Figure 1:
# # # Another useful Python package is # [pandas](https://pandas.pydata.org/), which is an open source library # providing high-performance, easy-to-use data structures and data # 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. # **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. # **pandas** allows you also to perform mathematical operations on the data, spanning from simple reshapings of vectors and matrices to statistical operations. # # 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. # In[17]: import pandas as pd from IPython.display import display data = {'First Name': ["Frodo", "Bilbo", "Aragorn II", "Samwise"], 'Last Name': ["Baggins", "Baggins","Elessar","Gamgee"], 'Place of birth': ["Shire", "Shire", "Eriador", "Shire"], 'Date of Birth T.A.': [2968, 2890, 2931, 2980] } data_pandas = pd.DataFrame(data) display(data_pandas) # 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 # 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*. # Displaying these results, we see that the indices are given by the default numbers from zero to three. # **pandas** is extremely flexible and we can easily change the above indices by defining a new type of indexing as # In[18]: data_pandas = pd.DataFrame(data,index=['Frodo','Bilbo','Aragorn','Sam']) display(data_pandas) # Thereafter we display the content of the row which begins with the index **Aragorn** # In[19]: display(data_pandas.loc['Aragorn']) # We can easily append data to this, for example # In[20]: new_hobbit = {'First Name': ["Peregrin"], 'Last Name': ["Took"], 'Place of birth': ["Shire"], 'Date of Birth T.A.': [2990] } data_pandas=data_pandas.append(pd.DataFrame(new_hobbit, index=['Pippin'])) display(data_pandas) # 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 # 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. # In[21]: import numpy as np import pandas as pd from IPython.display import display np.random.seed(100) # setting up a 10 x 5 matrix rows = 10 cols = 5 a = np.random.randn(rows,cols) df = pd.DataFrame(a) display(df) print(df.mean()) print(df.std()) display(df**2) # Thereafter we can select specific columns only and plot final results # In[22]: df.columns = ['First', 'Second', 'Third', 'Fourth', 'Fifth'] df.index = np.arange(10) display(df) print(df['Second'].mean() ) print(df.info()) print(df.describe()) from pylab import plt, mpl plt.style.use('seaborn') mpl.rcParams['font.family'] = 'serif' df.cumsum().plot(lw=2.0, figsize=(10,6)) plt.show() df.plot.bar(figsize=(10,6), rot=15) plt.show() # We can produce a $4\times 4$ matrix # In[23]: b = np.arange(16).reshape((4,4)) print(b) df1 = pd.DataFrame(b) print(df1) # and many other operations. # # The **Series** class is another important class included in # **pandas**. You can view it as a specialization of **DataFrame** but where # we have just a single column of data. It shares many of the same features as **DataFrame**. As with **DataFrame**, # most operations are vectorized, achieving thereby a high performance when dealing with computations of arrays, in particular labeled arrays. # As we will see below it leads also to a very concice code close to the mathematical operations we may be interested in. # 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**. # ## Pandas AI # # Try out [Pandas AI](https://pandas-ai.com/) # ### Simple linear regression model using **scikit-learn** # # We start with perhaps our simplest possible example, using **Scikit-Learn** to perform linear regression analysis on a data set produced by us. # # What follows is a simple Python code where we have defined a function # $y$ in terms of the variable $x$. Both are defined as vectors with $100$ entries. # The numbers in the vector $\boldsymbol{x}$ are given # by random numbers generated with a uniform distribution with entries # $x_i \in [0,1]$ (more about probability distribution functions # later). These values are then used to define a function $y(x)$ # (tabulated again as a vector) with a linear dependence on $x$ plus a # random noise added via the normal distribution. # # The Numpy functions are imported used the **import numpy as np** # statement and the random number generator for the uniform distribution # is called using the function **np.random.rand()**, where we specificy # that we want $100$ random variables. Using Numpy we define # automatically an array with the specified number of elements, $100$ in # our case. With the Numpy function **randn()** we can compute random # numbers with the normal distribution (mean value $\mu$ equal to zero and # variance $\sigma^2$ set to one) and produce the values of $y$ assuming a linear # dependence as function of $x$ # $$ # y = 2x+N(0,1), # $$ # where $N(0,1)$ represents random numbers generated by the normal # distribution. From **Scikit-Learn** we import then the # **LinearRegression** functionality and make a prediction $\tilde{y} = # \alpha + \beta x$ using the function **fit(x,y)**. We call the set of # data $(\boldsymbol{x},\boldsymbol{y})$ for our training data. The Python package # **scikit-learn** has also a functionality which extracts the above # fitting parameters $\alpha$ and $\beta$ (see below). Later we will # distinguish between training data and test data. # # For plotting we use the Python package # [matplotlib](https://matplotlib.org/) which produces publication # quality figures. Feel free to explore the extensive # [gallery](https://matplotlib.org/gallery/index.html) of examples. In # this example we plot our original values of $x$ and $y$ as well as the # prediction **ypredict** ($\tilde{y}$), which attempts at fitting our # data with a straight line. # # The Python code follows here. # In[24]: # Importing various packages import numpy as np import matplotlib.pyplot as plt from sklearn.linear_model import LinearRegression x = np.random.rand(100,1) y = 2*x+np.random.randn(100,1) linreg = LinearRegression() linreg.fit(x,y) xnew = np.array([[0],[1]]) ypredict = linreg.predict(xnew) plt.plot(xnew, ypredict, "r-") plt.plot(x, y ,'ro') plt.axis([0,1.0,0, 5.0]) plt.xlabel(r'$x$') plt.ylabel(r'$y$') plt.title(r'Simple Linear Regression') plt.show() # This example serves several aims. It allows us to demonstrate several # aspects of data analysis and later machine learning algorithms. The # immediate visualization shows that our linear fit is not # impressive. It goes through the data points, but there are many # outliers which are not reproduced by our linear regression. We could # now play around with this small program and change for example the # factor in front of $x$ and the normal distribution. Try to change the # function $y$ to # $$ # y = 10x+0.01 \times N(0,1), # $$ # where $x$ is defined as before. Does the fit look better? Indeed, by # reducing the role of the noise given by the normal distribution we see immediately that # our linear prediction seemingly reproduces better the training # set. However, this testing 'by the eye' is obviouly not satisfactory in the # long run. Here we have only defined the training data and our model, and # have not discussed a more rigorous approach to the **cost** function. # # We need more rigorous criteria in defining whether we have succeeded or # not in modeling our training data. You will be surprised to see that # many scientists seldomly venture beyond this 'by the eye' approach. A # standard approach for the *cost* function is the so-called $\chi^2$ # function (a variant of the mean-squared error (MSE)) # $$ # \chi^2 = \frac{1}{n} # \sum_{i=0}^{n-1}\frac{(y_i-\tilde{y}_i)^2}{\sigma_i^2}, # $$ # where $\sigma_i^2$ is the variance (to be defined later) of the entry # $y_i$. We may not know the explicit value of $\sigma_i^2$, it serves # however the aim of scaling the equations and make the cost function # dimensionless. # # Minimizing the cost function is a central aspect of # our discussions to come. Finding its minima as function of the model # parameters ($\alpha$ and $\beta$ in our case) will be a recurring # theme in these series of lectures. Essentially all machine learning # algorithms we will discuss center around the minimization of the # chosen cost function. This depends in turn on our specific # model for describing the data, a typical situation in supervised # learning. Automatizing the search for the minima of the cost function is a # central ingredient in all algorithms. Typical methods which are # employed are various variants of **gradient** methods. These will be # discussed in more detail later. Again, you'll be surprised to hear that # many practitioners minimize the above function ''by the eye', popularly dubbed as # 'chi by the eye'. That is, change a parameter and see (visually and numerically) that # the $\chi^2$ function becomes smaller. # # 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 # the relative error (why would we prefer the MSE instead of the relative error?) as # $$ # \epsilon_{\mathrm{relative}}= \frac{\vert \boldsymbol{y} -\boldsymbol{\tilde{y}}\vert}{\vert \boldsymbol{y}\vert}. # $$ # The squared cost function results in an arithmetic mean-unbiased # estimator, and the absolute-value cost function results in a # median-unbiased estimator (in the one-dimensional case, and a # geometric median-unbiased estimator for the multi-dimensional # case). The squared cost function has the disadvantage that it has the tendency # to be dominated by outliers. # # We can modify easily the above Python code and plot the relative error instead # In[25]: import numpy as np import matplotlib.pyplot as plt from sklearn.linear_model import LinearRegression x = np.random.rand(100,1) y = 5*x+0.01*np.random.randn(100,1) linreg = LinearRegression() linreg.fit(x,y) ypredict = linreg.predict(x) plt.plot(x, np.abs(ypredict-y)/abs(y), "ro") plt.axis([0,1.0,0.0, 0.5]) plt.xlabel(r'$x$') plt.ylabel(r'$\epsilon_{\mathrm{relative}}$') plt.title(r'Relative error') plt.show() # Depending on the parameter in front of the normal distribution, we may # have a small or larger relative error. Try to play around with # different training data sets and study (graphically) the value of the # relative error. # # As mentioned above, **Scikit-Learn** has an impressive functionality. # We can for example extract the values of $\alpha$ and $\beta$ and # their error estimates, or the variance and standard deviation and many # other properties from the statistical data analysis. # # Here we show an # example of the functionality of **Scikit-Learn**. # In[26]: import numpy as np import matplotlib.pyplot as plt from sklearn.linear_model import LinearRegression from sklearn.metrics import mean_squared_error, r2_score, mean_squared_log_error, mean_absolute_error x = np.random.rand(100,1) y = 2.0+ 5*x+0.5*np.random.randn(100,1) linreg = LinearRegression() linreg.fit(x,y) ypredict = linreg.predict(x) print('The intercept alpha: \n', linreg.intercept_) print('Coefficient beta : \n', linreg.coef_) # The mean squared error print("Mean squared error: %.2f" % mean_squared_error(y, ypredict)) # Explained variance score: 1 is perfect prediction print('Variance score: %.2f' % r2_score(y, ypredict)) # Mean squared log error print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) ) # Mean absolute error print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict)) plt.plot(x, ypredict, "r-") plt.plot(x, y ,'ro') plt.axis([0.0,1.0,1.5, 7.0]) plt.xlabel(r'$x$') plt.ylabel(r'$y$') plt.title(r'Linear Regression fit ') plt.show() # The function **coef** gives us the parameter $\beta$ of our fit while **intercept** yields # $\alpha$. 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 # $$ # MSE(\boldsymbol{y},\boldsymbol{\tilde{y}}) = \frac{1}{n} # \sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, # $$ # The smaller the value, the better the fit. Ideally we would like to # have an MSE equal zero. The attentive reader has probably recognized # this function as being similar to the $\chi^2$ function defined above. # # The **r2score** function computes $R^2$, the coefficient of # determination. It provides a measure of how well future samples are # likely to be predicted by the model. Best possible score is 1.0 and it # can be negative (because the model can be arbitrarily worse). A # constant model that always predicts the expected value of $\boldsymbol{y}$, # disregarding the input features, would get a $R^2$ score of $0.0$. # # If $\tilde{\boldsymbol{y}}_i$ 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 # $$ # R^2(\boldsymbol{y}, \tilde{\boldsymbol{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2}, # $$ # where we have defined the mean value of $\boldsymbol{y}$ as # $$ # \bar{y} = \frac{1}{n} \sum_{i=0}^{n - 1} y_i. # $$ # Another quantity taht we will meet again in our discussions of regression analysis is # 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. # The MAE is defined as follows # $$ # \text{MAE}(\boldsymbol{y}, \boldsymbol{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n-1} \left| y_i - \tilde{y}_i \right|. # $$ # We present the # squared logarithmic (quadratic) error # $$ # \text{MSLE}(\boldsymbol{y}, \boldsymbol{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n - 1} (\log_e (1 + y_i) - \log_e (1 + \tilde{y}_i) )^2, # $$ # where $\log_e (x)$ stands for the natural logarithm of $x$. This error # estimate is best to use when targets having exponential growth, such # as population counts, average sales of a commodity over a span of # years etc. # # Finally, another cost function is the Huber cost function used in robust regression. # # The rationale behind this possible cost function is its reduced # sensitivity to outliers in the data set. In our discussions on # dimensionality reduction and normalization of data we will meet other # ways of dealing with outliers. # # The Huber cost function is defined as # $$ # H_{\delta}(\boldsymbol{a})=\left\{\begin{array}{cc}\frac{1}{2} \boldsymbol{a}^{2}& \text{for }|\boldsymbol{a}|\leq \delta\\ \delta (|\boldsymbol{a}|-\frac{1}{2}\delta ),&\text{otherwise}.\end{array}\right. # $$ # Here $\boldsymbol{a}=\boldsymbol{y} - \boldsymbol{\tilde{y}}$. # # We will discuss in more detail these and other functions in the # various lectures and lab sessions. # ### To our real data: nuclear binding energies. Brief reminder on masses and binding energies # # Let us now dive into nuclear physics and remind ourselves briefly about some basic features about binding # energies. A basic quantity which can be measured for the ground # states of nuclei is the atomic mass $M(N, Z)$ of the neutral atom with # 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). # # Atomic masses are usually tabulated in terms of the mass excess defined by # $$ # \Delta M(N, Z) = M(N, Z) - uA, # $$ # where $u$ is the Atomic Mass Unit # $$ # u = M(^{12}\mathrm{C})/12 = 931.4940954(57) \hspace{0.1cm} \mathrm{MeV}/c^2. # $$ # The nucleon masses are # $$ # m_p = 1.00727646693(9)u, # $$ # and # $$ # m_n = 939.56536(8)\hspace{0.1cm} \mathrm{MeV}/c^2 = 1.0086649156(6)u. # $$ # 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) # there are data on masses and decays of 3437 nuclei. # # The nuclear binding energy is defined as the energy required to break # up a given nucleus into its constituent parts of $N$ neutrons and $Z$ # protons. In terms of the atomic masses $M(N, Z)$ the binding energy is # defined by # $$ # BE(N, Z) = ZM_H c^2 + Nm_n c^2 - M(N, Z)c^2 , # $$ # where $M_H$ is the mass of the hydrogen atom and $m_n$ is the mass of the neutron. # In terms of the mass excess the binding energy is given by # $$ # BE(N, Z) = Z\Delta_H c^2 + N\Delta_n c^2 -\Delta(N, Z)c^2 , # $$ # where $\Delta_H c^2 = 7.2890$ MeV and $\Delta_n c^2 = 8.0713$ MeV. # # A popular and physically intuitive model which can be used to parametrize # the experimental binding energies as function of $A$, is the so-called # **liquid drop model**. The ansatz is based on the following expression # $$ # BE(N,Z) = a_1A-a_2A^{2/3}-a_3\frac{Z^2}{A^{1/3}}-a_4\frac{(N-Z)^2}{A}, # $$ # where $A$ stands for the number of nucleons and the $a_i$s are parameters which are determined by a fit # to the experimental data. # # To arrive at the above expression we have assumed that we can make the following assumptions: # # * 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. # # * 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. # # * 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. # # * 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. # # We could also add a so-called pairing term, which is a correction term that # arises from the tendency of proton pairs and neutron pairs to # occur. An even number of particles is more stable than an odd number. # ### Organizing our data # # Let us start with reading and organizing our data. # We start with the compilation of masses and binding energies from 2016. # After having downloaded this file to our own computer, we are now ready to read the file and start structuring our data. # # 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**. # In[27]: # Common imports import numpy as np import pandas as pd import matplotlib.pyplot as plt import sklearn.linear_model as skl from sklearn.model_selection import train_test_split from sklearn.metrics import mean_squared_error, r2_score, mean_absolute_error import os # Where to save the figures and data files PROJECT_ROOT_DIR = "Results" FIGURE_ID = "Results/FigureFiles" DATA_ID = "DataFiles/" if not os.path.exists(PROJECT_ROOT_DIR): os.mkdir(PROJECT_ROOT_DIR) if not os.path.exists(FIGURE_ID): os.makedirs(FIGURE_ID) if not os.path.exists(DATA_ID): os.makedirs(DATA_ID) def image_path(fig_id): return os.path.join(FIGURE_ID, fig_id) def data_path(dat_id): return os.path.join(DATA_ID, dat_id) def save_fig(fig_id): plt.savefig(image_path(fig_id) + ".png", format='png') infile = open(data_path("MassEval2016.dat"),'r') # 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. # In[28]: from pylab import plt, mpl plt.style.use('seaborn') mpl.rcParams['font.family'] = 'serif' def MakePlot(x,y, styles, labels, axlabels): plt.figure(figsize=(10,6)) for i in range(len(x)): plt.plot(x[i], y[i], styles[i], label = labels[i]) plt.xlabel(axlabels[0]) plt.ylabel(axlabels[1]) plt.legend(loc=0) # Our next step is to read the data on experimental binding energies and # reorganize them as functions of the mass number $A$, the number of # protons $Z$ and neutrons $N$ using **pandas**. Before we do this it is # always useful (unless you have a binary file or other types of compressed # data) to actually open the file and simply take a look at it! # # 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. # In[29]: """ This is taken from the data file of the mass 2016 evaluation. All files are 3436 lines long with 124 character per line. Headers are 39 lines long. col 1 : Fortran character control: 1 = page feed 0 = line feed 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 These formats are reflected in the pandas widths variable below, see the statement widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1), Pandas has also a variable header, with length 39 in this case. """ # The data we are interested in are in columns 2, 3, 4 and 11, giving us # the number of neutrons, protons, mass numbers and binding energies, # respectively. We add also for the sake of completeness the element name. The data are in fixed-width formatted lines and we will # covert them into the **pandas** DataFrame structure. # In[30]: # Read the experimental data with Pandas Masses = pd.read_fwf(infile, usecols=(2,3,4,6,11), names=('N', 'Z', 'A', 'Element', 'Ebinding'), widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1), header=39, index_col=False) # Extrapolated values are indicated by '#' in place of the decimal place, so # the Ebinding column won't be numeric. Coerce to float and drop these entries. Masses['Ebinding'] = pd.to_numeric(Masses['Ebinding'], errors='coerce') Masses = Masses.dropna() # Convert from keV to MeV. Masses['Ebinding'] /= 1000 # Group the DataFrame by nucleon number, A. Masses = Masses.groupby('A') # Find the rows of the grouped DataFrame with the maximum binding energy. Masses = Masses.apply(lambda t: t[t.Ebinding==t.Ebinding.max()]) # We have now read in the data, grouped them according to the variables we are interested in. # We see how easy it is to reorganize the data using **pandas**. If we # were to do these operations in C/C++ or Fortran, we would have had to # write various functions/subroutines which perform the above # reorganizations for us. Having reorganized the data, we can now start # to make some simple fits using both the functionalities in **numpy** and # **Scikit-Learn** afterwards. # # Now we define five variables which contain # the number of nucleons $A$, the number of protons $Z$ and the number of neutrons $N$, the element name and finally the energies themselves. # In[31]: A = Masses['A'] Z = Masses['Z'] N = Masses['N'] Element = Masses['Element'] Energies = Masses['Ebinding'] print(Masses) # 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 $\boldsymbol{X}$. # 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. # In[32]: # Now we set up the design matrix X X = np.zeros((len(A),5)) X[:,0] = 1 X[:,1] = A X[:,2] = A**(2.0/3.0) X[:,3] = A**(-1.0/3.0) X[:,4] = A**(-1.0) # With **scikitlearn** we are now ready to use linear regression and fit our data. # In[33]: clf = skl.LinearRegression().fit(X, Energies) fity = clf.predict(X) # Pretty simple! # 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. # In[34]: # The mean squared error print("Mean squared error: %.2f" % mean_squared_error(Energies, fity)) # Explained variance score: 1 is perfect prediction print('Variance score: %.2f' % r2_score(Energies, fity)) # Mean absolute error print('Mean absolute error: %.2f' % mean_absolute_error(Energies, fity)) print(clf.coef_, clf.intercept_) Masses['Eapprox'] = fity # Generate a plot comparing the experimental with the fitted values values. fig, ax = plt.subplots() ax.set_xlabel(r'$A = N + Z$') ax.set_ylabel(r'$E_\mathrm{bind}\,/\mathrm{MeV}$') ax.plot(Masses['A'], Masses['Ebinding'], alpha=0.7, lw=2, label='Ame2016') ax.plot(Masses['A'], Masses['Eapprox'], alpha=0.7, lw=2, c='m', label='Fit') ax.legend() save_fig("Masses2016") plt.show() # ### And what about using neural networks? # # 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) # functionality. # In[35]: from sklearn.neural_network import MLPRegressor from sklearn.metrics import accuracy_score import seaborn as sns X_train = X Y_train = Energies n_hidden_neurons = 50 epochs = 100 # store models for later use eta_vals = np.logspace(-3, 0, 4) lmbd_vals = np.logspace(-3, 0, 4) # store the models for later use DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object) train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals))) sns.set() for i, eta in enumerate(eta_vals): for j, lmbd in enumerate(lmbd_vals): dnn = MLPRegressor(hidden_layer_sizes=(n_hidden_neurons), activation='relu', solver='adam', alpha=lmbd, learning_rate_init=eta, max_iter=epochs) dnn.fit(X_train, Y_train) DNN_scikit[i][j] = dnn train_accuracy[i][j] = dnn.score(X_train, Y_train) fity = dnn.predict(X_train) MSE = mean_squared_error(Y_train, fity) print("Mean squared error: %.2f" % mean_squared_error(Y_train, fity)) train_accuracy[i][j] = MSE fig, ax = plt.subplots(figsize = (10, 10)) sns.heatmap(train_accuracy, annot=True, ax=ax, cmap="viridis") ax.set_title("Training Accuracy") ax.set_ylabel("$\eta$") ax.set_xlabel("$\lambda$") plt.show() print(train_accuracy) # ## A first summary # # The aim behind these introductory words was to present to you various # Python libraries and their functionalities, in particular libraries like # **numpy**, **pandas**, **xarray** and **matplotlib** and other that make our life much easier # in handling various data sets and visualizing data. # # Furthermore, # **Scikit-Learn** allows us with few lines of code to implement popular # Machine Learning algorithms for supervised learning. Later we will meet **Tensorflow**, a powerful library for deep learning. # 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. # ## Why Linear Regression (aka Ordinary Least Squares and family) # # Fitting a continuous function with linear parameterization in terms of the parameters $\boldsymbol{\beta}$. # * Method of choice for fitting a continuous function! # # * Gives an excellent introduction to central Machine Learning features with **understandable pedagogical** links to other methods like **Neural Networks**, **Support Vector Machines** etc # # * Analytical expression for the fitting parameters $\boldsymbol{\beta}$ # # * Analytical expressions for statistical propertiers like mean values, variances, confidence intervals and more # # * Analytical relation with probabilistic interpretations # # * Easy to introduce basic concepts like bias-variance tradeoff, cross-validation, resampling and regularization techniques and many other ML topics # # * Easy to code! And links well with classification problems and logistic regression and neural networks # # * Allows for **easy** hands-on understanding of gradient descent methods # # * and many more features # # For more discussions of Ridge and Lasso regression, [Wessel van Wieringen's](https://arxiv.org/abs/1509.09169) article is highly recommended. # Similarly, [Mehta et al's article](https://arxiv.org/abs/1803.08823) is also recommended. # ## Regression analysis, overarching aims # # 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 $\boldsymbol{x} =[x_0, x_1,\dots, x_{n-1}]^T$. # 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, or simply just the **inputs**. # # A regression model aims at finding a likelihood function $p(\boldsymbol{y}\vert \boldsymbol{x})$ or in the more traditional sense a function $\boldsymbol{y}(\boldsymbol{x})$, 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 # * $n$ cases $i = 0, 1, 2, \dots, n-1$ # # * Response (target, dependent or outcome) variable $y_i$ with $i = 0, 1, 2, \dots, n-1$ # # * $p$ so-called explanatory (independent or predictor or feature) 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. # # The goal of the regression analysis is to extract/exploit relationship between $\boldsymbol{y}$ and $\boldsymbol{x}$ in order to infer specific dependencies, approximations to the likelihood functions, functional relationships and to make predictions, making fits and many other things. # ## Regression analysis, overarching aims II # # Consider an experiment in which $p$ characteristics/features of $n$ samples are # measured. The data from this experiment, for various explanatory variables $p$ are normally represented by a matrix # $\mathbf{X}$. # # The matrix $\mathbf{X}$ is called the *design # matrix*. Additional information of the samples is available in the # form of $\boldsymbol{y}$ (also as above). The variable $\boldsymbol{y}$ is # generally referred to as the *response variable*. The aim of # regression analysis is to explain $\boldsymbol{y}$ in terms of # $\boldsymbol{X}$ through a functional relationship like $y_i = # f(\mathbf{X}_{i,\ast})$. When no prior knowledge on the form of # $f(\cdot)$ is available, it is common to assume a linear relationship # between $\boldsymbol{X}$ and $\boldsymbol{y}$. This assumption gives rise to # the *linear regression model* where $\boldsymbol{\beta} = [\beta_0, \ldots, # \beta_{p-1}]^{T}$ are the *regression parameters*. # # Linear regression gives us a set of analytical equations for the parameters $\beta_j$. # ## Examples # In order to understand the relation among the predictors (or features or properties) $p$, the set of data $n$ and the target (outcome, output etc) $\boldsymbol{y}$, # consider the model we discussed for describing nuclear binding energies. # # There we assumed that we could parametrize the data using a polynomial approximation based on the liquid drop model. # Assuming # $$ # BE(A) = a_0+a_1A+a_2A^{2/3}+a_3A^{-1/3}+a_4A^{-1}, # $$ # 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. # This gives $p=0,1,2,3,4$. Furthermore we have $n$ entries for each predictor. It means that our design matrix is a # $p\times n$ matrix $\boldsymbol{X}$. # # Here the predictors are based on a model we have made. A popular data set which is widely encountered in ML applications is the # 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. # ## General linear models and linear algebra # Before we proceed let us study a case 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. # # 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 # $$ # y=y(x) \rightarrow y(x_i)=\tilde{y}_i+\epsilon_i=\sum_{j=0}^{n-1} \beta_j x_i^j+\epsilon_i, # $$ # where $\epsilon_i$ is the error in our approximation. # ## Rewriting the fitting procedure as a linear algebra problem # For every set of values $y_i,x_i$ we have thus the corresponding set of equations # $$ # \begin{align*} # y_0&=\beta_0+\beta_1x_0^1+\beta_2x_0^2+\dots+\beta_{n-1}x_0^{n-1}+\epsilon_0\\ # y_1&=\beta_0+\beta_1x_1^1+\beta_2x_1^2+\dots+\beta_{n-1}x_1^{n-1}+\epsilon_1\\ # y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2\\ # \dots & \dots \\ # y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_{n-1}x_{n-1}^{n-1}+\epsilon_{n-1}.\\ # \end{align*} # $$ # ## Rewriting the fitting procedure as a linear algebra problem, more details # Defining the vectors # $$ # \boldsymbol{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T, # $$ # and # $$ # \boldsymbol{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T, # $$ # and # $$ # \boldsymbol{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T, # $$ # and the design matrix # $$ # \boldsymbol{X}= # \begin{bmatrix} # 1& x_{0}^1 &x_{0}^2& \dots & \dots &x_{0}^{n-1}\\ # 1& x_{1}^1 &x_{1}^2& \dots & \dots &x_{1}^{n-1}\\ # 1& x_{2}^1 &x_{2}^2& \dots & \dots &x_{2}^{n-1}\\ # \dots& \dots &\dots& \dots & \dots &\dots\\ # 1& x_{n-1}^1 &x_{n-1}^2& \dots & \dots &x_{n-1}^{n-1}\\ # \end{bmatrix} # $$ # we can rewrite our equations as # $$ # \boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}. # $$ # The above design matrix is called a [Vandermonde matrix](https://en.wikipedia.org/wiki/Vandermonde_matrix). # ## Generalizing the fitting procedure as a linear algebra problem # # We are obviously not limited to the above polynomial expansions. We # could replace the various powers of $x$ with elements of Fourier # series or instead of $x_i^j$ we could have $\cos{(j x_i)}$ or $\sin{(j # x_i)}$, or time series or other orthogonal functions. For every set # of values $y_i,x_i$ we can then generalize the equations to # $$ # \begin{align*} # y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ # y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ # y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_2\\ # \dots & \dots \\ # y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_i\\ # \dots & \dots \\ # y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\ # \end{align*} # $$ # **Note that we have $p=n$ here. The matrix is symmetric. This is generally not the case!** # ## Generalizing the fitting procedure as a linear algebra problem # We redefine in turn the matrix $\boldsymbol{X}$ as # $$ # \boldsymbol{X}= # \begin{bmatrix} # x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\ # x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\ # x_{20}& x_{21} &x_{22}& \dots & \dots &x_{2,n-1}\\ # \dots& \dots &\dots& \dots & \dots &\dots\\ # x_{n-1,0}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\ # \end{bmatrix} # $$ # and without loss of generality we rewrite again our equations as # $$ # \boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}. # $$ # 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? # ## Optimizing our parameters # We have defined the matrix $\boldsymbol{X}$ via the equations # $$ # \begin{align*} # y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ # y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ # y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1\\ # \dots & \dots \\ # y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1\\ # \dots & \dots \\ # y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\ # \end{align*} # $$ # As we noted above, we stayed with a system with the design matrix # $\boldsymbol{X}\in {\mathbb{R}}^{n\times n}$, that is we have $p=n$. For reasons to come later (algorithmic arguments) we will hereafter define # 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. # ## Our model for the nuclear binding energies # # 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. # # We restate the parts of the code we are most interested in. # In[36]: # Common imports import numpy as np import pandas as pd import matplotlib.pyplot as plt from IPython.display import display import os # Where to save the figures and data files PROJECT_ROOT_DIR = "Results" FIGURE_ID = "Results/FigureFiles" DATA_ID = "DataFiles/" if not os.path.exists(PROJECT_ROOT_DIR): os.mkdir(PROJECT_ROOT_DIR) if not os.path.exists(FIGURE_ID): os.makedirs(FIGURE_ID) if not os.path.exists(DATA_ID): os.makedirs(DATA_ID) def image_path(fig_id): return os.path.join(FIGURE_ID, fig_id) def data_path(dat_id): return os.path.join(DATA_ID, dat_id) def save_fig(fig_id): plt.savefig(image_path(fig_id) + ".png", format='png') infile = open(data_path("MassEval2016.dat"),'r') # Read the experimental data with Pandas Masses = pd.read_fwf(infile, usecols=(2,3,4,6,11), names=('N', 'Z', 'A', 'Element', 'Ebinding'), widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1), header=39, index_col=False) # Extrapolated values are indicated by '#' in place of the decimal place, so # the Ebinding column won't be numeric. Coerce to float and drop these entries. Masses['Ebinding'] = pd.to_numeric(Masses['Ebinding'], errors='coerce') Masses = Masses.dropna() # Convert from keV to MeV. Masses['Ebinding'] /= 1000 # Group the DataFrame by nucleon number, A. Masses = Masses.groupby('A') # Find the rows of the grouped DataFrame with the maximum binding energy. Masses = Masses.apply(lambda t: t[t.Ebinding==t.Ebinding.max()]) A = Masses['A'] Z = Masses['Z'] N = Masses['N'] Element = Masses['Element'] Energies = Masses['Ebinding'] # Now we set up the design matrix X X = np.zeros((len(A),5)) X[:,0] = 1 X[:,1] = A X[:,2] = A**(2.0/3.0) X[:,3] = A**(-1.0/3.0) X[:,4] = A**(-1.0) # Then nice printout using pandas DesignMatrix = pd.DataFrame(X) DesignMatrix.index = A DesignMatrix.columns = ['1', 'A', 'A^(2/3)', 'A^(-1/3)', '1/A'] display(DesignMatrix) # With $\boldsymbol{\beta}\in {\mathbb{R}}^{p\times 1}$, it means that we will hereafter write our equations for the approximation as # $$ # \boldsymbol{\tilde{y}}= \boldsymbol{X}\boldsymbol{\beta}, # $$ # throughout these lectures. # ## Optimizing our parameters, more details # With the above we use the design matrix to define the approximation $\boldsymbol{\tilde{y}}$ via the unknown quantity $\boldsymbol{\beta}$ as # $$ # \boldsymbol{\tilde{y}}= \boldsymbol{X}\boldsymbol{\beta}, # $$ # 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 # $$ # C(\boldsymbol{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)^T\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)\right\}, # $$ # or using the matrix $\boldsymbol{X}$ and in a more compact matrix-vector notation as # $$ # C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}. # $$ # This function is one possible way to define the so-called cost function. # # It is also common to define # the function $C$ as # $$ # C(\boldsymbol{\beta})=\frac{1}{2n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2, # $$ # since when taking the first derivative with respect to the unknown parameters $\beta$, the factor of $2$ cancels out. # ## Interpretations and optimizing our parameters # # The function # $$ # C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}, # $$ # can be linked to the variance of the quantity $y_i$ if we interpret the latter as the mean value. # When linking (see the discussion below) with the maximum likelihood approach below, we will indeed interpret $y_i$ as a mean value # $$ # y_{i}=\langle y_i \rangle = \beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}+\epsilon_i, # $$ # where $\langle y_i \rangle$ is the mean value. Keep in mind also that # till now we have treated $y_i$ as the exact value. Normally, the # response (dependent or outcome) variable $y_i$ the outcome of a # numerical experiment or another type of experiment and is thus only an # approximation to the true value. It is then always accompanied by an # error estimate, often limited to a statistical error estimate given by # the standard deviation discussed earlier. In the discussion here we # will treat $y_i$ as our exact value for the response variable. # # 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 # $$ # {\displaystyle \min_{\boldsymbol{\beta}\in # {\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}. # $$ # In practical terms it means we will require # $$ # \frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0, # $$ # which results in # $$ # \frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0, # $$ # or in a matrix-vector form as # $$ # \frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right). # $$ # ## Interpretations and optimizing our parameters # We can rewrite # $$ # \frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right), # $$ # as # $$ # \boldsymbol{X}^T\boldsymbol{y} = \boldsymbol{X}^T\boldsymbol{X}\boldsymbol{\beta}, # $$ # and if the matrix $\boldsymbol{X}^T\boldsymbol{X}$ is invertible we have the solution # $$ # \boldsymbol{\beta} =\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}. # $$ # We note also that since our design matrix is defined as $\boldsymbol{X}\in # {\mathbb{R}}^{n\times p}$, the product $\boldsymbol{X}^T\boldsymbol{X} \in # {\mathbb{R}}^{p\times p}$. In the above case we have that $p \ll n$, # in our case $p=5$ meaning that we end up with inverting a small # $5\times 5$ matrix. This is a rather common situation, in many cases we end up with low-dimensional # matrices to invert. The methods discussed here and for many other # supervised learning algorithms like classification with logistic # regression or support vector machines, exhibit dimensionalities which # 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 # $\boldsymbol{X}^T\boldsymbol{X}$. # # **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? # ## Interpretations and optimizing our parameters # The residuals $\boldsymbol{\epsilon}$ are in turn given by # $$ # \boldsymbol{\epsilon} = \boldsymbol{y}-\boldsymbol{\tilde{y}} = \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}, # $$ # and with # $$ # \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)= 0, # $$ # we have # $$ # \boldsymbol{X}^T\boldsymbol{\epsilon}=\boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)= 0, # $$ # meaning that the solution for $\boldsymbol{\beta}$ is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach. # # Let us now return to our nuclear binding energies and simply code the above equations. # ## Own code for Ordinary Least Squares # # 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 # write # In[37]: # matrix inversion to find beta beta = np.linalg.inv(X.T.dot(X)).dot(X.T).dot(Energies) # and then make the prediction ytilde = X @ beta # Alternatively, you can use the least squares functionality in **Numpy** as # In[38]: fit = np.linalg.lstsq(X, Energies, rcond =None)[0] ytildenp = np.dot(fit,X.T) # And finally we plot our fit with and compare with data # In[39]: Masses['Eapprox'] = ytilde # Generate a plot comparing the experimental with the fitted values values. fig, ax = plt.subplots() ax.set_xlabel(r'$A = N + Z$') ax.set_ylabel(r'$E_\mathrm{bind}\,/\mathrm{MeV}$') ax.plot(Masses['A'], Masses['Ebinding'], alpha=0.7, lw=2, label='Ame2016') ax.plot(Masses['A'], Masses['Eapprox'], alpha=0.7, lw=2, c='m', label='Fit') ax.legend() save_fig("Masses2016OLS") plt.show() # ## Adding error analysis and training set up # # 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. # Since we are not using **Scikit-Learn** here we can define our own $R2$ function as # In[40]: def R2(y_data, y_model): return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2) # and we would be using it as # In[41]: print(R2(Energies,ytilde)) # We can easily add our **MSE** score as # In[42]: def MSE(y_data,y_model): n = np.size(y_model) return np.sum((y_data-y_model)**2)/n print(MSE(Energies,ytilde)) # and finally the relative error as # In[43]: def RelativeError(y_data,y_model): return abs((y_data-y_model)/y_data) print(RelativeError(Energies, ytilde)) # ## The $\chi^2$ function # # Normally, the response (dependent or outcome) variable $y_i$ is the # outcome of a numerical experiment or another type of experiment and is # thus only an approximation to the true value. It is then always # accompanied by an error estimate, often limited to a statistical error # estimate given by the standard deviation discussed earlier. In the # discussion here we will treat $y_i$ as our exact value for the # response variable. # # Introducing the standard deviation $\sigma_i$ for each measurement # $y_i$, we define now the $\chi^2$ function (omitting the $1/n$ term) # as # $$ # \chi^2(\boldsymbol{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)^T\frac{1}{\boldsymbol{\Sigma^2}}\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)\right\}, # $$ # where the matrix $\boldsymbol{\Sigma}$ is a diagonal matrix with $\sigma_i$ as matrix elements. # ## The $\chi^2$ function # # In order to find the parameters $\beta_i$ we will then minimize the spread of $\chi^2(\boldsymbol{\beta})$ by requiring # $$ # \frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0, # $$ # which results in # $$ # \frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0, # $$ # or in a matrix-vector form as # $$ # \frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{A}^T\left( \boldsymbol{b}-\boldsymbol{A}\boldsymbol{\beta}\right). # $$ # 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 $b_i = y_i/\sigma_i$. # ## The $\chi^2$ function # # We can rewrite # $$ # \frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{A}^T\left( \boldsymbol{b}-\boldsymbol{A}\boldsymbol{\beta}\right), # $$ # as # $$ # \boldsymbol{A}^T\boldsymbol{b} = \boldsymbol{A}^T\boldsymbol{A}\boldsymbol{\beta}, # $$ # and if the matrix $\boldsymbol{A}^T\boldsymbol{A}$ is invertible we have the solution # $$ # \boldsymbol{\beta} =\left(\boldsymbol{A}^T\boldsymbol{A}\right)^{-1}\boldsymbol{A}^T\boldsymbol{b}. # $$ # ## The $\chi^2$ function # # If we then introduce the matrix # $$ # \boldsymbol{H} = \left(\boldsymbol{A}^T\boldsymbol{A}\right)^{-1}, # $$ # we have then the following expression for the parameters $\beta_j$ (the matrix elements of $\boldsymbol{H}$ are $h_{ij}$) # $$ # \beta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik} # $$ # We state without proof the expression for the uncertainty in the parameters $\beta_j$ as (we leave this as an exercise) # $$ # \sigma^2(\beta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \beta_j}{\partial y_i}\right)^2, # $$ # resulting in # $$ # \sigma^2(\beta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}! # $$ # ## The $\chi^2$ function # The first step here is to approximate the function $y$ with a first-order polynomial, that is we write # $$ # y=y(x) \rightarrow y(x_i) \approx \beta_0+\beta_1 x_i. # $$ # By computing the derivatives of $\chi^2$ with respect to $\beta_0$ and $\beta_1$ show that these are given by # $$ # \frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_0} = -2\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0, # $$ # and # $$ # \frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_1} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0. # $$ # ## The $\chi^2$ function # # For a linear fit (a first-order polynomial) we don't need to invert a matrix!! # Defining # $$ # \gamma = \sum_{i=0}^{n-1}\frac{1}{\sigma_i^2}, # $$ # $$ # \gamma_x = \sum_{i=0}^{n-1}\frac{x_{i}}{\sigma_i^2}, # $$ # $$ # \gamma_y = \sum_{i=0}^{n-1}\left(\frac{y_i}{\sigma_i^2}\right), # $$ # $$ # \gamma_{xx} = \sum_{i=0}^{n-1}\frac{x_ix_{i}}{\sigma_i^2}, # $$ # $$ # \gamma_{xy} = \sum_{i=0}^{n-1}\frac{y_ix_{i}}{\sigma_i^2}, # $$ # we obtain # $$ # \beta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}, # $$ # $$ # \beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}. # $$ # This approach (different linear and non-linear regression) suffers # often from both being underdetermined and overdetermined in the # unknown coefficients $\beta_i$. A better approach is to use the # Singular Value Decomposition (SVD) method discussed next week. # ## Fitting an Equation of State for Dense Nuclear Matter # # Before we continue, let us introduce yet another example. We are going to fit the # nuclear equation of state using results from many-body calculations. # The equation of state we have made available here, as function of # density, has been derived using modern nucleon-nucleon potentials with # [the addition of three-body # forces](https://www.sciencedirect.com/science/article/pii/S0370157399001106). This # time the file is presented as a standard **csv** file. # # The beginning of the Python code here is similar to what you have seen # before, with the same initializations and declarations. We use also # **pandas** again, rather extensively in order to organize our data. # # The difference now is that we use **Scikit-Learn's** regression tools # instead of our own matrix inversion implementation. Furthermore, we # sneak in **Ridge** regression (to be discussed below) which includes a # hyperparameter $\lambda$, also to be explained below. # ## The code # In[44]: # Common imports import os import numpy as np import pandas as pd import matplotlib.pyplot as plt import matplotlib.pyplot as plt import sklearn.linear_model as skl from sklearn.metrics import mean_squared_error, r2_score, mean_absolute_error # Where to save the figures and data files PROJECT_ROOT_DIR = "Results" FIGURE_ID = "Results/FigureFiles" DATA_ID = "DataFiles/" if not os.path.exists(PROJECT_ROOT_DIR): os.mkdir(PROJECT_ROOT_DIR) if not os.path.exists(FIGURE_ID): os.makedirs(FIGURE_ID) if not os.path.exists(DATA_ID): os.makedirs(DATA_ID) def image_path(fig_id): return os.path.join(FIGURE_ID, fig_id) def data_path(dat_id): return os.path.join(DATA_ID, dat_id) def save_fig(fig_id): plt.savefig(image_path(fig_id) + ".png", format='png') infile = open(data_path("EoS.csv"),'r') # Read the EoS data as csv file and organize the data into two arrays with density and energies EoS = pd.read_csv(infile, names=('Density', 'Energy')) EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce') EoS = EoS.dropna() Energies = EoS['Energy'] Density = EoS['Density'] # The design matrix now as function of various polytrops X = np.zeros((len(Density),4)) X[:,3] = Density**(4.0/3.0) X[:,2] = Density X[:,1] = Density**(2.0/3.0) X[:,0] = 1 # We use now Scikit-Learn's linear regressor and ridge regressor # OLS part clf = skl.LinearRegression().fit(X, Energies) ytilde = clf.predict(X) EoS['Eols'] = ytilde # The mean squared error print("Mean squared error: %.2f" % mean_squared_error(Energies, ytilde)) # Explained variance score: 1 is perfect prediction print('Variance score: %.2f' % r2_score(Energies, ytilde)) # Mean absolute error print('Mean absolute error: %.2f' % mean_absolute_error(Energies, ytilde)) print(clf.coef_, clf.intercept_) # The Ridge regression with a hyperparameter lambda = 0.1 _lambda = 0.1 clf_ridge = skl.Ridge(alpha=_lambda).fit(X, Energies) yridge = clf_ridge.predict(X) EoS['Eridge'] = yridge # The mean squared error print("Mean squared error: %.2f" % mean_squared_error(Energies, yridge)) # Explained variance score: 1 is perfect prediction print('Variance score: %.2f' % r2_score(Energies, yridge)) # Mean absolute error print('Mean absolute error: %.2f' % mean_absolute_error(Energies, yridge)) print(clf_ridge.coef_, clf_ridge.intercept_) fig, ax = plt.subplots() ax.set_xlabel(r'$\rho[\mathrm{fm}^{-3}]$') ax.set_ylabel(r'Energy per particle') ax.plot(EoS['Density'], EoS['Energy'], alpha=0.7, lw=2, label='Theoretical data') ax.plot(EoS['Density'], EoS['Eols'], alpha=0.7, lw=2, c='m', label='OLS') ax.plot(EoS['Density'], EoS['Eridge'], alpha=0.7, lw=2, c='g', label='Ridge $\lambda = 0.1$') ax.legend() save_fig("EoSfitting") plt.show() # The above simple polynomial in density $\rho$ gives an excellent fit # to the data. # # We note also that there is a small deviation between the # standard OLS and the Ridge regression at higher densities. We discuss this in more detail # below. # ## Splitting our Data in Training and Test data # # It is normal in essentially all Machine Learning studies to split the # data in a training set and a test set (sometimes also an additional # validation set). **Scikit-Learn** has an own function for this. There # is no explicit recipe for how much data should be included as training # data and say test data. An accepted rule of thumb is to use # approximately $2/3$ to $4/5$ of the data as training data. We will # postpone a discussion of this splitting to the end of these notes and # our discussion of the so-called **bias-variance** tradeoff. Here we # limit ourselves to repeat the above equation of state fitting example # but now splitting the data into a training set and a test set. # In[45]: import os import numpy as np import pandas as pd import matplotlib.pyplot as plt from sklearn.model_selection import train_test_split # Where to save the figures and data files PROJECT_ROOT_DIR = "Results" FIGURE_ID = "Results/FigureFiles" DATA_ID = "DataFiles/" if not os.path.exists(PROJECT_ROOT_DIR): os.mkdir(PROJECT_ROOT_DIR) if not os.path.exists(FIGURE_ID): os.makedirs(FIGURE_ID) if not os.path.exists(DATA_ID): os.makedirs(DATA_ID) def image_path(fig_id): return os.path.join(FIGURE_ID, fig_id) def data_path(dat_id): return os.path.join(DATA_ID, dat_id) def save_fig(fig_id): plt.savefig(image_path(fig_id) + ".png", format='png') def R2(y_data, y_model): return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2) def MSE(y_data,y_model): n = np.size(y_model) return np.sum((y_data-y_model)**2)/n infile = open(data_path("EoS.csv"),'r') # Read the EoS data as csv file and organized into two arrays with density and energies EoS = pd.read_csv(infile, names=('Density', 'Energy')) EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce') EoS = EoS.dropna() Energies = EoS['Energy'] Density = EoS['Density'] # The design matrix now as function of various polytrops X = np.zeros((len(Density),5)) X[:,0] = 1 X[:,1] = Density**(2.0/3.0) X[:,2] = Density X[:,3] = Density**(4.0/3.0) X[:,4] = Density**(5.0/3.0) # We split the data in test and training data X_train, X_test, y_train, y_test = train_test_split(X, Energies, test_size=0.2) # matrix inversion to find beta beta = np.linalg.inv(X_train.T.dot(X_train)).dot(X_train.T).dot(y_train) # and then make the prediction ytilde = X_train @ beta print("Training R2") print(R2(y_train,ytilde)) print("Training MSE") print(MSE(y_train,ytilde)) ypredict = X_test @ beta print("Test R2") print(R2(y_test,ypredict)) print("Test MSE") print(MSE(y_test,ypredict)) # ## Exercises # # Here are three possible exercises for week 34 # ## Exercise 1: Setting up various Python environments # # The first exercise here is of a mere technical art. We want you to have # * git as a version control software and to establish a user account on a provider like GitHub. Other providers like GitLab etc are equally fine. You can also use the University of Oslo [GitHub facilities](https://www.uio.no/tjenester/it/maskin/filer/versjonskontroll/github.html). # # * Install various Python packages # # We will make extensive use of Python as programming language and its # myriad of available libraries. You will find # IPython/Jupyter notebooks invaluable in your work. You can run **R** # codes in the Jupyter/IPython notebooks, with the immediate benefit of # visualizing your data. You can also use compiled languages like C++, # Rust, Fortran etc if you prefer. The focus in these lectures will be # on Python. # # If you have Python installed (we recommend Python3) and you feel # pretty familiar with installing different packages, we recommend that # you install the following Python packages via **pip** as # # 1. pip install numpy scipy matplotlib ipython scikit-learn sympy pandas pillow # # For **Tensorflow**, we recommend following the instructions in the text of # [Aurelien Geron, Hands‑On Machine Learning with Scikit‑Learn and TensorFlow, O'Reilly](http://shop.oreilly.com/product/0636920052289.do) # # We will come back to **tensorflow** later. # # For Python3, replace **pip** with **pip3**. # # For OSX users we recommend, after having installed Xcode, to # install **brew**. Brew allows for a seamless installation of additional # software via for example # # 1. brew install python3 # # For Linux users, with its variety of distributions like for example the widely popular Ubuntu distribution, # you can use **pip** as well and simply install Python as # # 1. sudo apt-get install python3 (or python for Python2.7) # # If you don't want to perform these operations separately and venture # into the hassle of exploring how to set up dependencies and paths, we # recommend two widely used distrubutions which set up all relevant # dependencies for Python, namely # # * [Anaconda](https://docs.anaconda.com/), # # which is an open source # distribution of the Python and R programming languages for large-scale # data processing, predictive analytics, and scientific computing, that # aims to simplify package management and deployment. Package versions # are managed by the package management system **conda**. # # * [Enthought canopy](https://www.enthought.com/product/canopy/) # # is a Python # distribution for scientific and analytic computing distribution and # analysis environment, available for free and under a commercial # license. # # We recommend using **Anaconda** if you are not too familiar with setting paths in a terminal environment. # ## Exercise 2: making your own data and exploring scikit-learn # # We will generate our own dataset for a function $y(x)$ where $x \in [0,1]$ and defined by random numbers computed with the uniform distribution. The function $y$ is a quadratic polynomial in $x$ with added stochastic noise according to the normal distribution $\cal {N}(0,1)$. # The following simple Python instructions define our $x$ and $y$ values (with 100 data points). # In[46]: x = np.random.rand(100,1) y = 2.0+5*x*x+0.1*np.random.randn(100,1) # 1. Write your own code (following the examples under the [regression notes](https://compphysics.github.io/MachineLearning/doc/LectureNotes/_build/html/chapter1.html)) for computing the parametrization of the data set fitting a second-order polynomial. # # 2. Use thereafter **scikit-learn** (see again the examples in the regression slides) and compare with your own code. # # 3. Using scikit-learn, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as # $$ # MSE(\boldsymbol{y},\boldsymbol{\tilde{y}}) = \frac{1}{n} # \sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, # $$ # and the $R^2$ score function. # If $\tilde{\boldsymbol{y}}_i$ 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 # $$ # R^2(\boldsymbol{y}, \tilde{\boldsymbol{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2}, # $$ # where we have defined the mean value of $\boldsymbol{y}$ as # $$ # \bar{y} = \frac{1}{n} \sum_{i=0}^{n - 1} y_i. # $$ # You can use the functionality included in scikit-learn. If you feel for it, you can use your own program and define functions which compute the above two functions. # Discuss the meaning of these results. Try also to vary the coefficient in front of the added stochastic noise term and discuss the quality of the fits. # ## Exercise 3: Split data in test and training data # # In this exercise we want you to to compute the MSE for the training # data and the test data as function of the complexity of a polynomial, # that is the degree of a given polynomial. # # The aim is to reproduce Figure 2.11 of [Hastie et al](https://github.com/CompPhysics/MLErasmus/blob/master/doc/Textbooks/elementsstat.pdf). # # Our data is defined by $x\in [-3,3]$ with a total of for example $n=100$ data points. You should try to vary the number of data points $n$ in your analysis. # In[47]: np.random.seed() n = 100 # Make data set. x = np.linspace(-3, 3, n).reshape(-1, 1) y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.normal(0, 0.1, x.shape) # where $y$ is the function we want to fit with a given polynomial. # **a)** # Write a first code which sets up a design matrix $X$ defined by a fifth-order polynomial and split your data set in training and test data. # **b)** # Write thereafter (using either **scikit-learn** or your matrix inversion code using for example **numpy**) # and perform an ordinary least squares fitting and compute the mean squared error for the training data and the test data. These calculations should apply to a model given by a fifth-order polynomial. # **c)** # Add now a model which allows you to make polynomials up to degree $15$. Perform a standard OLS fitting of the training data and compute the MSE for the training and test data and plot both test and training data MSE as functions of the polynomial degree. Compare what you see with Figure 2.11 of Hastie et al. Comment your results. For which polynomial degree do you find an optimal MSE (smallest value)?