of an orthogonal/unitary transformation $\\boldsymbol{U}$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X} = \\boldsymbol{U}\\boldsymbol{\\Sigma}\\boldsymbol{V}^T,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "with $\\boldsymbol{U}\\boldsymbol{U}^T=\\boldsymbol{I}$ or $\\boldsymbol{U}\\boldsymbol{U}^{\\dagger}=\\boldsymbol{I}$.\n",
+ "\n",
+ "Not all square matrices are diagonalizable. A matrix like the one discussed above"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X} = \\begin{bmatrix} \n",
+ "1& -1 \\\\\n",
+ "1& -1\\\\\n",
+ "\\end{bmatrix}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "is not diagonalizable, it is a so-called [defective matrix](https://en.wikipedia.org/wiki/Defective_matrix). It is easy to see that the condition\n",
+ "$\\boldsymbol{X}\\boldsymbol{X}^T=\\boldsymbol{X}^T\\boldsymbol{X}$ is not fulfilled. \n",
+ "\n",
+ "\n",
+ "## The SVD, a Fantastic Algorithm\n",
+ "\n",
+ "\n",
+ "However, and this is the strength of the SVD algorithm, any general\n",
+ "matrix $\\boldsymbol{X}$ can be decomposed in terms of a diagonal matrix and\n",
+ "two orthogonal/unitary matrices. The [Singular Value Decompostion\n",
+ "(SVD) theorem](https://en.wikipedia.org/wiki/Singular_value_decomposition)\n",
+ "states that a general $m\\times n$ matrix $\\boldsymbol{X}$ can be written in\n",
+ "terms of a diagonal matrix $\\boldsymbol{\\Sigma}$ of dimensionality $m\\times n$\n",
+ "and two orthognal matrices $\\boldsymbol{U}$ and $\\boldsymbol{V}$, where the first has\n",
+ "dimensionality $m \\times m$ and the last dimensionality $n\\times n$.\n",
+ "We have then"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X} = \\boldsymbol{U}\\boldsymbol{\\Sigma}\\boldsymbol{V}^T\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "As an example, the above defective matrix can be decomposed as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X} = \\frac{1}{\\sqrt{2}}\\begin{bmatrix} 1& 1 \\\\ 1& -1\\\\ \\end{bmatrix} \\begin{bmatrix} 2& 0 \\\\ 0& 0\\\\ \\end{bmatrix} \\frac{1}{\\sqrt{2}}\\begin{bmatrix} 1& -1 \\\\ 1& 1\\\\ \\end{bmatrix}=\\boldsymbol{U}\\boldsymbol{\\Sigma}\\boldsymbol{V}^T,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "with eigenvalues $\\sigma_1=2$ and $\\sigma_2=0$. \n",
+ "The SVD exits always! \n",
+ "\n",
+ "The SVD\n",
+ "decomposition (singular values) gives eigenvalues \n",
+ "$\\sigma_i\\geq\\sigma_{i+1}$ for all $i$ and for dimensions larger than $i=p$, the\n",
+ "eigenvalues (singular values) are zero.\n",
+ "\n",
+ "In the general case, where our design matrix $\\boldsymbol{X}$ has dimension\n",
+ "$n\\times p$, the matrix is thus decomposed into an $n\\times n$\n",
+ "orthogonal matrix $\\boldsymbol{U}$, a $p\\times p$ orthogonal matrix $\\boldsymbol{V}$\n",
+ "and a diagonal matrix $\\boldsymbol{\\Sigma}$ with $r=\\mathrm{min}(n,p)$\n",
+ "singular values $\\sigma_i\\geq 0$ on the main diagonal and zeros filling\n",
+ "the rest of the matrix. There are at most $p$ singular values\n",
+ "assuming that $n > p$. In our regression examples for the nuclear\n",
+ "masses and the equation of state this is indeed the case, while for\n",
+ "the Ising model we have $p > n$. These are often cases that lead to\n",
+ "near singular or singular matrices.\n",
+ "\n",
+ "The columns of $\\boldsymbol{U}$ are called the left singular vectors while the columns of $\\boldsymbol{V}$ are the right singular vectors.\n",
+ "\n",
+ "## Economy-size SVD\n",
+ "\n",
+ "If we assume that $n > p$, then our matrix $\\boldsymbol{U}$ has dimension $n\n",
+ "\\times n$. The last $n-p$ columns of $\\boldsymbol{U}$ become however\n",
+ "irrelevant in our calculations since they are multiplied with the\n",
+ "zeros in $\\boldsymbol{\\Sigma}$.\n",
+ "\n",
+ "The economy-size decomposition removes extra rows or columns of zeros\n",
+ "from the diagonal matrix of singular values, $\\boldsymbol{\\Sigma}$, along with the columns\n",
+ "in either $\\boldsymbol{U}$ or $\\boldsymbol{V}$ that multiply those zeros in the expression. \n",
+ "Removing these zeros and columns can improve execution time\n",
+ "and reduce storage requirements without compromising the accuracy of\n",
+ "the decomposition.\n",
+ "\n",
+ "If $n > p$, we keep only the first $p$ columns of $\\boldsymbol{U}$ and $\\boldsymbol{\\Sigma}$ has dimension $p\\times p$. \n",
+ "If $p > n$, then only the first $n$ columns of $\\boldsymbol{V}$ are computed and $\\boldsymbol{\\Sigma}$ has dimension $n\\times n$.\n",
+ "The $n=p$ case is obvious, we retain the full SVD. \n",
+ "In general the economy-size SVD leads to less FLOPS and still conserving the desired accuracy.\n",
+ "\n",
+ "## Codes for the SVD"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 23,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "import numpy as np\n",
+ "# SVD inversion\n",
+ "def SVDinv(A):\n",
+ " ''' Takes as input a numpy matrix A and returns inv(A) based on singular value decomposition (SVD).\n",
+ " SVD is numerically more stable than the inversion algorithms provided by\n",
+ " numpy and scipy.linalg at the cost of being slower.\n",
+ " '''\n",
+ " U, s, VT = np.linalg.svd(A)\n",
+ "# print('test U')\n",
+ "# print( (np.transpose(U) @ U - U @np.transpose(U)))\n",
+ "# print('test VT')\n",
+ "# print( (np.transpose(VT) @ VT - VT @np.transpose(VT)))\n",
+ " print(U)\n",
+ " print(s)\n",
+ " print(VT)\n",
+ "\n",
+ " D = np.zeros((len(U),len(VT)))\n",
+ " for i in range(0,len(VT)):\n",
+ " D[i,i]=s[i]\n",
+ " UT = np.transpose(U); V = np.transpose(VT); invD = np.linalg.inv(D)\n",
+ " return np.matmul(V,np.matmul(invD,UT))\n",
+ "\n",
+ "\n",
+ "X = np.array([ [1.0, -1.0, 2.0], [1.0, 0.0, 1.0], [1.0, 2.0, -1.0], [1.0, 1.0, 0.0] ])\n",
+ "print(X)\n",
+ "A = np.transpose(X) @ X\n",
+ "print(A)\n",
+ "# Brute force inversion of super-collinear matrix\n",
+ "#B = np.linalg.inv(A)\n",
+ "#print(B)\n",
+ "C = SVDinv(A)\n",
+ "print(C)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The matrix $\\boldsymbol{X}$ has columns that are linearly dependent. The first\n",
+ "column is the row-wise sum of the other two columns. The rank of a\n",
+ "matrix (the column rank) is the dimension of space spanned by the\n",
+ "column vectors. The rank of the matrix is the number of linearly\n",
+ "independent columns, in this case just $2$. We see this from the\n",
+ "singular values when running the above code. Running the standard\n",
+ "inversion algorithm for matrix inversion with $\\boldsymbol{X}^T\\boldsymbol{X}$ results\n",
+ "in the program terminating due to a singular matrix.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Mathematical Properties\n",
+ "\n",
+ "There are several interesting mathematical properties which will be\n",
+ "relevant when we are going to discuss the differences between say\n",
+ "ordinary least squares (OLS) and **Ridge** regression.\n",
+ "\n",
+ "We have from OLS that the parameters of the linear approximation are given by"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{\\tilde{y}} = \\boldsymbol{X}\\boldsymbol{\\beta} = \\boldsymbol{X}\\left(\\boldsymbol{X}^T\\boldsymbol{X}\\right)^{-1}\\boldsymbol{X}^T\\boldsymbol{y}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The matrix to invert can be rewritten in terms of our SVD decomposition as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X}^T\\boldsymbol{X} = \\boldsymbol{V}\\boldsymbol{\\Sigma}^T\\boldsymbol{U}^T\\boldsymbol{U}\\boldsymbol{\\Sigma}\\boldsymbol{V}^T.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Using the orthogonality properties of $\\boldsymbol{U}$ we have"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X}^T\\boldsymbol{X} = \\boldsymbol{V}\\boldsymbol{\\Sigma}^T\\boldsymbol{\\Sigma}\\boldsymbol{V}^T = \\boldsymbol{V}\\boldsymbol{D}\\boldsymbol{V}^T,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "with $\\boldsymbol{D}$ being a diagonal matrix with values along the diagonal given by the singular values squared. \n",
+ "\n",
+ "This means that"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "(\\boldsymbol{X}^T\\boldsymbol{X})\\boldsymbol{V} = \\boldsymbol{V}\\boldsymbol{D},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "that is the eigenvectors of $(\\boldsymbol{X}^T\\boldsymbol{X})$ are given by the columns of the right singular matrix of $\\boldsymbol{X}$ and the eigenvalues are the squared singular values. It is easy to show (show this) that"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "(\\boldsymbol{X}\\boldsymbol{X}^T)\\boldsymbol{U} = \\boldsymbol{U}\\boldsymbol{D},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "that is, the eigenvectors of $(\\boldsymbol{X}\\boldsymbol{X})^T$ are the columns of the left singular matrix and the eigenvalues are the same. \n",
+ "\n",
+ "Going back to our OLS equation we have"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X}\\boldsymbol{\\beta} = \\boldsymbol{X}\\left(\\boldsymbol{V}\\boldsymbol{D}\\boldsymbol{V}^T \\right)^{-1}\\boldsymbol{X}^T\\boldsymbol{y}=\\boldsymbol{U\\Sigma V^T}\\left(\\boldsymbol{V}\\boldsymbol{D}\\boldsymbol{V}^T \\right)^{-1}(\\boldsymbol{U\\Sigma V^T})^T\\boldsymbol{y}=\\boldsymbol{U}\\boldsymbol{U}^T\\boldsymbol{y}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We will come back to this expression when we discuss Ridge regression. \n",
+ "\n",
+ "\n",
+ "$$ \\tilde{y}^{OLS}=\\boldsymbol{X}\\hat{\\beta}^{OLS}=\\sum_{j=1}^p \\boldsymbol{u}_j\\boldsymbol{u}_j^T\\boldsymbol{y}$$ and for Ridge we have \n",
+ "\n",
+ "$$ \\tilde{y}^{Ridge}=\\boldsymbol{X}\\hat{\\beta}^{Ridge}=\\sum_{j=1}^p \\boldsymbol{u}_j\\frac{\\sigma_j^2}{\\sigma_j^2+\\lambda}\\boldsymbol{u}_j^T\\boldsymbol{y}$$ . \n",
+ "\n",
+ "It is indeed the economy-sized SVD, note the summation runs up tp $$p$$ only and not $$n$$. \n",
+ "\n",
+ "Here we have that $$\\boldsymbol{X} = \\boldsymbol{U}\\boldsymbol{\\Sigma}\\boldsymbol{V}^T$$, with $$\\Sigma$$ being an $$ n\\times p$$ matrix and $$\\boldsymbol{V}$$ being a $$ p\\times p$$ matrix. We also have assumed here that $$ n > p$$. \n",
+ "\n",
+ "\n",
+ "## Ridge and LASSO Regression\n",
+ "\n",
+ "Let us remind ourselves about the expression for the standard Mean Squared Error (MSE) which we used to define our cost function and the equations for the ordinary least squares (OLS) method, that is \n",
+ "our optimization problem is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "{\\displaystyle \\min_{\\boldsymbol{\\beta}\\in {\\mathbb{R}}^{p}}}\\frac{1}{n}\\left\\{\\left(\\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta}\\right)^T\\left(\\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta}\\right)\\right\\}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "or we can state it as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "{\\displaystyle \\min_{\\boldsymbol{\\beta}\\in\n",
+ "{\\mathbb{R}}^{p}}}\\frac{1}{n}\\sum_{i=0}^{n-1}\\left(y_i-\\tilde{y}_i\\right)^2=\\frac{1}{n}\\vert\\vert \\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta}\\vert\\vert_2^2,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where we have used the definition of a norm-2 vector, that is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\vert\\vert \\boldsymbol{x}\\vert\\vert_2 = \\sqrt{\\sum_i x_i^2}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "By minimizing the above equation with respect to the parameters\n",
+ "$\\boldsymbol{\\beta}$ we could then obtain an analytical expression for the\n",
+ "parameters $\\boldsymbol{\\beta}$. We can add a regularization parameter $\\lambda$ by\n",
+ "defining a new cost function to be optimized, that is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "{\\displaystyle \\min_{\\boldsymbol{\\beta}\\in\n",
+ "{\\mathbb{R}}^{p}}}\\frac{1}{n}\\vert\\vert \\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta}\\vert\\vert_2^2+\\lambda\\vert\\vert \\boldsymbol{\\beta}\\vert\\vert_2^2\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which leads to the Ridge regression minimization problem where we\n",
+ "require that $\\vert\\vert \\boldsymbol{\\beta}\\vert\\vert_2^2\\le t$, where $t$ is\n",
+ "a finite number larger than zero. By defining"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "C(\\boldsymbol{X},\\boldsymbol{\\beta})=\\frac{1}{n}\\vert\\vert \\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta}\\vert\\vert_2^2+\\lambda\\vert\\vert \\boldsymbol{\\beta}\\vert\\vert_1,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "we have a new optimization equation"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "{\\displaystyle \\min_{\\boldsymbol{\\beta}\\in\n",
+ "{\\mathbb{R}}^{p}}}\\frac{1}{n}\\vert\\vert \\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta}\\vert\\vert_2^2+\\lambda\\vert\\vert \\boldsymbol{\\beta}\\vert\\vert_1\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which leads to Lasso regression. Lasso stands for least absolute shrinkage and selection operator. \n",
+ "\n",
+ "Here we have defined the norm-1 as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\vert\\vert \\boldsymbol{x}\\vert\\vert_1 = \\sum_i \\vert x_i\\vert.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## More on Ridge Regression\n",
+ "\n",
+ "Using the matrix-vector expression for Ridge regression,"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "C(\\boldsymbol{X},\\boldsymbol{\\beta})=\\frac{1}{n}\\left\\{(\\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta})^T(\\boldsymbol{y}-\\boldsymbol{X}\\boldsymbol{\\beta})\\right\\}+\\lambda\\boldsymbol{\\beta}^T\\boldsymbol{\\beta},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "by taking the derivatives with respect to $\\boldsymbol{\\beta}$ we obtain then\n",
+ "a slightly modified matrix inversion problem which for finite values\n",
+ "of $\\lambda$ does not suffer from singularity problems. We obtain"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{\\beta}^{\\mathrm{Ridge}} = \\left(\\boldsymbol{X}^T\\boldsymbol{X}+\\lambda\\boldsymbol{I}\\right)^{-1}\\boldsymbol{X}^T\\boldsymbol{y},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "with $\\boldsymbol{I}$ being a $p\\times p$ identity matrix with the constraint that"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\sum_{i=0}^{p-1} \\beta_i^2 \\leq t,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "with $t$ a finite positive number. \n",
+ "\n",
+ "We see that Ridge regression is nothing but the standard\n",
+ "OLS with a modified diagonal term added to $\\boldsymbol{X}^T\\boldsymbol{X}$. The\n",
+ "consequences, in particular for our discussion of the bias-variance tradeoff \n",
+ "are rather interesting.\n",
+ "\n",
+ "Furthermore, if we use the result above in terms of the SVD decomposition (our analysis was done for the OLS method), we had"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "(\\boldsymbol{X}\\boldsymbol{X}^T)\\boldsymbol{U} = \\boldsymbol{U}\\boldsymbol{D}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We can analyse the OLS solutions in terms of the eigenvectors (the columns) of the right singular value matrix $\\boldsymbol{U}$ as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X}\\boldsymbol{\\beta} = \\boldsymbol{X}\\left(\\boldsymbol{V}\\boldsymbol{D}\\boldsymbol{V}^T \\right)^{-1}\\boldsymbol{X}^T\\boldsymbol{y}=\\boldsymbol{U\\Sigma V^T}\\left(\\boldsymbol{V}\\boldsymbol{D}\\boldsymbol{V}^T \\right)^{-1}(\\boldsymbol{U\\Sigma V^T})^T\\boldsymbol{y}=\\boldsymbol{U}\\boldsymbol{U}^T\\boldsymbol{y}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "For Ridge regression this becomes"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X}\\boldsymbol{\\beta}^{\\mathrm{Ridge}} = \\boldsymbol{U\\Sigma V^T}\\left(\\boldsymbol{V}\\boldsymbol{D}\\boldsymbol{V}^T+\\lambda\\boldsymbol{I} \\right)^{-1}(\\boldsymbol{U\\Sigma V^T})^T\\boldsymbol{y}=\\sum_{j=0}^{p-1}\\boldsymbol{u}_j\\boldsymbol{u}_j^T\\frac{\\sigma_j^2}{\\sigma_j^2+\\lambda}\\boldsymbol{y},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "with the vectors $\\boldsymbol{u}_j$ being the columns of $\\boldsymbol{U}$. \n",
+ "\n",
+ "## Interpreting the Ridge results\n",
+ "\n",
+ "Since $\\lambda \\geq 0$, it means that compared to OLS, we have"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\sigma_j^2}{\\sigma_j^2+\\lambda} \\leq 1.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Ridge regression finds the coordinates of $\\boldsymbol{y}$ with respect to the\n",
+ "orthonormal basis $\\boldsymbol{U}$, it then shrinks the coordinates by\n",
+ "$\\frac{\\sigma_j^2}{\\sigma_j^2+\\lambda}$. Recall that the SVD has\n",
+ "eigenvalues ordered in a descending way, that is $\\sigma_i \\geq\n",
+ "\\sigma_{i+1}$.\n",
+ "\n",
+ "For small eigenvalues $\\sigma_i$ it means that their contributions become less important, a fact which can be used to reduce the number of degrees of freedom.\n",
+ "Actually, calculating the variance of $\\boldsymbol{X}\\boldsymbol{v}_j$ shows that this quantity is equal to $\\sigma_j^2/n$.\n",
+ "With a parameter $\\lambda$ we can thus shrink the role of specific parameters. \n",
+ "\n",
+ "\n",
+ "## More interpretations\n",
+ "\n",
+ "For the sake of simplicity, let us assume that the design matrix is orthonormal, that is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X}^T\\boldsymbol{X}=(\\boldsymbol{X}^T\\boldsymbol{X})^{-1} =\\boldsymbol{I}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "In this case the standard OLS results in"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{\\beta}^{\\mathrm{OLS}} = \\boldsymbol{X}^T\\boldsymbol{y}=\\sum_{i=0}^{p-1}\\boldsymbol{u}_j\\boldsymbol{u}_j^T\\boldsymbol{y},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{\\beta}^{\\mathrm{Ridge}} = \\left(\\boldsymbol{I}+\\lambda\\boldsymbol{I}\\right)^{-1}\\boldsymbol{X}^T\\boldsymbol{y}=\\left(1+\\lambda\\right)^{-1}\\boldsymbol{\\beta}^{\\mathrm{OLS}},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "that is the Ridge estimator scales the OLS estimator by the inverse of a factor $1+\\lambda$, and\n",
+ "the Ridge estimator converges to zero when the hyperparameter goes to\n",
+ "infinity.\n",
+ "\n",
+ "We will come back to more interpreations after we have gone through some of the statistical analysis part. \n",
+ "\n",
+ "For more discussions of Ridge and Lasso regression, [Wessel van Wieringen's](https://arxiv.org/abs/1509.09169) article is highly recommended.\n",
+ "Similarly, [Mehta et al's article](https://arxiv.org/abs/1803.08823) is also recommended.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## A better understanding of regularization\n",
+ "\n",
+ "The parameter $\\lambda$ that we have introduced in the Ridge (and\n",
+ "Lasso as well) regression is often called a regularization parameter\n",
+ "or shrinkage parameter. It is common to call it a hyperparameter. What does it mean mathemtically?\n",
+ "\n",
+ "Here we will first look at how to analyze the difference between the\n",
+ "standard OLS equations and the Ridge expressions in terms of a linear\n",
+ "algebra analysis using the SVD algorithm. Thereafter, we will link\n",
+ "(see the material on the bias-variance tradeoff below) these\n",
+ "observation to the statisical analysis of the results. In particular\n",
+ "we consider how the variance of the parameters $\\boldsymbol{\\beta}$ is\n",
+ "affected by changing the parameter $\\lambda$.\n",
+ "\n",
+ "## Decomposing the OLS and Ridge expressions\n",
+ "\n",
+ "We have our design matrix\n",
+ " $\\boldsymbol{X}\\in {\\mathbb{R}}^{n\\times p}$. With the SVD we decompose it as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X} = \\boldsymbol{U\\Sigma V^T},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "with $\\boldsymbol{U}\\in {\\mathbb{R}}^{n\\times n}$, $\\boldsymbol{\\Sigma}\\in {\\mathbb{R}}^{n\\times p}$\n",
+ "and $\\boldsymbol{V}\\in {\\mathbb{R}}^{p\\times p}$.\n",
+ "\n",
+ "The matrices $\\boldsymbol{U}$ and $\\boldsymbol{V}$ are unitary/orthonormal matrices, that is in case the matrices are real we have $\\boldsymbol{U}^T\\boldsymbol{U}=\\boldsymbol{U}\\boldsymbol{U}^T=\\boldsymbol{I}$ and $\\boldsymbol{V}^T\\boldsymbol{V}=\\boldsymbol{V}\\boldsymbol{V}^T=\\boldsymbol{I}$.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Introducing the Covariance and Correlation functions\n",
+ "\n",
+ "Before we discuss the link between for example Ridge regression and the singular value decomposition, we need to remind ourselves about\n",
+ "the definition of the covariance and the correlation function. These are quantities \n",
+ "\n",
+ "Suppose we have defined two vectors\n",
+ "$\\hat{x}$ and $\\hat{y}$ with $n$ elements each. The covariance matrix $\\boldsymbol{C}$ is defined as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{C}[\\boldsymbol{x},\\boldsymbol{y}] = \\begin{bmatrix} \\mathrm{cov}[\\boldsymbol{x},\\boldsymbol{x}] & \\mathrm{cov}[\\boldsymbol{x},\\boldsymbol{y}] \\\\\n",
+ " \\mathrm{cov}[\\boldsymbol{y},\\boldsymbol{x}] & \\mathrm{cov}[\\boldsymbol{y},\\boldsymbol{y}] \\\\\n",
+ " \\end{bmatrix},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where for example"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathrm{cov}[\\boldsymbol{x},\\boldsymbol{y}] =\\frac{1}{n} \\sum_{i=0}^{n-1}(x_i- \\overline{x})(y_i- \\overline{y}).\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "With this definition and recalling that the variance is defined as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathrm{var}[\\boldsymbol{x}]=\\frac{1}{n} \\sum_{i=0}^{n-1}(x_i- \\overline{x})^2,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "we can rewrite the covariance matrix as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{C}[\\boldsymbol{x},\\boldsymbol{y}] = \\begin{bmatrix} \\mathrm{var}[\\boldsymbol{x}] & \\mathrm{cov}[\\boldsymbol{x},\\boldsymbol{y}] \\\\\n",
+ " \\mathrm{cov}[\\boldsymbol{x},\\boldsymbol{y}] & \\mathrm{var}[\\boldsymbol{y}] \\\\\n",
+ " \\end{bmatrix}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The covariance takes values between zero and infinity and may thus\n",
+ "lead to problems with loss of numerical precision for particularly\n",
+ "large values. It is common to scale the covariance matrix by\n",
+ "introducing instead the correlation matrix defined via the so-called\n",
+ "correlation function"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathrm{corr}[\\boldsymbol{x},\\boldsymbol{y}]=\\frac{\\mathrm{cov}[\\boldsymbol{x},\\boldsymbol{y}]}{\\sqrt{\\mathrm{var}[\\boldsymbol{x}] \\mathrm{var}[\\boldsymbol{y}]}}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The correlation function is then given by values $\\mathrm{corr}[\\boldsymbol{x},\\boldsymbol{y}]\n",
+ "\\in [-1,1]$. This avoids eventual problems with too large values. We\n",
+ "can then define the correlation matrix for the two vectors $\\boldsymbol{x}$\n",
+ "and $\\boldsymbol{y}$ as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{K}[\\boldsymbol{x},\\boldsymbol{y}] = \\begin{bmatrix} 1 & \\mathrm{corr}[\\boldsymbol{x},\\boldsymbol{y}] \\\\\n",
+ " \\mathrm{corr}[\\boldsymbol{y},\\boldsymbol{x}] & 1 \\\\\n",
+ " \\end{bmatrix},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "In the above example this is the function we constructed using **pandas**.\n",
+ "\n",
+ "## Correlation Function and Design/Feature Matrix\n",
+ "\n",
+ "In our derivation of the various regression algorithms like **Ordinary Least Squares** or **Ridge regression**\n",
+ "we defined the design/feature matrix $\\boldsymbol{X}$ as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X}=\\begin{bmatrix}\n",
+ "x_{0,0} & x_{0,1} & x_{0,2}& \\dots & \\dots x_{0,p-1}\\\\\n",
+ "x_{1,0} & x_{1,1} & x_{1,2}& \\dots & \\dots x_{1,p-1}\\\\\n",
+ "x_{2,0} & x_{2,1} & x_{2,2}& \\dots & \\dots x_{2,p-1}\\\\\n",
+ "\\dots & \\dots & \\dots & \\dots \\dots & \\dots \\\\\n",
+ "x_{n-2,0} & x_{n-2,1} & x_{n-2,2}& \\dots & \\dots x_{n-2,p-1}\\\\\n",
+ "x_{n-1,0} & x_{n-1,1} & x_{n-1,2}& \\dots & \\dots x_{n-1,p-1}\\\\\n",
+ "\\end{bmatrix},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "with $\\boldsymbol{X}\\in {\\mathbb{R}}^{n\\times p}$, with the predictors/features $p$ refering to the column numbers and the\n",
+ "entries $n$ being the row elements.\n",
+ "We can rewrite the design/feature matrix in terms of its column vectors as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X}=\\begin{bmatrix} \\boldsymbol{x}_0 & \\boldsymbol{x}_1 & \\boldsymbol{x}_2 & \\dots & \\dots & \\boldsymbol{x}_{p-1}\\end{bmatrix},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "with a given vector"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{x}_i^T = \\begin{bmatrix}x_{0,i} & x_{1,i} & x_{2,i}& \\dots & \\dots x_{n-1,i}\\end{bmatrix}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "With these definitions, we can now rewrite our $2\\times 2$\n",
+ "correaltion/covariance matrix in terms of a moe general design/feature\n",
+ "matrix $\\boldsymbol{X}\\in {\\mathbb{R}}^{n\\times p}$. This leads to a $p\\times p$\n",
+ "covariance matrix for the vectors $\\boldsymbol{x}_i$ with $i=0,1,\\dots,p-1$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{C}[\\boldsymbol{x}] = \\begin{bmatrix}\n",
+ "\\mathrm{var}[\\boldsymbol{x}_0] & \\mathrm{cov}[\\boldsymbol{x}_0,\\boldsymbol{x}_1] & \\mathrm{cov}[\\boldsymbol{x}_0,\\boldsymbol{x}_2] & \\dots & \\dots & \\mathrm{cov}[\\boldsymbol{x}_0,\\boldsymbol{x}_{p-1}]\\\\\n",
+ "\\mathrm{cov}[\\boldsymbol{x}_1,\\boldsymbol{x}_0] & \\mathrm{var}[\\boldsymbol{x}_1] & \\mathrm{cov}[\\boldsymbol{x}_1,\\boldsymbol{x}_2] & \\dots & \\dots & \\mathrm{cov}[\\boldsymbol{x}_1,\\boldsymbol{x}_{p-1}]\\\\\n",
+ "\\mathrm{cov}[\\boldsymbol{x}_2,\\boldsymbol{x}_0] & \\mathrm{cov}[\\boldsymbol{x}_2,\\boldsymbol{x}_1] & \\mathrm{var}[\\boldsymbol{x}_2] & \\dots & \\dots & \\mathrm{cov}[\\boldsymbol{x}_2,\\boldsymbol{x}_{p-1}]\\\\\n",
+ "\\dots & \\dots & \\dots & \\dots & \\dots & \\dots \\\\\n",
+ "\\dots & \\dots & \\dots & \\dots & \\dots & \\dots \\\\\n",
+ "\\mathrm{cov}[\\boldsymbol{x}_{p-1},\\boldsymbol{x}_0] & \\mathrm{cov}[\\boldsymbol{x}_{p-1},\\boldsymbol{x}_1] & \\mathrm{cov}[\\boldsymbol{x}_{p-1},\\boldsymbol{x}_{2}] & \\dots & \\dots & \\mathrm{var}[\\boldsymbol{x}_{p-1}]\\\\\n",
+ "\\end{bmatrix},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and the correlation matrix"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{K}[\\boldsymbol{x}] = \\begin{bmatrix}\n",
+ "1 & \\mathrm{corr}[\\boldsymbol{x}_0,\\boldsymbol{x}_1] & \\mathrm{corr}[\\boldsymbol{x}_0,\\boldsymbol{x}_2] & \\dots & \\dots & \\mathrm{corr}[\\boldsymbol{x}_0,\\boldsymbol{x}_{p-1}]\\\\\n",
+ "\\mathrm{corr}[\\boldsymbol{x}_1,\\boldsymbol{x}_0] & 1 & \\mathrm{corr}[\\boldsymbol{x}_1,\\boldsymbol{x}_2] & \\dots & \\dots & \\mathrm{corr}[\\boldsymbol{x}_1,\\boldsymbol{x}_{p-1}]\\\\\n",
+ "\\mathrm{corr}[\\boldsymbol{x}_2,\\boldsymbol{x}_0] & \\mathrm{corr}[\\boldsymbol{x}_2,\\boldsymbol{x}_1] & 1 & \\dots & \\dots & \\mathrm{corr}[\\boldsymbol{x}_2,\\boldsymbol{x}_{p-1}]\\\\\n",
+ "\\dots & \\dots & \\dots & \\dots & \\dots & \\dots \\\\\n",
+ "\\dots & \\dots & \\dots & \\dots & \\dots & \\dots \\\\\n",
+ "\\mathrm{corr}[\\boldsymbol{x}_{p-1},\\boldsymbol{x}_0] & \\mathrm{corr}[\\boldsymbol{x}_{p-1},\\boldsymbol{x}_1] & \\mathrm{corr}[\\boldsymbol{x}_{p-1},\\boldsymbol{x}_{2}] & \\dots & \\dots & 1\\\\\n",
+ "\\end{bmatrix},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Covariance Matrix Examples\n",
+ "\n",
+ "\n",
+ "The Numpy function **np.cov** calculates the covariance elements using\n",
+ "the factor $1/(n-1)$ instead of $1/n$ since it assumes we do not have\n",
+ "the exact mean values. The following simple function uses the\n",
+ "**np.vstack** function which takes each vector of dimension $1\\times n$\n",
+ "and produces a $2\\times n$ matrix $\\boldsymbol{W}$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{W} = \\begin{bmatrix} x_0 & y_0 \\\\\n",
+ " x_1 & y_1 \\\\\n",
+ " x_2 & y_2\\\\\n",
+ " \\dots & \\dots \\\\\n",
+ " x_{n-2} & y_{n-2}\\\\\n",
+ " x_{n-1} & y_{n-1} & \n",
+ " \\end{bmatrix},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which in turn is converted into into the $2\\times 2$ covariance matrix\n",
+ "$\\boldsymbol{C}$ via the Numpy function **np.cov()**. We note that we can also calculate\n",
+ "the mean value of each set of samples $\\boldsymbol{x}$ etc using the Numpy\n",
+ "function **np.mean(x)**. We can also extract the eigenvalues of the\n",
+ "covariance matrix through the **np.linalg.eig()** function."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 24,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# Importing various packages\n",
+ "import numpy as np\n",
+ "n = 100\n",
+ "x = np.random.normal(size=n)\n",
+ "print(np.mean(x))\n",
+ "y = 4+3*x+np.random.normal(size=n)\n",
+ "print(np.mean(y))\n",
+ "W = np.vstack((x, y))\n",
+ "C = np.cov(W)\n",
+ "print(C)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Correlation Matrix\n",
+ "\n",
+ "The previous example can be converted into the correlation matrix by\n",
+ "simply scaling the matrix elements with the variances. We should also\n",
+ "subtract the mean values for each column. This leads to the following\n",
+ "code which sets up the correlations matrix for the previous example in\n",
+ "a more brute force way. Here we scale the mean values for each column of the design matrix, calculate the relevant mean values and variances and then finally set up the $2\\times 2$ correlation matrix (since we have only two vectors)."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 25,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "import numpy as np\n",
+ "n = 100\n",
+ "# define two vectors \n",
+ "x = np.random.random(size=n)\n",
+ "y = 4+3*x+np.random.normal(size=n)\n",
+ "#scaling the x and y vectors \n",
+ "x = x - np.mean(x)\n",
+ "y = y - np.mean(y)\n",
+ "variance_x = np.sum(x@x)/n\n",
+ "variance_y = np.sum(y@y)/n\n",
+ "print(variance_x)\n",
+ "print(variance_y)\n",
+ "cov_xy = np.sum(x@y)/n\n",
+ "cov_xx = np.sum(x@x)/n\n",
+ "cov_yy = np.sum(y@y)/n\n",
+ "C = np.zeros((2,2))\n",
+ "C[0,0]= cov_xx/variance_x\n",
+ "C[1,1]= cov_yy/variance_y\n",
+ "C[0,1]= cov_xy/np.sqrt(variance_y*variance_x)\n",
+ "C[1,0]= C[0,1]\n",
+ "print(C)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We see that the matrix elements along the diagonal are one as they\n",
+ "should be and that the matrix is symmetric. Furthermore, diagonalizing\n",
+ "this matrix we easily see that it is a positive definite matrix.\n",
+ "\n",
+ "The above procedure with **numpy** can be made more compact if we use **pandas**.\n",
+ "\n",
+ "## Correlation Matrix with Pandas\n",
+ "\n",
+ "We whow here how we can set up the correlation matrix using **pandas**, as done in this simple code"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 26,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "import numpy as np\n",
+ "import pandas as pd\n",
+ "n = 10\n",
+ "x = np.random.normal(size=n)\n",
+ "x = x - np.mean(x)\n",
+ "y = 4+3*x+np.random.normal(size=n)\n",
+ "y = y - np.mean(y)\n",
+ "X = (np.vstack((x, y))).T\n",
+ "print(X)\n",
+ "Xpd = pd.DataFrame(X)\n",
+ "print(Xpd)\n",
+ "correlation_matrix = Xpd.corr()\n",
+ "print(correlation_matrix)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We expand this model to the Franke function discussed above.\n",
+ "\n",
+ "## Correlation Matrix with Pandas and the Franke function"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 27,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# Common imports\n",
+ "import numpy as np\n",
+ "import pandas as pd\n",
+ "\n",
+ "\n",
+ "def FrankeFunction(x,y):\n",
+ "\tterm1 = 0.75*np.exp(-(0.25*(9*x-2)**2) - 0.25*((9*y-2)**2))\n",
+ "\tterm2 = 0.75*np.exp(-((9*x+1)**2)/49.0 - 0.1*(9*y+1))\n",
+ "\tterm3 = 0.5*np.exp(-(9*x-7)**2/4.0 - 0.25*((9*y-3)**2))\n",
+ "\tterm4 = -0.2*np.exp(-(9*x-4)**2 - (9*y-7)**2)\n",
+ "\treturn term1 + term2 + term3 + term4\n",
+ "\n",
+ "\n",
+ "def create_X(x, y, n ):\n",
+ "\tif len(x.shape) > 1:\n",
+ "\t\tx = np.ravel(x)\n",
+ "\t\ty = np.ravel(y)\n",
+ "\n",
+ "\tN = len(x)\n",
+ "\tl = int((n+1)*(n+2)/2)\t\t# Number of elements in beta\n",
+ "\tX = np.ones((N,l))\n",
+ "\n",
+ "\tfor i in range(1,n+1):\n",
+ "\t\tq = int((i)*(i+1)/2)\n",
+ "\t\tfor k in range(i+1):\n",
+ "\t\t\tX[:,q+k] = (x**(i-k))*(y**k)\n",
+ "\n",
+ "\treturn X\n",
+ "\n",
+ "\n",
+ "# Making meshgrid of datapoints and compute Franke's function\n",
+ "n = 4\n",
+ "N = 100\n",
+ "x = np.sort(np.random.uniform(0, 1, N))\n",
+ "y = np.sort(np.random.uniform(0, 1, N))\n",
+ "z = FrankeFunction(x, y)\n",
+ "X = create_X(x, y, n=n) \n",
+ "\n",
+ "Xpd = pd.DataFrame(X)\n",
+ "# subtract the mean values and set up the covariance matrix\n",
+ "Xpd = Xpd - Xpd.mean()\n",
+ "covariance_matrix = Xpd.cov()\n",
+ "print(covariance_matrix)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We note here that the covariance is zero for the first rows and\n",
+ "columns since all matrix elements in the design matrix were set to one\n",
+ "(we are fitting the function in terms of a polynomial of degree $n$).\n",
+ "\n",
+ "This means that the variance for these elements will be zero and will\n",
+ "cause problems when we set up the correlation matrix. We can simply\n",
+ "drop these elements and construct a correlation\n",
+ "matrix without these elements. \n",
+ "\n",
+ "\n",
+ "## Rewriting the Covariance and/or Correlation Matrix\n",
+ "\n",
+ "We can rewrite the covariance matrix in a more compact form in terms of the design/feature matrix $\\boldsymbol{X}$ as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{C}[\\boldsymbol{x}] = \\frac{1}{n}\\boldsymbol{X}^T\\boldsymbol{X}= \\mathbb{E}[\\boldsymbol{X}^T\\boldsymbol{X}].\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "To see this let us simply look at a design matrix $\\boldsymbol{X}\\in {\\mathbb{R}}^{2\\times 2}$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X}=\\begin{bmatrix}\n",
+ "x_{00} & x_{01}\\\\\n",
+ "x_{10} & x_{11}\\\\\n",
+ "\\end{bmatrix}=\\begin{bmatrix}\n",
+ "\\boldsymbol{x}_{0} & \\boldsymbol{x}_{1}\\\\\n",
+ "\\end{bmatrix}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "If we then compute the expectation value"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathbb{E}[\\boldsymbol{X}^T\\boldsymbol{X}] = \\frac{1}{n}\\boldsymbol{X}^T\\boldsymbol{X}=\\begin{bmatrix}\n",
+ "x_{00}^2+x_{01}^2 & x_{00}x_{10}+x_{01}x_{11}\\\\\n",
+ "x_{10}x_{00}+x_{11}x_{01} & x_{10}^2+x_{11}^2\\\\\n",
+ "\\end{bmatrix},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which is just"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{C}[\\boldsymbol{x}_0,\\boldsymbol{x}_1] = \\boldsymbol{C}[\\boldsymbol{x}]=\\begin{bmatrix} \\mathrm{var}[\\boldsymbol{x}_0] & \\mathrm{cov}[\\boldsymbol{x}_0,\\boldsymbol{x}_1] \\\\\n",
+ " \\mathrm{cov}[\\boldsymbol{x}_1,\\boldsymbol{x}_0] & \\mathrm{var}[\\boldsymbol{x}_1] \\\\\n",
+ " \\end{bmatrix},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where we wrote $$\\boldsymbol{C}[\\boldsymbol{x}_0,\\boldsymbol{x}_1] = \\boldsymbol{C}[\\boldsymbol{x}]$$ to indicate that this the covariance of the vectors $\\boldsymbol{x}$ of the design/feature matrix $\\boldsymbol{X}$.\n",
+ "\n",
+ "It is easy to generalize this to a matrix $\\boldsymbol{X}\\in {\\mathbb{R}}^{n\\times p}$.\n",
+ "\n",
+ "\n",
+ "## Linking with SVD\n",
+ "\n",
+ "See lecture september 11. More text to be added here soon.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Where are we going?\n",
+ "\n",
+ "Before we proceed, we need to rethink what we have been doing. In our\n",
+ "eager to fit the data, we have omitted several important elements in\n",
+ "our regression analysis. In what follows we will\n",
+ "1. look at statistical properties, including a discussion of mean values, variance and the so-called bias-variance tradeoff\n",
+ "\n",
+ "2. introduce resampling techniques like cross-validation, bootstrapping and jackknife and more\n",
+ "\n",
+ "This will allow us to link the standard linear algebra methods we have discussed above to a statistical interpretation of the methods. \n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Resampling methods\n",
+ "Resampling methods are an indispensable tool in modern\n",
+ "statistics. They involve repeatedly drawing samples from a training\n",
+ "set and refitting a model of interest on each sample in order to\n",
+ "obtain additional information about the fitted model. For example, in\n",
+ "order to estimate the variability of a linear regression fit, we can\n",
+ "repeatedly draw different samples from the training data, fit a linear\n",
+ "regression to each new sample, and then examine the extent to which\n",
+ "the resulting fits differ. Such an approach may allow us to obtain\n",
+ "information that would not be available from fitting the model only\n",
+ "once using the original training sample.\n",
+ "\n",
+ "Two resampling methods are often used in Machine Learning analyses,\n",
+ "1. The **bootstrap method**\n",
+ "\n",
+ "2. and **Cross-Validation**\n",
+ "\n",
+ "In addition there are several other methods such as the Jackknife and the Blocking methods. We will discuss in particular\n",
+ "cross-validation and the bootstrap method.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Resampling approaches can be computationally expensive\n",
+ "\n",
+ "Resampling approaches can be computationally expensive, because they\n",
+ "involve fitting the same statistical method multiple times using\n",
+ "different subsets of the training data. However, due to recent\n",
+ "advances in computing power, the computational requirements of\n",
+ "resampling methods generally are not prohibitive. In this chapter, we\n",
+ "discuss two of the most commonly used resampling methods,\n",
+ "cross-validation and the bootstrap. Both methods are important tools\n",
+ "in the practical application of many statistical learning\n",
+ "procedures. For example, cross-validation can be used to estimate the\n",
+ "test error associated with a given statistical learning method in\n",
+ "order to evaluate its performance, or to select the appropriate level\n",
+ "of flexibility. The process of evaluating a model’s performance is\n",
+ "known as model assessment, whereas the process of selecting the proper\n",
+ "level of flexibility for a model is known as model selection. The\n",
+ "bootstrap is widely used.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Why resampling methods ?\n",
+ "**Statistical analysis.**\n",
+ "\n",
+ "\n",
+ "* Our simulations can be treated as *computer experiments*. This is particularly the case for Monte Carlo methods\n",
+ "\n",
+ "* The results can be analysed with the same statistical tools as we would use analysing experimental data.\n",
+ "\n",
+ "* As in all experiments, we are looking for expectation values and an estimate of how accurate they are, i.e., possible sources for errors.\n",
+ "\n",
+ " \n",
+ "\n",
+ "## Statistical analysis\n",
+ "\n",
+ "* As in other experiments, many numerical experiments have two classes of errors:\n",
+ "\n",
+ " * Statistical errors\n",
+ "\n",
+ " * Systematical errors\n",
+ "\n",
+ "\n",
+ "* Statistical errors can be estimated using standard tools from statistics\n",
+ "\n",
+ "* Systematical errors are method specific and must be treated differently from case to case.\n",
+ "\n",
+ " \n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Linking the regression analysis with a statistical interpretation\n",
+ "\n",
+ "\n",
+ "The\n",
+ "advantage of doing linear regression is that we actually end up with\n",
+ "analytical expressions for several statistical quantities. \n",
+ "Standard least squares and Ridge regression allow us to\n",
+ "derive quantities like the variance and other expectation values in a\n",
+ "rather straightforward way.\n",
+ "\n",
+ "\n",
+ "It is assumed that $\\varepsilon_i\n",
+ "\\sim \\mathcal{N}(0, \\sigma^2)$ and the $\\varepsilon_{i}$ are\n",
+ "independent, i.e.:"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\begin{align*} \n",
+ "\\mbox{Cov}(\\varepsilon_{i_1},\n",
+ "\\varepsilon_{i_2}) & = \\left\\{ \\begin{array}{lcc} \\sigma^2 & \\mbox{if}\n",
+ "& i_1 = i_2, \\\\ 0 & \\mbox{if} & i_1 \\not= i_2. \\end{array} \\right.\n",
+ "\\end{align*}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The randomness of $\\varepsilon_i$ implies that\n",
+ "$\\mathbf{y}_i$ is also a random variable. In particular,\n",
+ "$\\mathbf{y}_i$ is normally distributed, because $\\varepsilon_i \\sim\n",
+ "\\mathcal{N}(0, \\sigma^2)$ and $\\mathbf{X}_{i,\\ast} \\, \\boldsymbol{\\beta}$ is a\n",
+ "non-random scalar. To specify the parameters of the distribution of\n",
+ "$\\mathbf{y}_i$ we need to calculate its first two moments. \n",
+ "\n",
+ "Recall that $\\boldsymbol{X}$ is a matrix of dimensionality $n\\times p$. The\n",
+ "notation above $\\mathbf{X}_{i,\\ast}$ means that we are looking at the\n",
+ "row number $i$ and perform a sum over all values $p$.\n",
+ "\n",
+ "\n",
+ "## Assumptions made\n",
+ "\n",
+ "The assumption we have made here can be summarized as (and this is going to be useful when we discuss the bias-variance trade off)\n",
+ "that there exists a function $f(\\boldsymbol{x})$ and a normal distributed error $\\boldsymbol{\\varepsilon}\\sim \\mathcal{N}(0, \\sigma^2)$\n",
+ "which describe our data"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{y} = f(\\boldsymbol{x})+\\boldsymbol{\\varepsilon}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We approximate this function with our model from the solution of the linear regression equations, that is our\n",
+ "function $f$ is approximated by $\\boldsymbol{\\tilde{y}}$ where we want to minimize $(\\boldsymbol{y}-\\boldsymbol{\\tilde{y}})^2$, our MSE, with"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{\\tilde{y}} = \\boldsymbol{X}\\boldsymbol{\\beta}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Expectation value and variance\n",
+ "\n",
+ "We can calculate the expectation value of $\\boldsymbol{y}$ for a given element $i$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\begin{align*} \n",
+ "\\mathbb{E}(y_i) & =\n",
+ "\\mathbb{E}(\\mathbf{X}_{i, \\ast} \\, \\boldsymbol{\\beta}) + \\mathbb{E}(\\varepsilon_i)\n",
+ "\\, \\, \\, = \\, \\, \\, \\mathbf{X}_{i, \\ast} \\, \\beta, \n",
+ "\\end{align*}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "while\n",
+ "its variance is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\begin{align*} \\mbox{Var}(y_i) & = \\mathbb{E} \\{ [y_i\n",
+ "- \\mathbb{E}(y_i)]^2 \\} \\, \\, \\, = \\, \\, \\, \\mathbb{E} ( y_i^2 ) -\n",
+ "[\\mathbb{E}(y_i)]^2 \\\\ & = \\mathbb{E} [ ( \\mathbf{X}_{i, \\ast} \\,\n",
+ "\\beta + \\varepsilon_i )^2] - ( \\mathbf{X}_{i, \\ast} \\, \\boldsymbol{\\beta})^2 \\\\ &\n",
+ "= \\mathbb{E} [ ( \\mathbf{X}_{i, \\ast} \\, \\boldsymbol{\\beta})^2 + 2 \\varepsilon_i\n",
+ "\\mathbf{X}_{i, \\ast} \\, \\boldsymbol{\\beta} + \\varepsilon_i^2 ] - ( \\mathbf{X}_{i,\n",
+ "\\ast} \\, \\beta)^2 \\\\ & = ( \\mathbf{X}_{i, \\ast} \\, \\boldsymbol{\\beta})^2 + 2\n",
+ "\\mathbb{E}(\\varepsilon_i) \\mathbf{X}_{i, \\ast} \\, \\boldsymbol{\\beta} +\n",
+ "\\mathbb{E}(\\varepsilon_i^2 ) - ( \\mathbf{X}_{i, \\ast} \\, \\boldsymbol{\\beta})^2 \n",
+ "\\\\ & = \\mathbb{E}(\\varepsilon_i^2 ) \\, \\, \\, = \\, \\, \\,\n",
+ "\\mbox{Var}(\\varepsilon_i) \\, \\, \\, = \\, \\, \\, \\sigma^2. \n",
+ "\\end{align*}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Hence, $y_i \\sim \\mathcal{N}( \\mathbf{X}_{i, \\ast} \\, \\boldsymbol{\\beta}, \\sigma^2)$, that is $\\boldsymbol{y}$ follows a normal distribution with \n",
+ "mean value $\\boldsymbol{X}\\boldsymbol{\\beta}$ and variance $\\sigma^2$ (not be confused with the singular values of the SVD). \n",
+ "\n",
+ "## Expectation value and variance for $\\boldsymbol{\\beta}$\n",
+ "\n",
+ "With the OLS expressions for the parameters $\\boldsymbol{\\beta}$ we can evaluate the expectation value"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathbb{E}(\\boldsymbol{\\beta}) = \\mathbb{E}[ (\\mathbf{X}^{\\top} \\mathbf{X})^{-1}\\mathbf{X}^{T} \\mathbf{Y}]=(\\mathbf{X}^{T} \\mathbf{X})^{-1}\\mathbf{X}^{T} \\mathbb{E}[ \\mathbf{Y}]=(\\mathbf{X}^{T} \\mathbf{X})^{-1} \\mathbf{X}^{T}\\mathbf{X}\\boldsymbol{\\beta}=\\boldsymbol{\\beta}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "This means that the estimator of the regression parameters is unbiased.\n",
+ "\n",
+ "We can also calculate the variance\n",
+ "\n",
+ "The variance of $\\boldsymbol{\\beta}$ is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\begin{eqnarray*}\n",
+ "\\mbox{Var}(\\boldsymbol{\\beta}) & = & \\mathbb{E} \\{ [\\boldsymbol{\\beta} - \\mathbb{E}(\\boldsymbol{\\beta})] [\\boldsymbol{\\beta} - \\mathbb{E}(\\boldsymbol{\\beta})]^{T} \\}\n",
+ "\\\\\n",
+ "& = & \\mathbb{E} \\{ [(\\mathbf{X}^{T} \\mathbf{X})^{-1} \\, \\mathbf{X}^{T} \\mathbf{Y} - \\boldsymbol{\\beta}] \\, [(\\mathbf{X}^{T} \\mathbf{X})^{-1} \\, \\mathbf{X}^{T} \\mathbf{Y} - \\boldsymbol{\\beta}]^{T} \\}\n",
+ "\\\\\n",
+ "% & = & \\mathbb{E} \\{ [(\\mathbf{X}^{T} \\mathbf{X})^{-1} \\, \\mathbf{X}^{T} \\mathbf{Y}] \\, [(\\mathbf{X}^{T} \\mathbf{X})^{-1} \\, \\mathbf{X}^{T} \\mathbf{Y}]^{T} \\} - \\boldsymbol{\\beta} \\, \\boldsymbol{\\beta}^{T}\n",
+ "% \\\\\n",
+ "% & = & \\mathbb{E} \\{ (\\mathbf{X}^{T} \\mathbf{X})^{-1} \\, \\mathbf{X}^{T} \\mathbf{Y} \\, \\mathbf{Y}^{T} \\, \\mathbf{X} \\, (\\mathbf{X}^{T} \\mathbf{X})^{-1} \\} - \\boldsymbol{\\beta} \\, \\boldsymbol{\\beta}^{T}\n",
+ "% \\\\\n",
+ "& = & (\\mathbf{X}^{T} \\mathbf{X})^{-1} \\, \\mathbf{X}^{T} \\, \\mathbb{E} \\{ \\mathbf{Y} \\, \\mathbf{Y}^{T} \\} \\, \\mathbf{X} \\, (\\mathbf{X}^{T} \\mathbf{X})^{-1} - \\boldsymbol{\\beta} \\, \\boldsymbol{\\beta}^{T}\n",
+ "\\\\\n",
+ "& = & (\\mathbf{X}^{T} \\mathbf{X})^{-1} \\, \\mathbf{X}^{T} \\, \\{ \\mathbf{X} \\, \\boldsymbol{\\beta} \\, \\boldsymbol{\\beta}^{T} \\, \\mathbf{X}^{T} + \\sigma^2 \\} \\, \\mathbf{X} \\, (\\mathbf{X}^{T} \\mathbf{X})^{-1} - \\boldsymbol{\\beta} \\, \\boldsymbol{\\beta}^{T}\n",
+ "% \\\\\n",
+ "% & = & (\\mathbf{X}^T \\mathbf{X})^{-1} \\, \\mathbf{X}^T \\, \\mathbf{X} \\, \\boldsymbol{\\beta} \\, \\boldsymbol{\\beta}^T \\, \\mathbf{X}^T \\, \\mathbf{X} \\, (\\mathbf{X}^T % \\mathbf{X})^{-1}\n",
+ "% \\\\\n",
+ "% & & + \\, \\, \\sigma^2 \\, (\\mathbf{X}^T \\mathbf{X})^{-1} \\, \\mathbf{X}^T \\, \\mathbf{X} \\, (\\mathbf{X}^T \\mathbf{X})^{-1} - \\boldsymbol{\\beta} \\boldsymbol{\\beta}^T\n",
+ "\\\\\n",
+ "& = & \\boldsymbol{\\beta} \\, \\boldsymbol{\\beta}^{T} + \\sigma^2 \\, (\\mathbf{X}^{T} \\mathbf{X})^{-1} - \\boldsymbol{\\beta} \\, \\boldsymbol{\\beta}^{T}\n",
+ "\\, \\, \\, = \\, \\, \\, \\sigma^2 \\, (\\mathbf{X}^{T} \\mathbf{X})^{-1},\n",
+ "\\end{eqnarray*}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where we have used that $\\mathbb{E} (\\mathbf{Y} \\mathbf{Y}^{T}) =\n",
+ "\\mathbf{X} \\, \\boldsymbol{\\beta} \\, \\boldsymbol{\\beta}^{T} \\, \\mathbf{X}^{T} +\n",
+ "\\sigma^2 \\, \\mathbf{I}_{nn}$. From $\\mbox{Var}(\\boldsymbol{\\beta}) = \\sigma^2\n",
+ "\\, (\\mathbf{X}^{T} \\mathbf{X})^{-1}$, one obtains an estimate of the\n",
+ "variance of the estimate of the $j$-th regression coefficient:\n",
+ "$\\boldsymbol{\\sigma}^2 (\\boldsymbol{\\beta}_j ) = \\boldsymbol{\\sigma}^2 \\sqrt{\n",
+ "[(\\mathbf{X}^{T} \\mathbf{X})^{-1}]_{jj} }$. This may be used to\n",
+ "construct a confidence interval for the estimates.\n",
+ "\n",
+ "\n",
+ "In a similar way, we can obtain analytical expressions for say the\n",
+ "expectation values of the parameters $\\boldsymbol{\\beta}$ and their variance\n",
+ "when we employ Ridge regression, allowing us again to define a confidence interval. \n",
+ "\n",
+ "It is rather straightforward to show that"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathbb{E} \\big[ \\boldsymbol{\\beta}^{\\mathrm{Ridge}} \\big]=(\\mathbf{X}^{T} \\mathbf{X} + \\lambda \\mathbf{I}_{pp})^{-1} (\\mathbf{X}^{\\top} \\mathbf{X})\\boldsymbol{\\beta}^{\\mathrm{OLS}}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We see clearly that \n",
+ "$\\mathbb{E} \\big[ \\boldsymbol{\\beta}^{\\mathrm{Ridge}} \\big] \\not= \\boldsymbol{\\beta}^{\\mathrm{OLS}}$ for any $\\lambda > 0$. We say then that the ridge estimator is biased.\n",
+ "\n",
+ "We can also compute the variance as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mbox{Var}[\\boldsymbol{\\beta}^{\\mathrm{Ridge}}]=\\sigma^2[ \\mathbf{X}^{T} \\mathbf{X} + \\lambda \\mathbf{I} ]^{-1} \\mathbf{X}^{T} \\mathbf{X} \\{ [ \\mathbf{X}^{\\top} \\mathbf{X} + \\lambda \\mathbf{I} ]^{-1}\\}^{T},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and it is easy to see that if the parameter $\\lambda$ goes to infinity then the variance of Ridge parameters $\\boldsymbol{\\beta}$ goes to zero. \n",
+ "\n",
+ "With this, we can compute the difference"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mbox{Var}[\\boldsymbol{\\beta}^{\\mathrm{OLS}}]-\\mbox{Var}(\\boldsymbol{\\beta}^{\\mathrm{Ridge}})=\\sigma^2 [ \\mathbf{X}^{T} \\mathbf{X} + \\lambda \\mathbf{I} ]^{-1}[ 2\\lambda\\mathbf{I} + \\lambda^2 (\\mathbf{X}^{T} \\mathbf{X})^{-1} ] \\{ [ \\mathbf{X}^{T} \\mathbf{X} + \\lambda \\mathbf{I} ]^{-1}\\}^{T}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The difference is non-negative definite since each component of the\n",
+ "matrix product is non-negative definite. \n",
+ "This means the variance we obtain with the standard OLS will always for $\\lambda > 0$ be larger than the variance of $\\boldsymbol{\\beta}$ obtained with the Ridge estimator. This has interesting consequences when we discuss the so-called bias-variance trade-off below. \n",
+ "\n",
+ "\n",
+ "## Resampling methods\n",
+ "\n",
+ "With all these analytical equations for both the OLS and Ridge\n",
+ "regression, we will now outline how to assess a given model. This will\n",
+ "lead us to a discussion of the so-called bias-variance tradeoff (see\n",
+ "below) and so-called resampling methods.\n",
+ "\n",
+ "One of the quantities we have discussed as a way to measure errors is\n",
+ "the mean-squared error (MSE), mainly used for fitting of continuous\n",
+ "functions. Another choice is the absolute error.\n",
+ "\n",
+ "In the discussions below we will focus on the MSE and in particular since we will split the data into test and training data,\n",
+ "we discuss the\n",
+ "1. prediction error or simply the **test error** $\\mathrm{Err_{Test}}$, where we have a fixed training set and the test error is the MSE arising from the data reserved for testing. We discuss also the \n",
+ "\n",
+ "2. training error $\\mathrm{Err_{Train}}$, which is the average loss over the training data.\n",
+ "\n",
+ "As our model becomes more and more complex, more of the training data tends to used. The training may thence adapt to more complicated structures in the data. This may lead to a decrease in the bias (see below for code example) and a slight increase of the variance for the test error.\n",
+ "For a certain level of complexity the test error will reach minimum, before starting to increase again. The\n",
+ "training error reaches a saturation.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Resampling methods: Jackknife and Bootstrap\n",
+ "\n",
+ "Two famous\n",
+ "resampling methods are the **independent bootstrap** and **the jackknife**. \n",
+ "\n",
+ "The jackknife is a special case of the independent bootstrap. Still, the jackknife was made\n",
+ "popular prior to the independent bootstrap. And as the popularity of\n",
+ "the independent bootstrap soared, new variants, such as **the dependent bootstrap**.\n",
+ "\n",
+ "The Jackknife and independent bootstrap work for\n",
+ "independent, identically distributed random variables.\n",
+ "If these conditions are not\n",
+ "satisfied, the methods will fail. Yet, it should be said that if the data are\n",
+ "independent, identically distributed, and we only want to estimate the\n",
+ "variance of $\\overline{X}$ (which often is the case), then there is no\n",
+ "need for bootstrapping. \n",
+ "\n",
+ "## Resampling methods: Jackknife\n",
+ "\n",
+ "The Jackknife works by making many replicas of the estimator $\\widehat{\\theta}$. \n",
+ "The jackknife is a resampling method where we systematically leave out one observation from the vector of observed values $\\boldsymbol{x} = (x_1,x_2,\\cdots,X_n)$. \n",
+ "Let $\\boldsymbol{x}_i$ denote the vector"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{x}_i = (x_1,x_2,\\cdots,x_{i-1},x_{i+1},\\cdots,x_n),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which equals the vector $\\boldsymbol{x}$ with the exception that observation\n",
+ "number $i$ is left out. Using this notation, define\n",
+ "$\\widehat{\\theta}_i$ to be the estimator\n",
+ "$\\widehat{\\theta}$ computed using $\\vec{X}_i$. \n",
+ "\n",
+ "\n",
+ "## Jackknife code example"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 28,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "from numpy import *\n",
+ "from numpy.random import randint, randn\n",
+ "from time import time\n",
+ "\n",
+ "def jackknife(data, stat):\n",
+ " n = len(data);t = zeros(n); inds = arange(n); t0 = time()\n",
+ " ## 'jackknifing' by leaving out an observation for each i \n",
+ " for i in range(n):\n",
+ " t[i] = stat(delete(data,i) )\n",
+ "\n",
+ " # analysis \n",
+ " print(\"Runtime: %g sec\" % (time()-t0)); print(\"Jackknife Statistics :\")\n",
+ " print(\"original bias std. error\")\n",
+ " print(\"%8g %14g %15g\" % (stat(data),(n-1)*mean(t)/n, (n*var(t))**.5))\n",
+ "\n",
+ " return t\n",
+ "\n",
+ "\n",
+ "# Returns mean of data samples \n",
+ "def stat(data):\n",
+ " return mean(data)\n",
+ "\n",
+ "\n",
+ "mu, sigma = 100, 15\n",
+ "datapoints = 10000\n",
+ "x = mu + sigma*random.randn(datapoints)\n",
+ "# jackknife returns the data sample \n",
+ "t = jackknife(x, stat)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Resampling methods: Bootstrap\n",
+ "Bootstrapping is a nonparametric approach to statistical inference\n",
+ "that substitutes computation for more traditional distributional\n",
+ "assumptions and asymptotic results. Bootstrapping offers a number of\n",
+ "advantages: \n",
+ "1. The bootstrap is quite general, although there are some cases in which it fails. \n",
+ "\n",
+ "2. Because it does not require distributional assumptions (such as normally distributed errors), the bootstrap can provide more accurate inferences when the data are not well behaved or when the sample size is small. \n",
+ "\n",
+ "3. It is possible to apply the bootstrap to statistics with sampling distributions that are difficult to derive, even asymptotically. \n",
+ "\n",
+ "4. It is relatively simple to apply the bootstrap to complex data-collection plans (such as stratified and clustered samples).\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Resampling methods: Bootstrap background\n",
+ "\n",
+ "Since $\\widehat{\\theta} = \\widehat{\\theta}(\\boldsymbol{X})$ is a function of random variables,\n",
+ "$\\widehat{\\theta}$ itself must be a random variable. Thus it has\n",
+ "a pdf, call this function $p(\\boldsymbol{t})$. The aim of the bootstrap is to\n",
+ "estimate $p(\\boldsymbol{t})$ by the relative frequency of\n",
+ "$\\widehat{\\theta}$. You can think of this as using a histogram\n",
+ "in the place of $p(\\boldsymbol{t})$. If the relative frequency closely\n",
+ "resembles $p(\\vec{t})$, then using numerics, it is straight forward to\n",
+ "estimate all the interesting parameters of $p(\\boldsymbol{t})$ using point\n",
+ "estimators. \n",
+ "\n",
+ "\n",
+ "## Resampling methods: More Bootstrap background\n",
+ "\n",
+ "In the case that $\\widehat{\\theta}$ has\n",
+ "more than one component, and the components are independent, we use the\n",
+ "same estimator on each component separately. If the probability\n",
+ "density function of $X_i$, $p(x)$, had been known, then it would have\n",
+ "been straight forward to do this by: \n",
+ "1. Drawing lots of numbers from $p(x)$, suppose we call one such set of numbers $(X_1^*, X_2^*, \\cdots, X_n^*)$. \n",
+ "\n",
+ "2. Then using these numbers, we could compute a replica of $\\widehat{\\theta}$ called $\\widehat{\\theta}^*$. \n",
+ "\n",
+ "By repeated use of (1) and (2), many\n",
+ "estimates of $\\widehat{\\theta}$ could have been obtained. The\n",
+ "idea is to use the relative frequency of $\\widehat{\\theta}^*$\n",
+ "(think of a histogram) as an estimate of $p(\\boldsymbol{t})$.\n",
+ "\n",
+ "## Resampling methods: Bootstrap approach\n",
+ "\n",
+ "But\n",
+ "unless there is enough information available about the process that\n",
+ "generated $X_1,X_2,\\cdots,X_n$, $p(x)$ is in general\n",
+ "unknown. Therefore, [Efron in 1979](https://projecteuclid.org/euclid.aos/1176344552) asked the\n",
+ "question: What if we replace $p(x)$ by the relative frequency\n",
+ "of the observation $X_i$; if we draw observations in accordance with\n",
+ "the relative frequency of the observations, will we obtain the same\n",
+ "result in some asymptotic sense? The answer is yes.\n",
+ "\n",
+ "\n",
+ "Instead of generating the histogram for the relative\n",
+ "frequency of the observation $X_i$, just draw the values\n",
+ "$(X_1^*,X_2^*,\\cdots,X_n^*)$ with replacement from the vector\n",
+ "$\\boldsymbol{X}$. \n",
+ "\n",
+ "## Resampling methods: Bootstrap steps\n",
+ "\n",
+ "The independent bootstrap works like this: \n",
+ "\n",
+ "1. Draw with replacement $n$ numbers for the observed variables $\\boldsymbol{x} = (x_1,x_2,\\cdots,x_n)$. \n",
+ "\n",
+ "2. Define a vector $\\boldsymbol{x}^*$ containing the values which were drawn from $\\boldsymbol{x}$. \n",
+ "\n",
+ "3. Using the vector $\\boldsymbol{x}^*$ compute $\\widehat{\\theta}^*$ by evaluating $\\widehat \\theta$ under the observations $\\boldsymbol{x}^*$. \n",
+ "\n",
+ "4. Repeat this process $k$ times. \n",
+ "\n",
+ "When you are done, you can draw a histogram of the relative frequency\n",
+ "of $\\widehat \\theta^*$. This is your estimate of the probability\n",
+ "distribution $p(t)$. Using this probability distribution you can\n",
+ "estimate any statistics thereof. In principle you never draw the\n",
+ "histogram of the relative frequency of $\\widehat{\\theta}^*$. Instead\n",
+ "you use the estimators corresponding to the statistic of interest. For\n",
+ "example, if you are interested in estimating the variance of $\\widehat\n",
+ "\\theta$, apply the etsimator $\\widehat \\sigma^2$ to the values\n",
+ "$\\widehat \\theta ^*$.\n",
+ "\n",
+ "\n",
+ "## Code example for the Bootstrap method\n",
+ "\n",
+ "The following code starts with a Gaussian distribution with mean value\n",
+ "$\\mu =100$ and variance $\\sigma=15$. We use this to generate the data\n",
+ "used in the bootstrap analysis. The bootstrap analysis returns a data\n",
+ "set after a given number of bootstrap operations (as many as we have\n",
+ "data points). This data set consists of estimated mean values for each\n",
+ "bootstrap operation. The histogram generated by the bootstrap method\n",
+ "shows that the distribution for these mean values is also a Gaussian,\n",
+ "centered around the mean value $\\mu=100$ but with standard deviation\n",
+ "$\\sigma/\\sqrt{n}$, where $n$ is the number of bootstrap samples (in\n",
+ "this case the same as the number of original data points). The value\n",
+ "of the standard deviation is what we expect from the central limit\n",
+ "theorem."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 29,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "from numpy import *\n",
+ "from numpy.random import randint, randn\n",
+ "from time import time\n",
+ "import matplotlib.mlab as mlab\n",
+ "import matplotlib.pyplot as plt\n",
+ "\n",
+ "# Returns mean of bootstrap samples \n",
+ "def stat(data):\n",
+ " return mean(data)\n",
+ "\n",
+ "# Bootstrap algorithm\n",
+ "def bootstrap(data, statistic, R):\n",
+ " t = zeros(R); n = len(data); inds = arange(n); t0 = time()\n",
+ " # non-parametric bootstrap \n",
+ " for i in range(R):\n",
+ " t[i] = statistic(data[randint(0,n,n)])\n",
+ "\n",
+ " # analysis \n",
+ " print(\"Runtime: %g sec\" % (time()-t0)); print(\"Bootstrap Statistics :\")\n",
+ " print(\"original bias std. error\")\n",
+ " print(\"%8g %8g %14g %15g\" % (statistic(data), std(data),mean(t),std(t)))\n",
+ " return t\n",
+ "\n",
+ "\n",
+ "mu, sigma = 100, 15\n",
+ "datapoints = 10000\n",
+ "x = mu + sigma*random.randn(datapoints)\n",
+ "# bootstrap returns the data sample \n",
+ "t = bootstrap(x, stat, datapoints)\n",
+ "# the histogram of the bootstrapped data \n",
+ "n, binsboot, patches = plt.hist(t, 50, normed=1, facecolor='red', alpha=0.75)\n",
+ "\n",
+ "# add a 'best fit' line \n",
+ "y = mlab.normpdf( binsboot, mean(t), std(t))\n",
+ "lt = plt.plot(binsboot, y, 'r--', linewidth=1)\n",
+ "plt.xlabel('Smarts')\n",
+ "plt.ylabel('Probability')\n",
+ "plt.axis([99.5, 100.6, 0, 3.0])\n",
+ "plt.grid(True)\n",
+ "\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "## Various steps in cross-validation\n",
+ "\n",
+ "When the repetitive splitting of the data set is done randomly,\n",
+ "samples may accidently end up in a fast majority of the splits in\n",
+ "either training or test set. Such samples may have an unbalanced\n",
+ "influence on either model building or prediction evaluation. To avoid\n",
+ "this $k$-fold cross-validation structures the data splitting. The\n",
+ "samples are divided into $k$ more or less equally sized exhaustive and\n",
+ "mutually exclusive subsets. In turn (at each split) one of these\n",
+ "subsets plays the role of the test set while the union of the\n",
+ "remaining subsets constitutes the training set. Such a splitting\n",
+ "warrants a balanced representation of each sample in both training and\n",
+ "test set over the splits. Still the division into the $k$ subsets\n",
+ "involves a degree of randomness. This may be fully excluded when\n",
+ "choosing $k=n$. This particular case is referred to as leave-one-out\n",
+ "cross-validation (LOOCV). \n",
+ "\n",
+ "\n",
+ "## How to set up the cross-validation for Ridge and/or Lasso\n",
+ "\n",
+ "* Define a range of interest for the penalty parameter.\n",
+ "\n",
+ "* Divide the data set into training and test set comprising samples $\\{1, \\ldots, n\\} \\setminus i$ and $\\{ i \\}$, respectively.\n",
+ "\n",
+ "* Fit the linear regression model by means of ridge estimation for each $\\lambda$ in the grid using the training set, and the corresponding estimate of the error variance $\\boldsymbol{\\sigma}_{-i}^2(\\lambda)$, as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\begin{align*}\n",
+ "\\boldsymbol{\\beta}_{-i}(\\lambda) & = ( \\boldsymbol{X}_{-i, \\ast}^{T}\n",
+ "\\boldsymbol{X}_{-i, \\ast} + \\lambda \\boldsymbol{I}_{pp})^{-1}\n",
+ "\\boldsymbol{X}_{-i, \\ast}^{T} \\boldsymbol{y}_{-i}\n",
+ "\\end{align*}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "* Evaluate the prediction performance of these models on the test set by $\\log\\{L[y_i, \\boldsymbol{X}_{i, \\ast}; \\boldsymbol{\\beta}_{-i}(\\lambda), \\boldsymbol{\\sigma}_{-i}^2(\\lambda)]\\}$. Or, by the prediction error $|y_i - \\boldsymbol{X}_{i, \\ast} \\boldsymbol{\\beta}_{-i}(\\lambda)|$, the relative error, the error squared or the R2 score function.\n",
+ "\n",
+ "* Repeat the first three steps such that each sample plays the role of the test set once.\n",
+ "\n",
+ "* Average the prediction performances of the test sets at each grid point of the penalty bias/parameter. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. It is defined as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\begin{align*}\n",
+ "\\frac{1}{n} \\sum_{i = 1}^n \\log\\{L[y_i, \\mathbf{X}_{i, \\ast}; \\boldsymbol{\\beta}_{-i}(\\lambda), \\boldsymbol{\\sigma}_{-i}^2(\\lambda)]\\}.\n",
+ "\\end{align*}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Cross-validation in brief\n",
+ "\n",
+ "For the various values of $k$\n",
+ "\n",
+ "1. shuffle the dataset randomly.\n",
+ "\n",
+ "2. Split the dataset into $k$ groups.\n",
+ "\n",
+ "3. For each unique group:\n",
+ "\n",
+ "a. Decide which group to use as set for test data\n",
+ "\n",
+ "b. Take the remaining groups as a training data set\n",
+ "\n",
+ "c. Fit a model on the training set and evaluate it on the test set\n",
+ "\n",
+ "d. Retain the evaluation score and discard the model\n",
+ "\n",
+ "\n",
+ "5. Summarize the model using the sample of model evaluation scores\n",
+ "\n",
+ "## Code Example for Cross-validation and $k$-fold Cross-validation\n",
+ "\n",
+ "The code here uses Ridge regression with cross-validation (CV) resampling and $k$-fold CV in order to fit a specific polynomial."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 30,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "import numpy as np\n",
+ "import matplotlib.pyplot as plt\n",
+ "from sklearn.model_selection import KFold\n",
+ "from sklearn.linear_model import Ridge\n",
+ "from sklearn.model_selection import cross_val_score\n",
+ "from sklearn.preprocessing import PolynomialFeatures\n",
+ "\n",
+ "# A seed just to ensure that the random numbers are the same for every run.\n",
+ "# Useful for eventual debugging.\n",
+ "np.random.seed(3155)\n",
+ "\n",
+ "# Generate the data.\n",
+ "nsamples = 100\n",
+ "x = np.random.randn(nsamples)\n",
+ "y = 3*x**2 + np.random.randn(nsamples)\n",
+ "\n",
+ "## Cross-validation on Ridge regression using KFold only\n",
+ "\n",
+ "# Decide degree on polynomial to fit\n",
+ "poly = PolynomialFeatures(degree = 6)\n",
+ "\n",
+ "# Decide which values of lambda to use\n",
+ "nlambdas = 500\n",
+ "lambdas = np.logspace(-3, 5, nlambdas)\n",
+ "\n",
+ "# Initialize a KFold instance\n",
+ "k = 5\n",
+ "kfold = KFold(n_splits = k)\n",
+ "\n",
+ "# Perform the cross-validation to estimate MSE\n",
+ "scores_KFold = np.zeros((nlambdas, k))\n",
+ "\n",
+ "i = 0\n",
+ "for lmb in lambdas:\n",
+ " ridge = Ridge(alpha = lmb)\n",
+ " j = 0\n",
+ " for train_inds, test_inds in kfold.split(x):\n",
+ " xtrain = x[train_inds]\n",
+ " ytrain = y[train_inds]\n",
+ "\n",
+ " xtest = x[test_inds]\n",
+ " ytest = y[test_inds]\n",
+ "\n",
+ " Xtrain = poly.fit_transform(xtrain[:, np.newaxis])\n",
+ " ridge.fit(Xtrain, ytrain[:, np.newaxis])\n",
+ "\n",
+ " Xtest = poly.fit_transform(xtest[:, np.newaxis])\n",
+ " ypred = ridge.predict(Xtest)\n",
+ "\n",
+ " scores_KFold[i,j] = np.sum((ypred - ytest[:, np.newaxis])**2)/np.size(ypred)\n",
+ "\n",
+ " j += 1\n",
+ " i += 1\n",
+ "\n",
+ "\n",
+ "estimated_mse_KFold = np.mean(scores_KFold, axis = 1)\n",
+ "\n",
+ "## Cross-validation using cross_val_score from sklearn along with KFold\n",
+ "\n",
+ "# kfold is an instance initialized above as:\n",
+ "# kfold = KFold(n_splits = k)\n",
+ "\n",
+ "estimated_mse_sklearn = np.zeros(nlambdas)\n",
+ "i = 0\n",
+ "for lmb in lambdas:\n",
+ " ridge = Ridge(alpha = lmb)\n",
+ "\n",
+ " X = poly.fit_transform(x[:, np.newaxis])\n",
+ " estimated_mse_folds = cross_val_score(ridge, X, y[:, np.newaxis], scoring='neg_mean_squared_error', cv=kfold)\n",
+ "\n",
+ " # cross_val_score return an array containing the estimated negative mse for every fold.\n",
+ " # we have to the the mean of every array in order to get an estimate of the mse of the model\n",
+ " estimated_mse_sklearn[i] = np.mean(-estimated_mse_folds)\n",
+ "\n",
+ " i += 1\n",
+ "\n",
+ "## Plot and compare the slightly different ways to perform cross-validation\n",
+ "\n",
+ "plt.figure()\n",
+ "\n",
+ "plt.plot(np.log10(lambdas), estimated_mse_sklearn, label = 'cross_val_score')\n",
+ "plt.plot(np.log10(lambdas), estimated_mse_KFold, 'r--', label = 'KFold')\n",
+ "\n",
+ "plt.xlabel('log10(lambda)')\n",
+ "plt.ylabel('mse')\n",
+ "\n",
+ "plt.legend()\n",
+ "\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## The bias-variance tradeoff\n",
+ "\n",
+ "\n",
+ "We will discuss the bias-variance tradeoff in the context of\n",
+ "continuous predictions such as regression. However, many of the\n",
+ "intuitions and ideas discussed here also carry over to classification\n",
+ "tasks. Consider a dataset $\\mathcal{L}$ consisting of the data\n",
+ "$\\mathbf{X}_\\mathcal{L}=\\{(y_j, \\boldsymbol{x}_j), j=0\\ldots n-1\\}$. \n",
+ "\n",
+ "Let us assume that the true data is generated from a noisy model"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{y}=f(\\boldsymbol{x}) + \\boldsymbol{\\epsilon}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $\\epsilon$ is normally distributed with mean zero and standard deviation $\\sigma^2$.\n",
+ "\n",
+ "In our derivation of the ordinary least squares method we defined then\n",
+ "an approximation to the function $f$ in terms of the parameters\n",
+ "$\\boldsymbol{\\beta}$ and the design matrix $\\boldsymbol{X}$ which embody our model,\n",
+ "that is $\\boldsymbol{\\tilde{y}}=\\boldsymbol{X}\\boldsymbol{\\beta}$. \n",
+ "\n",
+ "Thereafter we found the parameters $\\boldsymbol{\\beta}$ by optimizing the means squared error via the so-called cost function"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "C(\\boldsymbol{X},\\boldsymbol{\\beta}) =\\frac{1}{n}\\sum_{i=0}^{n-1}(y_i-\\tilde{y}_i)^2=\\mathbb{E}\\left[(\\boldsymbol{y}-\\boldsymbol{\\tilde{y}})^2\\right].\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We can rewrite this as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathbb{E}\\left[(\\boldsymbol{y}-\\boldsymbol{\\tilde{y}})^2\\right]=\\frac{1}{n}\\sum_i(f_i-\\mathbb{E}\\left[\\boldsymbol{\\tilde{y}}\\right])^2+\\frac{1}{n}\\sum_i(\\tilde{y}_i-\\mathbb{E}\\left[\\boldsymbol{\\tilde{y}}\\right])^2+\\sigma^2.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The three terms represent the square of the bias of the learning\n",
+ "method, which can be thought of as the error caused by the simplifying\n",
+ "assumptions built into the method. The second term represents the\n",
+ "variance of the chosen model and finally the last terms is variance of\n",
+ "the error $\\boldsymbol{\\epsilon}$.\n",
+ "\n",
+ "To derive this equation, we need to recall that the variance of $\\boldsymbol{y}$ and $\\boldsymbol{\\epsilon}$ are both equal to $\\sigma^2$. The mean value of $\\boldsymbol{\\epsilon}$ is by definition equal to zero. Furthermore, the function $f$ is not a stochastics variable, idem for $\\boldsymbol{\\tilde{y}}$.\n",
+ "We use a more compact notation in terms of the expectation value"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathbb{E}\\left[(\\boldsymbol{y}-\\boldsymbol{\\tilde{y}})^2\\right]=\\mathbb{E}\\left[(\\boldsymbol{f}+\\boldsymbol{\\epsilon}-\\boldsymbol{\\tilde{y}})^2\\right],\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and adding and subtracting $\\mathbb{E}\\left[\\boldsymbol{\\tilde{y}}\\right]$ we get"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathbb{E}\\left[(\\boldsymbol{y}-\\boldsymbol{\\tilde{y}})^2\\right]=\\mathbb{E}\\left[(\\boldsymbol{f}+\\boldsymbol{\\epsilon}-\\boldsymbol{\\tilde{y}}+\\mathbb{E}\\left[\\boldsymbol{\\tilde{y}}\\right]-\\mathbb{E}\\left[\\boldsymbol{\\tilde{y}}\\right])^2\\right],\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which, using the abovementioned expectation values can be rewritten as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathbb{E}\\left[(\\boldsymbol{y}-\\boldsymbol{\\tilde{y}})^2\\right]=\\mathbb{E}\\left[(\\boldsymbol{y}-\\mathbb{E}\\left[\\boldsymbol{\\tilde{y}}\\right])^2\\right]+\\mathrm{Var}\\left[\\boldsymbol{\\tilde{y}}\\right]+\\sigma^2,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "that is the rewriting in terms of the so-called bias, the variance of the model $\\boldsymbol{\\tilde{y}}$ and the variance of $\\boldsymbol{\\epsilon}$.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Example code for Bias-Variance tradeoff"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 31,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "import matplotlib.pyplot as plt\n",
+ "import numpy as np\n",
+ "from sklearn.linear_model import LinearRegression, Ridge, Lasso\n",
+ "from sklearn.preprocessing import PolynomialFeatures\n",
+ "from sklearn.model_selection import train_test_split\n",
+ "from sklearn.pipeline import make_pipeline\n",
+ "from sklearn.utils import resample\n",
+ "\n",
+ "np.random.seed(2018)\n",
+ "\n",
+ "n = 500\n",
+ "n_boostraps = 100\n",
+ "degree = 18 # A quite high value, just to show.\n",
+ "noise = 0.1\n",
+ "\n",
+ "# Make data set.\n",
+ "x = np.linspace(-1, 3, n).reshape(-1, 1)\n",
+ "y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2) + np.random.normal(0, 0.1, x.shape)\n",
+ "\n",
+ "# Hold out some test data that is never used in training.\n",
+ "x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.2)\n",
+ "\n",
+ "# Combine x transformation and model into one operation.\n",
+ "# Not neccesary, but convenient.\n",
+ "model = make_pipeline(PolynomialFeatures(degree=degree), LinearRegression(fit_intercept=False))\n",
+ "\n",
+ "# The following (m x n_bootstraps) matrix holds the column vectors y_pred\n",
+ "# for each bootstrap iteration.\n",
+ "y_pred = np.empty((y_test.shape[0], n_boostraps))\n",
+ "for i in range(n_boostraps):\n",
+ " x_, y_ = resample(x_train, y_train)\n",
+ "\n",
+ " # Evaluate the new model on the same test data each time.\n",
+ " y_pred[:, i] = model.fit(x_, y_).predict(x_test).ravel()\n",
+ "\n",
+ "# Note: Expectations and variances taken w.r.t. different training\n",
+ "# data sets, hence the axis=1. Subsequent means are taken across the test data\n",
+ "# set in order to obtain a total value, but before this we have error/bias/variance\n",
+ "# calculated per data point in the test set.\n",
+ "# Note 2: The use of keepdims=True is important in the calculation of bias as this \n",
+ "# maintains the column vector form. Dropping this yields very unexpected results.\n",
+ "error = np.mean( np.mean((y_test - y_pred)**2, axis=1, keepdims=True) )\n",
+ "bias = np.mean( (y_test - np.mean(y_pred, axis=1, keepdims=True))**2 )\n",
+ "variance = np.mean( np.var(y_pred, axis=1, keepdims=True) )\n",
+ "print('Error:', error)\n",
+ "print('Bias^2:', bias)\n",
+ "print('Var:', variance)\n",
+ "print('{} >= {} + {} = {}'.format(error, bias, variance, bias+variance))\n",
+ "\n",
+ "plt.plot(x[::5, :], y[::5, :], label='f(x)')\n",
+ "plt.scatter(x_test, y_test, label='Data points')\n",
+ "plt.scatter(x_test, np.mean(y_pred, axis=1), label='Pred')\n",
+ "plt.legend()\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Understanding what happens"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 32,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "import matplotlib.pyplot as plt\n",
+ "import numpy as np\n",
+ "from sklearn.linear_model import LinearRegression, Ridge, Lasso\n",
+ "from sklearn.preprocessing import PolynomialFeatures\n",
+ "from sklearn.model_selection import train_test_split\n",
+ "from sklearn.pipeline import make_pipeline\n",
+ "from sklearn.utils import resample\n",
+ "\n",
+ "np.random.seed(2018)\n",
+ "\n",
+ "n = 40\n",
+ "n_boostraps = 100\n",
+ "maxdegree = 14\n",
+ "\n",
+ "\n",
+ "# Make data set.\n",
+ "x = np.linspace(-3, 3, n).reshape(-1, 1)\n",
+ "y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.normal(0, 0.1, x.shape)\n",
+ "error = np.zeros(maxdegree)\n",
+ "bias = np.zeros(maxdegree)\n",
+ "variance = np.zeros(maxdegree)\n",
+ "polydegree = np.zeros(maxdegree)\n",
+ "x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.2)\n",
+ "\n",
+ "for degree in range(maxdegree):\n",
+ " model = make_pipeline(PolynomialFeatures(degree=degree), LinearRegression(fit_intercept=False))\n",
+ " y_pred = np.empty((y_test.shape[0], n_boostraps))\n",
+ " for i in range(n_boostraps):\n",
+ " x_, y_ = resample(x_train, y_train)\n",
+ " y_pred[:, i] = model.fit(x_, y_).predict(x_test).ravel()\n",
+ "\n",
+ " polydegree[degree] = degree\n",
+ " error[degree] = np.mean( np.mean((y_test - y_pred)**2, axis=1, keepdims=True) )\n",
+ " bias[degree] = np.mean( (y_test - np.mean(y_pred, axis=1, keepdims=True))**2 )\n",
+ " variance[degree] = np.mean( np.var(y_pred, axis=1, keepdims=True) )\n",
+ " print('Polynomial degree:', degree)\n",
+ " print('Error:', error[degree])\n",
+ " print('Bias^2:', bias[degree])\n",
+ " print('Var:', variance[degree])\n",
+ " print('{} >= {} + {} = {}'.format(error[degree], bias[degree], variance[degree], bias[degree]+variance[degree]))\n",
+ "\n",
+ "plt.plot(polydegree, error, label='Error')\n",
+ "plt.plot(polydegree, bias, label='bias')\n",
+ "plt.plot(polydegree, variance, label='Variance')\n",
+ "plt.legend()\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "## Summing up\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "The bias-variance tradeoff summarizes the fundamental tension in\n",
+ "machine learning, particularly supervised learning, between the\n",
+ "complexity of a model and the amount of training data needed to train\n",
+ "it. Since data is often limited, in practice it is often useful to\n",
+ "use a less-complex model with higher bias, that is a model whose asymptotic\n",
+ "performance is worse than another model because it is easier to\n",
+ "train and less sensitive to sampling noise arising from having a\n",
+ "finite-sized training dataset (smaller variance). \n",
+ "\n",
+ "\n",
+ "\n",
+ "The above equations tell us that in\n",
+ "order to minimize the expected test error, we need to select a\n",
+ "statistical learning method that simultaneously achieves low variance\n",
+ "and low bias. Note that variance is inherently a nonnegative quantity,\n",
+ "and squared bias is also nonnegative. Hence, we see that the expected\n",
+ "test MSE can never lie below $Var(\\epsilon)$, the irreducible error.\n",
+ "\n",
+ "\n",
+ "What do we mean by the variance and bias of a statistical learning\n",
+ "method? The variance refers to the amount by which our model would change if we\n",
+ "estimated it using a different training data set. Since the training\n",
+ "data are used to fit the statistical learning method, different\n",
+ "training data sets will result in a different estimate. But ideally the\n",
+ "estimate for our model should not vary too much between training\n",
+ "sets. However, if a method has high variance then small changes in\n",
+ "the training data can result in large changes in the model. In general, more\n",
+ "flexible statistical methods have higher variance.\n",
+ "\n",
+ "\n",
+ "You may also find this recent [article](https://www.pnas.org/content/116/32/15849) of interest.\n",
+ "\n",
+ "## Another Example from Scikit-Learn's Repository"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 33,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "\"\"\"\n",
+ "============================\n",
+ "Underfitting vs. Overfitting\n",
+ "============================\n",
+ "\n",
+ "This example demonstrates the problems of underfitting and overfitting and\n",
+ "how we can use linear regression with polynomial features to approximate\n",
+ "nonlinear functions. The plot shows the function that we want to approximate,\n",
+ "which is a part of the cosine function. In addition, the samples from the\n",
+ "real function and the approximations of different models are displayed. The\n",
+ "models have polynomial features of different degrees. We can see that a\n",
+ "linear function (polynomial with degree 1) is not sufficient to fit the\n",
+ "training samples. This is called **underfitting**. A polynomial of degree 4\n",
+ "approximates the true function almost perfectly. However, for higher degrees\n",
+ "the model will **overfit** the training data, i.e. it learns the noise of the\n",
+ "training data.\n",
+ "We evaluate quantitatively **overfitting** / **underfitting** by using\n",
+ "cross-validation. We calculate the mean squared error (MSE) on the validation\n",
+ "set, the higher, the less likely the model generalizes correctly from the\n",
+ "training data.\n",
+ "\"\"\"\n",
+ "\n",
+ "print(__doc__)\n",
+ "\n",
+ "import numpy as np\n",
+ "import matplotlib.pyplot as plt\n",
+ "from sklearn.pipeline import Pipeline\n",
+ "from sklearn.preprocessing import PolynomialFeatures\n",
+ "from sklearn.linear_model import LinearRegression\n",
+ "from sklearn.model_selection import cross_val_score\n",
+ "\n",
+ "\n",
+ "def true_fun(X):\n",
+ " return np.cos(1.5 * np.pi * X)\n",
+ "\n",
+ "np.random.seed(0)\n",
+ "\n",
+ "n_samples = 30\n",
+ "degrees = [1, 4, 15]\n",
+ "\n",
+ "X = np.sort(np.random.rand(n_samples))\n",
+ "y = true_fun(X) + np.random.randn(n_samples) * 0.1\n",
+ "\n",
+ "plt.figure(figsize=(14, 5))\n",
+ "for i in range(len(degrees)):\n",
+ " ax = plt.subplot(1, len(degrees), i + 1)\n",
+ " plt.setp(ax, xticks=(), yticks=())\n",
+ "\n",
+ " polynomial_features = PolynomialFeatures(degree=degrees[i],\n",
+ " include_bias=False)\n",
+ " linear_regression = LinearRegression()\n",
+ " pipeline = Pipeline([(\"polynomial_features\", polynomial_features),\n",
+ " (\"linear_regression\", linear_regression)])\n",
+ " pipeline.fit(X[:, np.newaxis], y)\n",
+ "\n",
+ " # Evaluate the models using crossvalidation\n",
+ " scores = cross_val_score(pipeline, X[:, np.newaxis], y,\n",
+ " scoring=\"neg_mean_squared_error\", cv=10)\n",
+ "\n",
+ " X_test = np.linspace(0, 1, 100)\n",
+ " plt.plot(X_test, pipeline.predict(X_test[:, np.newaxis]), label=\"Model\")\n",
+ " plt.plot(X_test, true_fun(X_test), label=\"True function\")\n",
+ " plt.scatter(X, y, edgecolor='b', s=20, label=\"Samples\")\n",
+ " plt.xlabel(\"x\")\n",
+ " plt.ylabel(\"y\")\n",
+ " plt.xlim((0, 1))\n",
+ " plt.ylim((-2, 2))\n",
+ " plt.legend(loc=\"best\")\n",
+ " plt.title(\"Degree {}\\nMSE = {:.2e}(+/- {:.2e})\".format(\n",
+ " degrees[i], -scores.mean(), scores.std()))\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## More examples on bootstrap and cross-validation and errors"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 34,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# Common imports\n",
+ "import os\n",
+ "import numpy as np\n",
+ "import pandas as pd\n",
+ "import matplotlib.pyplot as plt\n",
+ "from sklearn.linear_model import LinearRegression, Ridge, Lasso\n",
+ "from sklearn.model_selection import train_test_split\n",
+ "from sklearn.utils import resample\n",
+ "from sklearn.metrics import mean_squared_error\n",
+ "# Where to save the figures and data files\n",
+ "PROJECT_ROOT_DIR = \"Results\"\n",
+ "FIGURE_ID = \"Results/FigureFiles\"\n",
+ "DATA_ID = \"DataFiles/\"\n",
+ "\n",
+ "if not os.path.exists(PROJECT_ROOT_DIR):\n",
+ " os.mkdir(PROJECT_ROOT_DIR)\n",
+ "\n",
+ "if not os.path.exists(FIGURE_ID):\n",
+ " os.makedirs(FIGURE_ID)\n",
+ "\n",
+ "if not os.path.exists(DATA_ID):\n",
+ " os.makedirs(DATA_ID)\n",
+ "\n",
+ "def image_path(fig_id):\n",
+ " return os.path.join(FIGURE_ID, fig_id)\n",
+ "\n",
+ "def data_path(dat_id):\n",
+ " return os.path.join(DATA_ID, dat_id)\n",
+ "\n",
+ "def save_fig(fig_id):\n",
+ " plt.savefig(image_path(fig_id) + \".png\", format='png')\n",
+ "\n",
+ "infile = open(data_path(\"EoS.csv\"),'r')\n",
+ "\n",
+ "# Read the EoS data as csv file and organize the data into two arrays with density and energies\n",
+ "EoS = pd.read_csv(infile, names=('Density', 'Energy'))\n",
+ "EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce')\n",
+ "EoS = EoS.dropna()\n",
+ "Energies = EoS['Energy']\n",
+ "Density = EoS['Density']\n",
+ "# The design matrix now as function of various polytrops\n",
+ "\n",
+ "Maxpolydegree = 30\n",
+ "X = np.zeros((len(Density),Maxpolydegree))\n",
+ "X[:,0] = 1.0\n",
+ "testerror = np.zeros(Maxpolydegree)\n",
+ "trainingerror = np.zeros(Maxpolydegree)\n",
+ "polynomial = np.zeros(Maxpolydegree)\n",
+ "\n",
+ "trials = 100\n",
+ "for polydegree in range(1, Maxpolydegree):\n",
+ " polynomial[polydegree] = polydegree\n",
+ " for degree in range(polydegree):\n",
+ " X[:,degree] = Density**(degree/3.0)\n",
+ "\n",
+ "# loop over trials in order to estimate the expectation value of the MSE\n",
+ " testerror[polydegree] = 0.0\n",
+ " trainingerror[polydegree] = 0.0\n",
+ " for samples in range(trials):\n",
+ " x_train, x_test, y_train, y_test = train_test_split(X, Energies, test_size=0.2)\n",
+ " model = LinearRegression(fit_intercept=True).fit(x_train, y_train)\n",
+ " ypred = model.predict(x_train)\n",
+ " ytilde = model.predict(x_test)\n",
+ " testerror[polydegree] += mean_squared_error(y_test, ytilde)\n",
+ " trainingerror[polydegree] += mean_squared_error(y_train, ypred) \n",
+ "\n",
+ " testerror[polydegree] /= trials\n",
+ " trainingerror[polydegree] /= trials\n",
+ " print(\"Degree of polynomial: %3d\"% polynomial[polydegree])\n",
+ " print(\"Mean squared error on training data: %.8f\" % trainingerror[polydegree])\n",
+ " print(\"Mean squared error on test data: %.8f\" % testerror[polydegree])\n",
+ "\n",
+ "plt.plot(polynomial, np.log10(trainingerror), label='Training Error')\n",
+ "plt.plot(polynomial, np.log10(testerror), label='Test Error')\n",
+ "plt.xlabel('Polynomial degree')\n",
+ "plt.ylabel('log10[MSE]')\n",
+ "plt.legend()\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "## The same example but now with cross-validation"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 35,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# Common imports\n",
+ "import os\n",
+ "import numpy as np\n",
+ "import pandas as pd\n",
+ "import matplotlib.pyplot as plt\n",
+ "from sklearn.linear_model import LinearRegression, Ridge, Lasso\n",
+ "from sklearn.metrics import mean_squared_error\n",
+ "from sklearn.model_selection import KFold\n",
+ "from sklearn.model_selection import cross_val_score\n",
+ "\n",
+ "\n",
+ "# Where to save the figures and data files\n",
+ "PROJECT_ROOT_DIR = \"Results\"\n",
+ "FIGURE_ID = \"Results/FigureFiles\"\n",
+ "DATA_ID = \"DataFiles/\"\n",
+ "\n",
+ "if not os.path.exists(PROJECT_ROOT_DIR):\n",
+ " os.mkdir(PROJECT_ROOT_DIR)\n",
+ "\n",
+ "if not os.path.exists(FIGURE_ID):\n",
+ " os.makedirs(FIGURE_ID)\n",
+ "\n",
+ "if not os.path.exists(DATA_ID):\n",
+ " os.makedirs(DATA_ID)\n",
+ "\n",
+ "def image_path(fig_id):\n",
+ " return os.path.join(FIGURE_ID, fig_id)\n",
+ "\n",
+ "def data_path(dat_id):\n",
+ " return os.path.join(DATA_ID, dat_id)\n",
+ "\n",
+ "def save_fig(fig_id):\n",
+ " plt.savefig(image_path(fig_id) + \".png\", format='png')\n",
+ "\n",
+ "infile = open(data_path(\"EoS.csv\"),'r')\n",
+ "\n",
+ "# Read the EoS data as csv file and organize the data into two arrays with density and energies\n",
+ "EoS = pd.read_csv(infile, names=('Density', 'Energy'))\n",
+ "EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce')\n",
+ "EoS = EoS.dropna()\n",
+ "Energies = EoS['Energy']\n",
+ "Density = EoS['Density']\n",
+ "# The design matrix now as function of various polytrops\n",
+ "\n",
+ "Maxpolydegree = 30\n",
+ "X = np.zeros((len(Density),Maxpolydegree))\n",
+ "X[:,0] = 1.0\n",
+ "estimated_mse_sklearn = np.zeros(Maxpolydegree)\n",
+ "polynomial = np.zeros(Maxpolydegree)\n",
+ "k =5\n",
+ "kfold = KFold(n_splits = k)\n",
+ "\n",
+ "for polydegree in range(1, Maxpolydegree):\n",
+ " polynomial[polydegree] = polydegree\n",
+ " for degree in range(polydegree):\n",
+ " X[:,degree] = Density**(degree/3.0)\n",
+ " OLS = LinearRegression()\n",
+ "# loop over trials in order to estimate the expectation value of the MSE\n",
+ " estimated_mse_folds = cross_val_score(OLS, X, Energies, scoring='neg_mean_squared_error', cv=kfold)\n",
+ "#[:, np.newaxis]\n",
+ " estimated_mse_sklearn[polydegree] = np.mean(-estimated_mse_folds)\n",
+ "\n",
+ "plt.plot(polynomial, np.log10(estimated_mse_sklearn), label='Test Error')\n",
+ "plt.xlabel('Polynomial degree')\n",
+ "plt.ylabel('log10[MSE]')\n",
+ "plt.legend()\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Cross-validation with Ridge"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 36,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "import numpy as np\n",
+ "import matplotlib.pyplot as plt\n",
+ "from sklearn.model_selection import KFold\n",
+ "from sklearn.linear_model import Ridge\n",
+ "from sklearn.model_selection import cross_val_score\n",
+ "from sklearn.preprocessing import PolynomialFeatures\n",
+ "\n",
+ "# A seed just to ensure that the random numbers are the same for every run.\n",
+ "np.random.seed(3155)\n",
+ "# Generate the data.\n",
+ "n = 100\n",
+ "x = np.linspace(-3, 3, n).reshape(-1, 1)\n",
+ "y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.normal(0, 0.1, x.shape)\n",
+ "# Decide degree on polynomial to fit\n",
+ "poly = PolynomialFeatures(degree = 10)\n",
+ "\n",
+ "# Decide which values of lambda to use\n",
+ "nlambdas = 500\n",
+ "lambdas = np.logspace(-3, 5, nlambdas)\n",
+ "# Initialize a KFold instance\n",
+ "k = 5\n",
+ "kfold = KFold(n_splits = k)\n",
+ "estimated_mse_sklearn = np.zeros(nlambdas)\n",
+ "i = 0\n",
+ "for lmb in lambdas:\n",
+ " ridge = Ridge(alpha = lmb)\n",
+ " estimated_mse_folds = cross_val_score(ridge, x, y, scoring='neg_mean_squared_error', cv=kfold)\n",
+ " estimated_mse_sklearn[i] = np.mean(-estimated_mse_folds)\n",
+ " i += 1\n",
+ "plt.figure()\n",
+ "plt.plot(np.log10(lambdas), estimated_mse_sklearn, label = 'cross_val_score')\n",
+ "plt.xlabel('log10(lambda)')\n",
+ "plt.ylabel('MSE')\n",
+ "plt.legend()\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## The Ising model\n",
+ "\n",
+ "The one-dimensional Ising model with nearest neighbor interaction, no\n",
+ "external field and a constant coupling constant $J$ is given by"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " H = -J \\sum_{k}^L s_k s_{k + 1},\n",
+ "\\label{_auto2} \\tag{2}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $s_i \\in \\{-1, 1\\}$ and $s_{N + 1} = s_1$. The number of spins\n",
+ "in the system is determined by $L$. For the one-dimensional system\n",
+ "there is no phase transition.\n",
+ "\n",
+ "We will look at a system of $L = 40$ spins with a coupling constant of\n",
+ "$J = 1$. To get enough training data we will generate 10000 states\n",
+ "with their respective energies."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 37,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "import numpy as np\n",
+ "import matplotlib.pyplot as plt\n",
+ "from mpl_toolkits.axes_grid1 import make_axes_locatable\n",
+ "import seaborn as sns\n",
+ "import scipy.linalg as scl\n",
+ "from sklearn.model_selection import train_test_split\n",
+ "import tqdm\n",
+ "sns.set(color_codes=True)\n",
+ "cmap_args=dict(vmin=-1., vmax=1., cmap='seismic')\n",
+ "\n",
+ "L = 40\n",
+ "n = int(1e4)\n",
+ "\n",
+ "spins = np.random.choice([-1, 1], size=(n, L))\n",
+ "J = 1.0\n",
+ "\n",
+ "energies = np.zeros(n)\n",
+ "\n",
+ "for i in range(n):\n",
+ " energies[i] = - J * np.dot(spins[i], np.roll(spins[i], 1))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Here we use ordinary least squares\n",
+ "regression to predict the energy for the nearest neighbor\n",
+ "one-dimensional Ising model on a ring, i.e., the endpoints wrap\n",
+ "around. We will use linear regression to fit a value for\n",
+ "the coupling constant to achieve this.\n",
+ "\n",
+ "## Reformulating the problem to suit regression\n",
+ "\n",
+ "A more general form for the one-dimensional Ising model is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " H = - \\sum_j^L \\sum_k^L s_j s_k J_{jk}.\n",
+ "\\label{_auto3} \\tag{3}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Here we allow for interactions beyond the nearest neighbors and a state dependent\n",
+ "coupling constant. This latter expression can be formulated as\n",
+ "a matrix-product"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " \\boldsymbol{H} = \\boldsymbol{X} J,\n",
+ "\\label{_auto4} \\tag{4}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $X_{jk} = s_j s_k$ and $J$ is a matrix which consists of the\n",
+ "elements $-J_{jk}$. This form of writing the energy fits perfectly\n",
+ "with the form utilized in linear regression, that is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " \\boldsymbol{y} = \\boldsymbol{X}\\boldsymbol{\\beta} + \\boldsymbol{\\epsilon},\n",
+ "\\label{_auto5} \\tag{5}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We split the data in training and test data as discussed in the previous example"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 38,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "X = np.zeros((n, L ** 2))\n",
+ "for i in range(n):\n",
+ " X[i] = np.outer(spins[i], spins[i]).ravel()\n",
+ "y = energies\n",
+ "X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Linear regression\n",
+ "\n",
+ "In the ordinary least squares method we choose the cost function"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " C(\\boldsymbol{X}, \\boldsymbol{\\beta})= \\frac{1}{n}\\left\\{(\\boldsymbol{X}\\boldsymbol{\\beta} - \\boldsymbol{y})^T(\\boldsymbol{X}\\boldsymbol{\\beta} - \\boldsymbol{y})\\right\\}.\n",
+ "\\label{_auto6} \\tag{6}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We then find the extremal point of $C$ by taking the derivative with respect to $\\boldsymbol{\\beta}$ as discussed above.\n",
+ "This yields the expression for $\\boldsymbol{\\beta}$ to be"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{\\beta} = \\frac{\\boldsymbol{X}^T \\boldsymbol{y}}{\\boldsymbol{X}^T \\boldsymbol{X}},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which immediately imposes some requirements on $\\boldsymbol{X}$ as there must exist\n",
+ "an inverse of $\\boldsymbol{X}^T \\boldsymbol{X}$. If the expression we are modeling contains an\n",
+ "intercept, i.e., a constant term, we must make sure that the\n",
+ "first column of $\\boldsymbol{X}$ consists of $1$. We do this here"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 39,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "X_train_own = np.concatenate(\n",
+ " (np.ones(len(X_train))[:, np.newaxis], X_train),\n",
+ " axis=1\n",
+ ")\n",
+ "X_test_own = np.concatenate(\n",
+ " (np.ones(len(X_test))[:, np.newaxis], X_test),\n",
+ " axis=1\n",
+ ")"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 40,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "def ols_inv(x: np.ndarray, y: np.ndarray) -> np.ndarray:\n",
+ " return scl.inv(x.T @ x) @ (x.T @ y)\n",
+ "beta = ols_inv(X_train_own, y_train)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Singular Value decomposition\n",
+ "\n",
+ "Doing the inversion directly turns out to be a bad idea since the matrix\n",
+ "$\\boldsymbol{X}^T\\boldsymbol{X}$ is singular. An alternative approach is to use the **singular\n",
+ "value decomposition**. Using the definition of the Moore-Penrose\n",
+ "pseudoinverse we can write the equation for $\\boldsymbol{\\beta}$ as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{\\beta} = \\boldsymbol{X}^{+}\\boldsymbol{y},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where the pseudoinverse of $\\boldsymbol{X}$ is given by"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\boldsymbol{X}^{+} = \\frac{\\boldsymbol{X}^T}{\\boldsymbol{X}^T\\boldsymbol{X}}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Using singular value decomposition we can decompose the matrix $\\boldsymbol{X} = \\boldsymbol{U}\\boldsymbol{\\Sigma} \\boldsymbol{V}^T$,\n",
+ "where $\\boldsymbol{U}$ and $\\boldsymbol{V}$ are orthogonal(unitary) matrices and $\\boldsymbol{\\Sigma}$ contains the singular values (more details below).\n",
+ "where $X^{+} = V\\Sigma^{+} U^T$. This reduces the equation for\n",
+ "$\\omega$ to"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " \\boldsymbol{\\beta} = \\boldsymbol{V}\\boldsymbol{\\Sigma}^{+} \\boldsymbol{U}^T \\boldsymbol{y}.\n",
+ "\\label{_auto7} \\tag{7}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Note that solving this equation by actually doing the pseudoinverse\n",
+ "(which is what we will do) is not a good idea as this operation scales\n",
+ "as $\\mathcal{O}(n^3)$, where $n$ is the number of elements in a\n",
+ "general matrix. Instead, doing $QR$-factorization and solving the\n",
+ "linear system as an equation would reduce this down to\n",
+ "$\\mathcal{O}(n^2)$ operations."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 41,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "def ols_svd(x: np.ndarray, y: np.ndarray) -> np.ndarray:\n",
+ " u, s, v = scl.svd(x)\n",
+ " return v.T @ scl.pinv(scl.diagsvd(s, u.shape[0], v.shape[0])) @ u.T @ y"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 42,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "beta = ols_svd(X_train_own,y_train)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "When extracting the $J$-matrix we need to make sure that we remove the intercept, as is done here"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 43,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "J = beta[1:].reshape(L, L)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "A way of looking at the coefficients in $J$ is to plot the matrices as images."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 44,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "fig = plt.figure(figsize=(20, 14))\n",
+ "im = plt.imshow(J, **cmap_args)\n",
+ "plt.title(\"OLS\", fontsize=18)\n",
+ "plt.xticks(fontsize=18)\n",
+ "plt.yticks(fontsize=18)\n",
+ "cb = fig.colorbar(im)\n",
+ "cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18)\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "It is interesting to note that OLS\n",
+ "considers both $J_{j, j + 1} = -0.5$ and $J_{j, j - 1} = -0.5$ as\n",
+ "valid matrix elements for $J$.\n",
+ "In our discussion below on hyperparameters and Ridge and Lasso regression we will see that\n",
+ "this problem can be removed, partly and only with Lasso regression. \n",
+ "\n",
+ "In this case our matrix inversion was actually possible. The obvious question now is what is the mathematics behind the SVD?\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## The one-dimensional Ising model\n",
+ "\n",
+ "Let us bring back the Ising model again, but now with an additional\n",
+ "focus on Ridge and Lasso regression as well. We repeat some of the\n",
+ "basic parts of the Ising model and the setup of the training and test\n",
+ "data. The one-dimensional Ising model with nearest neighbor\n",
+ "interaction, no external field and a constant coupling constant $J$ is\n",
+ "given by"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " H = -J \\sum_{k}^L s_k s_{k + 1},\n",
+ "\\label{_auto8} \\tag{8}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $s_i \\in \\{-1, 1\\}$ and $s_{N + 1} = s_1$. The number of spins in the system is determined by $L$. For the one-dimensional system there is no phase transition.\n",
+ "\n",
+ "We will look at a system of $L = 40$ spins with a coupling constant of $J = 1$. To get enough training data we will generate 10000 states with their respective energies."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 45,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "import numpy as np\n",
+ "import matplotlib.pyplot as plt\n",
+ "from mpl_toolkits.axes_grid1 import make_axes_locatable\n",
+ "import seaborn as sns\n",
+ "import scipy.linalg as scl\n",
+ "from sklearn.model_selection import train_test_split\n",
+ "import sklearn.linear_model as skl\n",
+ "import tqdm\n",
+ "sns.set(color_codes=True)\n",
+ "cmap_args=dict(vmin=-1., vmax=1., cmap='seismic')\n",
+ "\n",
+ "L = 40\n",
+ "n = int(1e4)\n",
+ "\n",
+ "spins = np.random.choice([-1, 1], size=(n, L))\n",
+ "J = 1.0\n",
+ "\n",
+ "energies = np.zeros(n)\n",
+ "\n",
+ "for i in range(n):\n",
+ " energies[i] = - J * np.dot(spins[i], np.roll(spins[i], 1))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "A more general form for the one-dimensional Ising model is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " H = - \\sum_j^L \\sum_k^L s_j s_k J_{jk}.\n",
+ "\\label{_auto9} \\tag{9}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Here we allow for interactions beyond the nearest neighbors and a more\n",
+ "adaptive coupling matrix. This latter expression can be formulated as\n",
+ "a matrix-product on the form"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " H = X J,\n",
+ "\\label{_auto10} \\tag{10}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $X_{jk} = s_j s_k$ and $J$ is the matrix consisting of the\n",
+ "elements $-J_{jk}$. This form of writing the energy fits perfectly\n",
+ "with the form utilized in linear regression, viz."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " \\boldsymbol{y} = \\boldsymbol{X}\\boldsymbol{\\beta} + \\boldsymbol{\\epsilon}.\n",
+ "\\label{_auto11} \\tag{11}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We organize the data as we did above"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 46,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "X = np.zeros((n, L ** 2))\n",
+ "for i in range(n):\n",
+ " X[i] = np.outer(spins[i], spins[i]).ravel()\n",
+ "y = energies\n",
+ "X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.96)\n",
+ "\n",
+ "X_train_own = np.concatenate(\n",
+ " (np.ones(len(X_train))[:, np.newaxis], X_train),\n",
+ " axis=1\n",
+ ")\n",
+ "\n",
+ "X_test_own = np.concatenate(\n",
+ " (np.ones(len(X_test))[:, np.newaxis], X_test),\n",
+ " axis=1\n",
+ ")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We will do all fitting with **Scikit-Learn**,"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 47,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "clf = skl.LinearRegression().fit(X_train, y_train)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "When extracting the $J$-matrix we make sure to remove the intercept"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 48,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "J_sk = clf.coef_.reshape(L, L)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "And then we plot the results"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 49,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "fig = plt.figure(figsize=(20, 14))\n",
+ "im = plt.imshow(J_sk, **cmap_args)\n",
+ "plt.title(\"LinearRegression from Scikit-learn\", fontsize=18)\n",
+ "plt.xticks(fontsize=18)\n",
+ "plt.yticks(fontsize=18)\n",
+ "cb = fig.colorbar(im)\n",
+ "cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18)\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The results perfectly with our previous discussion where we used our own code.\n",
+ "\n",
+ "## Ridge regression\n",
+ "\n",
+ "Having explored the ordinary least squares we move on to ridge\n",
+ "regression. In ridge regression we include a **regularizer**. This\n",
+ "involves a new cost function which leads to a new estimate for the\n",
+ "weights $\\boldsymbol{\\beta}$. This results in a penalized regression problem. The\n",
+ "cost function is given by"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "1\n",
+ "3\n",
+ "6\n",
+ " \n",
+ "<\n",
+ "<\n",
+ "<\n",
+ "!\n",
+ "!\n",
+ "M\n",
+ "A\n",
+ "T\n",
+ "H\n",
+ "_\n",
+ "B\n",
+ "L\n",
+ "O\n",
+ "C\n",
+ "K"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 50,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "_lambda = 0.1\n",
+ "clf_ridge = skl.Ridge(alpha=_lambda).fit(X_train, y_train)\n",
+ "J_ridge_sk = clf_ridge.coef_.reshape(L, L)\n",
+ "fig = plt.figure(figsize=(20, 14))\n",
+ "im = plt.imshow(J_ridge_sk, **cmap_args)\n",
+ "plt.title(\"Ridge from Scikit-learn\", fontsize=18)\n",
+ "plt.xticks(fontsize=18)\n",
+ "plt.yticks(fontsize=18)\n",
+ "cb = fig.colorbar(im)\n",
+ "cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18)\n",
+ "\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## LASSO regression\n",
+ "\n",
+ "In the **Least Absolute Shrinkage and Selection Operator** (LASSO)-method we get a third cost function."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " C(\\boldsymbol{X}, \\boldsymbol{\\beta}; \\lambda) = (\\boldsymbol{X}\\boldsymbol{\\beta} - \\boldsymbol{y})^T(\\boldsymbol{X}\\boldsymbol{\\beta} - \\boldsymbol{y}) + \\lambda \\sqrt{\\boldsymbol{\\beta}^T\\boldsymbol{\\beta}}.\n",
+ "\\label{_auto13} \\tag{13}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Finding the extremal point of this cost function is not so straight-forward as in least squares and ridge. We will therefore rely solely on the function ``Lasso`` from **Scikit-Learn**."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 51,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "clf_lasso = skl.Lasso(alpha=_lambda).fit(X_train, y_train)\n",
+ "J_lasso_sk = clf_lasso.coef_.reshape(L, L)\n",
+ "fig = plt.figure(figsize=(20, 14))\n",
+ "im = plt.imshow(J_lasso_sk, **cmap_args)\n",
+ "plt.title(\"Lasso from Scikit-learn\", fontsize=18)\n",
+ "plt.xticks(fontsize=18)\n",
+ "plt.yticks(fontsize=18)\n",
+ "cb = fig.colorbar(im)\n",
+ "cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18)\n",
+ "\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "It is quite striking how LASSO breaks the symmetry of the coupling\n",
+ "constant as opposed to ridge and OLS. We get a sparse solution with\n",
+ "$J_{j, j + 1} = -1$.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Performance as function of the regularization parameter\n",
+ "\n",
+ "We see how the different models perform for a different set of values for $\\lambda$."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 52,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "lambdas = np.logspace(-4, 5, 10)\n",
+ "\n",
+ "train_errors = {\n",
+ " \"ols_sk\": np.zeros(lambdas.size),\n",
+ " \"ridge_sk\": np.zeros(lambdas.size),\n",
+ " \"lasso_sk\": np.zeros(lambdas.size)\n",
+ "}\n",
+ "\n",
+ "test_errors = {\n",
+ " \"ols_sk\": np.zeros(lambdas.size),\n",
+ " \"ridge_sk\": np.zeros(lambdas.size),\n",
+ " \"lasso_sk\": np.zeros(lambdas.size)\n",
+ "}\n",
+ "\n",
+ "plot_counter = 1\n",
+ "\n",
+ "fig = plt.figure(figsize=(32, 54))\n",
+ "\n",
+ "for i, _lambda in enumerate(tqdm.tqdm(lambdas)):\n",
+ " for key, method in zip(\n",
+ " [\"ols_sk\", \"ridge_sk\", \"lasso_sk\"],\n",
+ " [skl.LinearRegression(), skl.Ridge(alpha=_lambda), skl.Lasso(alpha=_lambda)]\n",
+ " ):\n",
+ " method = method.fit(X_train, y_train)\n",
+ "\n",
+ " train_errors[key][i] = method.score(X_train, y_train)\n",
+ " test_errors[key][i] = method.score(X_test, y_test)\n",
+ "\n",
+ " omega = method.coef_.reshape(L, L)\n",
+ "\n",
+ " plt.subplot(10, 5, plot_counter)\n",
+ " plt.imshow(omega, **cmap_args)\n",
+ " plt.title(r\"%s, $\\lambda = %.4f$\" % (key, _lambda))\n",
+ " plot_counter += 1\n",
+ "\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We see that LASSO reaches a good solution for low\n",
+ "values of $\\lambda$, but will \"wither\" when we increase $\\lambda$ too\n",
+ "much. Ridge is more stable over a larger range of values for\n",
+ "$\\lambda$, but eventually also fades away.\n",
+ "\n",
+ "## Finding the optimal value of $\\lambda$\n",
+ "\n",
+ "To determine which value of $\\lambda$ is best we plot the accuracy of\n",
+ "the models when predicting the training and the testing set. We expect\n",
+ "the accuracy of the training set to be quite good, but if the accuracy\n",
+ "of the testing set is much lower this tells us that we might be\n",
+ "subject to an overfit model. The ideal scenario is an accuracy on the\n",
+ "testing set that is close to the accuracy of the training set."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 53,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "fig = plt.figure(figsize=(20, 14))\n",
+ "\n",
+ "colors = {\n",
+ " \"ols_sk\": \"r\",\n",
+ " \"ridge_sk\": \"y\",\n",
+ " \"lasso_sk\": \"c\"\n",
+ "}\n",
+ "\n",
+ "for key in train_errors:\n",
+ " plt.semilogx(\n",
+ " lambdas,\n",
+ " train_errors[key],\n",
+ " colors[key],\n",
+ " label=\"Train {0}\".format(key),\n",
+ " linewidth=4.0\n",
+ " )\n",
+ "\n",
+ "for key in test_errors:\n",
+ " plt.semilogx(\n",
+ " lambdas,\n",
+ " test_errors[key],\n",
+ " colors[key] + \"--\",\n",
+ " label=\"Test {0}\".format(key),\n",
+ " linewidth=4.0\n",
+ " )\n",
+ "plt.legend(loc=\"best\", fontsize=18)\n",
+ "plt.xlabel(r\"$\\lambda$\", fontsize=18)\n",
+ "plt.ylabel(r\"$R^2$\", fontsize=18)\n",
+ "plt.tick_params(labelsize=18)\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "From the above figure we can see that LASSO with $\\lambda = 10^{-2}$\n",
+ "achieves a very good accuracy on the test set. This by far surpasses the\n",
+ "other models for all values of $\\lambda$."
+ ]
+ }
+ ],
+ "metadata": {},
+ "nbformat": 4,
+ "nbformat_minor": 4
+}
diff --git a/doc/src/LectureNotes/chapter5.ipynb b/doc/src/LectureNotes/chapter5.ipynb
new file mode 100644
index 000000000..8372ef384
--- /dev/null
+++ b/doc/src/LectureNotes/chapter5.ipynb
@@ -0,0 +1,583 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Logistic Regression\n",
+ "\n",
+ "## Introduction\n",
+ "In linear regression our main interest was centered on learning the\n",
+ "coefficients of a functional fit (say a polynomial) in order to be\n",
+ "able to predict the response of a continuous variable on some unseen\n",
+ "data. The fit to the continuous variable $y_i$ is based on some\n",
+ "independent variables $\\hat{x}_i$. Linear regression resulted in\n",
+ "analytical expressions for standard ordinary Least Squares or Ridge\n",
+ "regression (in terms of matrices to invert) for several quantities,\n",
+ "ranging from the variance and thereby the confidence intervals of the\n",
+ "parameters $\\hat{\\beta}$ to the mean squared error. If we can invert\n",
+ "the product of the design matrices, linear regression gives then a\n",
+ "simple recipe for fitting our data.\n",
+ "\n",
+ "\n",
+ "Classification problems, however, are concerned with outcomes taking\n",
+ "the form of discrete variables (i.e. categories). We may for example,\n",
+ "on the basis of DNA sequencing for a number of patients, like to find\n",
+ "out which mutations are important for a certain disease; or based on\n",
+ "scans of various patients' brains, figure out if there is a tumor or\n",
+ "not; or given a specific physical system, we'd like to identify its\n",
+ "state, say whether it is an ordered or disordered system (typical\n",
+ "situation in solid state physics); or classify the status of a\n",
+ "patient, whether she/he has a stroke or not and many other similar\n",
+ "situations.\n",
+ "\n",
+ "The most common situation we encounter when we apply logistic\n",
+ "regression is that of two possible outcomes, normally denoted as a\n",
+ "binary outcome, true or false, positive or negative, success or\n",
+ "failure etc.\n",
+ "\n",
+ "Logistic regression will also serve as our stepping stone towards\n",
+ "neural network algorithms and supervised deep learning. For logistic\n",
+ "learning, the minimization of the cost function leads to a non-linear\n",
+ "equation in the parameters $\\hat{\\beta}$. The optimization of the\n",
+ "problem calls therefore for minimization algorithms. This forms the\n",
+ "bottle neck of all machine learning algorithms, namely how to find\n",
+ "reliable minima of a multi-variable function. This leads us to the\n",
+ "family of gradient descent methods. The latter are the working horses\n",
+ "of basically all modern machine learning algorithms.\n",
+ "\n",
+ "We note also that many of the topics discussed here on logistic \n",
+ "regression are also commonly used in modern supervised Deep Learning\n",
+ "models, as we will see later.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Basics\n",
+ "\n",
+ "We consider the case where the dependent variables, also called the\n",
+ "responses or the outcomes, $y_i$ are discrete and only take values\n",
+ "from $k=0,\\dots,K-1$ (i.e. $K$ classes).\n",
+ "\n",
+ "The goal is to predict the\n",
+ "output classes from the design matrix $\\hat{X}\\in\\mathbb{R}^{n\\times p}$\n",
+ "made of $n$ samples, each of which carries $p$ features or predictors. The\n",
+ "primary goal is to identify the classes to which new unseen samples\n",
+ "belong.\n",
+ "\n",
+ "Let us specialize to the case of two classes only, with outputs\n",
+ "$y_i=0$ and $y_i=1$. Our outcomes could represent the status of a\n",
+ "credit card user that could default or not on her/his credit card\n",
+ "debt. That is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "y_i = \\begin{bmatrix} 0 & \\mathrm{no}\\\\ 1 & \\mathrm{yes} \\end{bmatrix}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Before moving to the logistic model, let us try to use our linear\n",
+ "regression model to classify these two outcomes. We could for example\n",
+ "fit a linear model to the default case if $y_i > 0.5$ and the no\n",
+ "default case $y_i \\leq 0.5$.\n",
+ "\n",
+ "We would then have our \n",
+ "weighted linear combination, namely"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ "\\hat{y} = \\hat{X}^T\\hat{\\beta} + \\hat{\\epsilon},\n",
+ "\\label{_auto1} \\tag{1}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $\\hat{y}$ is a vector representing the possible outcomes, $\\hat{X}$ is our\n",
+ "$n\\times p$ design matrix and $\\hat{\\beta}$ represents our estimators/predictors.\n",
+ "\n",
+ "\n",
+ "The main problem with our function is that it takes values on the\n",
+ "entire real axis. In the case of logistic regression, however, the\n",
+ "labels $y_i$ are discrete variables. A typical example is the credit\n",
+ "card data discussed below here, where we can set the state of\n",
+ "defaulting the debt to $y_i=1$ and not to $y_i=0$ for one the persons\n",
+ "in the data set (see the full example below).\n",
+ "\n",
+ "One simple way to get a discrete output is to have sign\n",
+ "functions that map the output of a linear regressor to values $\\{0,1\\}$,\n",
+ "$f(s_i)=sign(s_i)=1$ if $s_i\\ge 0$ and 0 if otherwise. \n",
+ "We will encounter this model in our first demonstration of neural networks. Historically it is called the \"perceptron\" model in the machine learning\n",
+ "literature. This model is extremely simple. However, in many cases it is more\n",
+ "favorable to use a ``soft\" classifier that outputs\n",
+ "the probability of a given category. This leads us to the logistic function.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## The logistic function\n",
+ "\n",
+ "The perceptron is an example of a ``hard classification\" model. We\n",
+ "will encounter this model when we discuss neural networks as\n",
+ "well. Each datapoint is deterministically assigned to a category (i.e\n",
+ "$y_i=0$ or $y_i=1$). In many cases, it is favorable to have a \"soft\"\n",
+ "classifier that outputs the probability of a given category rather\n",
+ "than a single value. For example, given $x_i$, the classifier\n",
+ "outputs the probability of being in a category $k$. Logistic regression\n",
+ "is the most common example of a so-called soft classifier. In logistic\n",
+ "regression, the probability that a data point $x_i$\n",
+ "belongs to a category $y_i=\\{0,1\\}$ is given by the so-called logit function (or Sigmoid) which is meant to represent the likelihood for a given event,"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "p(t) = \\frac{1}{1+\\mathrm \\exp{-t}}=\\frac{\\exp{t}}{1+\\mathrm \\exp{t}}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Note that $1-p(t)= p(-t)$.\n",
+ "\n",
+ "\n",
+ "The following code plots the logistic function, the step function and other functions we will encounter from here and on."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "%matplotlib inline\n",
+ "\n",
+ "\"\"\"The sigmoid function (or the logistic curve) is a\n",
+ "function that takes any real number, z, and outputs a number (0,1).\n",
+ "It is useful in neural networks for assigning weights on a relative scale.\n",
+ "The value z is the weighted sum of parameters involved in the learning algorithm.\"\"\"\n",
+ "\n",
+ "import numpy\n",
+ "import matplotlib.pyplot as plt\n",
+ "import math as mt\n",
+ "\n",
+ "z = numpy.arange(-5, 5, .1)\n",
+ "sigma_fn = numpy.vectorize(lambda z: 1/(1+numpy.exp(-z)))\n",
+ "sigma = sigma_fn(z)\n",
+ "\n",
+ "fig = plt.figure()\n",
+ "ax = fig.add_subplot(111)\n",
+ "ax.plot(z, sigma)\n",
+ "ax.set_ylim([-0.1, 1.1])\n",
+ "ax.set_xlim([-5,5])\n",
+ "ax.grid(True)\n",
+ "ax.set_xlabel('z')\n",
+ "ax.set_title('sigmoid function')\n",
+ "\n",
+ "plt.show()\n",
+ "\n",
+ "\"\"\"Step Function\"\"\"\n",
+ "z = numpy.arange(-5, 5, .02)\n",
+ "step_fn = numpy.vectorize(lambda z: 1.0 if z >= 0.0 else 0.0)\n",
+ "step = step_fn(z)\n",
+ "\n",
+ "fig = plt.figure()\n",
+ "ax = fig.add_subplot(111)\n",
+ "ax.plot(z, step)\n",
+ "ax.set_ylim([-0.5, 1.5])\n",
+ "ax.set_xlim([-5,5])\n",
+ "ax.grid(True)\n",
+ "ax.set_xlabel('z')\n",
+ "ax.set_title('step function')\n",
+ "\n",
+ "plt.show()\n",
+ "\n",
+ "\"\"\"tanh Function\"\"\"\n",
+ "z = numpy.arange(-2*mt.pi, 2*mt.pi, 0.1)\n",
+ "t = numpy.tanh(z)\n",
+ "\n",
+ "fig = plt.figure()\n",
+ "ax = fig.add_subplot(111)\n",
+ "ax.plot(z, t)\n",
+ "ax.set_ylim([-1.0, 1.0])\n",
+ "ax.set_xlim([-2*mt.pi,2*mt.pi])\n",
+ "ax.grid(True)\n",
+ "ax.set_xlabel('z')\n",
+ "ax.set_title('tanh function')\n",
+ "\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Two parameters\n",
+ "\n",
+ "We assume now that we have two classes with $y_i$ either $0$ or $1$. Furthermore we assume also that we have only two parameters $\\beta$ in our fitting of the Sigmoid function, that is we define probabilities"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\begin{align*}\n",
+ "p(y_i=1|x_i,\\hat{\\beta}) &= \\frac{\\exp{(\\beta_0+\\beta_1x_i)}}{1+\\exp{(\\beta_0+\\beta_1x_i)}},\\nonumber\\\\\n",
+ "p(y_i=0|x_i,\\hat{\\beta}) &= 1 - p(y_i=1|x_i,\\hat{\\beta}),\n",
+ "\\end{align*}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $\\hat{\\beta}$ are the weights we wish to extract from data, in our case $\\beta_0$ and $\\beta_1$. \n",
+ "\n",
+ "Note that we used"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "p(y_i=0\\vert x_i, \\hat{\\beta}) = 1-p(y_i=1\\vert x_i, \\hat{\\beta}).\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Maximum likelihood\n",
+ "\n",
+ "In order to define the total likelihood for all possible outcomes from a \n",
+ "dataset $\\mathcal{D}=\\{(y_i,x_i)\\}$, with the binary labels\n",
+ "$y_i\\in\\{0,1\\}$ and where the data points are drawn independently, we use the so-called [Maximum Likelihood Estimation](https://en.wikipedia.org/wiki/Maximum_likelihood_estimation) (MLE) principle. \n",
+ "We aim thus at maximizing \n",
+ "the probability of seeing the observed data. We can then approximate the \n",
+ "likelihood in terms of the product of the individual probabilities of a specific outcome $y_i$, that is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\begin{align*}\n",
+ "P(\\mathcal{D}|\\hat{\\beta})& = \\prod_{i=1}^n \\left[p(y_i=1|x_i,\\hat{\\beta})\\right]^{y_i}\\left[1-p(y_i=1|x_i,\\hat{\\beta}))\\right]^{1-y_i}\\nonumber \\\\\n",
+ "\\end{align*}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "from which we obtain the log-likelihood and our **cost/loss** function"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathcal{C}(\\hat{\\beta}) = \\sum_{i=1}^n \\left( y_i\\log{p(y_i=1|x_i,\\hat{\\beta})} + (1-y_i)\\log\\left[1-p(y_i=1|x_i,\\hat{\\beta}))\\right]\\right).\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Reordering the logarithms, we can rewrite the **cost/loss** function as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathcal{C}(\\hat{\\beta}) = \\sum_{i=1}^n \\left(y_i(\\beta_0+\\beta_1x_i) -\\log{(1+\\exp{(\\beta_0+\\beta_1x_i)})}\\right).\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The maximum likelihood estimator is defined as the set of parameters that maximize the log-likelihood where we maximize with respect to $\\beta$.\n",
+ "Since the cost (error) function is just the negative log-likelihood, for logistic regression we have that"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathcal{C}(\\hat{\\beta})=-\\sum_{i=1}^n \\left(y_i(\\beta_0+\\beta_1x_i) -\\log{(1+\\exp{(\\beta_0+\\beta_1x_i)})}\\right).\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "This equation is known in statistics as the **cross entropy**. Finally, we note that just as in linear regression, \n",
+ "in practice we often supplement the cross-entropy with additional regularization terms, usually $L_1$ and $L_2$ regularization as we did for Ridge and Lasso regression.\n",
+ "\n",
+ "\n",
+ "The cross entropy is a convex function of the weights $\\hat{\\beta}$ and,\n",
+ "therefore, any local minimizer is a global minimizer. \n",
+ "\n",
+ "\n",
+ "Minimizing this\n",
+ "cost function with respect to the two parameters $\\beta_0$ and $\\beta_1$ we obtain"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial \\mathcal{C}(\\hat{\\beta})}{\\partial \\beta_0} = -\\sum_{i=1}^n \\left(y_i -\\frac{\\exp{(\\beta_0+\\beta_1x_i)}}{1+\\exp{(\\beta_0+\\beta_1x_i)}}\\right),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial \\mathcal{C}(\\hat{\\beta})}{\\partial \\beta_1} = -\\sum_{i=1}^n \\left(y_ix_i -x_i\\frac{\\exp{(\\beta_0+\\beta_1x_i)}}{1+\\exp{(\\beta_0+\\beta_1x_i)}}\\right).\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Let us now define a vector $\\hat{y}$ with $n$ elements $y_i$, an\n",
+ "$n\\times p$ matrix $\\hat{X}$ which contains the $x_i$ values and a\n",
+ "vector $\\hat{p}$ of fitted probabilities $p(y_i\\vert x_i,\\hat{\\beta})$. We can rewrite in a more compact form the first\n",
+ "derivative of cost function as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial \\mathcal{C}(\\hat{\\beta})}{\\partial \\hat{\\beta}} = -\\hat{X}^T\\left(\\hat{y}-\\hat{p}\\right).\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "If we in addition define a diagonal matrix $\\hat{W}$ with elements \n",
+ "$p(y_i\\vert x_i,\\hat{\\beta})(1-p(y_i\\vert x_i,\\hat{\\beta})$, we can obtain a compact expression of the second derivative as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial^2 \\mathcal{C}(\\hat{\\beta})}{\\partial \\hat{\\beta}\\partial \\hat{\\beta}^T} = \\hat{X}^T\\hat{W}\\hat{X}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Within a binary classification problem, we can easily expand our model to include multiple predictors. Our ratio between likelihoods is then with $p$ predictors"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\log{ \\frac{p(\\hat{\\beta}\\hat{x})}{1-p(\\hat{\\beta}\\hat{x})}} = \\beta_0+\\beta_1x_1+\\beta_2x_2+\\dots+\\beta_px_p.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Here we defined $\\hat{x}=[1,x_1,x_2,\\dots,x_p]$ and $\\hat{\\beta}=[\\beta_0, \\beta_1, \\dots, \\beta_p]$ leading to"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "p(\\hat{\\beta}\\hat{x})=\\frac{ \\exp{(\\beta_0+\\beta_1x_1+\\beta_2x_2+\\dots+\\beta_px_p)}}{1+\\exp{(\\beta_0+\\beta_1x_1+\\beta_2x_2+\\dots+\\beta_px_p)}}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Including more classes\n",
+ "\n",
+ "Till now we have mainly focused on two classes, the so-called binary\n",
+ "system. Suppose we wish to extend to $K$ classes. Let us for the sake\n",
+ "of simplicity assume we have only two predictors. We have then\n",
+ "following model"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "1\n",
+ "5\n",
+ " \n",
+ "<\n",
+ "<\n",
+ "<\n",
+ "!\n",
+ "!\n",
+ "M\n",
+ "A\n",
+ "T\n",
+ "H\n",
+ "_\n",
+ "B\n",
+ "L\n",
+ "O\n",
+ "C\n",
+ "K"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\log{\\frac{p(C=2\\vert x)}{p(K\\vert x)}} = \\beta_{20}+\\beta_{21}x_1,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and so on till the class $C=K-1$ class"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\log{\\frac{p(C=K-1\\vert x)}{p(K\\vert x)}} = \\beta_{(K-1)0}+\\beta_{(K-1)1}x_1,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and the model is specified in term of $K-1$ so-called log-odds or\n",
+ "**logit** transformations.\n",
+ "\n",
+ "\n",
+ "\n",
+ "In our discussion of neural networks we will encounter the above again\n",
+ "in terms of a slightly modified function, the so-called **Softmax** function.\n",
+ "\n",
+ "The softmax function is used in various multiclass classification\n",
+ "methods, such as multinomial logistic regression (also known as\n",
+ "softmax regression), multiclass linear discriminant analysis, naive\n",
+ "Bayes classifiers, and artificial neural networks. Specifically, in\n",
+ "multinomial logistic regression and linear discriminant analysis, the\n",
+ "input to the function is the result of $K$ distinct linear functions,\n",
+ "and the predicted probability for the $k$-th class given a sample\n",
+ "vector $\\hat{x}$ and a weighting vector $\\hat{\\beta}$ is (with two\n",
+ "predictors):"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "p(C=k\\vert \\mathbf {x} )=\\frac{\\exp{(\\beta_{k0}+\\beta_{k1}x_1)}}{1+\\sum_{l=1}^{K-1}\\exp{(\\beta_{l0}+\\beta_{l1}x_1)}}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "It is easy to extend to more predictors. The final class is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "p(C=K\\vert \\mathbf {x} )=\\frac{1}{1+\\sum_{l=1}^{K-1}\\exp{(\\beta_{l0}+\\beta_{l1}x_1)}},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and they sum to one. Our earlier discussions were all specialized to\n",
+ "the case with two classes only. It is easy to see from the above that\n",
+ "what we derived earlier is compatible with these equations.\n",
+ "\n",
+ "To find the optimal parameters we would typically use a gradient\n",
+ "descent method. Newton's method and gradient descent methods are\n",
+ "discussed in the material on [optimization\n",
+ "methods](https://compphysics.github.io/MachineLearning/doc/pub/Splines/html/Splines-bs.html)."
+ ]
+ }
+ ],
+ "metadata": {},
+ "nbformat": 4,
+ "nbformat_minor": 4
+}
diff --git a/doc/src/LectureNotes/chapter6.ipynb b/doc/src/LectureNotes/chapter6.ipynb
new file mode 100644
index 000000000..26a8daed9
--- /dev/null
+++ b/doc/src/LectureNotes/chapter6.ipynb
@@ -0,0 +1,3351 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Neural networks, from the simple perceptron to deep learning\n",
+ "\n",
+ "## To do list\n",
+ "\n",
+ "* write code for single perceptron model and make link with linear regression\n",
+ "\n",
+ "* revise initial info and add references\n",
+ "\n",
+ "* Update tensorflow material, with keras\n",
+ "\n",
+ "* think of adding material about pytorch\n",
+ "\n",
+ "* rework pulsar example and breast cancer example\n",
+ "\n",
+ "* add ising model example for both regression and classification\n",
+ "\n",
+ "* make data on gravitational problem, add reference to articles on uncovering physical laws from ML\n",
+ "\n",
+ "* think of genetic data\n",
+ "\n",
+ "## Neural networks\n",
+ "\n",
+ "Artificial neural networks are computational systems that can learn to\n",
+ "perform tasks by considering examples, generally without being\n",
+ "programmed with any task-specific rules. It is supposed to mimic a\n",
+ "biological system, wherein neurons interact by sending signals in the\n",
+ "form of mathematical functions between layers. All layers can contain\n",
+ "an arbitrary number of neurons, and each connection is represented by\n",
+ "a weight variable.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Artificial neurons\n",
+ "\n",
+ "The field of artificial neural networks has a long history of\n",
+ "development, and is closely connected with the advancement of computer\n",
+ "science and computers in general. A model of artificial neurons was\n",
+ "first developed by McCulloch and Pitts in 1943 to study signal\n",
+ "processing in the brain and has later been refined by others. The\n",
+ "general idea is to mimic neural networks in the human brain, which is\n",
+ "composed of billions of neurons that communicate with each other by\n",
+ "sending electrical signals. Each neuron accumulates its incoming\n",
+ "signals, which must exceed an activation threshold to yield an\n",
+ "output. If the threshold is not overcome, the neuron remains inactive,\n",
+ "i.e. has zero output.\n",
+ "\n",
+ "This behaviour has inspired a simple mathematical model for an artificial neuron."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " y = f\\left(\\sum_{i=1}^n w_ix_i\\right) = f(u)\n",
+ "\\label{artificialNeuron} \\tag{1}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Here, the output $y$ of the neuron is the value of its activation function, which have as input\n",
+ "a weighted sum of signals $x_i, \\dots ,x_n$ received by $n$ other neurons.\n",
+ "\n",
+ "Conceptually, it is helpful to divide neural networks into four\n",
+ "categories:\n",
+ "1. general purpose neural networks for supervised learning,\n",
+ "\n",
+ "2. neural networks designed specifically for image processing, the most prominent example of this class being Convolutional Neural Networks (CNNs),\n",
+ "\n",
+ "3. neural networks for sequential data such as Recurrent Neural Networks (RNNs), and\n",
+ "\n",
+ "4. neural networks for unsupervised learning such as Deep Boltzmann Machines.\n",
+ "\n",
+ "In natural science, DNNs and CNNs have already found numerous\n",
+ "applications. In statistical physics, they have been applied to detect\n",
+ "phase transitions in 2D Ising and Potts models, lattice gauge\n",
+ "theories, and different phases of polymers, or solving the\n",
+ "Navier-Stokes equation in weather forecasting. Deep learning has also\n",
+ "found interesting applications in quantum physics. Various quantum\n",
+ "phase transitions can be detected and studied using DNNs and CNNs,\n",
+ "topological phases, and even non-equilibrium many-body\n",
+ "localization. Representing quantum states as DNNs quantum state\n",
+ "tomography are among some of the impressive achievements to reveal the\n",
+ "potential of DNNs to facilitate the study of quantum systems.\n",
+ "\n",
+ "In quantum information theory, it has been shown that one can perform\n",
+ "gate decompositions with the help of neural. \n",
+ "\n",
+ "The applications are not limited to the natural sciences. There is a\n",
+ "plethora of applications in essentially all disciplines, from the\n",
+ "humanities to life science and medicine.\n",
+ "\n",
+ "\n",
+ "## Neural network types\n",
+ "\n",
+ "An artificial neural network (ANN), is a computational model that\n",
+ "consists of layers of connected neurons, or nodes or units. We will\n",
+ "refer to these interchangeably as units or nodes, and sometimes as\n",
+ "neurons.\n",
+ "\n",
+ "It is supposed to mimic a biological nervous system by letting each\n",
+ "neuron interact with other neurons by sending signals in the form of\n",
+ "mathematical functions between layers. A wide variety of different\n",
+ "ANNs have been developed, but most of them consist of an input layer,\n",
+ "an output layer and eventual layers in-between, called *hidden\n",
+ "layers*. All layers can contain an arbitrary number of nodes, and each\n",
+ "connection between two nodes is associated with a weight variable.\n",
+ "\n",
+ "Neural networks (also called neural nets) are neural-inspired\n",
+ "nonlinear models for supervised learning. As we will see, neural nets\n",
+ "can be viewed as natural, more powerful extensions of supervised\n",
+ "learning methods such as linear and logistic regression and soft-max\n",
+ "methods we discussed earlier.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Feed-forward neural networks\n",
+ "\n",
+ "The feed-forward neural network (FFNN) was the first and simplest type\n",
+ "of ANNs that were devised. In this network, the information moves in\n",
+ "only one direction: forward through the layers.\n",
+ "\n",
+ "Nodes are represented by circles, while the arrows display the\n",
+ "connections between the nodes, including the direction of information\n",
+ "flow. Additionally, each arrow corresponds to a weight variable\n",
+ "(figure to come). We observe that each node in a layer is connected\n",
+ "to *all* nodes in the subsequent layer, making this a so-called\n",
+ "*fully-connected* FFNN.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Convolutional Neural Network\n",
+ "\n",
+ "A different variant of FFNNs are *convolutional neural networks*\n",
+ "(CNNs), which have a connectivity pattern inspired by the animal\n",
+ "visual cortex. Individual neurons in the visual cortex only respond to\n",
+ "stimuli from small sub-regions of the visual field, called a receptive\n",
+ "field. This makes the neurons well-suited to exploit the strong\n",
+ "spatially local correlation present in natural images. The response of\n",
+ "each neuron can be approximated mathematically as a convolution\n",
+ "operation. (figure to come)\n",
+ "\n",
+ "Convolutional neural networks emulate the behaviour of neurons in the\n",
+ "visual cortex by enforcing a *local* connectivity pattern between\n",
+ "nodes of adjacent layers: Each node in a convolutional layer is\n",
+ "connected only to a subset of the nodes in the previous layer, in\n",
+ "contrast to the fully-connected FFNN. Often, CNNs consist of several\n",
+ "convolutional layers that learn local features of the input, with a\n",
+ "fully-connected layer at the end, which gathers all the local data and\n",
+ "produces the outputs. They have wide applications in image and video\n",
+ "recognition.\n",
+ "\n",
+ "\n",
+ "## Recurrent neural networks\n",
+ "\n",
+ "So far we have only mentioned ANNs where information flows in one\n",
+ "direction: forward. *Recurrent neural networks* on the other hand,\n",
+ "have connections between nodes that form directed *cycles*. This\n",
+ "creates a form of internal memory which are able to capture\n",
+ "information on what has been calculated before; the output is\n",
+ "dependent on the previous computations. Recurrent NNs make use of\n",
+ "sequential information by performing the same task for every element\n",
+ "in a sequence, where each element depends on previous elements. An\n",
+ "example of such information is sentences, making recurrent NNs\n",
+ "especially well-suited for handwriting and speech recognition.\n",
+ "\n",
+ "\n",
+ "## Other types of networks\n",
+ "\n",
+ "There are many other kinds of ANNs that have been developed. One type\n",
+ "that is specifically designed for interpolation in multidimensional\n",
+ "space is the radial basis function (RBF) network. RBFs are typically\n",
+ "made up of three layers: an input layer, a hidden layer with\n",
+ "non-linear radial symmetric activation functions and a linear output\n",
+ "layer (''linear'' here means that each node in the output layer has a\n",
+ "linear activation function). The layers are normally fully-connected\n",
+ "and there are no cycles, thus RBFs can be viewed as a type of\n",
+ "fully-connected FFNN. They are however usually treated as a separate\n",
+ "type of NN due the unusual activation functions.\n",
+ "\n",
+ "\n",
+ "## Multilayer perceptrons\n",
+ "\n",
+ "One uses often so-called fully-connected feed-forward neural networks\n",
+ "with three or more layers (an input layer, one or more hidden layers\n",
+ "and an output layer) consisting of neurons that have non-linear\n",
+ "activation functions.\n",
+ "\n",
+ "Such networks are often called *multilayer perceptrons* (MLPs).\n",
+ "\n",
+ "\n",
+ "## Why multilayer perceptrons?\n",
+ "\n",
+ "According to the *Universal approximation theorem*, a feed-forward\n",
+ "neural network with just a single hidden layer containing a finite\n",
+ "number of neurons can approximate a continuous multidimensional\n",
+ "function to arbitrary accuracy, assuming the activation function for\n",
+ "the hidden layer is a **non-constant, bounded and\n",
+ "monotonically-increasing continuous function**.\n",
+ "\n",
+ "Note that the requirements on the activation function only applies to\n",
+ "the hidden layer, the output nodes are always assumed to be linear, so\n",
+ "as to not restrict the range of output values.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Mathematical model\n",
+ "\n",
+ "The output $y$ is produced via the activation function $f$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "y = f\\left(\\sum_{i=1}^n w_ix_i + b_i\\right) = f(z),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "This function receives $x_i$ as inputs.\n",
+ "Here the activation $z=(\\sum_{i=1}^n w_ix_i+b_i)$. \n",
+ "In an FFNN of such neurons, the *inputs* $x_i$ are the *outputs* of\n",
+ "the neurons in the preceding layer. Furthermore, an MLP is\n",
+ "fully-connected, which means that each neuron receives a weighted sum\n",
+ "of the outputs of *all* neurons in the previous layer.\n",
+ "\n",
+ "\n",
+ "## Mathematical model\n",
+ "\n",
+ "First, for each node $i$ in the first hidden layer, we calculate a weighted sum $z_i^1$ of the input coordinates $x_j$,"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation} z_i^1 = \\sum_{j=1}^{M} w_{ij}^1 x_j + b_i^1\n",
+ "\\label{_auto1} \\tag{2}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Here $b_i$ is the so-called bias which is normally needed in\n",
+ "case of zero activation weights or inputs. How to fix the biases and\n",
+ "the weights will be discussed below. The value of $z_i^1$ is the\n",
+ "argument to the activation function $f_i$ of each node $i$, The\n",
+ "variable $M$ stands for all possible inputs to a given node $i$ in the\n",
+ "first layer. We define the output $y_i^1$ of all neurons in layer 1 as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " y_i^1 = f(z_i^1) = f\\left(\\sum_{j=1}^M w_{ij}^1 x_j + b_i^1\\right)\n",
+ "\\label{outputLayer1} \\tag{3}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where we assume that all nodes in the same layer have identical\n",
+ "activation functions, hence the notation $f$. In general, we could assume in the more general case that different layers have different activation functions.\n",
+ "In this case we would identify these functions with a superscript $l$ for the $l$-th layer,"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " y_i^l = f^l(u_i^l) = f^l\\left(\\sum_{j=1}^{N_{l-1}} w_{ij}^l y_j^{l-1} + b_i^l\\right)\n",
+ "\\label{generalLayer} \\tag{4}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $N_l$ is the number of nodes in layer $l$. When the output of\n",
+ "all the nodes in the first hidden layer are computed, the values of\n",
+ "the subsequent layer can be calculated and so forth until the output\n",
+ "is obtained.\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Mathematical model\n",
+ "\n",
+ "The output of neuron $i$ in layer 2 is thus,"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " y_i^2 = f^2\\left(\\sum_{j=1}^N w_{ij}^2 y_j^1 + b_i^2\\right) \n",
+ "\\label{_auto2} \\tag{5}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation} \n",
+ " = f^2\\left[\\sum_{j=1}^N w_{ij}^2f^1\\left(\\sum_{k=1}^M w_{jk}^1 x_k + b_j^1\\right) + b_i^2\\right]\n",
+ "\\label{outputLayer2} \\tag{6}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where we have substituted $y_k^1$ with the inputs $x_k$. Finally, the ANN output reads"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " y_i^3 = f^3\\left(\\sum_{j=1}^N w_{ij}^3 y_j^2 + b_i^3\\right) \n",
+ "\\label{_auto3} \\tag{7}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation} \n",
+ " = f_3\\left[\\sum_{j} w_{ij}^3 f^2\\left(\\sum_{k} w_{jk}^2 f^1\\left(\\sum_{m} w_{km}^1 x_m + b_k^1\\right) + b_j^2\\right)\n",
+ " + b_1^3\\right]\n",
+ "\\label{_auto4} \\tag{8}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Mathematical model\n",
+ "\n",
+ "We can generalize this expression to an MLP with $l$ hidden\n",
+ "layers. The complete functional form is,"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ "y^{l+1}_i = f^{l+1}\\left[\\!\\sum_{j=1}^{N_l} w_{ij}^3 f^l\\left(\\sum_{k=1}^{N_{l-1}}w_{jk}^{l-1}\\left(\\dots f^1\\left(\\sum_{n=1}^{N_0} w_{mn}^1 x_n+ b_m^1\\right)\\dots\\right)+b_k^2\\right)+b_1^3\\right] \n",
+ "\\label{completeNN} \\tag{9}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which illustrates a basic property of MLPs: The only independent\n",
+ "variables are the input values $x_n$.\n",
+ "\n",
+ "\n",
+ "## Mathematical model\n",
+ "\n",
+ "This confirms that an MLP, despite its quite convoluted mathematical\n",
+ "form, is nothing more than an analytic function, specifically a\n",
+ "mapping of real-valued vectors $\\hat{x} \\in \\mathbb{R}^n \\rightarrow\n",
+ "\\hat{y} \\in \\mathbb{R}^m$.\n",
+ "\n",
+ "Furthermore, the flexibility and universality of an MLP can be\n",
+ "illustrated by realizing that the expression is essentially a nested\n",
+ "sum of scaled activation functions of the form"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " f(x) = c_1 f(c_2 x + c_3) + c_4\n",
+ "\\label{_auto5} \\tag{10}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where the parameters $c_i$ are weights and biases. By adjusting these\n",
+ "parameters, the activation functions can be shifted up and down or\n",
+ "left and right, change slope or be rescaled which is the key to the\n",
+ "flexibility of a neural network.\n",
+ "\n",
+ "\n",
+ "### Matrix-vector notation\n",
+ "\n",
+ "We can introduce a more convenient notation for the activations in an A NN. \n",
+ "\n",
+ "Additionally, we can represent the biases and activations\n",
+ "as layer-wise column vectors $\\hat{b}_l$ and $\\hat{y}_l$, so that the $i$-th element of each vector \n",
+ "is the bias $b_i^l$ and activation $y_i^l$ of node $i$ in layer $l$ respectively. \n",
+ "\n",
+ "We have that $\\mathrm{W}_l$ is an $N_{l-1} \\times N_l$ matrix, while $\\hat{b}_l$ and $\\hat{y}_l$ are $N_l \\times 1$ column vectors. \n",
+ "With this notation, the sum becomes a matrix-vector multiplication, and we can write\n",
+ "the equation for the activations of hidden layer 2 (assuming three nodes for simplicity) as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " \\hat{y}_2 = f_2(\\mathrm{W}_2 \\hat{y}_{1} + \\hat{b}_{2}) = \n",
+ " f_2\\left(\\left[\\begin{array}{ccc}\n",
+ " w^2_{11} &w^2_{12} &w^2_{13} \\\\\n",
+ " w^2_{21} &w^2_{22} &w^2_{23} \\\\\n",
+ " w^2_{31} &w^2_{32} &w^2_{33} \\\\\n",
+ " \\end{array} \\right] \\cdot\n",
+ " \\left[\\begin{array}{c}\n",
+ " y^1_1 \\\\\n",
+ " y^1_2 \\\\\n",
+ " y^1_3 \\\\\n",
+ " \\end{array}\\right] + \n",
+ " \\left[\\begin{array}{c}\n",
+ " b^2_1 \\\\\n",
+ " b^2_2 \\\\\n",
+ " b^2_3 \\\\\n",
+ " \\end{array}\\right]\\right).\n",
+ "\\label{_auto6} \\tag{11}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "### Matrix-vector notation and activation\n",
+ "\n",
+ "The activation of node $i$ in layer 2 is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ " y^2_i = f_2\\Bigr(w^2_{i1}y^1_1 + w^2_{i2}y^1_2 + w^2_{i3}y^1_3 + b^2_i\\Bigr) = \n",
+ " f_2\\left(\\sum_{j=1}^3 w^2_{ij} y_j^1 + b^2_i\\right).\n",
+ "\\label{_auto7} \\tag{12}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "This is not just a convenient and compact notation, but also a useful\n",
+ "and intuitive way to think about MLPs: The output is calculated by a\n",
+ "series of matrix-vector multiplications and vector additions that are\n",
+ "used as input to the activation functions. For each operation\n",
+ "$\\mathrm{W}_l \\hat{y}_{l-1}$ we move forward one layer.\n",
+ "\n",
+ "\n",
+ "\n",
+ "### Activation functions\n",
+ "\n",
+ "A property that characterizes a neural network, other than its\n",
+ "connectivity, is the choice of activation function(s). As described\n",
+ "in, the following restrictions are imposed on an activation function\n",
+ "for a FFNN to fulfill the universal approximation theorem\n",
+ "\n",
+ " * Non-constant\n",
+ "\n",
+ " * Bounded\n",
+ "\n",
+ " * Monotonically-increasing\n",
+ "\n",
+ " * Continuous\n",
+ "\n",
+ "### Activation functions, Logistic and Hyperbolic ones\n",
+ "\n",
+ "The second requirement excludes all linear functions. Furthermore, in\n",
+ "a MLP with only linear activation functions, each layer simply\n",
+ "performs a linear transformation of its inputs.\n",
+ "\n",
+ "Regardless of the number of layers, the output of the NN will be\n",
+ "nothing but a linear function of the inputs. Thus we need to introduce\n",
+ "some kind of non-linearity to the NN to be able to fit non-linear\n",
+ "functions Typical examples are the logistic *Sigmoid*"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "f(x) = \\frac{1}{1 + e^{-x}},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and the *hyperbolic tangent* function"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "f(x) = \\tanh(x)\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "### Relevance\n",
+ "\n",
+ "The *sigmoid* function are more biologically plausible because the\n",
+ "output of inactive neurons are zero. Such activation function are\n",
+ "called *one-sided*. However, it has been shown that the hyperbolic\n",
+ "tangent performs better than the sigmoid for training MLPs. has\n",
+ "become the most popular for *deep neural networks*"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "%matplotlib inline\n",
+ "\n",
+ "\"\"\"The sigmoid function (or the logistic curve) is a \n",
+ "function that takes any real number, z, and outputs a number (0,1).\n",
+ "It is useful in neural networks for assigning weights on a relative scale.\n",
+ "The value z is the weighted sum of parameters involved in the learning algorithm.\"\"\"\n",
+ "\n",
+ "import numpy\n",
+ "import matplotlib.pyplot as plt\n",
+ "import math as mt\n",
+ "\n",
+ "z = numpy.arange(-5, 5, .1)\n",
+ "sigma_fn = numpy.vectorize(lambda z: 1/(1+numpy.exp(-z)))\n",
+ "sigma = sigma_fn(z)\n",
+ "\n",
+ "fig = plt.figure()\n",
+ "ax = fig.add_subplot(111)\n",
+ "ax.plot(z, sigma)\n",
+ "ax.set_ylim([-0.1, 1.1])\n",
+ "ax.set_xlim([-5,5])\n",
+ "ax.grid(True)\n",
+ "ax.set_xlabel('z')\n",
+ "ax.set_title('sigmoid function')\n",
+ "\n",
+ "plt.show()\n",
+ "\n",
+ "\"\"\"Step Function\"\"\"\n",
+ "z = numpy.arange(-5, 5, .02)\n",
+ "step_fn = numpy.vectorize(lambda z: 1.0 if z >= 0.0 else 0.0)\n",
+ "step = step_fn(z)\n",
+ "\n",
+ "fig = plt.figure()\n",
+ "ax = fig.add_subplot(111)\n",
+ "ax.plot(z, step)\n",
+ "ax.set_ylim([-0.5, 1.5])\n",
+ "ax.set_xlim([-5,5])\n",
+ "ax.grid(True)\n",
+ "ax.set_xlabel('z')\n",
+ "ax.set_title('step function')\n",
+ "\n",
+ "plt.show()\n",
+ "\n",
+ "\"\"\"Sine Function\"\"\"\n",
+ "z = numpy.arange(-2*mt.pi, 2*mt.pi, 0.1)\n",
+ "t = numpy.sin(z)\n",
+ "\n",
+ "fig = plt.figure()\n",
+ "ax = fig.add_subplot(111)\n",
+ "ax.plot(z, t)\n",
+ "ax.set_ylim([-1.0, 1.0])\n",
+ "ax.set_xlim([-2*mt.pi,2*mt.pi])\n",
+ "ax.grid(True)\n",
+ "ax.set_xlabel('z')\n",
+ "ax.set_title('sine function')\n",
+ "\n",
+ "plt.show()\n",
+ "\n",
+ "\"\"\"Plots a graph of the squashing function used by a rectified linear\n",
+ "unit\"\"\"\n",
+ "z = numpy.arange(-2, 2, .1)\n",
+ "zero = numpy.zeros(len(z))\n",
+ "y = numpy.max([zero, z], axis=0)\n",
+ "\n",
+ "fig = plt.figure()\n",
+ "ax = fig.add_subplot(111)\n",
+ "ax.plot(z, y)\n",
+ "ax.set_ylim([-2.0, 2.0])\n",
+ "ax.set_xlim([-2.0, 2.0])\n",
+ "ax.grid(True)\n",
+ "ax.set_xlabel('z')\n",
+ "ax.set_title('Rectified linear unit')\n",
+ "\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## The multilayer perceptron (MLP)\n",
+ "\n",
+ "The multilayer perceptron is a very popular, and easy to implement approach, to deep learning. It consists of\n",
+ "1. A neural network with one or more layers of nodes between the input and the output nodes.\n",
+ "\n",
+ "2. The multilayer network structure, or architecture, or topology, consists of an input layer, one or more hidden layers, and one output layer.\n",
+ "\n",
+ "3. The input nodes pass values to the first hidden layer, its nodes pass the information on to the second and so on till we reach the output layer.\n",
+ "\n",
+ "As a convention it is normal to call a network with one layer of input units, one layer of hidden\n",
+ "units and one layer of output units as a two-layer network. A network with two layers of hidden units is called a three-layer network etc etc.\n",
+ "\n",
+ "For an MLP network there is no direct connection between the output nodes/neurons/units and the input nodes/neurons/units.\n",
+ "Hereafter we will call the various entities of a layer for nodes.\n",
+ "There are also no connections within a single layer.\n",
+ "\n",
+ "The number of input nodes does not need to equal the number of output\n",
+ "nodes. This applies also to the hidden layers. Each layer may have its\n",
+ "own number of nodes and activation functions.\n",
+ "\n",
+ "The hidden layers have their name from the fact that they are not\n",
+ "linked to observables and as we will see below when we define the\n",
+ "so-called activation $\\hat{z}$, we can think of this as a basis\n",
+ "expansion of the original inputs $\\hat{x}$. The difference however\n",
+ "between neural networks and say linear regression is that now these\n",
+ "basis functions (which will correspond to the weights in the network)\n",
+ "are learned from data. This results in an important difference between\n",
+ "neural networks and deep learning approaches on one side and methods\n",
+ "like logistic regression or linear regression and their modifications on the other side.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## From one to many layers, the universal approximation theorem\n",
+ "\n",
+ "\n",
+ "A neural network with only one layer, what we called the simple\n",
+ "perceptron, is best suited if we have a standard binary model with\n",
+ "clear (linear) boundaries between the outcomes. As such it could\n",
+ "equally well be replaced by standard linear regression or logistic\n",
+ "regression. Networks with one or more hidden layers approximate\n",
+ "systems with more complex boundaries.\n",
+ "\n",
+ "As stated earlier, \n",
+ "an important theorem in studies of neural networks, restated without\n",
+ "proof here, is the [universal approximation\n",
+ "theorem](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.441.7873&rep=rep1&type=pdf).\n",
+ "\n",
+ "It states that a feed-forward network with a single hidden layer\n",
+ "containing a finite number of neurons can approximate continuous\n",
+ "functions on compact subsets of real functions. The theorem thus\n",
+ "states that simple neural networks can represent a wide variety of\n",
+ "interesting functions when given appropriate parameters. It is the\n",
+ "multilayer feedforward architecture itself which gives neural networks\n",
+ "the potential of being universal approximators.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Deriving the back propagation code for a multilayer perceptron model\n",
+ "\n",
+ "\n",
+ "**Note: figures will be inserted later!**\n",
+ "\n",
+ "As we have seen now in a feed forward network, we can express the final output of our network in terms of basic matrix-vector multiplications.\n",
+ "The unknowwn quantities are our weights $w_{ij}$ and we need to find an algorithm for changing them so that our errors are as small as possible.\n",
+ "This leads us to the famous [back propagation algorithm](https://www.nature.com/articles/323533a0).\n",
+ "\n",
+ "The questions we want to ask are how do changes in the biases and the\n",
+ "weights in our network change the cost function and how can we use the\n",
+ "final output to modify the weights?\n",
+ "\n",
+ "To derive these equations let us start with a plain regression problem\n",
+ "and define our cost function as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "{\\cal C}(\\hat{W}) = \\frac{1}{2}\\sum_{i=1}^n\\left(y_i - t_i\\right)^2,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where the $t_i$s are our $n$ targets (the values we want to\n",
+ "reproduce), while the outputs of the network after having propagated\n",
+ "all inputs $\\hat{x}$ are given by $y_i$. Below we will demonstrate\n",
+ "how the basic equations arising from the back propagation algorithm\n",
+ "can be modified in order to study classification problems with $K$\n",
+ "classes.\n",
+ "\n",
+ "\n",
+ "## Definitions\n",
+ "\n",
+ "With our definition of the targets $\\hat{t}$, the outputs of the\n",
+ "network $\\hat{y}$ and the inputs $\\hat{x}$ we\n",
+ "define now the activation $z_j^l$ of node/neuron/unit $j$ of the\n",
+ "$l$-th layer as a function of the bias, the weights which add up from\n",
+ "the previous layer $l-1$ and the forward passes/outputs\n",
+ "$\\hat{a}^{l-1}$ from the previous layer as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "z_j^l = \\sum_{i=1}^{M_{l-1}}w_{ij}^la_i^{l-1}+b_j^l,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $b_k^l$ are the biases from layer $l$. Here $M_{l-1}$\n",
+ "represents the total number of nodes/neurons/units of layer $l-1$. The\n",
+ "figure here illustrates this equation. We can rewrite this in a more\n",
+ "compact form as the matrix-vector products we discussed earlier,"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\hat{z}^l = \\left(\\hat{W}^l\\right)^T\\hat{a}^{l-1}+\\hat{b}^l.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "With the activation values $\\hat{z}^l$ we can in turn define the\n",
+ "output of layer $l$ as $\\hat{a}^l = f(\\hat{z}^l)$ where $f$ is our\n",
+ "activation function. In the examples here we will use the sigmoid\n",
+ "function discussed in our logistic regression lectures. We will also use the same activation function $f$ for all layers\n",
+ "and their nodes. It means we have"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "a_j^l = f(z_j^l) = \\frac{1}{1+\\exp{-(z_j^l)}}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Derivatives and the chain rule\n",
+ "\n",
+ "From the definition of the activation $z_j^l$ we have"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial z_j^l}{\\partial w_{ij}^l} = a_i^{l-1},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial z_j^l}{\\partial a_i^{l-1}} = w_{ji}^l.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "With our definition of the activation function we have that (note that this function depends only on $z_j^l$)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial a_j^l}{\\partial z_j^{l}} = a_j^l(1-a_j^l)=f(z_j^l)(1-f(z_j^l)).\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Derivative of the cost function\n",
+ "\n",
+ "With these definitions we can now compute the derivative of the cost function in terms of the weights.\n",
+ "\n",
+ "Let us specialize to the output layer $l=L$. Our cost function is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "{\\cal C}(\\hat{W^L}) = \\frac{1}{2}\\sum_{i=1}^n\\left(y_i - t_i\\right)^2=\\frac{1}{2}\\sum_{i=1}^n\\left(a_i^L - t_i\\right)^2,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The derivative of this function with respect to the weights is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial{\\cal C}(\\hat{W^L})}{\\partial w_{jk}^L} = \\left(a_j^L - t_j\\right)\\frac{\\partial a_j^L}{\\partial w_{jk}^{L}},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The last partial derivative can easily be computed and reads (by applying the chain rule)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial a_j^L}{\\partial w_{jk}^{L}} = \\frac{\\partial a_j^L}{\\partial z_{j}^{L}}\\frac{\\partial z_j^L}{\\partial w_{jk}^{L}}=a_j^L(1-a_j^L)a_k^{L-1},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Bringing it together, first back propagation equation\n",
+ "\n",
+ "We have thus"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial{\\cal C}(\\hat{W^L})}{\\partial w_{jk}^L} = \\left(a_j^L - t_j\\right)a_j^L(1-a_j^L)a_k^{L-1},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Defining"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\delta_j^L = a_j^L(1-a_j^L)\\left(a_j^L - t_j\\right) = f'(z_j^L)\\frac{\\partial {\\cal C}}{\\partial (a_j^L)},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and using the Hadamard product of two vectors we can write this as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\hat{\\delta}^L = f'(\\hat{z}^L)\\circ\\frac{\\partial {\\cal C}}{\\partial (\\hat{a}^L)}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "This is an important expression. The second term on the right handside\n",
+ "measures how fast the cost function is changing as a function of the $j$th\n",
+ "output activation. If, for example, the cost function doesn't depend\n",
+ "much on a particular output node $j$, then $\\delta_j^L$ will be small,\n",
+ "which is what we would expect. The first term on the right, measures\n",
+ "how fast the activation function $f$ is changing at a given activation\n",
+ "value $z_j^L$.\n",
+ "\n",
+ "Notice that everything in the above equations is easily computed. In\n",
+ "particular, we compute $z_j^L$ while computing the behaviour of the\n",
+ "network, and it is only a small additional overhead to compute\n",
+ "$f'(z^L_j)$. The exact form of the derivative with respect to the\n",
+ "output depends on the form of the cost function.\n",
+ "However, provided the cost function is known there should be little\n",
+ "trouble in calculating"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial {\\cal C}}{\\partial (a_j^L)}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "With the definition of $\\delta_j^L$ we have a more compact definition of the derivative of the cost function in terms of the weights, namely"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial{\\cal C}(\\hat{W^L})}{\\partial w_{jk}^L} = \\delta_j^La_k^{L-1}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Derivatives in terms of $z_j^L$\n",
+ "\n",
+ "It is also easy to see that our previous equation can be written as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\delta_j^L =\\frac{\\partial {\\cal C}}{\\partial z_j^L}= \\frac{\\partial {\\cal C}}{\\partial a_j^L}\\frac{\\partial a_j^L}{\\partial z_j^L},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which can also be interpreted as the partial derivative of the cost function with respect to the biases $b_j^L$, namely"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\delta_j^L = \\frac{\\partial {\\cal C}}{\\partial b_j^L}\\frac{\\partial b_j^L}{\\partial z_j^L}=\\frac{\\partial {\\cal C}}{\\partial b_j^L},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "That is, the error $\\delta_j^L$ is exactly equal to the rate of change of the cost function as a function of the bias. \n",
+ "\n",
+ "## Bringing it together\n",
+ "\n",
+ "We have now three equations that are essential for the computations of the derivatives of the cost function at the output layer. These equations are needed to start the algorithm and they are\n",
+ "\n",
+ "**The starting equations.**"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ "\\frac{\\partial{\\cal C}(\\hat{W^L})}{\\partial w_{jk}^L} = \\delta_j^La_k^{L-1},\n",
+ "\\label{_auto8} \\tag{13}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ "\\delta_j^L = f'(z_j^L)\\frac{\\partial {\\cal C}}{\\partial (a_j^L)},\n",
+ "\\label{_auto9} \\tag{14}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ "\\delta_j^L = \\frac{\\partial {\\cal C}}{\\partial b_j^L},\n",
+ "\\label{_auto10} \\tag{15}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "An interesting consequence of the above equations is that when the\n",
+ "activation $a_k^{L-1}$ is small, the gradient term, that is the\n",
+ "derivative of the cost function with respect to the weights, will also\n",
+ "tend to be small. We say then that the weight learns slowly, meaning\n",
+ "that it changes slowly when we minimize the weights via say gradient\n",
+ "descent. In this case we say the system learns slowly.\n",
+ "\n",
+ "Another interesting feature is that is when the activation function,\n",
+ "represented by the sigmoid function here, is rather flat when we move towards\n",
+ "its end values $0$ and $1$ (see the above Python codes). In these\n",
+ "cases, the derivatives of the activation function will also be close\n",
+ "to zero, meaning again that the gradients will be small and the\n",
+ "network learns slowly again.\n",
+ "\n",
+ "\n",
+ "\n",
+ "We need a fourth equation and we are set. We are going to propagate\n",
+ "backwards in order to the determine the weights and biases. In order\n",
+ "to do so we need to represent the error in the layer before the final\n",
+ "one $L-1$ in terms of the errors in the final output layer.\n",
+ "\n",
+ "\n",
+ "## Final back propagating equation\n",
+ "\n",
+ "We have that (replacing $L$ with a general layer $l$)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\delta_j^l =\\frac{\\partial {\\cal C}}{\\partial z_j^l}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We want to express this in terms of the equations for layer $l+1$. Using the chain rule and summing over all $k$ entries we have"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\delta_j^l =\\sum_k \\frac{\\partial {\\cal C}}{\\partial z_k^{l+1}}\\frac{\\partial z_k^{l+1}}{\\partial z_j^{l}}=\\sum_k \\delta_k^{l+1}\\frac{\\partial z_k^{l+1}}{\\partial z_j^{l}},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and recalling that"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "z_j^{l+1} = \\sum_{i=1}^{M_{l}}w_{ij}^{l+1}a_i^{l}+b_j^{l+1},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "with $M_l$ being the number of nodes in layer $l$, we obtain"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\delta_j^l =\\sum_k \\delta_k^{l+1}w_{kj}^{l+1}f'(z_j^l),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "This is our final equation.\n",
+ "\n",
+ "We are now ready to set up the algorithm for back propagation and learning the weights and biases.\n",
+ "\n",
+ "\n",
+ "## Setting up the Back propagation algorithm\n",
+ "\n",
+ "\n",
+ "\n",
+ "The four equations provide us with a way of computing the gradient of the cost function. Let us write this out in the form of an algorithm.\n",
+ "\n",
+ "First, we set up the input data $\\hat{x}$ and the activations\n",
+ "$\\hat{z}_1$ of the input layer and compute the activation function and\n",
+ "the pertinent outputs $\\hat{a}^1$.\n",
+ "\n",
+ "\n",
+ "\n",
+ "Secondly, we perform then the feed forward till we reach the output\n",
+ "layer and compute all $\\hat{z}_l$ of the input layer and compute the\n",
+ "activation function and the pertinent outputs $\\hat{a}^l$ for\n",
+ "$l=2,3,\\dots,L$.\n",
+ "\n",
+ "\n",
+ "\n",
+ "Thereafter we compute the ouput error $\\hat{\\delta}^L$ by computing all"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\delta_j^L = f'(z_j^L)\\frac{\\partial {\\cal C}}{\\partial (a_j^L)}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Then we compute the back propagate error for each $l=L-1,L-2,\\dots,2$ as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\delta_j^l = \\sum_k \\delta_k^{l+1}w_{kj}^{l+1}f'(z_j^l).\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Finally, we update the weights and the biases using gradient descent for each $l=L-1,L-2,\\dots,2$ and update the weights and biases according to the rules"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "w_{jk}^l\\leftarrow = w_{jk}^l- \\eta \\delta_j^la_k^{l-1},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "b_j^l \\leftarrow b_j^l-\\eta \\frac{\\partial {\\cal C}}{\\partial b_j^l}=b_j^l-\\eta \\delta_j^l,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The parameter $\\eta$ is the learning parameter discussed in connection with the gradient descent methods.\n",
+ "Here it is convenient to use stochastic gradient descent (see the examples below) with mini-batches with an outer loop that steps through multiple epochs of training.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Setting up a Multi-layer perceptron model for classification\n",
+ "\n",
+ "We are now gong to develop an example based on the MNIST data\n",
+ "base. This is a classification problem and we need to use our\n",
+ "cross-entropy function we discussed in connection with logistic\n",
+ "regression. The cross-entropy defines our cost function for the\n",
+ "classificaton problems with neural networks.\n",
+ "\n",
+ "In binary classification with two classes $(0, 1)$ we define the\n",
+ "logistic/sigmoid function as the probability that a particular input\n",
+ "is in class $0$ or $1$. This is possible because the logistic\n",
+ "function takes any input from the real numbers and inputs a number\n",
+ "between 0 and 1, and can therefore be interpreted as a probability. It\n",
+ "also has other nice properties, such as a derivative that is simple to\n",
+ "calculate.\n",
+ "\n",
+ "For an input $\\boldsymbol{a}$ from the hidden layer, the probability that the input $\\boldsymbol{x}$\n",
+ "is in class 0 or 1 is just. We let $\\theta$ represent the unknown weights and biases to be adjusted by our equations). The variable $x$\n",
+ "represents our activation values $z$. We have"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "P(y = 0 \\mid \\hat{x}, \\hat{\\theta}) = \\frac{1}{1 + \\exp{(- \\hat{x}})} ,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "P(y = 1 \\mid \\hat{x}, \\hat{\\theta}) = 1 - P(y = 0 \\mid \\hat{x}, \\hat{\\theta}) ,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where $y \\in \\{0, 1\\}$ and $\\hat{\\theta}$ represents the weights and biases\n",
+ "of our network.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Defining the cost function\n",
+ "\n",
+ "Our cost function is given as (see the Logistic regression lectures)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathcal{C}(\\hat{\\theta}) = - \\ln P(\\mathcal{D} \\mid \\hat{\\theta}) = - \\sum_{i=1}^n\n",
+ "y_i \\ln[P(y_i = 0)] + (1 - y_i) \\ln [1 - P(y_i = 0)] = \\sum_{i=1}^n \\mathcal{L}_i(\\hat{\\theta}) .\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "This last equality means that we can interpret our *cost* function as a sum over the *loss* function\n",
+ "for each point in the dataset $\\mathcal{L}_i(\\hat{\\theta})$. \n",
+ "The negative sign is just so that we can think about our algorithm as minimizing a positive number, rather\n",
+ "than maximizing a negative number. \n",
+ "\n",
+ "In *multiclass* classification it is common to treat each integer label as a so called *one-hot* vector: \n",
+ "\n",
+ "$y = 5 \\quad \\rightarrow \\quad \\hat{y} = (0, 0, 0, 0, 0, 1, 0, 0, 0, 0) ,$ and\n",
+ "\n",
+ "\n",
+ "$y = 1 \\quad \\rightarrow \\quad \\hat{y} = (0, 1, 0, 0, 0, 0, 0, 0, 0, 0) ,$ \n",
+ "\n",
+ "\n",
+ "i.e. a binary bit string of length $C$, where $C = 10$ is the number of classes in the MNIST dataset (numbers from $0$ to $9$).. \n",
+ "\n",
+ "If $\\hat{x}_i$ is the $i$-th input (image), $y_{ic}$ refers to the $c$-th component of the $i$-th\n",
+ "output vector $\\hat{y}_i$. \n",
+ "The probability of $\\hat{x}_i$ being in class $c$ will be given by the softmax function:"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "P(y_{ic} = 1 \\mid \\hat{x}_i, \\hat{\\theta}) = \\frac{\\exp{((\\hat{a}_i^{hidden})^T \\hat{w}_c)}}\n",
+ "{\\sum_{c'=0}^{C-1} \\exp{((\\hat{a}_i^{hidden})^T \\hat{w}_{c'})}} ,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which reduces to the logistic function in the binary case. \n",
+ "The likelihood of this $C$-class classifier\n",
+ "is now given as:"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "P(\\mathcal{D} \\mid \\hat{\\theta}) = \\prod_{i=1}^n \\prod_{c=0}^{C-1} [P(y_{ic} = 1)]^{y_{ic}} .\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Again we take the negative log-likelihood to define our cost function:"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathcal{C}(\\hat{\\theta}) = - \\log{P(\\mathcal{D} \\mid \\hat{\\theta})}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "See the logistic regression lectures for a full definition of the cost function.\n",
+ "\n",
+ "The back propagation equations need now only a small change, namely the definition of a new cost function. We are thus ready to use the same equations as before!\n",
+ "\n",
+ "\n",
+ "## Example: binary classification problem\n",
+ "\n",
+ "As an example of the above, relevant for project 2 as well, let us consider a binary class. As discussed in our logistic regression lectures, we defined a cost function in terms of the parameters $\\beta$ as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathcal{C}(\\hat{\\beta}) = - \\sum_{i=1}^n \\left(y_i\\log{p(y_i \\vert x_i,\\hat{\\beta})}+(1-y_i)\\log{1-p(y_i \\vert x_i,\\hat{\\beta})}\\right),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where we had defined the logistic (sigmoid) function"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "p(y_i =1\\vert x_i,\\hat{\\beta})=\\frac{\\exp{(\\beta_0+\\beta_1 x_i)}}{1+\\exp{(\\beta_0+\\beta_1 x_i)}},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "p(y_i =0\\vert x_i,\\hat{\\beta})=1-p(y_i =1\\vert x_i,\\hat{\\beta}).\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The parameters $\\hat{\\beta}$ were defined using a minimization method like gradient descent or Newton-Raphson's method. \n",
+ "\n",
+ "Now we replace $x_i$ with the activation $z_i^l$ for a given layer $l$ and the outputs as $y_i=a_i^l=f(z_i^l)$, with $z_i^l$ now being a function of the weights $w_{ij}^l$ and biases $b_i^l$. \n",
+ "We have then"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "a_i^l = y_i = \\frac{\\exp{(z_i^l)}}{1+\\exp{(z_i^l)}},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "with"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "z_i^l = \\sum_{j}w_{ij}^l a_j^{l-1}+b_i^l,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where the superscript $l-1$ indicates that these are the outputs from layer $l-1$.\n",
+ "Our cost function at the final layer $l=L$ is now"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\mathcal{C}(\\hat{W}) = - \\sum_{i=1}^n \\left(t_i\\log{a_i^L}+(1-t_i)\\log{(1-a_i^L)}\\right),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "where we have defined the targets $t_i$. The derivatives of the cost function with respect to the output $a_i^L$ are then easily calculated and we get"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial \\mathcal{C}(\\hat{W})}{\\partial a_i^L} = \\frac{a_i^L-t_i}{a_i^L(1-a_i^L)}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "In case we use another activation function than the logistic one, we need to evaluate other derivatives. \n",
+ "\n",
+ "\n",
+ "\n",
+ "## The Softmax function\n",
+ "In case we employ the more general case given by the Softmax equation, we need to evaluate the derivative of the activation function with respect to the activation $z_i^l$, that is we need"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial f(z_i^l)}{\\partial w_{jk}^l} =\n",
+ "\\frac{\\partial f(z_i^l)}{\\partial z_j^l} \\frac{\\partial z_j^l}{\\partial w_{jk}^l}= \\frac{\\partial f(z_i^l)}{\\partial z_j^l}a_k^{l-1}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "For the Softmax function we have"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "f(z_i^l) = \\frac{\\exp{(z_i^l)}}{\\sum_{m=1}^K\\exp{(z_m^l)}}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Its derivative with respect to $z_j^l$ gives"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "\\frac{\\partial f(z_i^l)}{\\partial z_j^l}= f(z_i^l)\\left(\\delta_{ij}-f(z_j^l)\\right),\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "which in case of the simply binary model reduces to having $i=j$. \n",
+ "\n",
+ "\n",
+ "## Developing a code for doing neural networks with back propagation\n",
+ "\n",
+ "\n",
+ "One can identify a set of key steps when using neural networks to solve supervised learning problems: \n",
+ "\n",
+ "1. Collect and pre-process data \n",
+ "\n",
+ "2. Define model and architecture \n",
+ "\n",
+ "3. Choose cost function and optimizer \n",
+ "\n",
+ "4. Train the model \n",
+ "\n",
+ "5. Evaluate model performance on test data \n",
+ "\n",
+ "6. Adjust hyperparameters (if necessary, network architecture)\n",
+ "\n",
+ "## Collect and pre-process data\n",
+ "\n",
+ "Here we will be using the MNIST dataset, which is readily available through the **scikit-learn**\n",
+ "package. You may also find it for example [here](http://yann.lecun.com/exdb/mnist/). \n",
+ "The *MNIST* (Modified National Institute of Standards and Technology) database is a large database\n",
+ "of handwritten digits that is commonly used for training various image processing systems. \n",
+ "The MNIST dataset consists of 70 000 images of size $28\\times 28$ pixels, each labeled from 0 to 9. \n",
+ "The scikit-learn dataset we will use consists of a selection of 1797 images of size $8\\times 8$ collected and processed from this database. \n",
+ "\n",
+ "To feed data into a feed-forward neural network we need to represent\n",
+ "the inputs as a design/feature matrix $X = (n_{inputs}, n_{features})$. Each\n",
+ "row represents an *input*, in this case a handwritten digit, and\n",
+ "each column represents a *feature*, in this case a pixel. The\n",
+ "correct answers, also known as *labels* or *targets* are\n",
+ "represented as a 1D array of integers \n",
+ "$Y = (n_{inputs}) = (5, 3, 1, 8,...)$.\n",
+ "\n",
+ "As an example, say we want to build a neural network using supervised learning to predict Body-Mass Index (BMI) from\n",
+ "measurements of height (in m) \n",
+ "and weight (in kg). If we have measurements of 5 people the design/feature matrix could be for example: \n",
+ "\n",
+ "$$ X = \\begin{bmatrix}\n",
+ "1.85 & 81\\\\\n",
+ "1.71 & 65\\\\\n",
+ "1.95 & 103\\\\\n",
+ "1.55 & 42\\\\\n",
+ "1.63 & 56\n",
+ "\\end{bmatrix} ,$$ \n",
+ "\n",
+ "and the targets would be: \n",
+ "\n",
+ "$$ Y = (23.7, 22.2, 27.1, 17.5, 21.1) $$ \n",
+ "\n",
+ "Since each input image is a 2D matrix, we need to flatten the image\n",
+ "(i.e. \"unravel\" the 2D matrix into a 1D array) to turn the data into a\n",
+ "design/feature matrix. This means we lose all spatial information in the\n",
+ "image, such as locality and translational invariance. More complicated\n",
+ "architectures such as Convolutional Neural Networks can take advantage\n",
+ "of such information, and are most commonly applied when analyzing\n",
+ "images."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# import necessary packages\n",
+ "import numpy as np\n",
+ "import matplotlib.pyplot as plt\n",
+ "from sklearn import datasets\n",
+ "\n",
+ "\n",
+ "# ensure the same random numbers appear every time\n",
+ "np.random.seed(0)\n",
+ "\n",
+ "# display images in notebook\n",
+ "%matplotlib inline\n",
+ "plt.rcParams['figure.figsize'] = (12,12)\n",
+ "\n",
+ "\n",
+ "# download MNIST dataset\n",
+ "digits = datasets.load_digits()\n",
+ "\n",
+ "# define inputs and labels\n",
+ "inputs = digits.images\n",
+ "labels = digits.target\n",
+ "\n",
+ "print(\"inputs = (n_inputs, pixel_width, pixel_height) = \" + str(inputs.shape))\n",
+ "print(\"labels = (n_inputs) = \" + str(labels.shape))\n",
+ "\n",
+ "\n",
+ "# flatten the image\n",
+ "# the value -1 means dimension is inferred from the remaining dimensions: 8x8 = 64\n",
+ "n_inputs = len(inputs)\n",
+ "inputs = inputs.reshape(n_inputs, -1)\n",
+ "print(\"X = (n_inputs, n_features) = \" + str(inputs.shape))\n",
+ "\n",
+ "\n",
+ "# choose some random images to display\n",
+ "indices = np.arange(n_inputs)\n",
+ "random_indices = np.random.choice(indices, size=5)\n",
+ "\n",
+ "for i, image in enumerate(digits.images[random_indices]):\n",
+ " plt.subplot(1, 5, i+1)\n",
+ " plt.axis('off')\n",
+ " plt.imshow(image, cmap=plt.cm.gray_r, interpolation='nearest')\n",
+ " plt.title(\"Label: %d\" % digits.target[random_indices[i]])\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Train and test datasets\n",
+ "\n",
+ "Performing analysis before partitioning the dataset is a major error, that can lead to incorrect conclusions. \n",
+ "\n",
+ "We will reserve $80 \\%$ of our dataset for training and $20 \\%$ for testing. \n",
+ "\n",
+ "It is important that the train and test datasets are drawn randomly from our dataset, to ensure\n",
+ "no bias in the sampling. \n",
+ "Say you are taking measurements of weather data to predict the weather in the coming 5 days.\n",
+ "You don't want to train your model on measurements taken from the hours 00.00 to 12.00, and then test it on data\n",
+ "collected from 12.00 to 24.00."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "from sklearn.model_selection import train_test_split\n",
+ "\n",
+ "# one-liner from scikit-learn library\n",
+ "train_size = 0.8\n",
+ "test_size = 1 - train_size\n",
+ "X_train, X_test, Y_train, Y_test = train_test_split(inputs, labels, train_size=train_size,\n",
+ " test_size=test_size)\n",
+ "\n",
+ "# equivalently in numpy\n",
+ "def train_test_split_numpy(inputs, labels, train_size, test_size):\n",
+ " n_inputs = len(inputs)\n",
+ " inputs_shuffled = inputs.copy()\n",
+ " labels_shuffled = labels.copy()\n",
+ " \n",
+ " np.random.shuffle(inputs_shuffled)\n",
+ " np.random.shuffle(labels_shuffled)\n",
+ " \n",
+ " train_end = int(n_inputs*train_size)\n",
+ " X_train, X_test = inputs_shuffled[:train_end], inputs_shuffled[train_end:]\n",
+ " Y_train, Y_test = labels_shuffled[:train_end], labels_shuffled[train_end:]\n",
+ " \n",
+ " return X_train, X_test, Y_train, Y_test\n",
+ "\n",
+ "#X_train, X_test, Y_train, Y_test = train_test_split_numpy(inputs, labels, train_size, test_size)\n",
+ "\n",
+ "print(\"Number of training images: \" + str(len(X_train)))\n",
+ "print(\"Number of test images: \" + str(len(X_test)))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Define model and architecture\n",
+ "\n",
+ "Our simple feed-forward neural network will consist of an *input* layer, a single *hidden* layer and an *output* layer. The activation $y$ of each neuron is a weighted sum of inputs, passed through an activation function. In case of the simple perceptron model we have \n",
+ "\n",
+ "$$ z = \\sum_{i=1}^n w_i a_i ,$$\n",
+ "\n",
+ "$$ y = f(z) ,$$\n",
+ "\n",
+ "where $f$ is the activation function, $a_i$ represents input from neuron $i$ in the preceding layer\n",
+ "and $w_i$ is the weight to input $i$. \n",
+ "The activation of the neurons in the input layer is just the features (e.g. a pixel value). \n",
+ "\n",
+ "The simplest activation function for a neuron is the *Heaviside* function:\n",
+ "\n",
+ "$$ f(z) = \n",
+ "\\begin{cases}\n",
+ "1, & z > 0\\\\\n",
+ "0, & \\text{otherwise}\n",
+ "\\end{cases}\n",
+ "$$\n",
+ "\n",
+ "A feed-forward neural network with this activation is known as a *perceptron*. \n",
+ "For a binary classifier (i.e. two classes, 0 or 1, dog or not-dog) we can also use this in our output layer. \n",
+ "This activation can be generalized to $k$ classes (using e.g. the *one-against-all* strategy), \n",
+ "and we call these architectures *multiclass perceptrons*. \n",
+ "\n",
+ "However, it is now common to use the terms Single Layer Perceptron (SLP) (1 hidden layer) and \n",
+ "Multilayer Perceptron (MLP) (2 or more hidden layers) to refer to feed-forward neural networks with any activation function. \n",
+ "\n",
+ "Typical choices for activation functions include the sigmoid function, hyperbolic tangent, and Rectified Linear Unit (ReLU). \n",
+ "We will be using the sigmoid function $\\sigma(x)$: \n",
+ "\n",
+ "$$ f(x) = \\sigma(x) = \\frac{1}{1 + e^{-x}} ,$$\n",
+ "\n",
+ "which is inspired by probability theory (see logistic regression) and was most commonly used until about 2011. See the discussion below concerning other activation functions.\n",
+ "\n",
+ "\n",
+ "## Layers\n",
+ "\n",
+ "* Input \n",
+ "\n",
+ "Since each input image has 8x8 = 64 pixels or features, we have an input layer of 64 neurons. \n",
+ "\n",
+ "* Hidden layer\n",
+ "\n",
+ "We will use 50 neurons in the hidden layer receiving input from the neurons in the input layer. \n",
+ "Since each neuron in the hidden layer is connected to the 64 inputs we have 64x50 = 3200 weights to the hidden layer. \n",
+ "\n",
+ "* Output\n",
+ "\n",
+ "If we were building a binary classifier, it would be sufficient with a single neuron in the output layer,\n",
+ "which could output 0 or 1 according to the Heaviside function. This would be an example of a *hard* classifier, meaning it outputs the class of the input directly. However, if we are dealing with noisy data it is often beneficial to use a *soft* classifier, which outputs the probability of being in class 0 or 1. \n",
+ "\n",
+ "For a soft binary classifier, we could use a single neuron and interpret the output as either being the probability of being in class 0 or the probability of being in class 1. Alternatively we could use 2 neurons, and interpret each neuron as the probability of being in each class. \n",
+ "\n",
+ "Since we are doing multiclass classification, with 10 categories, it is natural to use 10 neurons in the output layer. We number the neurons $j = 0,1,...,9$. The activation of each output neuron $j$ will be according to the *softmax* function: \n",
+ "\n",
+ "$$ P(\\text{class $j$} \\mid \\text{input $\\hat{a}$}) = \\frac{\\exp{(\\hat{a}^T \\hat{w}_j)}}\n",
+ "{\\sum_{c=0}^{9} \\exp{(\\hat{a}^T \\hat{w}_c)}} ,$$ \n",
+ "\n",
+ "i.e. each neuron $j$ outputs the probability of being in class $j$ given an input from the hidden layer $\\hat{a}$, with $\\hat{w}_j$ the weights of neuron $j$ to the inputs. \n",
+ "The denominator is a normalization factor to ensure the outputs (probabilities) sum up to 1. \n",
+ "The exponent is just the weighted sum of inputs as before: \n",
+ "\n",
+ "$$ z_j = \\sum_{i=1}^n w_ {ij} a_i+b_j.$$ \n",
+ "\n",
+ "Since each neuron in the output layer is connected to the 50 inputs from the hidden layer we have 50x10 = 500\n",
+ "weights to the output layer.\n",
+ "\n",
+ "\n",
+ "## Weights and biases\n",
+ "\n",
+ "Typically weights are initialized with small values distributed around zero, drawn from a uniform\n",
+ "or normal distribution. Setting all weights to zero means all neurons give the same output, making the network useless. \n",
+ "\n",
+ "Adding a bias value to the weighted sum of inputs allows the neural network to represent a greater range\n",
+ "of values. Without it, any input with the value 0 will be mapped to zero (before being passed through the activation). The bias unit has an output of 1, and a weight to each neuron $j$, $b_j$: \n",
+ "\n",
+ "$$ z_j = \\sum_{i=1}^n w_ {ij} a_i + b_j.$$ \n",
+ "\n",
+ "The bias weights $\\hat{b}$ are often initialized to zero, but a small value like $0.01$ ensures all neurons have some output which can be backpropagated in the first training cycle."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# building our neural network\n",
+ "\n",
+ "n_inputs, n_features = X_train.shape\n",
+ "n_hidden_neurons = 50\n",
+ "n_categories = 10\n",
+ "\n",
+ "# we make the weights normally distributed using numpy.random.randn\n",
+ "\n",
+ "# weights and bias in the hidden layer\n",
+ "hidden_weights = np.random.randn(n_features, n_hidden_neurons)\n",
+ "hidden_bias = np.zeros(n_hidden_neurons) + 0.01\n",
+ "\n",
+ "# weights and bias in the output layer\n",
+ "output_weights = np.random.randn(n_hidden_neurons, n_categories)\n",
+ "output_bias = np.zeros(n_categories) + 0.01"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Feed-forward pass\n",
+ "\n",
+ "Denote $F$ the number of features, $H$ the number of hidden neurons and $C$ the number of categories. \n",
+ "For each input image we calculate a weighted sum of input features (pixel values) to each neuron $j$ in the hidden layer $l$: \n",
+ "\n",
+ "$$ z_{j}^{l} = \\sum_{i=1}^{F} w_{ij}^{l} x_i + b_{j}^{l},$$\n",
+ "\n",
+ "this is then passed through our activation function \n",
+ "\n",
+ "$$ a_{j}^{l} = f(z_{j}^{l}) .$$ \n",
+ "\n",
+ "We calculate a weighted sum of inputs (activations in the hidden layer) to each neuron $j$ in the output layer: \n",
+ "\n",
+ "$$ z_{j}^{L} = \\sum_{i=1}^{H} w_{ij}^{L} a_{i}^{l} + b_{j}^{L}.$$ \n",
+ "\n",
+ "Finally we calculate the output of neuron $j$ in the output layer using the softmax function: \n",
+ "\n",
+ "$$ a_{j}^{L} = \\frac{\\exp{(z_j^{L})}}\n",
+ "{\\sum_{c=0}^{C-1} \\exp{(z_c^{L})}} .$$ \n",
+ "\n",
+ "\n",
+ "## Matrix multiplications\n",
+ "\n",
+ "Since our data has the dimensions $X = (n_{inputs}, n_{features})$ and our weights to the hidden\n",
+ "layer have the dimensions \n",
+ "$W_{hidden} = (n_{features}, n_{hidden})$,\n",
+ "we can easily feed the network all our training data in one go by taking the matrix product \n",
+ "\n",
+ "$$ X W^{h} = (n_{inputs}, n_{hidden}),$$ \n",
+ "\n",
+ "and obtain a matrix that holds the weighted sum of inputs to the hidden layer\n",
+ "for each input image and each hidden neuron. \n",
+ "We also add the bias to obtain a matrix of weighted sums to the hidden layer $Z^{h}$: \n",
+ "\n",
+ "$$ \\hat{z}^{l} = \\hat{X} \\hat{W}^{l} + \\hat{b}^{l} ,$$\n",
+ "\n",
+ "meaning the same bias (1D array with size equal number of hidden neurons) is added to each input image. \n",
+ "This is then passed through the activation: \n",
+ "\n",
+ "$$ \\hat{a}^{l} = f(\\hat{z}^l) .$$ \n",
+ "\n",
+ "This is fed to the output layer: \n",
+ "\n",
+ "$$ \\hat{z}^{L} = \\hat{a}^{L} \\hat{W}^{L} + \\hat{b}^{L} .$$\n",
+ "\n",
+ "Finally we receive our output values for each image and each category by passing it through the softmax function: \n",
+ "\n",
+ "$$ output = softmax (\\hat{z}^{L}) = (n_{inputs}, n_{categories}) .$$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# setup the feed-forward pass, subscript h = hidden layer\n",
+ "\n",
+ "def sigmoid(x):\n",
+ " return 1/(1 + np.exp(-x))\n",
+ "\n",
+ "def feed_forward(X):\n",
+ " # weighted sum of inputs to the hidden layer\n",
+ " z_h = np.matmul(X, hidden_weights) + hidden_bias\n",
+ " # activation in the hidden layer\n",
+ " a_h = sigmoid(z_h)\n",
+ " \n",
+ " # weighted sum of inputs to the output layer\n",
+ " z_o = np.matmul(a_h, output_weights) + output_bias\n",
+ " # softmax output\n",
+ " # axis 0 holds each input and axis 1 the probabilities of each category\n",
+ " exp_term = np.exp(z_o)\n",
+ " probabilities = exp_term / np.sum(exp_term, axis=1, keepdims=True)\n",
+ " \n",
+ " return probabilities\n",
+ "\n",
+ "probabilities = feed_forward(X_train)\n",
+ "print(\"probabilities = (n_inputs, n_categories) = \" + str(probabilities.shape))\n",
+ "print(\"probability that image 0 is in category 0,1,2,...,9 = \\n\" + str(probabilities[0]))\n",
+ "print(\"probabilities sum up to: \" + str(probabilities[0].sum()))\n",
+ "print()\n",
+ "\n",
+ "# we obtain a prediction by taking the class with the highest likelihood\n",
+ "def predict(X):\n",
+ " probabilities = feed_forward(X)\n",
+ " return np.argmax(probabilities, axis=1)\n",
+ "\n",
+ "predictions = predict(X_train)\n",
+ "print(\"predictions = (n_inputs) = \" + str(predictions.shape))\n",
+ "print(\"prediction for image 0: \" + str(predictions[0]))\n",
+ "print(\"correct label for image 0: \" + str(Y_train[0]))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Choose cost function and optimizer\n",
+ "\n",
+ "To measure how well our neural network is doing we need to introduce a cost function. \n",
+ "We will call the function that gives the error of a single sample output the *loss* function, and the function\n",
+ "that gives the total error of our network across all samples the *cost* function.\n",
+ "A typical choice for multiclass classification is the *cross-entropy* loss, also known as the negative log likelihood. \n",
+ "\n",
+ "In *multiclass* classification it is common to treat each integer label as a so called *one-hot* vector: \n",
+ "\n",
+ "$$ y = 5 \\quad \\rightarrow \\quad \\hat{y} = (0, 0, 0, 0, 0, 1, 0, 0, 0, 0) ,$$ \n",
+ "\n",
+ "\n",
+ "$$ y = 1 \\quad \\rightarrow \\quad \\hat{y} = (0, 1, 0, 0, 0, 0, 0, 0, 0, 0) ,$$ \n",
+ "\n",
+ "\n",
+ "i.e. a binary bit string of length $C$, where $C = 10$ is the number of classes in the MNIST dataset. \n",
+ "\n",
+ "Let $y_{ic}$ denote the $c$-th component of the $i$-th one-hot vector. \n",
+ "We define the cost function $\\mathcal{C}$ as a sum over the cross-entropy loss for each point $\\hat{x}_i$ in the dataset.\n",
+ "\n",
+ "In the one-hot representation only one of the terms in the loss function is non-zero, namely the\n",
+ "probability of the correct category $c'$ \n",
+ "(i.e. the category $c'$ such that $y_{ic'} = 1$). This means that the cross entropy loss only punishes you for how wrong\n",
+ "you got the correct label. The probability of category $c$ is given by the softmax function. The vector $\\hat{\\theta}$ represents the parameters of our network, i.e. all the weights and biases. \n",
+ "\n",
+ "\n",
+ "\n",
+ "## Optimizing the cost function\n",
+ "\n",
+ "The network is trained by finding the weights and biases that minimize the cost function. One of the most widely used classes of methods is *gradient descent* and its generalizations. The idea behind gradient descent\n",
+ "is simply to adjust the weights in the direction where the gradient of the cost function is large and negative. This ensures we flow toward a *local* minimum of the cost function. \n",
+ "Each parameter $\\theta$ is iteratively adjusted according to the rule \n",
+ "\n",
+ "$$ \\theta_{i+1} = \\theta_i - \\eta \\nabla \\mathcal{C}(\\theta_i) ,$$\n",
+ "\n",
+ "where $\\eta$ is known as the *learning rate*, which controls how big a step we take towards the minimum. \n",
+ "This update can be repeated for any number of iterations, or until we are satisfied with the result. \n",
+ "\n",
+ "A simple and effective improvement is a variant called *Batch Gradient Descent*. \n",
+ "Instead of calculating the gradient on the whole dataset, we calculate an approximation of the gradient\n",
+ "on a subset of the data called a *minibatch*. \n",
+ "If there are $N$ data points and we have a minibatch size of $M$, the total number of batches\n",
+ "is $N/M$. \n",
+ "We denote each minibatch $B_k$, with $k = 1, 2,...,N/M$. The gradient then becomes: \n",
+ "\n",
+ "$$ \\nabla \\mathcal{C}(\\theta) = \\frac{1}{N} \\sum_{i=1}^N \\nabla \\mathcal{L}_i(\\theta) \\quad \\rightarrow \\quad\n",
+ "\\frac{1}{M} \\sum_{i \\in B_k} \\nabla \\mathcal{L}_i(\\theta) ,$$\n",
+ "\n",
+ "i.e. instead of averaging the loss over the entire dataset, we average over a minibatch. \n",
+ "\n",
+ "This has two important benefits: \n",
+ "1. Introducing stochasticity decreases the chance that the algorithm becomes stuck in a local minima. \n",
+ "\n",
+ "2. It significantly speeds up the calculation, since we do not have to use the entire dataset to calculate the gradient. \n",
+ "\n",
+ "The various optmization methods, with codes and algorithms, are discussed in our lectures on [Gradient descent approaches](https://compphysics.github.io/MachineLearning/doc/pub/Splines/html/Splines-bs.html).\n",
+ "\n",
+ "\n",
+ "## Regularization\n",
+ "\n",
+ "It is common to add an extra term to the cost function, proportional\n",
+ "to the size of the weights. This is equivalent to constraining the\n",
+ "size of the weights, so that they do not grow out of control.\n",
+ "Constraining the size of the weights means that the weights cannot\n",
+ "grow arbitrarily large to fit the training data, and in this way\n",
+ "reduces *overfitting*.\n",
+ "\n",
+ "We will measure the size of the weights using the so called *L2-norm*, meaning our cost function becomes: \n",
+ "\n",
+ "$$ \\mathcal{C}(\\theta) = \\frac{1}{N} \\sum_{i=1}^N \\mathcal{L}_i(\\theta) \\quad \\rightarrow \\quad\n",
+ "\\frac{1}{N} \\sum_{i=1}^N \\mathcal{L}_i(\\theta) + \\lambda \\lvert \\lvert \\hat{w} \\rvert \\rvert_2^2 \n",
+ "= \\frac{1}{N} \\sum_{i=1}^N \\mathcal{L}(\\theta) + \\lambda \\sum_{ij} w_{ij}^2,$$ \n",
+ "\n",
+ "i.e. we sum up all the weights squared. The factor $\\lambda$ is known as a regularization parameter.\n",
+ "\n",
+ "\n",
+ "In order to train the model, we need to calculate the derivative of\n",
+ "the cost function with respect to every bias and weight in the\n",
+ "network. In total our network has $(64 + 1)\\times 50=3250$ weights in\n",
+ "the hidden layer and $(50 + 1)\\times 10=510$ weights to the output\n",
+ "layer ($+1$ for the bias), and the gradient must be calculated for\n",
+ "every parameter. We use the *backpropagation* algorithm discussed\n",
+ "above. This is a clever use of the chain rule that allows us to\n",
+ "calculate the gradient efficently. \n",
+ "\n",
+ "\n",
+ "\n",
+ "## Matrix multiplication\n",
+ "\n",
+ "To more efficently train our network these equations are implemented using matrix operations. \n",
+ "The error in the output layer is calculated simply as, with $\\hat{t}$ being our targets, \n",
+ "\n",
+ "$$ \\delta_L = \\hat{t} - \\hat{y} = (n_{inputs}, n_{categories}) .$$ \n",
+ "\n",
+ "The gradient for the output weights is calculated as \n",
+ "\n",
+ "$$ \\nabla W_{L} = \\hat{a}^T \\delta_L = (n_{hidden}, n_{categories}) ,$$\n",
+ "\n",
+ "where $\\hat{a} = (n_{inputs}, n_{hidden})$. This simply means that we are summing up the gradients for each input. \n",
+ "Since we are going backwards we have to transpose the activation matrix. \n",
+ "\n",
+ "The gradient with respect to the output bias is then \n",
+ "\n",
+ "$$ \\nabla \\hat{b}_{L} = \\sum_{i=1}^{n_{inputs}} \\delta_L = (n_{categories}) .$$ \n",
+ "\n",
+ "The error in the hidden layer is \n",
+ "\n",
+ "$$ \\Delta_h = \\delta_L W_{L}^T \\circ f'(z_{h}) = \\delta_L W_{L}^T \\circ a_{h} \\circ (1 - a_{h}) = (n_{inputs}, n_{hidden}) ,$$ \n",
+ "\n",
+ "where $f'(a_{h})$ is the derivative of the activation in the hidden layer. The matrix products mean\n",
+ "that we are summing up the products for each neuron in the output layer. The symbol $\\circ$ denotes\n",
+ "the *Hadamard product*, meaning element-wise multiplication. \n",
+ "\n",
+ "This again gives us the gradients in the hidden layer: \n",
+ "\n",
+ "$$ \\nabla W_{h} = X^T \\delta_h = (n_{features}, n_{hidden}) ,$$ \n",
+ "\n",
+ "$$ \\nabla b_{h} = \\sum_{i=1}^{n_{inputs}} \\delta_h = (n_{hidden}) .$$"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# to categorical turns our integer vector into a onehot representation\n",
+ "from sklearn.metrics import accuracy_score\n",
+ "\n",
+ "# one-hot in numpy\n",
+ "def to_categorical_numpy(integer_vector):\n",
+ " n_inputs = len(integer_vector)\n",
+ " n_categories = np.max(integer_vector) + 1\n",
+ " onehot_vector = np.zeros((n_inputs, n_categories))\n",
+ " onehot_vector[range(n_inputs), integer_vector] = 1\n",
+ " \n",
+ " return onehot_vector\n",
+ "\n",
+ "#Y_train_onehot, Y_test_onehot = to_categorical(Y_train), to_categorical(Y_test)\n",
+ "Y_train_onehot, Y_test_onehot = to_categorical_numpy(Y_train), to_categorical_numpy(Y_test)\n",
+ "\n",
+ "def feed_forward_train(X):\n",
+ " # weighted sum of inputs to the hidden layer\n",
+ " z_h = np.matmul(X, hidden_weights) + hidden_bias\n",
+ " # activation in the hidden layer\n",
+ " a_h = sigmoid(z_h)\n",
+ " \n",
+ " # weighted sum of inputs to the output layer\n",
+ " z_o = np.matmul(a_h, output_weights) + output_bias\n",
+ " # softmax output\n",
+ " # axis 0 holds each input and axis 1 the probabilities of each category\n",
+ " exp_term = np.exp(z_o)\n",
+ " probabilities = exp_term / np.sum(exp_term, axis=1, keepdims=True)\n",
+ " \n",
+ " # for backpropagation need activations in hidden and output layers\n",
+ " return a_h, probabilities\n",
+ "\n",
+ "def backpropagation(X, Y):\n",
+ " a_h, probabilities = feed_forward_train(X)\n",
+ " \n",
+ " # error in the output layer\n",
+ " error_output = probabilities - Y\n",
+ " # error in the hidden layer\n",
+ " error_hidden = np.matmul(error_output, output_weights.T) * a_h * (1 - a_h)\n",
+ " \n",
+ " # gradients for the output layer\n",
+ " output_weights_gradient = np.matmul(a_h.T, error_output)\n",
+ " output_bias_gradient = np.sum(error_output, axis=0)\n",
+ " \n",
+ " # gradient for the hidden layer\n",
+ " hidden_weights_gradient = np.matmul(X.T, error_hidden)\n",
+ " hidden_bias_gradient = np.sum(error_hidden, axis=0)\n",
+ "\n",
+ " return output_weights_gradient, output_bias_gradient, hidden_weights_gradient, hidden_bias_gradient\n",
+ "\n",
+ "print(\"Old accuracy on training data: \" + str(accuracy_score(predict(X_train), Y_train)))\n",
+ "\n",
+ "eta = 0.01\n",
+ "lmbd = 0.01\n",
+ "for i in range(1000):\n",
+ " # calculate gradients\n",
+ " dWo, dBo, dWh, dBh = backpropagation(X_train, Y_train_onehot)\n",
+ " \n",
+ " # regularization term gradients\n",
+ " dWo += lmbd * output_weights\n",
+ " dWh += lmbd * hidden_weights\n",
+ " \n",
+ " # update weights and biases\n",
+ " output_weights -= eta * dWo\n",
+ " output_bias -= eta * dBo\n",
+ " hidden_weights -= eta * dWh\n",
+ " hidden_bias -= eta * dBh\n",
+ "\n",
+ "print(\"New accuracy on training data: \" + str(accuracy_score(predict(X_train), Y_train)))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Improving performance\n",
+ "\n",
+ "As we can see the network does not seem to be learning at all. It seems to be just guessing the label for each image. \n",
+ "In order to obtain a network that does something useful, we will have to do a bit more work. \n",
+ "\n",
+ "The choice of *hyperparameters* such as learning rate and regularization parameter is hugely influential for the performance of the network. Typically a *grid-search* is performed, wherein we test different hyperparameters separated by orders of magnitude. For example we could test the learning rates $\\eta = 10^{-6}, 10^{-5},...,10^{-1}$ with different regularization parameters $\\lambda = 10^{-6},...,10^{-0}$. \n",
+ "\n",
+ "Next, we haven't implemented minibatching yet, which introduces stochasticity and is though to act as an important regularizer on the weights. We call a feed-forward + backward pass with a minibatch an *iteration*, and a full training period\n",
+ "going through the entire dataset ($n/M$ batches) an *epoch*.\n",
+ "\n",
+ "If this does not improve network performance, you may want to consider altering the network architecture, adding more neurons or hidden layers. \n",
+ "Andrew Ng goes through some of these considerations in this [video](https://youtu.be/F1ka6a13S9I). You can find a summary of the video [here](https://kevinzakka.github.io/2016/09/26/applying-deep-learning/). \n",
+ "\n",
+ "\n",
+ "## Full object-oriented implementation\n",
+ "\n",
+ "It is very natural to think of the network as an object, with specific instances of the network\n",
+ "being realizations of this object with different hyperparameters. An implementation using Python classes provides a clean structure and interface, and the full implementation of our neural network is given below."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "class NeuralNetwork:\n",
+ " def __init__(\n",
+ " self,\n",
+ " X_data,\n",
+ " Y_data,\n",
+ " n_hidden_neurons=50,\n",
+ " n_categories=10,\n",
+ " epochs=10,\n",
+ " batch_size=100,\n",
+ " eta=0.1,\n",
+ " lmbd=0.0):\n",
+ "\n",
+ " self.X_data_full = X_data\n",
+ " self.Y_data_full = Y_data\n",
+ "\n",
+ " self.n_inputs = X_data.shape[0]\n",
+ " self.n_features = X_data.shape[1]\n",
+ " self.n_hidden_neurons = n_hidden_neurons\n",
+ " self.n_categories = n_categories\n",
+ "\n",
+ " self.epochs = epochs\n",
+ " self.batch_size = batch_size\n",
+ " self.iterations = self.n_inputs // self.batch_size\n",
+ " self.eta = eta\n",
+ " self.lmbd = lmbd\n",
+ "\n",
+ " self.create_biases_and_weights()\n",
+ "\n",
+ " def create_biases_and_weights(self):\n",
+ " self.hidden_weights = np.random.randn(self.n_features, self.n_hidden_neurons)\n",
+ " self.hidden_bias = np.zeros(self.n_hidden_neurons) + 0.01\n",
+ "\n",
+ " self.output_weights = np.random.randn(self.n_hidden_neurons, self.n_categories)\n",
+ " self.output_bias = np.zeros(self.n_categories) + 0.01\n",
+ "\n",
+ " def feed_forward(self):\n",
+ " # feed-forward for training\n",
+ " self.z_h = np.matmul(self.X_data, self.hidden_weights) + self.hidden_bias\n",
+ " self.a_h = sigmoid(self.z_h)\n",
+ "\n",
+ " self.z_o = np.matmul(self.a_h, self.output_weights) + self.output_bias\n",
+ "\n",
+ " exp_term = np.exp(self.z_o)\n",
+ " self.probabilities = exp_term / np.sum(exp_term, axis=1, keepdims=True)\n",
+ "\n",
+ " def feed_forward_out(self, X):\n",
+ " # feed-forward for output\n",
+ " z_h = np.matmul(X, self.hidden_weights) + self.hidden_bias\n",
+ " a_h = sigmoid(z_h)\n",
+ "\n",
+ " z_o = np.matmul(a_h, self.output_weights) + self.output_bias\n",
+ " \n",
+ " exp_term = np.exp(z_o)\n",
+ " probabilities = exp_term / np.sum(exp_term, axis=1, keepdims=True)\n",
+ " return probabilities\n",
+ "\n",
+ " def backpropagation(self):\n",
+ " error_output = self.probabilities - self.Y_data\n",
+ " error_hidden = np.matmul(error_output, self.output_weights.T) * self.a_h * (1 - self.a_h)\n",
+ "\n",
+ " self.output_weights_gradient = np.matmul(self.a_h.T, error_output)\n",
+ " self.output_bias_gradient = np.sum(error_output, axis=0)\n",
+ "\n",
+ " self.hidden_weights_gradient = np.matmul(self.X_data.T, error_hidden)\n",
+ " self.hidden_bias_gradient = np.sum(error_hidden, axis=0)\n",
+ "\n",
+ " if self.lmbd > 0.0:\n",
+ " self.output_weights_gradient += self.lmbd * self.output_weights\n",
+ " self.hidden_weights_gradient += self.lmbd * self.hidden_weights\n",
+ "\n",
+ " self.output_weights -= self.eta * self.output_weights_gradient\n",
+ " self.output_bias -= self.eta * self.output_bias_gradient\n",
+ " self.hidden_weights -= self.eta * self.hidden_weights_gradient\n",
+ " self.hidden_bias -= self.eta * self.hidden_bias_gradient\n",
+ "\n",
+ " def predict(self, X):\n",
+ " probabilities = self.feed_forward_out(X)\n",
+ " return np.argmax(probabilities, axis=1)\n",
+ "\n",
+ " def predict_probabilities(self, X):\n",
+ " probabilities = self.feed_forward_out(X)\n",
+ " return probabilities\n",
+ "\n",
+ " def train(self):\n",
+ " data_indices = np.arange(self.n_inputs)\n",
+ "\n",
+ " for i in range(self.epochs):\n",
+ " for j in range(self.iterations):\n",
+ " # pick datapoints with replacement\n",
+ " chosen_datapoints = np.random.choice(\n",
+ " data_indices, size=self.batch_size, replace=False\n",
+ " )\n",
+ "\n",
+ " # minibatch training data\n",
+ " self.X_data = self.X_data_full[chosen_datapoints]\n",
+ " self.Y_data = self.Y_data_full[chosen_datapoints]\n",
+ "\n",
+ " self.feed_forward()\n",
+ " self.backpropagation()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Evaluate model performance on test data\n",
+ "\n",
+ "To measure the performance of our network we evaluate how well it does it data it has never seen before, i.e. the test data. \n",
+ "We measure the performance of the network using the *accuracy* score. \n",
+ "The accuracy is as you would expect just the number of images correctly labeled divided by the total number of images. A perfect classifier will have an accuracy score of $1$. \n",
+ "\n",
+ "$$ \\text{Accuracy} = \\frac{\\sum_{i=1}^n I(\\hat{y}_i = y_i)}{n} ,$$ \n",
+ "\n",
+ "where $I$ is the indicator function, $1$ if $\\hat{y}_i = y_i$ and $0$ otherwise."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "epochs = 100\n",
+ "batch_size = 100\n",
+ "\n",
+ "dnn = NeuralNetwork(X_train, Y_train_onehot, eta=eta, lmbd=lmbd, epochs=epochs, batch_size=batch_size,\n",
+ " n_hidden_neurons=n_hidden_neurons, n_categories=n_categories)\n",
+ "dnn.train()\n",
+ "test_predict = dnn.predict(X_test)\n",
+ "\n",
+ "# accuracy score from scikit library\n",
+ "print(\"Accuracy score on test set: \", accuracy_score(Y_test, test_predict))\n",
+ "\n",
+ "# equivalent in numpy\n",
+ "def accuracy_score_numpy(Y_test, Y_pred):\n",
+ " return np.sum(Y_test == Y_pred) / len(Y_test)\n",
+ "\n",
+ "#print(\"Accuracy score on test set: \", accuracy_score_numpy(Y_test, test_predict))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Adjust hyperparameters\n",
+ "\n",
+ "We now perform a grid search to find the optimal hyperparameters for the network. \n",
+ "Note that we are only using 1 layer with 50 neurons, and human performance is estimated to be around $98\\%$ ($2\\%$ error rate)."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "eta_vals = np.logspace(-5, 1, 7)\n",
+ "lmbd_vals = np.logspace(-5, 1, 7)\n",
+ "# store the models for later use\n",
+ "DNN_numpy = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)\n",
+ "\n",
+ "# grid search\n",
+ "for i, eta in enumerate(eta_vals):\n",
+ " for j, lmbd in enumerate(lmbd_vals):\n",
+ " dnn = NeuralNetwork(X_train, Y_train_onehot, eta=eta, lmbd=lmbd, epochs=epochs, batch_size=batch_size,\n",
+ " n_hidden_neurons=n_hidden_neurons, n_categories=n_categories)\n",
+ " dnn.train()\n",
+ " \n",
+ " DNN_numpy[i][j] = dnn\n",
+ " \n",
+ " test_predict = dnn.predict(X_test)\n",
+ " \n",
+ " print(\"Learning rate = \", eta)\n",
+ " print(\"Lambda = \", lmbd)\n",
+ " print(\"Accuracy score on test set: \", accuracy_score(Y_test, test_predict))\n",
+ " print()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Visualization"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# visual representation of grid search\n",
+ "# uses seaborn heatmap, you can also do this with matplotlib imshow\n",
+ "import seaborn as sns\n",
+ "\n",
+ "sns.set()\n",
+ "\n",
+ "train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n",
+ "test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n",
+ "\n",
+ "for i in range(len(eta_vals)):\n",
+ " for j in range(len(lmbd_vals)):\n",
+ " dnn = DNN_numpy[i][j]\n",
+ " \n",
+ " train_pred = dnn.predict(X_train) \n",
+ " test_pred = dnn.predict(X_test)\n",
+ "\n",
+ " train_accuracy[i][j] = accuracy_score(Y_train, train_pred)\n",
+ " test_accuracy[i][j] = accuracy_score(Y_test, test_pred)\n",
+ "\n",
+ " \n",
+ "fig, ax = plt.subplots(figsize = (10, 10))\n",
+ "sns.heatmap(train_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n",
+ "ax.set_title(\"Training Accuracy\")\n",
+ "ax.set_ylabel(\"$\\eta$\")\n",
+ "ax.set_xlabel(\"$\\lambda$\")\n",
+ "plt.show()\n",
+ "\n",
+ "fig, ax = plt.subplots(figsize = (10, 10))\n",
+ "sns.heatmap(test_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n",
+ "ax.set_title(\"Test Accuracy\")\n",
+ "ax.set_ylabel(\"$\\eta$\")\n",
+ "ax.set_xlabel(\"$\\lambda$\")\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## scikit-learn implementation\n",
+ "\n",
+ "**scikit-learn** focuses more\n",
+ "on traditional machine learning methods, such as regression,\n",
+ "clustering, decision trees, etc. As such, it has only two types of\n",
+ "neural networks: Multi Layer Perceptron outputting continuous values,\n",
+ "*MPLRegressor*, and Multi Layer Perceptron outputting labels,\n",
+ "*MLPClassifier*. We will see how simple it is to use these classes.\n",
+ "\n",
+ "**scikit-learn** implements a few improvements from our neural network,\n",
+ "such as early stopping, a varying learning rate, different\n",
+ "optimization methods, etc. We would therefore expect a better\n",
+ "performance overall."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "from sklearn.neural_network import MLPClassifier\n",
+ "# store models for later use\n",
+ "DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)\n",
+ "\n",
+ "for i, eta in enumerate(eta_vals):\n",
+ " for j, lmbd in enumerate(lmbd_vals):\n",
+ " dnn = MLPClassifier(hidden_layer_sizes=(n_hidden_neurons), activation='logistic',\n",
+ " alpha=lmbd, learning_rate_init=eta, max_iter=epochs)\n",
+ " dnn.fit(X_train, Y_train)\n",
+ " \n",
+ " DNN_scikit[i][j] = dnn\n",
+ " \n",
+ " print(\"Learning rate = \", eta)\n",
+ " print(\"Lambda = \", lmbd)\n",
+ " print(\"Accuracy score on test set: \", dnn.score(X_test, Y_test))\n",
+ " print()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Visualization"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# optional\n",
+ "# visual representation of grid search\n",
+ "# uses seaborn heatmap, could probably do this in matplotlib\n",
+ "import seaborn as sns\n",
+ "\n",
+ "sns.set()\n",
+ "\n",
+ "train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n",
+ "test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n",
+ "\n",
+ "for i in range(len(eta_vals)):\n",
+ " for j in range(len(lmbd_vals)):\n",
+ " dnn = DNN_scikit[i][j]\n",
+ " \n",
+ " train_pred = dnn.predict(X_train) \n",
+ " test_pred = dnn.predict(X_test)\n",
+ "\n",
+ " train_accuracy[i][j] = accuracy_score(Y_train, train_pred)\n",
+ " test_accuracy[i][j] = accuracy_score(Y_test, test_pred)\n",
+ "\n",
+ " \n",
+ "fig, ax = plt.subplots(figsize = (10, 10))\n",
+ "sns.heatmap(train_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n",
+ "ax.set_title(\"Training Accuracy\")\n",
+ "ax.set_ylabel(\"$\\eta$\")\n",
+ "ax.set_xlabel(\"$\\lambda$\")\n",
+ "plt.show()\n",
+ "\n",
+ "fig, ax = plt.subplots(figsize = (10, 10))\n",
+ "sns.heatmap(test_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n",
+ "ax.set_title(\"Test Accuracy\")\n",
+ "ax.set_ylabel(\"$\\eta$\")\n",
+ "ax.set_xlabel(\"$\\lambda$\")\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Building neural networks in Tensorflow and Keras\n",
+ "\n",
+ "Now we want to build on the experience gained from our neural network implementation in NumPy and scikit-learn\n",
+ "and use it to construct a neural network in Tensorflow. Once we have constructed a neural network in NumPy\n",
+ "and Tensorflow, building one in Keras is really quite trivial, though the performance may suffer. \n",
+ "\n",
+ "In our previous example we used only one hidden layer, and in this we will use two. From this it should be quite\n",
+ "clear how to build one using an arbitrary number of hidden layers, using data structures such as Python lists or\n",
+ "NumPy arrays.\n",
+ "\n",
+ "\n",
+ "## Tensorflow\n",
+ "\n",
+ "Tensorflow is an open source library machine learning library\n",
+ "developed by the Google Brain team for internal use. It was released\n",
+ "under the Apache 2.0 open source license in November 9, 2015.\n",
+ "\n",
+ "Tensorflow is a computational framework that allows you to construct\n",
+ "machine learning models at different levels of abstraction, from\n",
+ "high-level, object-oriented APIs like Keras, down to the C++ kernels\n",
+ "that Tensorflow is built upon. The higher levels of abstraction are\n",
+ "simpler to use, but less flexible, and our choice of implementation\n",
+ "should reflect the problems we are trying to solve.\n",
+ "\n",
+ "[Tensorflow uses](https://www.tensorflow.org/guide/graphs) so-called graphs to represent your computation\n",
+ "in terms of the dependencies between individual operations, such that you first build a Tensorflow *graph*\n",
+ "to represent your model, and then create a Tensorflow *session* to run the graph.\n",
+ "\n",
+ "In this guide we will analyze the same data as we did in our NumPy and\n",
+ "scikit-learn tutorial, gathered from the MNIST database of images. We\n",
+ "will give an introduction to the lower level Python Application\n",
+ "Program Interfaces (APIs), and see how we use them to build our graph.\n",
+ "Then we will build (effectively) the same graph in Keras, to see just\n",
+ "how simple solving a machine learning problem can be.\n",
+ "\n",
+ "To install tensorflow on Unix/Linux systems, use pip as"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "pip3 install tensorflow"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "and/or if you use **anaconda**, just write (or install from the graphical user interface)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "conda install tensorflow"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Collect and pre-process data"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# import necessary packages\n",
+ "import numpy as np\n",
+ "import matplotlib.pyplot as plt\n",
+ "from sklearn import datasets\n",
+ "\n",
+ "\n",
+ "# ensure the same random numbers appear every time\n",
+ "np.random.seed(0)\n",
+ "\n",
+ "# display images in notebook\n",
+ "%matplotlib inline\n",
+ "plt.rcParams['figure.figsize'] = (12,12)\n",
+ "\n",
+ "\n",
+ "# download MNIST dataset\n",
+ "digits = datasets.load_digits()\n",
+ "\n",
+ "# define inputs and labels\n",
+ "inputs = digits.images\n",
+ "labels = digits.target\n",
+ "\n",
+ "print(\"inputs = (n_inputs, pixel_width, pixel_height) = \" + str(inputs.shape))\n",
+ "print(\"labels = (n_inputs) = \" + str(labels.shape))\n",
+ "\n",
+ "\n",
+ "# flatten the image\n",
+ "# the value -1 means dimension is inferred from the remaining dimensions: 8x8 = 64\n",
+ "n_inputs = len(inputs)\n",
+ "inputs = inputs.reshape(n_inputs, -1)\n",
+ "print(\"X = (n_inputs, n_features) = \" + str(inputs.shape))\n",
+ "\n",
+ "\n",
+ "# choose some random images to display\n",
+ "indices = np.arange(n_inputs)\n",
+ "random_indices = np.random.choice(indices, size=5)\n",
+ "\n",
+ "for i, image in enumerate(digits.images[random_indices]):\n",
+ " plt.subplot(1, 5, i+1)\n",
+ " plt.axis('off')\n",
+ " plt.imshow(image, cmap=plt.cm.gray_r, interpolation='nearest')\n",
+ " plt.title(\"Label: %d\" % digits.target[random_indices[i]])\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "from keras.utils import to_categorical\n",
+ "from sklearn.model_selection import train_test_split\n",
+ "\n",
+ "# one-hot representation of labels\n",
+ "labels = to_categorical(labels)\n",
+ "\n",
+ "# split into train and test data\n",
+ "train_size = 0.8\n",
+ "test_size = 1 - train_size\n",
+ "X_train, X_test, Y_train, Y_test = train_test_split(inputs, labels, train_size=train_size,\n",
+ " test_size=test_size)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Using TensorFlow backend\n",
+ "\n",
+ "1. Define model and architecture\n",
+ "\n",
+ "2. Choose cost function and optimizer"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "import tensorflow as tf\n",
+ "\n",
+ "class NeuralNetworkTensorflow:\n",
+ " def __init__(\n",
+ " self,\n",
+ " X_train,\n",
+ " Y_train,\n",
+ " X_test,\n",
+ " Y_test,\n",
+ " n_neurons_layer1=100,\n",
+ " n_neurons_layer2=50,\n",
+ " n_categories=2,\n",
+ " epochs=10,\n",
+ " batch_size=100,\n",
+ " eta=0.1,\n",
+ " lmbd=0.0):\n",
+ " \n",
+ " # keep track of number of steps\n",
+ " self.global_step = tf.Variable(0, dtype=tf.int32, trainable=False, name='global_step')\n",
+ " \n",
+ " self.X_train = X_train\n",
+ " self.Y_train = Y_train\n",
+ " self.X_test = X_test\n",
+ " self.Y_test = Y_test\n",
+ " \n",
+ " self.n_inputs = X_train.shape[0]\n",
+ " self.n_features = X_train.shape[1]\n",
+ " self.n_neurons_layer1 = n_neurons_layer1\n",
+ " self.n_neurons_layer2 = n_neurons_layer2\n",
+ " self.n_categories = n_categories\n",
+ " \n",
+ " self.epochs = epochs\n",
+ " self.batch_size = batch_size\n",
+ " self.iterations = self.n_inputs // self.batch_size\n",
+ " self.eta = eta\n",
+ " self.lmbd = lmbd\n",
+ " \n",
+ " # build network piece by piece\n",
+ " # name scopes (with) are used to enforce creation of new variables\n",
+ " # https://www.tensorflow.org/guide/variables\n",
+ " self.create_placeholders()\n",
+ " self.create_DNN()\n",
+ " self.create_loss()\n",
+ " self.create_optimiser()\n",
+ " self.create_accuracy()\n",
+ " \n",
+ " def create_placeholders(self):\n",
+ " # placeholders are fine here, but \"Datasets\" are the preferred method\n",
+ " # of streaming data into a model\n",
+ " with tf.name_scope('data'):\n",
+ " self.X = tf.placeholder(tf.float32, shape=(None, self.n_features), name='X_data')\n",
+ " self.Y = tf.placeholder(tf.float32, shape=(None, self.n_categories), name='Y_data')\n",
+ " \n",
+ " def create_DNN(self):\n",
+ " with tf.name_scope('DNN'):\n",
+ " # the weights are stored to calculate regularization loss later\n",
+ " \n",
+ " # Fully connected layer 1\n",
+ " self.W_fc1 = self.weight_variable([self.n_features, self.n_neurons_layer1], name='fc1', dtype=tf.float32)\n",
+ " b_fc1 = self.bias_variable([self.n_neurons_layer1], name='fc1', dtype=tf.float32)\n",
+ " a_fc1 = tf.nn.sigmoid(tf.matmul(self.X, self.W_fc1) + b_fc1)\n",
+ " \n",
+ " # Fully connected layer 2\n",
+ " self.W_fc2 = self.weight_variable([self.n_neurons_layer1, self.n_neurons_layer2], name='fc2', dtype=tf.float32)\n",
+ " b_fc2 = self.bias_variable([self.n_neurons_layer2], name='fc2', dtype=tf.float32)\n",
+ " a_fc2 = tf.nn.sigmoid(tf.matmul(a_fc1, self.W_fc2) + b_fc2)\n",
+ " \n",
+ " # Output layer\n",
+ " self.W_out = self.weight_variable([self.n_neurons_layer2, self.n_categories], name='out', dtype=tf.float32)\n",
+ " b_out = self.bias_variable([self.n_categories], name='out', dtype=tf.float32)\n",
+ " self.z_out = tf.matmul(a_fc2, self.W_out) + b_out\n",
+ " \n",
+ " def create_loss(self):\n",
+ " with tf.name_scope('loss'):\n",
+ " softmax_loss = tf.reduce_mean(tf.nn.softmax_cross_entropy_with_logits_v2(labels=self.Y, logits=self.z_out))\n",
+ " \n",
+ " regularizer_loss_fc1 = tf.nn.l2_loss(self.W_fc1)\n",
+ " regularizer_loss_fc2 = tf.nn.l2_loss(self.W_fc2)\n",
+ " regularizer_loss_out = tf.nn.l2_loss(self.W_out)\n",
+ " regularizer_loss = self.lmbd*(regularizer_loss_fc1 + regularizer_loss_fc2 + regularizer_loss_out)\n",
+ " \n",
+ " self.loss = softmax_loss + regularizer_loss\n",
+ "\n",
+ " def create_accuracy(self):\n",
+ " with tf.name_scope('accuracy'):\n",
+ " probabilities = tf.nn.softmax(self.z_out)\n",
+ " predictions = tf.argmax(probabilities, axis=1)\n",
+ " labels = tf.argmax(self.Y, axis=1)\n",
+ " \n",
+ " correct_predictions = tf.equal(predictions, labels)\n",
+ " correct_predictions = tf.cast(correct_predictions, tf.float32)\n",
+ " self.accuracy = tf.reduce_mean(correct_predictions)\n",
+ " \n",
+ " def create_optimiser(self):\n",
+ " with tf.name_scope('optimizer'):\n",
+ " self.optimizer = tf.train.GradientDescentOptimizer(learning_rate=self.eta).minimize(self.loss, global_step=self.global_step)\n",
+ " \n",
+ " def weight_variable(self, shape, name='', dtype=tf.float32):\n",
+ " initial = tf.truncated_normal(shape, stddev=0.1)\n",
+ " return tf.Variable(initial, name=name, dtype=dtype)\n",
+ " \n",
+ " def bias_variable(self, shape, name='', dtype=tf.float32):\n",
+ " initial = tf.constant(0.1, shape=shape)\n",
+ " return tf.Variable(initial, name=name, dtype=dtype)\n",
+ " \n",
+ " def fit(self):\n",
+ " data_indices = np.arange(self.n_inputs)\n",
+ "\n",
+ " with tf.Session() as sess:\n",
+ " sess.run(tf.global_variables_initializer())\n",
+ " for i in range(self.epochs):\n",
+ " for j in range(self.iterations):\n",
+ " chosen_datapoints = np.random.choice(data_indices, size=self.batch_size, replace=False)\n",
+ " batch_X, batch_Y = self.X_train[chosen_datapoints], self.Y_train[chosen_datapoints]\n",
+ " \n",
+ " sess.run([DNN.loss, DNN.optimizer],\n",
+ " feed_dict={DNN.X: batch_X,\n",
+ " DNN.Y: batch_Y})\n",
+ " accuracy = sess.run(DNN.accuracy,\n",
+ " feed_dict={DNN.X: batch_X,\n",
+ " DNN.Y: batch_Y})\n",
+ " step = sess.run(DNN.global_step)\n",
+ " \n",
+ " self.train_loss, self.train_accuracy = sess.run([DNN.loss, DNN.accuracy],\n",
+ " feed_dict={DNN.X: self.X_train,\n",
+ " DNN.Y: self.Y_train})\n",
+ " \n",
+ " self.test_loss, self.test_accuracy = sess.run([DNN.loss, DNN.accuracy],\n",
+ " feed_dict={DNN.X: self.X_test,\n",
+ " DNN.Y: self.Y_test})"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Optimizing and using gradient descent"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 18,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "epochs = 100\n",
+ "batch_size = 100\n",
+ "n_neurons_layer1 = 100\n",
+ "n_neurons_layer2 = 50\n",
+ "n_categories = 10\n",
+ "eta_vals = np.logspace(-5, 1, 7)\n",
+ "lmbd_vals = np.logspace(-5, 1, 7)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 19,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "DNN_tf = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)\n",
+ " \n",
+ "for i, eta in enumerate(eta_vals):\n",
+ " for j, lmbd in enumerate(lmbd_vals):\n",
+ " DNN = NeuralNetworkTensorflow(X_train, Y_train, X_test, Y_test,\n",
+ " n_neurons_layer1, n_neurons_layer2, n_categories,\n",
+ " epochs=epochs, batch_size=batch_size, eta=eta, lmbd=lmbd)\n",
+ " DNN.fit()\n",
+ " \n",
+ " DNN_tf[i][j] = DNN\n",
+ " \n",
+ " print(\"Learning rate = \", eta)\n",
+ " print(\"Lambda = \", lmbd)\n",
+ " print(\"Test accuracy: %.3f\" % DNN.test_accuracy)\n",
+ " print()"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 20,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# optional\n",
+ "# visual representation of grid search\n",
+ "# uses seaborn heatmap, could probably do this in matplotlib\n",
+ "import seaborn as sns\n",
+ "\n",
+ "sns.set()\n",
+ "\n",
+ "train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n",
+ "test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n",
+ "\n",
+ "for i in range(len(eta_vals)):\n",
+ " for j in range(len(lmbd_vals)):\n",
+ " DNN = DNN_tf[i][j]\n",
+ "\n",
+ " train_accuracy[i][j] = DNN.train_accuracy\n",
+ " test_accuracy[i][j] = DNN.test_accuracy\n",
+ "\n",
+ " \n",
+ "fig, ax = plt.subplots(figsize = (10, 10))\n",
+ "sns.heatmap(train_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n",
+ "ax.set_title(\"Training Accuracy\")\n",
+ "ax.set_ylabel(\"$\\eta$\")\n",
+ "ax.set_xlabel(\"$\\lambda$\")\n",
+ "plt.show()\n",
+ "\n",
+ "fig, ax = plt.subplots(figsize = (10, 10))\n",
+ "sns.heatmap(test_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n",
+ "ax.set_title(\"Test Accuracy\")\n",
+ "ax.set_ylabel(\"$\\eta$\")\n",
+ "ax.set_xlabel(\"$\\lambda$\")\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 21,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# optional\n",
+ "# we can use log files to visualize our graph in Tensorboard\n",
+ "writer = tf.summary.FileWriter('logs/')\n",
+ "writer.add_graph(tf.get_default_graph())"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Using Keras\n",
+ "\n",
+ "Keras is a high level [neural network](https://en.wikipedia.org/wiki/Application_programming_interface)\n",
+ "that supports Tensorflow, CTNK and Theano as backends. \n",
+ "If you have Tensorflow installed Keras is available through the *tf.keras* module. \n",
+ "If you have Anaconda installed you may run the following command"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 22,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "conda install keras"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Alternatively, if you have Tensorflow or one of the other supported backends install you may use the pip package manager:"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 23,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "pip3 install keras"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "or look up the [instructions here](https://keras.io/)."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 24,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "from keras.models import Sequential\n",
+ "from keras.layers import Dense\n",
+ "from keras.regularizers import l2\n",
+ "from keras.optimizers import SGD\n",
+ "\n",
+ "def create_neural_network_keras(n_neurons_layer1, n_neurons_layer2, n_categories, eta, lmbd):\n",
+ " model = Sequential()\n",
+ " model.add(Dense(n_neurons_layer1, activation='sigmoid', kernel_regularizer=l2(lmbd)))\n",
+ " model.add(Dense(n_neurons_layer2, activation='sigmoid', kernel_regularizer=l2(lmbd)))\n",
+ " model.add(Dense(n_categories, activation='softmax'))\n",
+ " \n",
+ " sgd = SGD(lr=eta)\n",
+ " model.compile(loss='categorical_crossentropy', optimizer=sgd, metrics=['accuracy'])\n",
+ " \n",
+ " return model"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 25,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "DNN_keras = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)\n",
+ " \n",
+ "for i, eta in enumerate(eta_vals):\n",
+ " for j, lmbd in enumerate(lmbd_vals):\n",
+ " DNN = create_neural_network_keras(n_neurons_layer1, n_neurons_layer2, n_categories,\n",
+ " eta=eta, lmbd=lmbd)\n",
+ " DNN.fit(X_train, Y_train, epochs=epochs, batch_size=batch_size, verbose=0)\n",
+ " scores = DNN.evaluate(X_test, Y_test)\n",
+ " \n",
+ " DNN_keras[i][j] = DNN\n",
+ " \n",
+ " print(\"Learning rate = \", eta)\n",
+ " print(\"Lambda = \", lmbd)\n",
+ " print(\"Test accuracy: %.3f\" % scores[1])\n",
+ " print()"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 26,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [],
+ "source": [
+ "# optional\n",
+ "# visual representation of grid search\n",
+ "# uses seaborn heatmap, could probably do this in matplotlib\n",
+ "import seaborn as sns\n",
+ "\n",
+ "sns.set()\n",
+ "\n",
+ "train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n",
+ "test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n",
+ "\n",
+ "for i in range(len(eta_vals)):\n",
+ " for j in range(len(lmbd_vals)):\n",
+ " DNN = DNN_keras[i][j]\n",
+ "\n",
+ " train_accuracy[i][j] = DNN.evaluate(X_train, Y_train)[1]\n",
+ " test_accuracy[i][j] = DNN.evaluate(X_test, Y_test)[1]\n",
+ "\n",
+ " \n",
+ "fig, ax = plt.subplots(figsize = (10, 10))\n",
+ "sns.heatmap(train_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n",
+ "ax.set_title(\"Training Accuracy\")\n",
+ "ax.set_ylabel(\"$\\eta$\")\n",
+ "ax.set_xlabel(\"$\\lambda$\")\n",
+ "plt.show()\n",
+ "\n",
+ "fig, ax = plt.subplots(figsize = (10, 10))\n",
+ "sns.heatmap(test_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n",
+ "ax.set_title(\"Test Accuracy\")\n",
+ "ax.set_ylabel(\"$\\eta$\")\n",
+ "ax.set_xlabel(\"$\\lambda$\")\n",
+ "plt.show()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Which activation function should I use?\n",
+ "\n",
+ "The Back propagation algorithm we derived above works by going from\n",
+ "the output layer to the input layer, propagating the error gradient on\n",
+ "the way. Once the algorithm has computed the gradient of the cost\n",
+ "function with regards to each parameter in the network, it uses these\n",
+ "gradients to update each parameter with a Gradient Descent (GD) step.\n",
+ "\n",
+ "\n",
+ "Unfortunately for us, the gradients often get smaller and smaller as the\n",
+ "algorithm progresses down to the first hidden layers. As a result, the\n",
+ "GD update leaves the lower layer connection weights\n",
+ "virtually unchanged, and training never converges to a good\n",
+ "solution. This is known in the literature as \n",
+ "**the vanishing gradients problem**. \n",
+ "\n",
+ "In other cases, the opposite can happen, namely the the gradients can grow bigger and\n",
+ "bigger. The result is that many of the layers get large updates of the \n",
+ "weights the\n",
+ "algorithm diverges. This is the **exploding gradients problem**, which is\n",
+ "mostly encountered in recurrent neural networks. More generally, deep\n",
+ "neural networks suffer from unstable gradients, different layers may\n",
+ "learn at widely different speeds\n",
+ "\n",
+ "\n",
+ "## Is the Logistic activation function (Sigmoid) our choice?\n",
+ "\n",
+ "Although this unfortunate behavior has been empirically observed for\n",
+ "quite a while (it was one of the reasons why deep neural networks were\n",
+ "mostly abandoned for a long time), it is only around 2010 that\n",
+ "significant progress was made in understanding it.\n",
+ "\n",
+ "A paper titled [Understanding the Difficulty of Training Deep\n",
+ "Feedforward Neural Networks by Xavier Glorot and Yoshua Bengio](http://proceedings.mlr.press/v9/glorot10a.html) found that\n",
+ "the problems with the popular logistic\n",
+ "sigmoid activation function and the weight initialization technique\n",
+ "that was most popular at the time, namely random initialization using\n",
+ "a normal distribution with a mean of 0 and a standard deviation of\n",
+ "1. \n",
+ "\n",
+ "They showed that with this activation function and this\n",
+ "initialization scheme, the variance of the outputs of each layer is\n",
+ "much greater than the variance of its inputs. Going forward in the\n",
+ "network, the variance keeps increasing after each layer until the\n",
+ "activation function saturates at the top layers. This is actually made\n",
+ "worse by the fact that the logistic function has a mean of 0.5, not 0\n",
+ "(the hyperbolic tangent function has a mean of 0 and behaves slightly\n",
+ "better than the logistic function in deep networks).\n",
+ "\n",
+ "\n",
+ "\n",
+ "## The derivative of the Logistic funtion\n",
+ "\n",
+ "Looking at the logistic activation function, when inputs become large\n",
+ "(negative or positive), the function saturates at 0 or 1, with a\n",
+ "derivative extremely close to 0. Thus when backpropagation kicks in,\n",
+ "it has virtually no gradient to propagate back through the network,\n",
+ "and what little gradient exists keeps getting diluted as\n",
+ "backpropagation progresses down through the top layers, so there is\n",
+ "really nothing left for the lower layers.\n",
+ "\n",
+ "In their paper, Glorot and Bengio propose a way to significantly\n",
+ "alleviate this problem. We need the signal to flow properly in both\n",
+ "directions: in the forward direction when making predictions, and in\n",
+ "the reverse direction when backpropagating gradients. We don’t want\n",
+ "the signal to die out, nor do we want it to explode and saturate. For\n",
+ "the signal to flow properly, the authors argue that we need the\n",
+ "variance of the outputs of each layer to be equal to the variance of\n",
+ "its inputs, and we also need the gradients to have equal variance\n",
+ "before and after flowing through a layer in the reverse direction.\n",
+ "\n",
+ "\n",
+ "\n",
+ "One of the insights in the 2010 paper by Glorot and Bengio was that\n",
+ "the vanishing/exploding gradients problems were in part due to a poor\n",
+ "choice of activation function. Until then most people had assumed that\n",
+ "if Nature had chosen to use roughly sigmoid activation functions in\n",
+ "biological neurons, they must be an excellent choice. But it turns out\n",
+ "that other activation functions behave much better in deep neural\n",
+ "networks, in particular the ReLU activation function, mostly because\n",
+ "it does not saturate for positive values (and also because it is quite\n",
+ "fast to compute).\n",
+ "\n",
+ "\n",
+ "\n",
+ "## The RELU function family\n",
+ "\n",
+ "The ReLU activation function suffers from a problem known as the dying\n",
+ "ReLUs: during training, some neurons effectively die, meaning they\n",
+ "stop outputting anything other than 0.\n",
+ "\n",
+ "In some cases, you may find that half of your network’s neurons are\n",
+ "dead, especially if you used a large learning rate. During training,\n",
+ "if a neuron’s weights get updated such that the weighted sum of the\n",
+ "neuron’s inputs is negative, it will start outputting 0. When this\n",
+ "happen, the neuron is unlikely to come back to life since the gradient\n",
+ "of the ReLU function is 0 when its input is negative.\n",
+ "\n",
+ "To solve this problem, nowadays practitioners use a variant of the ReLU\n",
+ "function, such as the leaky ReLU discussed above or the so-called\n",
+ "exponential linear unit (ELU) function"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "$$\n",
+ "ELU(z) = \\left\\{\\begin{array}{cc} \\alpha\\left( \\exp{(z)}-1\\right) & z < 0,\\\\ z & z \\ge 0.\\end{array}\\right.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Which activation function should we use?\n",
+ "\n",
+ "In general it seems that the ELU activation function is better than\n",
+ "the leaky ReLU function (and its variants), which is better than\n",
+ "ReLU. ReLU performs better than $\\tanh$ which in turn performs better\n",
+ "than the logistic function. \n",
+ "\n",
+ "If runtime\n",
+ "performance is an issue, then you may opt for the leaky ReLU function over the \n",
+ "ELU function If you don’t\n",
+ "want to tweak yet another hyperparameter, you may just use the default\n",
+ "$\\alpha$ of $0.01$ for the leaky ReLU, and $1$ for ELU. If you have\n",
+ "spare time and computing power, you can use cross-validation or\n",
+ "bootstrap to evaluate other activation functions.\n",
+ "\n",
+ "\n",
+ "\n",
+ "## A top-down perspective on Neural networks\n",
+ "\n",
+ "\n",
+ "The first thing we would like to do is divide the data into two or three\n",
+ "parts. A training set, a validation or dev (development) set, and a\n",
+ "test set. The test set is the data on which we want to make\n",
+ "predictions. The dev set is a subset of the training data we use to\n",
+ "check how well we are doing out-of-sample, after training the model on\n",
+ "the training dataset. We use the validation error as a proxy for the\n",
+ "test error in order to make tweaks to our model. It is crucial that we\n",
+ "do not use any of the test data to train the algorithm. This is a\n",
+ "cardinal sin in ML. Then:\n",
+ "\n",
+ "\n",
+ "* Estimate optimal error rate\n",
+ "\n",
+ "* Minimize underfitting (bias) on training data set.\n",
+ "\n",
+ "* Make sure you are not overfitting.\n",
+ "\n",
+ "If the validation and test sets are drawn from the same distributions,\n",
+ "then a good performance on the validation set should lead to similarly\n",
+ "good performance on the test set. \n",
+ "\n",
+ "However, sometimes\n",
+ "the training data and test data differ in subtle ways because, for\n",
+ "example, they are collected using slightly different methods, or\n",
+ "because it is cheaper to collect data in one way versus another. In\n",
+ "this case, there can be a mismatch between the training and test\n",
+ "data. This can lead to the neural network overfitting these small\n",
+ "differences between the test and training sets, and a poor performance\n",
+ "on the test set despite having a good performance on the validation\n",
+ "set. To rectify this, Andrew Ng suggests making two validation or dev\n",
+ "sets, one constructed from the training data and one constructed from\n",
+ "the test data. The difference between the performance of the algorithm\n",
+ "on these two validation sets quantifies the train-test mismatch. This\n",
+ "can serve as another important diagnostic when using DNNs for\n",
+ "supervised learning.\n",
+ "\n",
+ "\n",
+ "## Limitations of supervised learning with deep networks\n",
+ "\n",
+ "Like all statistical methods, supervised learning using neural\n",
+ "networks has important limitations. This is especially important when\n",
+ "one seeks to apply these methods, especially to physics problems. Like\n",
+ "all tools, DNNs are not a universal solution. Often, the same or\n",
+ "better performance on a task can be achieved by using a few\n",
+ "hand-engineered features (or even a collection of random\n",
+ "features). \n",
+ "\n",
+ "Here we list some of the important limitations of supervised neural network based models. \n",
+ "\n",
+ "\n",
+ "\n",
+ "* **Need labeled data**. All supervised learning methods, DNNs for supervised learning require labeled data. Often, labeled data is harder to acquire than unlabeled data (e.g. one must pay for human experts to label images).\n",
+ "\n",
+ "* **Supervised neural networks are extremely data intensive.** DNNs are data hungry. They perform best when data is plentiful. This is doubly so for supervised methods where the data must also be labeled. The utility of DNNs is extremely limited if data is hard to acquire or the datasets are small (hundreds to a few thousand samples). In this case, the performance of other methods that utilize hand-engineered features can exceed that of DNNs.\n",
+ "\n",
+ "* **Homogeneous data.** Almost all DNNs deal with homogeneous data of one type. It is very hard to design architectures that mix and match data types (i.e. some continuous variables, some discrete variables, some time series). In applications beyond images, video, and language, this is often what is required. In contrast, ensemble models like random forests or gradient-boosted trees have no difficulty handling mixed data types.\n",
+ "\n",
+ "* **Many problems are not about prediction.** In natural science we are often interested in learning something about the underlying distribution that generates the data. In this case, it is often difficult to cast these ideas in a supervised learning setting. While the problems are related, it is possible to make good predictions with a *wrong* model. The model might or might not be useful for understanding the underlying science.\n",
+ "\n",
+ "Some of these remarks are particular to DNNs, others are shared by all supervised learning methods. This motivates the use of unsupervised methods which in part circumvent these problems."
+ ]
+ }
+ ],
+ "metadata": {},
+ "nbformat": 4,
+ "nbformat_minor": 4
+}
diff --git a/doc/src/Statistics/Statistics.do.txt b/doc/src/Statistics/Statistics.do.txt
index 9cadba238..db8d38f50 100644
--- a/doc/src/Statistics/Statistics.do.txt
+++ b/doc/src/Statistics/Statistics.do.txt
@@ -1,17 +1,13 @@
-TITLE: Data Analysis and Machine Learning: Elements of Probability Theory and Statistical Data Analysis
-AUTHOR: Morten Hjorth-Jensen {copyright, 1999-present|CC BY-NC} at Department of Physics, University of Oslo & Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University
-DATE: today
+======= Elements of Probability Theory and Statistical Data Analysis =======
-!split
-===== To do list =====
* add math about MVN and define MLE and other quantities
* rewrite about covariance matrix
* add KL theorem
-!split
+
===== Domains and probabilities =====
!bblock
Consider the following simple example, namely the tossing of two dice, resulting in the following possible values
@@ -29,7 +25,7 @@ To this domain we have the corresponding *probabilities*
!et
!eblock
-!split
+
===== Tossing the dice =====
!bblock
The numbers in the domain are the outcomes of the physical process of tossing say two dice.
@@ -55,14 +51,14 @@ appear in a random order. After 11 throws the results may look like
!et
!eblock
-!split
+
===== Stochastic variables =====
!bblock
_Random variables are characterized by a domain which contains all possible values that the random value may take. This domain has a corresponding probability distribution function(PDF)_.
!eblock
-!split
+
===== Stochastic variables and the main concepts, the discrete case =====
!bblock
There are two main concepts associated with a stochastic variable. The
@@ -83,7 +79,7 @@ p(x) = \mathrm{Prob}(X=x).
!eblock
-!split
+
===== Stochastic variables and the main concepts, the continuous case =====
!bblock
In the continuous case, the PDF does not directly depict the
@@ -104,7 +100,7 @@ numbers chosen as if by chance from some specified PDF so that the
selection of a large set of these numbers reproduces this PDF.
!eblock
-!split
+
===== The cumulative probability =====
!bblock
Of interest to us is the *cumulative probability
@@ -125,7 +121,7 @@ p(x) = \frac{d}{dx}P(x).
!et
!eblock
-!split
+
===== Properties of PDFs =====
!bblock
@@ -150,7 +146,7 @@ both discrete and continuous PDFs, this condition is
!et
!eblock
-!split
+
===== Important distributions, the uniform distribution =====
!bblock
The first one
@@ -172,7 +168,7 @@ p(x)dx = dx & \in [0,1].
The latter distribution is used to generate random numbers. For other PDFs, one needs normally a mapping from this distribution to say for example the exponential distribution.
!eblock
-!split
+
===== Gaussian distribution =====
!bblock
The second one is the Gaussian Distribution
@@ -231,7 +227,7 @@ plt.show()
!eblock
-!split
+
===== Exponential distribution =====
!bblock
Another important distribution in science is the exponential distribution
@@ -242,7 +238,7 @@ p(x) = \alpha\exp{-(\alpha x)}.
!et
!eblock
-!split
+
===== Expectation values =====
!bblock
Let $h(x)$ be an arbitrary continuous function on the domain of the stochastic
@@ -267,7 +263,7 @@ follows
!et
!eblock
-!split
+
===== Stochastic variables and the main concepts, mean values =====
!bblock
The zero-th moment $\langle 1\rangle$ is just the normalization condition of
@@ -289,7 +285,7 @@ Qualitatively it represents the centroid or the average value of the
PDF and is therefore simply called the expectation value of $p(x)$.
!eblock
-!split
+
===== Stochastic variables and the main concepts, central moments, the variance =====
!bblock
@@ -320,7 +316,7 @@ qualitatively as the ``spread'' of $p$ around its mean.
-!split
+
===== Probability Distribution Functions =====
!bblock
@@ -344,7 +340,7 @@ while $P(x)$ is the cumulative probability.
!eblock
-!split
+
===== Probability Distribution Functions =====
!bblock
With a PDF we can compute expectation values of selected quantities such as
@@ -365,7 +361,7 @@ in the case of a continuous PDF. We have already defined the mean value $\mu$
and the variance $\sigma^2$.
!eblock
-!split
+
===== The three famous Probability Distribution Functions =====
!bblock
@@ -399,7 +395,7 @@ with variance
\end{equation*}
!et
-!split
+
===== Probability Distribution Functions, the normal distribution =====
!bblock
Finally, we have the so-called univariate normal distribution, or just the _normal distribution_
@@ -425,7 +421,7 @@ which becomes with a suitable change of variables
!et
!eblock
-!split
+
===== Probability Distribution Functions, the normal distribution =====
!bblock
Similarly, the variance becomes
@@ -453,7 +449,7 @@ centered around zero and with a variance $\sigma^2=1$, leading to
!et
!eblock
-!split
+
===== Probability Distribution Functions, the cumulative distribution =====
!bblock
@@ -470,7 +466,7 @@ which is difficult to evaluate in a quick way.
!eblock
-!split
+
===== Probability Distribution Functions, other important distribution =====
!bblock
@@ -492,7 +488,7 @@ The sequence of binomial trials is characterized by the following definitions
* The probability for every outcome in a single trial has the same value, for example the outcome of tossing (either heads or tails) a coin is always $1/2$.
!eblock
-!split
+
===== Probability Distribution Functions, the binomial distribution =====
!bblock
@@ -535,7 +531,7 @@ which we rewrite as
!eblock
The variance is slightly trickier to get. It reads $\sigma^2=ny(1-y)$.
-!split
+
===== Probability Distribution Functions, Poisson's distribution =====
!bblock
@@ -559,7 +555,7 @@ and the variance is $\sigma^2=\lambda$.
-!split
+
===== Probability Distribution Functions, Poisson's distribution =====
!bblock
An example of applications of the Poisson distribution could be the counting
@@ -576,7 +572,7 @@ the binomial distribution we can show that
!eblock
-!split
+
===== Meet the covariance! =====
!bblock
An important quantity in a statistical analysis is the so-called covariance.
@@ -604,7 +600,7 @@ with
-!split
+
===== Meet the covariance in matrix disguise =====
!bblock
If we consider the above covariance as a matrix
@@ -619,7 +615,7 @@ all the off-diagonal elements are zero if the stochastic variables are
uncorrelated.
!eblock
-!split
+
===== Covariance =====
!bc pycod
# Importing various packages
@@ -651,7 +647,7 @@ print(c)
-!split
+
===== Meet the covariance, uncorrelated events =====
!bblock
@@ -683,7 +679,7 @@ Cov(X_i, X_j) = 0 \hspace{0.1cm} (i\neq j).
-!split
+
===== Numerical experiments and the covariance =====
!bblock
@@ -704,7 +700,7 @@ chain of values
!eblock
-!split
+
===== Numerical experiments and the covariance =====
!bblock
We will call these
@@ -722,7 +718,7 @@ $\mu_X^{\phantom X}$ and the variance $\sigma_X^{\phantom X}$.
!eblock
-!split
+
===== Numerical experiments and the covariance, actual situations =====
!bblock
In practical situations however, a sample is always of finite size. Let that
@@ -742,7 +738,7 @@ with its square root being the *standard deviation of the sample*.
!eblock
-!split
+
===== Numerical experiments and the covariance, our observables =====
!bblock
You can think of the above observables as a set of quantities which define
@@ -770,7 +766,7 @@ label{eq:exptvariance}
!et
!eblock
-!split
+
===== Numerical experiments and the covariance, the sample variance =====
!bblock
@@ -791,7 +787,7 @@ exact values for the mean value $\mu_X$, the variance $\mathrm{Var}(X)$
and the covariance $\mathrm{Cov}(X,Y)$.
!eblock
-!split
+
===== Numerical experiments and the covariance, central limit theorem =====
!bblock
@@ -813,7 +809,7 @@ standard deviation, given by
In many cases the above estimate for the standard deviation, in particular if correlations are strong, may be too simplistic. We need therefore a more precise defintion of the error and the variance in our results.
!eblock
-!split
+
===== Definition of Correlation Functions and Standard Deviation =====
!bblock
Our estimate of the true average $\mu_{X}$ is the sample mean $\langle X_m \rangle$
@@ -845,7 +841,7 @@ and the last term is nothing but the covariance which arises when $k\ne l$.
-!split
+
===== Definition of Correlation Functions and Standard Deviation =====
!bblock
Our estimate of the true average $\mu_{X}$ is the sample mean $\langle X_m \rangle$
@@ -863,7 +859,7 @@ measurements to calculate the cross terms. Therefore, all measurements
have to be stored throughout the experiment.
!eblock
-!split
+
===== Definition of Correlation Functions and Standard Deviation =====
!bblock
@@ -887,7 +883,7 @@ $f_d$
!eblock
-!split
+
===== Definition of Correlation Functions and Standard Deviation =====
!bblock
The value of $f_d$ reflects the correlation between measurements
@@ -905,7 +901,7 @@ which gives us a useful measure of the correlation pair correlation
starting always at $1$ for $d=0$.
!eblock
-!split
+
===== Definition of Correlation Functions and Standard Deviation, sample variance =====
!bblock
@@ -938,7 +934,7 @@ equals 1.
-!split
+
===== Definition of Correlation Functions and Standard Deviation =====
!bblock
From the point of view of
@@ -961,7 +957,7 @@ more practically oriented methods like the blocking technique.
!eblock
-!split
+
===== Code to compute the Covariance matrix and the Covariance =====
!bc pycod
# Importing various packages
@@ -1003,7 +999,7 @@ Eigvals, Eigvecs = np.linalg.eig(c)
print(Eigvals)
!ec
-!split
+
======= Random Numbers =======
!bblock
@@ -1020,7 +1016,7 @@ or Monte Carlo computer work.
!eblock
-!split
+
======= Random Numbers, better name: pseudo random numbers =======
!bblock
@@ -1040,7 +1036,7 @@ numbers, hopefully abiding to the following criteria:
!eblock
-!split
+
======= Random number generator RNG =======
!bblock
The most common random number generators are based on so-called
@@ -1067,7 +1063,7 @@ that is if we were to evaluate $(13)\mathrm{MOD}(9)$, the outcome is the remaind
of the division $13/9$, namely $4$.
!eblock
-!split
+
======= Random number generator RNG and periodic outputs =======
!bblock
@@ -1097,7 +1093,7 @@ which still, with $N_0=2$, results in $11,38,11,38,11,38,\dots$, a period of
just $2$.
!eblock
-!split
+
======= Random number generator RNG and its period =======
!bblock
Typical periods for the random generators provided in the program library
@@ -1115,7 +1111,7 @@ number is the sum of the $l-i$th and $l-j$th values with modulo $M$,
!et
!eblock
-!split
+
======= Random number generator RNG, other examples =======
!bblock
Such a generator again produces a sequence of pseudorandom numbers
@@ -1141,7 +1137,7 @@ followed by
which according to the authors has a period larger than $2^{94}$.
!eblock
-!split
+
======= Random number generator RNG, other examples =======
!bblock
Instead of using modular addition, we could use the bitwise
@@ -1165,7 +1161,7 @@ it is given by $m\wedge n$.
!eblock
-!split
+
======= Random number generator RNG, RAN0 =======
!bblock
@@ -1203,7 +1199,7 @@ $q$ and $r$ are chosen so that $r < q$.
-!split
+
======= Random number generator RNG, RAN0 =======
!bblock
@@ -1220,7 +1216,7 @@ $[N_{i-1}/q]$ just yields a constant which is multiplied with $M$.
!eblock
-!split
+
======= Random number generator RNG, RAN0 =======
!bblock
We can now rewrite Eq. (ref{eq:rntrick1}) as
@@ -1250,7 +1246,7 @@ label{eq:rntrick4}
!eblock
-!split
+
======= Random number generator RNG, RAN0 =======
!bblock
The term $[N_{i-1}/q]r$ is always smaller or equal $N_{i-1}(r/q)$ and with $r < q$ we obtain always a
@@ -1268,7 +1264,7 @@ set to $0$.
!eblock
-!split
+
======= Random number generator RNG, RAN0 code =======
!bblock
@@ -1302,7 +1298,7 @@ double ran0(long &idum)
!ec
!eblock
-!split
+
===== Properties of Selected Random Number Generators =====
!bblock
@@ -1321,7 +1317,7 @@ Two additional measures are the standard deviation $\sigma$ and the mean
$\mu=\langle x\rangle$.
!eblock
-!split
+
===== Properties of Selected Random Number Generators =====
!bblock
For the uniform distribution, the mean value $\mu$ is then
@@ -1340,7 +1336,7 @@ while the standard deviation is
!et
!eblock
-!split
+
===== Properties of Selected Random Number Generators =====
!bblock
The various random number generators produce results which agree rather well with
@@ -1365,7 +1361,7 @@ these limiting values.
!eblock
-!split
+
===== Simple demonstration of RNGs using python =====
!bblock
The following simple Python code plots the distribution of the produced random numbers using the linear congruential RNG employed by Python. The trend displayed in the previous table is seen rather clearly.
@@ -1396,7 +1392,7 @@ plt.show()
!eblock
-!split
+
===== Properties of Selected Random Number Generators =====
!bblock
Since our random numbers, which are typically generated via a linear congruential algorithm,
@@ -1427,7 +1423,7 @@ assumption for approximating $\sigma_N$ is no longer valid.
!eblock
-!split
+
===== Autocorrelation function =====
This program computes the autocorrelation function as discussed in the equation on the previous slide for random numbers generated with the normal distribution $N(0,1)$.
!bc pycod
@@ -1479,7 +1475,7 @@ For the remaining values we notice that there are still non-zero values for the
-!split
+
===== Correlation function and which random number generators should I use =====
!bblock
The program here computes the correlation function for one of the standard functions included with the c++ compiler.
@@ -1555,7 +1551,7 @@ int main(int argc, char* argv[])
-!split
+
======= Which RNG should I use? =======
!bblock
* C++ has a class called _random_. The "random class":"http://www.cplusplus.com/reference/random/" contains a large selection of RNGs and is highly recommended. Some of these RNGs have very large periods making it thereby very safe to use these RNGs in case one is performing large calculations. In particular, the "Mersenne twister random number engine":"http://www.cplusplus.com/reference/random/mersenne_twister_engine/" has a period of $2^{19937}$.
@@ -1564,7 +1560,7 @@ int main(int argc, char* argv[])
!eblock
-!split
+
===== How to use the Mersenne generator =====
!bblock
The following part of a c++ code (from project 4) sets up the uniform distribution for $x\in [0,1]$.
@@ -1588,7 +1584,7 @@ int ix = (int) (RandomNumberGenerator(gen)*NSpins);
-!split
+
===== Why blocking? =====
!bblock Statistical analysis
* Monte Carlo simulations can be treated as *computer experiments*
@@ -1601,7 +1597,7 @@ A very good article which explains blocking is H. Flyvbjerg and H. G. Petersen,
-!split
+
===== Why blocking? =====
!bblock Statistical analysis
* As in other experiments, Monte Carlo experiments have two classes of errors:
@@ -1612,7 +1608,7 @@ A very good article which explains blocking is H. Flyvbjerg and H. G. Petersen,
!eblock
-!split
+
===== Code to demonstrate the calculation of the autocorrelation function =====
The following code computes the autocorrelation function, the covariance and the standard deviation
for standard RNG.
@@ -1700,7 +1696,7 @@ int main(int argc, char* argv[])
!ec
-!split
+
===== What is blocking? =====
!bblock Blocking
* Say that we have a set of samples from a Monte Carlo experiment
@@ -1719,7 +1715,7 @@ int main(int argc, char* argv[])
where $\tau$ is the correlation time (the time between a sample and the next uncorrelated sample) and $\Delta t$ is time between each sample
!eblock
-!split
+
===== What is blocking? =====
!bblock Blocking
* If $\Delta t\gg\tau$ our first estimate of $\sigma$ still holds
@@ -1730,7 +1726,7 @@ int main(int argc, char* argv[])
* The correlation time $\tau$ would be a good choice
!eblock
-!split
+
===== What is blocking? =====
!bblock Blocking
* Problem: We don't know $\tau$ or it is too expensive to compute
@@ -1739,7 +1735,7 @@ int main(int argc, char* argv[])
* When the std. dev. stops increasing the blocks are uncorrelated
!eblock
-!split
+
===== Implementation =====
!bblock
* Do a Monte Carlo simulation, storing all samples to file
@@ -1756,7 +1752,7 @@ int main(int argc, char* argv[])
-!split
+
===== Actual implementation with code, main function =====
When the file gets large, it can be useful to write your data in binary mode instead of ascii characters.
The "following python file":"https://github.com/CompPhysics/MachineLearning/blob/master/doc/Programs/Sampling/analysis.py" reads data from file with the output from every Monte Carlo cycle.
@@ -1797,7 +1793,7 @@ The "following python file":"https://github.com/CompPhysics/MachineLearning/blob
-!split
+
===== The Bootstrap method =====
The Bootstrap resampling method is also very popular. It is very simple:
@@ -1811,7 +1807,7 @@ that it produces the correct standard deviation.
This method is very useful for small ensembles of data points.
-!split
+
===== Bootstrapping =====
Given a set of $N$ data, assume that we are interested in some
observable $\theta$ which may be estimated from that set. This observable can also be for example the result of a fit based on all $N$ raw data.
@@ -1820,7 +1816,7 @@ data set $\hat{\theta}$. One recreates from the sample repeatedly
other samples by choosing randomly $N$ data out of the original set.
This costs essentially nothing, since we just recycle the original data set for the building of new sets.
-!split
+
===== Bootstrapping, recipe =====
Let us assume we have done this $K$ times and thus have $K$ sets of $N$
data values each.
@@ -1844,7 +1840,7 @@ The difference is called bias and gives an idea on how far away the result may b
the true $\angle\theta\rangle$. As final result for the observable one quotes $\angle\theta\rangle = \tilde{\theta} \pm \sigma_{\tilde{\theta}}$ .
-!split
+
===== Bootstrapping, "code":"https://github.com/CompPhysics/MachineLearning/blob/master/doc/Programs/Sampling/analysis.py" =====
!bc
# Bootstrap
@@ -1858,7 +1854,7 @@ the true $\angle\theta\rangle$. As final result for the observable one quotes $\
self.bootStd = np.std(bootVec)
!ec
-!split
+
===== Jackknife, "code":"https://github.com/CompPhysics/MachineLearning/blob/master/doc/Programs/Sampling/analysis.py" =====
!bc
# Jackknife