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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<!-- dom:TITLE: Data Analysis and Machine Learning: Logistic Regression -->\n",
"# Data Analysis and Machine Learning: Logistic Regression\n",
"<!-- dom:AUTHOR: Morten Hjorth-Jensen at Department of Physics, University of Oslo & Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University -->\n",
"<!-- Author: --> \n",
"**Morten Hjorth-Jensen**, Department of Physics, University of Oslo and Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University\n",
"\n",
"Date: **Sep 17, 2020**\n",
"\n",
"Copyright 1999-2020, Morten Hjorth-Jensen. Released under CC Attribution-NonCommercial 4.0 license\n",
"\n",
"\n",
"\n",
"\n",
"## Plans for week 38\n",
"\n",
"* Thursday: Summary of regression methods and discussion of project 1. We revisit also cross-validation and bootstrap as resampling techniques with examples\n",
"\n",
"* Friday: Logistic Regression\n",
"\n",
"## Thursday September 17\n",
"\n",
"\n",
"## Ridge and LASSO Regression, reminder\n",
"\n",
"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": [
"<!-- !split -->\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",
"<!-- !split -->\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": 1,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"%matplotlib inline\n",
"\n",
"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": [
"## Bias-Variance tradeoff with Bootstrap"
]
},
{
"cell_type": "code",
"execution_count": 2,
"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": [
"## Another Example from Scikit-Learn's Repository"
]
},
{
"cell_type": "code",
"execution_count": 3,
"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": [
"## Cross-validation with Ridge"
]
},
{
"cell_type": "code",
"execution_count": 4,
"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": [
"<!-- Equation labels as ordinary links -->\n",
"<div id=\"_auto1\"></div>\n",
"\n",
"$$\n",
"\\begin{equation}\n",
" H = -J \\sum_{k}^L s_k s_{k + 1},\n",
"\\label{_auto1} \\tag{1}\n",
"\\end{equation}\n",
"$$"
]
},
{
"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": 5,
"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": [
"<!-- Equation labels as ordinary links -->\n",
"<div id=\"_auto2\"></div>\n",
"\n",
"$$\n",
"\\begin{equation}\n",
" H = - \\sum_j^L \\sum_k^L s_j s_k J_{jk}.\n",
"\\label{_auto2} \\tag{2}\n",
"\\end{equation}\n",
"$$"
]
},
{
"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": [
"<!-- Equation labels as ordinary links -->\n",
"<div id=\"_auto3\"></div>\n",
"\n",
"$$\n",
"\\begin{equation}\n",
" \\boldsymbol{H} = \\boldsymbol{X} J,\n",
"\\label{_auto3} \\tag{3}\n",
"\\end{equation}\n",
"$$"
]
},
{
"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": [
"<!-- Equation labels as ordinary links -->\n",
"<div id=\"_auto4\"></div>\n",
"\n",
"$$\n",
"\\begin{equation}\n",
" \\boldsymbol{y} = \\boldsymbol{X}\\boldsymbol{\\beta} + \\boldsymbol{\\epsilon},\n",
"\\label{_auto4} \\tag{4}\n",
"\\end{equation}\n",
"$$"
]
},
{
"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": 6,
"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": [
"<!-- Equation labels as ordinary links -->\n",
"<div id=\"_auto5\"></div>\n",
"\n",
"$$\n",
"\\begin{equation}\n",
" C(\\boldsymbol{X}, \\boldsymbol{\\beta})= \\frac{1}{n}\\left\\{(\\boldsymbol{X}\\boldsymbol{\\beta} - \\boldsymbol{y})^T(\\boldsymbol{X}\\boldsymbol{\\beta} - \\boldsymbol{y})\\right\\}.\n",
"\\label{_auto5} \\tag{5}\n",
"\\end{equation}\n",
"$$"
]
},
{
"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": 7,
"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": 8,
"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": [
"<!-- Equation labels as ordinary links -->\n",
"<div id=\"_auto6\"></div>\n",
"\n",
"$$\n",
"\\begin{equation}\n",
" \\boldsymbol{\\beta} = \\boldsymbol{V}\\boldsymbol{\\Sigma}^{+} \\boldsymbol{U}^T \\boldsymbol{y}.\n",
"\\label{_auto6} \\tag{6}\n",
"\\end{equation}\n",
"$$"
]
},
{
"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": 9,
"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": 10,
"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": 11,
"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": 12,
"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": [
"<!-- Equation labels as ordinary links -->\n",
"<div id=\"_auto7\"></div>\n",
"\n",
"$$\n",
"\\begin{equation}\n",
" H = -J \\sum_{k}^L s_k s_{k + 1},\n",
"\\label{_auto7} \\tag{7}\n",
"\\end{equation}\n",
"$$"
]
},
{
"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": 13,
"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": [
"<!-- Equation labels as ordinary links -->\n",
"<div id=\"_auto8\"></div>\n",
"\n",
"$$\n",
"\\begin{equation}\n",
" H = - \\sum_j^L \\sum_k^L s_j s_k J_{jk}.\n",
"\\label{_auto8} \\tag{8}\n",
"\\end{equation}\n",
"$$"
]
},
{
"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": [
"<!-- Equation labels as ordinary links -->\n",
"<div id=\"_auto9\"></div>\n",
"\n",
"$$\n",
"\\begin{equation}\n",
" H = X J,\n",
"\\label{_auto9} \\tag{9}\n",
"\\end{equation}\n",
"$$"
]
},
{
"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": [
"<!-- Equation labels as ordinary links -->\n",
"<div id=\"_auto10\"></div>\n",
"\n",
"$$\n",
"\\begin{equation}\n",
" \\boldsymbol{y} = \\boldsymbol{X}\\boldsymbol{\\beta} + \\boldsymbol{\\epsilon}.\n",
"\\label{_auto10} \\tag{10}\n",
"\\end{equation}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"We organize the data as we did above"
]
},
{
"cell_type": "code",
"execution_count": 14,
"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": 15,
"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": 16,
"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": 17,
"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": [
"2\n",
"2\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": 18,
"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": [
"<!-- Equation labels as ordinary links -->\n",
"<div id=\"_auto12\"></div>\n",
"\n",
"$$\n",
"\\begin{equation}\n",
" C(\\boldsymbol{X}, \\boldsymbol{\\beta}; \\lambda) = (\\boldsymbol{X}\\boldsymbol{\\beta} - \\boldsymbol{y})^T(\\boldsymbol{X}\\boldsymbol{\\beta} - \\boldsymbol{y}) + \\lambda \\sqrt{\\boldsymbol{\\beta}^T\\boldsymbol{\\beta}}.\n",
"\\label{_auto12} \\tag{12}\n",
"\\end{equation}\n",
"$$"
]
},
{
"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": 19,
"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": 20,
"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": 21,
"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 $\\lambda$.\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"## Friday September 18: Intro to Logistic Regression\n",
"\n",
"\n",
"<!-- !split -->\n",
"## Logistic Regression\n",
"\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",
"<!-- !split -->\n",
"## Classification problems\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",
"## Optimization and Deep learning\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",
"<!-- !split -->\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": [
"## Linear classifier\n",
"\n",
"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": [
"<!-- Equation labels as ordinary links -->\n",
"<div id=\"_auto13\"></div>\n",
"\n",
"$$\n",
"\\begin{equation}\n",
"\\hat{y} = \\hat{X}^T\\hat{\\beta} + \\hat{\\epsilon},\n",
"\\label{_auto13} \\tag{13}\n",
"\\end{equation}\n",
"$$"
]
},
{
"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",
"## Some selected properties\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",
"## 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",
"## Examples of likelihood functions used in logistic regression and nueral networks\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": 22,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"\"\"\"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": [
"<!-- !split -->\n",
"## 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": [
"## The cost function rewritten\n",
"\n",
"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",
"## Minimizing the cross entropy\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": [
"## A more compact expression\n",
"\n",
"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": [
"## Extending to more predictors\n",
"\n",
"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": [
"3\n",
"9\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",
"## More classes\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).\n",
"\n",
"\n",
"\n",
"\n",
"## A simple classification problem"
]
},
{
"cell_type": "code",
"execution_count": 23,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"import numpy as np\n",
"from sklearn import datasets, linear_model\n",
"import matplotlib.pyplot as plt\n",
"\n",
"\n",
"def generate_data():\n",
" np.random.seed(0)\n",
" X, y = datasets.make_moons(200, noise=0.20)\n",
" return X, y\n",
"\n",
"\n",
"def visualize(X, y, clf):\n",
" plot_decision_boundary(lambda x: clf.predict(x), X, y)\n",
"\n",
"def plot_decision_boundary(pred_func, X, y):\n",
" # Set min and max values and give it some padding\n",
" x_min, x_max = X[:, 0].min() - .5, X[:, 0].max() + .5\n",
" y_min, y_max = X[:, 1].min() - .5, X[:, 1].max() + .5\n",
" h = 0.01\n",
" # Generate a grid of points with distance h between them\n",
" xx, yy = np.meshgrid(np.arange(x_min, x_max, h), np.arange(y_min, y_max, h))\n",
" # Predict the function value for the whole gid\n",
" Z = pred_func(np.c_[xx.ravel(), yy.ravel()])\n",
" Z = Z.reshape(xx.shape)\n",
" # Plot the contour and training examples\n",
" plt.contourf(xx, yy, Z, cmap=plt.cm.Spectral)\n",
" plt.scatter(X[:, 0], X[:, 1], c=y, cmap=plt.cm.Spectral)\n",
" plt.show()\n",
"\n",
"\n",
"def classify(X, y):\n",
" clf = linear_model.LogisticRegressionCV()\n",
" clf.fit(X, y)\n",
" return clf\n",
"\n",
"\n",
"def main():\n",
" X, y = generate_data()\n",
" # visualize(X, y)\n",
" clf = classify(X, y)\n",
" visualize(X, y, clf)\n",
"\n",
"if __name__ == \"__main__\":\n",
" main()"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Cancer Data again now with Decision Trees and other Methods"
]
},
{
"cell_type": "code",
"execution_count": 24,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"import matplotlib.pyplot as plt\n",
"import numpy as np\n",
"from sklearn.model_selection import train_test_split \n",
"from sklearn.datasets import load_breast_cancer\n",
"from sklearn.linear_model import LogisticRegression\n",
"\n",
"# Load the data\n",
"cancer = load_breast_cancer()\n",
"\n",
"X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)\n",
"print(X_train.shape)\n",
"print(X_test.shape)\n",
"# Logistic Regression\n",
"logreg = LogisticRegression(solver='lbfgs')\n",
"logreg.fit(X_train, y_train)\n",
"print(\"Test set accuracy with Logistic Regression: {:.2f}\".format(logreg.score(X_test,y_test)))\n",
"#now scale the data\n",
"from sklearn.preprocessing import StandardScaler\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",
"# Logistic Regression\n",
"logreg.fit(X_train_scaled, y_train)\n",
"print(\"Test set accuracy Logistic Regression with scaled data: {:.2f}\".format(logreg.score(X_test_scaled,y_test)))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Other measures in classification studies: Cancer Data again"
]
},
{
"cell_type": "code",
"execution_count": 25,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"import matplotlib.pyplot as plt\n",
"import numpy as np\n",
"from sklearn.model_selection import train_test_split \n",
"from sklearn.datasets import load_breast_cancer\n",
"from sklearn.linear_model import LogisticRegression\n",
"\n",
"# Load the data\n",
"cancer = load_breast_cancer()\n",
"\n",
"X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)\n",
"print(X_train.shape)\n",
"print(X_test.shape)\n",
"# Logistic Regression\n",
"logreg = LogisticRegression(solver='lbfgs')\n",
"logreg.fit(X_train, y_train)\n",
"print(\"Test set accuracy with Logistic Regression: {:.2f}\".format(logreg.score(X_test,y_test)))\n",
"#now scale the data\n",
"from sklearn.preprocessing import StandardScaler\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",
"# Logistic Regression\n",
"logreg.fit(X_train_scaled, y_train)\n",
"print(\"Test set accuracy Logistic Regression with scaled data: {:.2f}\".format(logreg.score(X_test_scaled,y_test)))\n",
"\n",
"\n",
"from sklearn.preprocessing import LabelEncoder\n",
"from sklearn.model_selection import cross_validate\n",
"#Cross validation\n",
"accuracy = cross_validate(logreg,X_test_scaled,y_test,cv=10)['test_score']\n",
"print(accuracy)\n",
"print(\"Test set accuracy with Logistic Regression and scaled data: {:.2f}\".format(logreg.score(X_test_scaled,y_test)))\n",
"\n",
"\n",
"import scikitplot as skplt\n",
"y_pred = logreg.predict(X_test_scaled)\n",
"skplt.metrics.plot_confusion_matrix(y_test, y_pred, normalize=True)\n",
"plt.show()\n",
"y_probas = logreg.predict_proba(X_test_scaled)\n",
"skplt.metrics.plot_roc(y_test, y_probas)\n",
"plt.show()\n",
"skplt.metrics.plot_cumulative_gain(y_test, y_probas)\n",
"plt.show()"
]
}
],
"metadata": {},
"nbformat": 4,
"nbformat_minor": 2
}
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