Morten Hjorth-Jensen
1595 lines
50 KiB
Plaintext
1595 lines
50 KiB
Plaintext
TITLE: Data Analysis and Machine Learning: Logistic Regression
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AUTHOR: Morten Hjorth-Jensen {copyright, 1999-present|CC BY-NC} at Department of Physics, University of Oslo & Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University
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DATE: today
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!split
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===== Plans for week 38 =====
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* Thursday: Summary of regression methods and discussion of project 1. We revisit also cross-validation and bootstrap as resampling techniques with examples. Recommended reading: "Hastie et al":"https://www.springer.com/gp/book/9780387848570" chapters 3 and 7.1-7.6 and 7.10-7.12.
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* Friday: Logistic Regression. Recommended reading: "Hastie et al":"https://www.springer.com/gp/book/9780387848570" chapters 4.1-4.4 and "Murphy":"https://mitpress.mit.edu/books/machine-learning-1" chapter 8.1-8.2
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!split
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===== Thursday September 17 =====
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"Video of Lecture":"https://www.uio.no/studier/emner/matnat/fys/FYS-STK4155/h20/forelesningsvideoer/LectureSeptember17.mp4?vrtx=view-as-webpage" and "link to handwritten notes":"https://github.com/CompPhysics/MachineLearning/blob/master/doc/HandWrittenNotes/NotesSeptember17.pdf".
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!split
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===== Ridge and LASSO Regression, reminder =====
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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
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our optimization problem is
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!bt
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\[
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{\displaystyle \min_{\bm{\beta}\in {\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\bm{y}-\bm{X}\bm{\beta}\right)^T\left(\bm{y}-\bm{X}\bm{\beta}\right)\right\}.
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\]
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!et
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or we can state it as
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!bt
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\[
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{\displaystyle \min_{\bm{\beta}\in
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{\mathbb{R}}^{p}}}\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\vert\vert \bm{y}-\bm{X}\bm{\beta}\vert\vert_2^2,
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\]
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!et
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where we have used the definition of a norm-2 vector, that is
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!bt
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\[
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\vert\vert \bm{x}\vert\vert_2 = \sqrt{\sum_i x_i^2}.
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\]
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!et
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By minimizing the above equation with respect to the parameters
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$\bm{\beta}$ we could then obtain an analytical expression for the
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parameters $\bm{\beta}$. We can add a regularization parameter $\lambda$ by
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defining a new cost function to be optimized, that is
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!bt
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\[
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{\displaystyle \min_{\bm{\beta}\in
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{\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \bm{y}-\bm{X}\bm{\beta}\vert\vert_2^2+\lambda\vert\vert \bm{\beta}\vert\vert_2^2
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\]
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!et
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which leads to the Ridge regression minimization problem where we
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require that $\vert\vert \bm{\beta}\vert\vert_2^2\le t$, where $t$ is
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a finite number larger than zero. By defining
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!bt
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\[
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C(\bm{X},\bm{\beta})=\frac{1}{n}\vert\vert \bm{y}-\bm{X}\bm{\beta}\vert\vert_2^2+\lambda\vert\vert \bm{\beta}\vert\vert_1,
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\]
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!et
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we have a new optimization equation
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!bt
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\[
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{\displaystyle \min_{\bm{\beta}\in
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{\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \bm{y}-\bm{X}\bm{\beta}\vert\vert_2^2+\lambda\vert\vert \bm{\beta}\vert\vert_1
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\]
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!et
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which leads to Lasso regression. Lasso stands for least absolute shrinkage and selection operator.
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Here we have defined the norm-1 as
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!bt
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\[
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\vert\vert \bm{x}\vert\vert_1 = \sum_i \vert x_i\vert.
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\]
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!et
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!split
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===== Various steps in cross-validation =====
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When the repetitive splitting of the data set is done randomly,
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samples may accidently end up in a fast majority of the splits in
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either training or test set. Such samples may have an unbalanced
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influence on either model building or prediction evaluation. To avoid
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this $k$-fold cross-validation structures the data splitting. The
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samples are divided into $k$ more or less equally sized exhaustive and
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mutually exclusive subsets. In turn (at each split) one of these
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subsets plays the role of the test set while the union of the
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remaining subsets constitutes the training set. Such a splitting
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warrants a balanced representation of each sample in both training and
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test set over the splits. Still the division into the $k$ subsets
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involves a degree of randomness. This may be fully excluded when
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choosing $k=n$. This particular case is referred to as leave-one-out
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cross-validation (LOOCV).
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!split
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===== How to set up the cross-validation for Ridge and/or Lasso =====
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* Define a range of interest for the penalty parameter.
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* Divide the data set into training and test set comprising samples $\{1, \ldots, n\} \setminus i$ and $\{ i \}$, respectively.
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* 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 $\bm{\sigma}_{-i}^2(\lambda)$, as
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!bt
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\begin{align*}
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\bm{\beta}_{-i}(\lambda) & = ( \bm{X}_{-i, \ast}^{T}
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\bm{X}_{-i, \ast} + \lambda \bm{I}_{pp})^{-1}
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\bm{X}_{-i, \ast}^{T} \bm{y}_{-i}
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\end{align*}
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!et
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* Evaluate the prediction performance of these models on the test set by $\log\{L[y_i, \bm{X}_{i, \ast}; \bm{\beta}_{-i}(\lambda), \bm{\sigma}_{-i}^2(\lambda)]\}$. Or, by the prediction error $|y_i - \bm{X}_{i, \ast} \bm{\beta}_{-i}(\lambda)|$, the relative error, the error squared or the R2 score function.
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* Repeat the first three steps such that each sample plays the role of the test set once.
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* 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
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!bt
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\begin{align*}
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\frac{1}{n} \sum_{i = 1}^n \log\{L[y_i, \mathbf{X}_{i, \ast}; \bm{\beta}_{-i}(\lambda), \bm{\sigma}_{-i}^2(\lambda)]\}.
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\end{align*}
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!et
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!split
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===== Cross-validation in brief =====
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For the various values of $k$
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o shuffle the dataset randomly.
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o Split the dataset into $k$ groups.
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o For each unique group:
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o Decide which group to use as set for test data
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o Take the remaining groups as a training data set
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o Fit a model on the training set and evaluate it on the test set
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o Retain the evaluation score and discard the model
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o Summarize the model using the sample of model evaluation scores
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!split
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===== Code Example for Cross-validation and $k$-fold Cross-validation =====
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The code here uses Ridge regression with cross-validation (CV) resampling and $k$-fold CV in order to fit a specific polynomial.
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!bc pycod
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import numpy as np
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import matplotlib.pyplot as plt
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from sklearn.model_selection import KFold
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from sklearn.linear_model import Ridge
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from sklearn.model_selection import cross_val_score
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from sklearn.preprocessing import PolynomialFeatures
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# A seed just to ensure that the random numbers are the same for every run.
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# Useful for eventual debugging.
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np.random.seed(3155)
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# Generate the data.
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nsamples = 100
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x = np.random.randn(nsamples)
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y = 3*x**2 + np.random.randn(nsamples)
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## Cross-validation on Ridge regression using KFold only
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# Decide degree on polynomial to fit
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poly = PolynomialFeatures(degree = 6)
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# Decide which values of lambda to use
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nlambdas = 500
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lambdas = np.logspace(-3, 5, nlambdas)
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# Initialize a KFold instance
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k = 5
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kfold = KFold(n_splits = k)
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# Perform the cross-validation to estimate MSE
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scores_KFold = np.zeros((nlambdas, k))
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i = 0
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for lmb in lambdas:
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ridge = Ridge(alpha = lmb)
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j = 0
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for train_inds, test_inds in kfold.split(x):
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xtrain = x[train_inds]
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ytrain = y[train_inds]
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xtest = x[test_inds]
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ytest = y[test_inds]
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Xtrain = poly.fit_transform(xtrain[:, np.newaxis])
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ridge.fit(Xtrain, ytrain[:, np.newaxis])
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Xtest = poly.fit_transform(xtest[:, np.newaxis])
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ypred = ridge.predict(Xtest)
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scores_KFold[i,j] = np.sum((ypred - ytest[:, np.newaxis])**2)/np.size(ypred)
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j += 1
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i += 1
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estimated_mse_KFold = np.mean(scores_KFold, axis = 1)
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## Cross-validation using cross_val_score from sklearn along with KFold
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# kfold is an instance initialized above as:
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# kfold = KFold(n_splits = k)
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estimated_mse_sklearn = np.zeros(nlambdas)
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i = 0
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for lmb in lambdas:
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ridge = Ridge(alpha = lmb)
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X = poly.fit_transform(x[:, np.newaxis])
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estimated_mse_folds = cross_val_score(ridge, X, y[:, np.newaxis], scoring='neg_mean_squared_error', cv=kfold)
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# cross_val_score return an array containing the estimated negative mse for every fold.
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# we have to the the mean of every array in order to get an estimate of the mse of the model
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estimated_mse_sklearn[i] = np.mean(-estimated_mse_folds)
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i += 1
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## Plot and compare the slightly different ways to perform cross-validation
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plt.figure()
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plt.plot(np.log10(lambdas), estimated_mse_sklearn, label = 'cross_val_score')
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plt.plot(np.log10(lambdas), estimated_mse_KFold, 'r--', label = 'KFold')
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plt.xlabel('log10(lambda)')
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plt.ylabel('mse')
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plt.legend()
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plt.show()
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!ec
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!split
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===== Bias-Variance tradeoff with Bootstrap =====
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!bc pycod
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import matplotlib.pyplot as plt
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import numpy as np
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from sklearn.linear_model import LinearRegression, Ridge, Lasso
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from sklearn.preprocessing import PolynomialFeatures
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from sklearn.model_selection import train_test_split
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from sklearn.pipeline import make_pipeline
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from sklearn.utils import resample
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np.random.seed(2018)
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n = 40
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n_boostraps = 100
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maxdegree = 14
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# Make data set.
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x = np.linspace(-3, 3, n).reshape(-1, 1)
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y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.normal(0, 0.1, x.shape)
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error = np.zeros(maxdegree)
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bias = np.zeros(maxdegree)
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variance = np.zeros(maxdegree)
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polydegree = np.zeros(maxdegree)
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x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.2)
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for degree in range(maxdegree):
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model = make_pipeline(PolynomialFeatures(degree=degree), LinearRegression(fit_intercept=False))
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y_pred = np.empty((y_test.shape[0], n_boostraps))
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for i in range(n_boostraps):
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x_, y_ = resample(x_train, y_train)
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y_pred[:, i] = model.fit(x_, y_).predict(x_test).ravel()
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polydegree[degree] = degree
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error[degree] = np.mean( np.mean((y_test - y_pred)**2, axis=1, keepdims=True) )
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bias[degree] = np.mean( (y_test - np.mean(y_pred, axis=1, keepdims=True))**2 )
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variance[degree] = np.mean( np.var(y_pred, axis=1, keepdims=True) )
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print('Polynomial degree:', degree)
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print('Error:', error[degree])
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print('Bias^2:', bias[degree])
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print('Var:', variance[degree])
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print('{} >= {} + {} = {}'.format(error[degree], bias[degree], variance[degree], bias[degree]+variance[degree]))
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plt.plot(polydegree, error, label='Error')
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plt.plot(polydegree, bias, label='bias')
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plt.plot(polydegree, variance, label='Variance')
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plt.legend()
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plt.show()
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!ec
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!split
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===== Another Example from Scikit-Learn's Repository =====
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!bc pycod
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"""
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============================
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Underfitting vs. Overfitting
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============================
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This example demonstrates the problems of underfitting and overfitting and
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how we can use linear regression with polynomial features to approximate
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nonlinear functions. The plot shows the function that we want to approximate,
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which is a part of the cosine function. In addition, the samples from the
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real function and the approximations of different models are displayed. The
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models have polynomial features of different degrees. We can see that a
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linear function (polynomial with degree 1) is not sufficient to fit the
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training samples. This is called **underfitting**. A polynomial of degree 4
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approximates the true function almost perfectly. However, for higher degrees
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the model will **overfit** the training data, i.e. it learns the noise of the
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training data.
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We evaluate quantitatively **overfitting** / **underfitting** by using
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cross-validation. We calculate the mean squared error (MSE) on the validation
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set, the higher, the less likely the model generalizes correctly from the
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training data.
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"""
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print(__doc__)
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import numpy as np
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import matplotlib.pyplot as plt
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from sklearn.pipeline import Pipeline
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from sklearn.preprocessing import PolynomialFeatures
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from sklearn.linear_model import LinearRegression
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from sklearn.model_selection import cross_val_score
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def true_fun(X):
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return np.cos(1.5 * np.pi * X)
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np.random.seed(0)
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n_samples = 30
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degrees = [1, 4, 15]
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X = np.sort(np.random.rand(n_samples))
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y = true_fun(X) + np.random.randn(n_samples) * 0.1
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plt.figure(figsize=(14, 5))
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for i in range(len(degrees)):
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ax = plt.subplot(1, len(degrees), i + 1)
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plt.setp(ax, xticks=(), yticks=())
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polynomial_features = PolynomialFeatures(degree=degrees[i],
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include_bias=False)
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linear_regression = LinearRegression()
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pipeline = Pipeline([("polynomial_features", polynomial_features),
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("linear_regression", linear_regression)])
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pipeline.fit(X[:, np.newaxis], y)
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# Evaluate the models using crossvalidation
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scores = cross_val_score(pipeline, X[:, np.newaxis], y,
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scoring="neg_mean_squared_error", cv=10)
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X_test = np.linspace(0, 1, 100)
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plt.plot(X_test, pipeline.predict(X_test[:, np.newaxis]), label="Model")
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plt.plot(X_test, true_fun(X_test), label="True function")
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plt.scatter(X, y, edgecolor='b', s=20, label="Samples")
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plt.xlabel("x")
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plt.ylabel("y")
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plt.xlim((0, 1))
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plt.ylim((-2, 2))
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plt.legend(loc="best")
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plt.title("Degree {}\nMSE = {:.2e}(+/- {:.2e})".format(
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degrees[i], -scores.mean(), scores.std()))
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plt.show()
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!ec
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!split
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===== Cross-validation with Ridge =====
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!bc pycod
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import numpy as np
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import matplotlib.pyplot as plt
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from sklearn.model_selection import KFold
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from sklearn.linear_model import Ridge
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from sklearn.model_selection import cross_val_score
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from sklearn.preprocessing import PolynomialFeatures
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# A seed just to ensure that the random numbers are the same for every run.
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np.random.seed(3155)
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# Generate the data.
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n = 100
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x = np.linspace(-3, 3, n).reshape(-1, 1)
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y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.normal(0, 0.1, x.shape)
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# Decide degree on polynomial to fit
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poly = PolynomialFeatures(degree = 10)
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# Decide which values of lambda to use
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nlambdas = 500
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lambdas = np.logspace(-3, 5, nlambdas)
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# Initialize a KFold instance
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k = 5
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kfold = KFold(n_splits = k)
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estimated_mse_sklearn = np.zeros(nlambdas)
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i = 0
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for lmb in lambdas:
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ridge = Ridge(alpha = lmb)
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estimated_mse_folds = cross_val_score(ridge, x, y, scoring='neg_mean_squared_error', cv=kfold)
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estimated_mse_sklearn[i] = np.mean(-estimated_mse_folds)
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i += 1
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plt.figure()
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plt.plot(np.log10(lambdas), estimated_mse_sklearn, label = 'cross_val_score')
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plt.xlabel('log10(lambda)')
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plt.ylabel('MSE')
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plt.legend()
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plt.show()
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!ec
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!split
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===== The Ising model =====
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The one-dimensional Ising model with nearest neighbor interaction, no
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external field and a constant coupling constant $J$ is given by
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!bt
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\begin{align}
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H = -J \sum_{k}^L s_k s_{k + 1},
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\end{align}
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!et
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where $s_i \in \{-1, 1\}$ and $s_{N + 1} = s_1$. The number of spins
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in the system is determined by $L$. For the one-dimensional system
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there is no phase transition.
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We will look at a system of $L = 40$ spins with a coupling constant of
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$J = 1$. To get enough training data we will generate 10000 states
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with their respective energies.
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!bc pycod
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import numpy as np
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import matplotlib.pyplot as plt
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from mpl_toolkits.axes_grid1 import make_axes_locatable
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import seaborn as sns
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import scipy.linalg as scl
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from sklearn.model_selection import train_test_split
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import tqdm
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sns.set(color_codes=True)
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cmap_args=dict(vmin=-1., vmax=1., cmap='seismic')
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L = 40
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n = int(1e4)
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spins = np.random.choice([-1, 1], size=(n, L))
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J = 1.0
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energies = np.zeros(n)
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for i in range(n):
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energies[i] = - J * np.dot(spins[i], np.roll(spins[i], 1))
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!ec
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Here we use ordinary least squares
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regression to predict the energy for the nearest neighbor
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one-dimensional Ising model on a ring, i.e., the endpoints wrap
|
|
around. We will use linear regression to fit a value for
|
|
the coupling constant to achieve this.
|
|
|
|
!split
|
|
===== Reformulating the problem to suit regression =====
|
|
|
|
A more general form for the one-dimensional Ising model is
|
|
|
|
!bt
|
|
\begin{align}
|
|
H = - \sum_j^L \sum_k^L s_j s_k J_{jk}.
|
|
\end{align}
|
|
!et
|
|
|
|
Here we allow for interactions beyond the nearest neighbors and a state dependent
|
|
coupling constant. This latter expression can be formulated as
|
|
a matrix-product
|
|
!bt
|
|
\begin{align}
|
|
\bm{H} = \bm{X} J,
|
|
\end{align}
|
|
!et
|
|
|
|
where $X_{jk} = s_j s_k$ and $J$ is a matrix which consists of the
|
|
elements $-J_{jk}$. This form of writing the energy fits perfectly
|
|
with the form utilized in linear regression, that is
|
|
|
|
!bt
|
|
\begin{align}
|
|
\bm{y} = \bm{X}\bm{\beta} + \bm{\epsilon},
|
|
\end{align}
|
|
!et
|
|
|
|
We split the data in training and test data as discussed in the previous example
|
|
|
|
!bc pycod
|
|
X = np.zeros((n, L ** 2))
|
|
for i in range(n):
|
|
X[i] = np.outer(spins[i], spins[i]).ravel()
|
|
y = energies
|
|
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
|
|
!ec
|
|
|
|
!split
|
|
===== Linear regression =====
|
|
|
|
In the ordinary least squares method we choose the cost function
|
|
|
|
!bt
|
|
\begin{align}
|
|
C(\bm{X}, \bm{\beta})= \frac{1}{n}\left\{(\bm{X}\bm{\beta} - \bm{y})^T(\bm{X}\bm{\beta} - \bm{y})\right\}.
|
|
\end{align}
|
|
!et
|
|
|
|
We then find the extremal point of $C$ by taking the derivative with respect to $\bm{\beta}$ as discussed above.
|
|
This yields the expression for $\bm{\beta}$ to be
|
|
|
|
!bt
|
|
\[
|
|
\bm{\beta} = \frac{\bm{X}^T \bm{y}}{\bm{X}^T \bm{X}},
|
|
\]
|
|
!et
|
|
|
|
which immediately imposes some requirements on $\bm{X}$ as there must exist
|
|
an inverse of $\bm{X}^T \bm{X}$. If the expression we are modeling contains an
|
|
intercept, i.e., a constant term, we must make sure that the
|
|
first column of $\bm{X}$ consists of $1$. We do this here
|
|
|
|
!bc pycod
|
|
X_train_own = np.concatenate(
|
|
(np.ones(len(X_train))[:, np.newaxis], X_train),
|
|
axis=1
|
|
)
|
|
X_test_own = np.concatenate(
|
|
(np.ones(len(X_test))[:, np.newaxis], X_test),
|
|
axis=1
|
|
)
|
|
!ec
|
|
|
|
!bc pycod
|
|
def ols_inv(x: np.ndarray, y: np.ndarray) -> np.ndarray:
|
|
return scl.inv(x.T @ x) @ (x.T @ y)
|
|
beta = ols_inv(X_train_own, y_train)
|
|
!ec
|
|
|
|
|
|
!split
|
|
===== Singular Value decomposition =====
|
|
|
|
Doing the inversion directly turns out to be a bad idea since the matrix
|
|
$\bm{X}^T\bm{X}$ is singular. An alternative approach is to use the _singular
|
|
value decomposition_. Using the definition of the Moore-Penrose
|
|
pseudoinverse we can write the equation for $\bm{\beta}$ as
|
|
|
|
!bt
|
|
\[
|
|
\bm{\beta} = \bm{X}^{+}\bm{y},
|
|
\]
|
|
!et
|
|
|
|
where the pseudoinverse of $\bm{X}$ is given by
|
|
|
|
!bt
|
|
\[
|
|
\bm{X}^{+} = \frac{\bm{X}^T}{\bm{X}^T\bm{X}}.
|
|
\]
|
|
!et
|
|
|
|
Using singular value decomposition we can decompose the matrix $\bm{X} = \bm{U}\bm{\Sigma} \bm{V}^T$,
|
|
where $\bm{U}$ and $\bm{V}$ are orthogonal(unitary) matrices and $\bm{\Sigma}$ contains the singular values (more details below).
|
|
where $X^{+} = V\Sigma^{+} U^T$. This reduces the equation for
|
|
$\omega$ to
|
|
!bt
|
|
\begin{align}
|
|
\bm{\beta} = \bm{V}\bm{\Sigma}^{+} \bm{U}^T \bm{y}.
|
|
\end{align}
|
|
!et
|
|
|
|
Note that solving this equation by actually doing the pseudoinverse
|
|
(which is what we will do) is not a good idea as this operation scales
|
|
as $\mathcal{O}(n^3)$, where $n$ is the number of elements in a
|
|
general matrix. Instead, doing $QR$-factorization and solving the
|
|
linear system as an equation would reduce this down to
|
|
$\mathcal{O}(n^2)$ operations.
|
|
|
|
|
|
!bc pycod
|
|
def ols_svd(x: np.ndarray, y: np.ndarray) -> np.ndarray:
|
|
u, s, v = scl.svd(x)
|
|
return v.T @ scl.pinv(scl.diagsvd(s, u.shape[0], v.shape[0])) @ u.T @ y
|
|
!ec
|
|
|
|
!bc pycod
|
|
beta = ols_svd(X_train_own,y_train)
|
|
!ec
|
|
|
|
When extracting the $J$-matrix we need to make sure that we remove the intercept, as is done here
|
|
|
|
!bc pycod
|
|
J = beta[1:].reshape(L, L)
|
|
!ec
|
|
|
|
A way of looking at the coefficients in $J$ is to plot the matrices as images.
|
|
|
|
|
|
!bc pycod
|
|
fig = plt.figure(figsize=(20, 14))
|
|
im = plt.imshow(J, **cmap_args)
|
|
plt.title("OLS", fontsize=18)
|
|
plt.xticks(fontsize=18)
|
|
plt.yticks(fontsize=18)
|
|
cb = fig.colorbar(im)
|
|
cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18)
|
|
plt.show()
|
|
!ec
|
|
It is interesting to note that OLS
|
|
considers both $J_{j, j + 1} = -0.5$ and $J_{j, j - 1} = -0.5$ as
|
|
valid matrix elements for $J$.
|
|
In our discussion below on hyperparameters and Ridge and Lasso regression we will see that
|
|
this problem can be removed, partly and only with Lasso regression.
|
|
|
|
In this case our matrix inversion was actually possible. The obvious question now is what is the mathematics behind the SVD?
|
|
|
|
|
|
|
|
|
|
|
|
!split
|
|
===== The one-dimensional Ising model =====
|
|
|
|
Let us bring back the Ising model again, but now with an additional
|
|
focus on Ridge and Lasso regression as well. We repeat some of the
|
|
basic parts of the Ising model and the setup of the training and test
|
|
data. The one-dimensional Ising model with nearest neighbor
|
|
interaction, no external field and a constant coupling constant $J$ is
|
|
given by
|
|
|
|
!bt
|
|
\begin{align}
|
|
H = -J \sum_{k}^L s_k s_{k + 1},
|
|
\end{align}
|
|
!et
|
|
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.
|
|
|
|
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.
|
|
|
|
|
|
!bc pycod
|
|
import numpy as np
|
|
import matplotlib.pyplot as plt
|
|
from mpl_toolkits.axes_grid1 import make_axes_locatable
|
|
import seaborn as sns
|
|
import scipy.linalg as scl
|
|
from sklearn.model_selection import train_test_split
|
|
import sklearn.linear_model as skl
|
|
import tqdm
|
|
sns.set(color_codes=True)
|
|
cmap_args=dict(vmin=-1., vmax=1., cmap='seismic')
|
|
|
|
L = 40
|
|
n = int(1e4)
|
|
|
|
spins = np.random.choice([-1, 1], size=(n, L))
|
|
J = 1.0
|
|
|
|
energies = np.zeros(n)
|
|
|
|
for i in range(n):
|
|
energies[i] = - J * np.dot(spins[i], np.roll(spins[i], 1))
|
|
!ec
|
|
|
|
A more general form for the one-dimensional Ising model is
|
|
|
|
!bt
|
|
\begin{align}
|
|
H = - \sum_j^L \sum_k^L s_j s_k J_{jk}.
|
|
\end{align}
|
|
!et
|
|
|
|
Here we allow for interactions beyond the nearest neighbors and a more
|
|
adaptive coupling matrix. This latter expression can be formulated as
|
|
a matrix-product on the form
|
|
!bt
|
|
\begin{align}
|
|
H = X J,
|
|
\end{align}
|
|
!et
|
|
|
|
where $X_{jk} = s_j s_k$ and $J$ is the matrix consisting of the
|
|
elements $-J_{jk}$. This form of writing the energy fits perfectly
|
|
with the form utilized in linear regression, viz.
|
|
!bt
|
|
\begin{align}
|
|
\bm{y} = \bm{X}\bm{\beta} + \bm{\epsilon}.
|
|
\end{align}
|
|
!et
|
|
We organize the data as we did above
|
|
!bc pycod
|
|
X = np.zeros((n, L ** 2))
|
|
for i in range(n):
|
|
X[i] = np.outer(spins[i], spins[i]).ravel()
|
|
y = energies
|
|
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.96)
|
|
|
|
X_train_own = np.concatenate(
|
|
(np.ones(len(X_train))[:, np.newaxis], X_train),
|
|
axis=1
|
|
)
|
|
|
|
X_test_own = np.concatenate(
|
|
(np.ones(len(X_test))[:, np.newaxis], X_test),
|
|
axis=1
|
|
)
|
|
!ec
|
|
|
|
We will do all fitting with _Scikit-Learn_,
|
|
|
|
!bc pycod
|
|
clf = skl.LinearRegression().fit(X_train, y_train)
|
|
!ec
|
|
When extracting the $J$-matrix we make sure to remove the intercept
|
|
!bc pycod
|
|
J_sk = clf.coef_.reshape(L, L)
|
|
!ec
|
|
And then we plot the results
|
|
!bc pycod
|
|
fig = plt.figure(figsize=(20, 14))
|
|
im = plt.imshow(J_sk, **cmap_args)
|
|
plt.title("LinearRegression from Scikit-learn", fontsize=18)
|
|
plt.xticks(fontsize=18)
|
|
plt.yticks(fontsize=18)
|
|
cb = fig.colorbar(im)
|
|
cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18)
|
|
plt.show()
|
|
!ec
|
|
The results perfectly with our previous discussion where we used our own code.
|
|
|
|
!split
|
|
===== Ridge regression =====
|
|
|
|
Having explored the ordinary least squares we move on to ridge
|
|
regression. In ridge regression we include a _regularizer_. This
|
|
involves a new cost function which leads to a new estimate for the
|
|
weights $\bm{\beta}$. This results in a penalized regression problem. The
|
|
cost function is given by
|
|
|
|
!bt
|
|
\begin{align}
|
|
C(\bm{X}, \bm{\beta}; \lambda) = (\bm{X}\bm{\beta} - \bm{y})^T(\bm{X}\bm{\beta} - \bm{y}) + \lambda \bm{\beta}^T\bm{\beta}.
|
|
\end{align}
|
|
!et
|
|
!bc pycod
|
|
_lambda = 0.1
|
|
clf_ridge = skl.Ridge(alpha=_lambda).fit(X_train, y_train)
|
|
J_ridge_sk = clf_ridge.coef_.reshape(L, L)
|
|
fig = plt.figure(figsize=(20, 14))
|
|
im = plt.imshow(J_ridge_sk, **cmap_args)
|
|
plt.title("Ridge from Scikit-learn", fontsize=18)
|
|
plt.xticks(fontsize=18)
|
|
plt.yticks(fontsize=18)
|
|
cb = fig.colorbar(im)
|
|
cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18)
|
|
|
|
plt.show()
|
|
!ec
|
|
|
|
!split
|
|
===== LASSO regression =====
|
|
|
|
In the _Least Absolute Shrinkage and Selection Operator_ (LASSO)-method we get a third cost function.
|
|
|
|
!bt
|
|
\begin{align}
|
|
C(\bm{X}, \bm{\beta}; \lambda) = (\bm{X}\bm{\beta} - \bm{y})^T(\bm{X}\bm{\beta} - \bm{y}) + \lambda \sqrt{\bm{\beta}^T\bm{\beta}}.
|
|
\end{align}
|
|
!et
|
|
|
|
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_.
|
|
|
|
!bc pycod
|
|
clf_lasso = skl.Lasso(alpha=_lambda).fit(X_train, y_train)
|
|
J_lasso_sk = clf_lasso.coef_.reshape(L, L)
|
|
fig = plt.figure(figsize=(20, 14))
|
|
im = plt.imshow(J_lasso_sk, **cmap_args)
|
|
plt.title("Lasso from Scikit-learn", fontsize=18)
|
|
plt.xticks(fontsize=18)
|
|
plt.yticks(fontsize=18)
|
|
cb = fig.colorbar(im)
|
|
cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18)
|
|
|
|
plt.show()
|
|
!ec
|
|
|
|
It is quite striking how LASSO breaks the symmetry of the coupling
|
|
constant as opposed to ridge and OLS. We get a sparse solution with
|
|
$J_{j, j + 1} = -1$.
|
|
|
|
|
|
|
|
!split
|
|
===== Performance as function of the regularization parameter =====
|
|
|
|
We see how the different models perform for a different set of values for $\lambda$.
|
|
|
|
|
|
!bc pycod
|
|
lambdas = np.logspace(-4, 5, 10)
|
|
|
|
train_errors = {
|
|
"ols_sk": np.zeros(lambdas.size),
|
|
"ridge_sk": np.zeros(lambdas.size),
|
|
"lasso_sk": np.zeros(lambdas.size)
|
|
}
|
|
|
|
test_errors = {
|
|
"ols_sk": np.zeros(lambdas.size),
|
|
"ridge_sk": np.zeros(lambdas.size),
|
|
"lasso_sk": np.zeros(lambdas.size)
|
|
}
|
|
|
|
plot_counter = 1
|
|
|
|
fig = plt.figure(figsize=(32, 54))
|
|
|
|
for i, _lambda in enumerate(tqdm.tqdm(lambdas)):
|
|
for key, method in zip(
|
|
["ols_sk", "ridge_sk", "lasso_sk"],
|
|
[skl.LinearRegression(), skl.Ridge(alpha=_lambda), skl.Lasso(alpha=_lambda)]
|
|
):
|
|
method = method.fit(X_train, y_train)
|
|
|
|
train_errors[key][i] = method.score(X_train, y_train)
|
|
test_errors[key][i] = method.score(X_test, y_test)
|
|
|
|
omega = method.coef_.reshape(L, L)
|
|
|
|
plt.subplot(10, 5, plot_counter)
|
|
plt.imshow(omega, **cmap_args)
|
|
plt.title(r"%s, $\lambda = %.4f$" % (key, _lambda))
|
|
plot_counter += 1
|
|
|
|
plt.show()
|
|
!ec
|
|
|
|
We see that LASSO reaches a good solution for low
|
|
values of $\lambda$, but will "wither" when we increase $\lambda$ too
|
|
much. Ridge is more stable over a larger range of values for
|
|
$\lambda$, but eventually also fades away.
|
|
|
|
!split
|
|
===== Finding the optimal value of $\lambda$ =====
|
|
|
|
To determine which value of $\lambda$ is best we plot the accuracy of
|
|
the models when predicting the training and the testing set. We expect
|
|
the accuracy of the training set to be quite good, but if the accuracy
|
|
of the testing set is much lower this tells us that we might be
|
|
subject to an overfit model. The ideal scenario is an accuracy on the
|
|
testing set that is close to the accuracy of the training set.
|
|
|
|
|
|
!bc pycod
|
|
fig = plt.figure(figsize=(20, 14))
|
|
|
|
colors = {
|
|
"ols_sk": "r",
|
|
"ridge_sk": "y",
|
|
"lasso_sk": "c"
|
|
}
|
|
|
|
for key in train_errors:
|
|
plt.semilogx(
|
|
lambdas,
|
|
train_errors[key],
|
|
colors[key],
|
|
label="Train {0}".format(key),
|
|
linewidth=4.0
|
|
)
|
|
|
|
for key in test_errors:
|
|
plt.semilogx(
|
|
lambdas,
|
|
test_errors[key],
|
|
colors[key] + "--",
|
|
label="Test {0}".format(key),
|
|
linewidth=4.0
|
|
)
|
|
plt.legend(loc="best", fontsize=18)
|
|
plt.xlabel(r"$\lambda$", fontsize=18)
|
|
plt.ylabel(r"$R^2$", fontsize=18)
|
|
plt.tick_params(labelsize=18)
|
|
plt.show()
|
|
!ec
|
|
|
|
From the above figure we can see that LASSO with $\lambda = 10^{-2}$
|
|
achieves a very good accuracy on the test set. This by far surpasses the
|
|
other models for all values of $\lambda$.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
!split
|
|
===== Friday September 18: Intro to Logistic Regression =====
|
|
|
|
"Video of Lecture":"https://www.uio.no/studier/emner/matnat/fys/FYS-STK3155/h20/forelesningsvideoer/LectureSeptember18.mp4?vrtx=view-as-webpage" and "link to handwritten notes":"https://github.com/CompPhysics/MachineLearning/blob/master/doc/HandWrittenNotes/NotesSeptember18.pdf".
|
|
|
|
|
|
|
|
!split
|
|
===== Logistic Regression =====
|
|
|
|
In linear regression our main interest was centered on learning the
|
|
coefficients of a functional fit (say a polynomial) in order to be
|
|
able to predict the response of a continuous variable on some unseen
|
|
data. The fit to the continuous variable $y_i$ is based on some
|
|
independent variables $\hat{x}_i$. Linear regression resulted in
|
|
analytical expressions for standard ordinary Least Squares or Ridge
|
|
regression (in terms of matrices to invert) for several quantities,
|
|
ranging from the variance and thereby the confidence intervals of the
|
|
parameters $\hat{\beta}$ to the mean squared error. If we can invert
|
|
the product of the design matrices, linear regression gives then a
|
|
simple recipe for fitting our data.
|
|
|
|
!split
|
|
===== Classification problems =====
|
|
|
|
|
|
Classification problems, however, are concerned with outcomes taking
|
|
the form of discrete variables (i.e. categories). We may for example,
|
|
on the basis of DNA sequencing for a number of patients, like to find
|
|
out which mutations are important for a certain disease; or based on
|
|
scans of various patients' brains, figure out if there is a tumor or
|
|
not; or given a specific physical system, we'd like to identify its
|
|
state, say whether it is an ordered or disordered system (typical
|
|
situation in solid state physics); or classify the status of a
|
|
patient, whether she/he has a stroke or not and many other similar
|
|
situations.
|
|
|
|
The most common situation we encounter when we apply logistic
|
|
regression is that of two possible outcomes, normally denoted as a
|
|
binary outcome, true or false, positive or negative, success or
|
|
failure etc.
|
|
|
|
!split
|
|
===== Optimization and Deep learning =====
|
|
|
|
Logistic regression will also serve as our stepping stone towards
|
|
neural network algorithms and supervised deep learning. For logistic
|
|
learning, the minimization of the cost function leads to a non-linear
|
|
equation in the parameters $\hat{\beta}$. The optimization of the
|
|
problem calls therefore for minimization algorithms. This forms the
|
|
bottle neck of all machine learning algorithms, namely how to find
|
|
reliable minima of a multi-variable function. This leads us to the
|
|
family of gradient descent methods. The latter are the working horses
|
|
of basically all modern machine learning algorithms.
|
|
|
|
We note also that many of the topics discussed here on logistic
|
|
regression are also commonly used in modern supervised Deep Learning
|
|
models, as we will see later.
|
|
|
|
|
|
!split
|
|
===== Basics =====
|
|
|
|
We consider the case where the dependent variables, also called the
|
|
responses or the outcomes, $y_i$ are discrete and only take values
|
|
from $k=0,\dots,K-1$ (i.e. $K$ classes).
|
|
|
|
The goal is to predict the
|
|
output classes from the design matrix $\hat{X}\in\mathbb{R}^{n\times p}$
|
|
made of $n$ samples, each of which carries $p$ features or predictors. The
|
|
primary goal is to identify the classes to which new unseen samples
|
|
belong.
|
|
|
|
Let us specialize to the case of two classes only, with outputs
|
|
$y_i=0$ and $y_i=1$. Our outcomes could represent the status of a
|
|
credit card user that could default or not on her/his credit card
|
|
debt. That is
|
|
|
|
|
|
!bt
|
|
\[
|
|
y_i = \begin{bmatrix} 0 & \mathrm{no}\\ 1 & \mathrm{yes} \end{bmatrix}.
|
|
\]
|
|
!et
|
|
|
|
|
|
|
|
!split
|
|
===== Linear classifier =====
|
|
|
|
Before moving to the logistic model, let us try to use our linear
|
|
regression model to classify these two outcomes. We could for example
|
|
fit a linear model to the default case if $y_i > 0.5$ and the no
|
|
default case $y_i \leq 0.5$.
|
|
|
|
We would then have our
|
|
weighted linear combination, namely
|
|
!bt
|
|
\begin{equation}
|
|
\hat{y} = \hat{X}^T\hat{\beta} + \hat{\epsilon},
|
|
\end{equation}
|
|
!et
|
|
where $\hat{y}$ is a vector representing the possible outcomes, $\hat{X}$ is our
|
|
$n\times p$ design matrix and $\hat{\beta}$ represents our estimators/predictors.
|
|
|
|
!split
|
|
===== Some selected properties =====
|
|
|
|
The main problem with our function is that it takes values on the
|
|
entire real axis. In the case of logistic regression, however, the
|
|
labels $y_i$ are discrete variables. A typical example is the credit
|
|
card data discussed below here, where we can set the state of
|
|
defaulting the debt to $y_i=1$ and not to $y_i=0$ for one the persons
|
|
in the data set (see the full example below).
|
|
|
|
One simple way to get a discrete output is to have sign
|
|
functions that map the output of a linear regressor to values $\{0,1\}$,
|
|
$f(s_i)=sign(s_i)=1$ if $s_i\ge 0$ and 0 if otherwise.
|
|
We will encounter this model in our first demonstration of neural networks. Historically it is called the ``perceptron" model in the machine learning
|
|
literature. This model is extremely simple. However, in many cases it is more
|
|
favorable to use a ``soft" classifier that outputs
|
|
the probability of a given category. This leads us to the logistic function.
|
|
|
|
!split
|
|
===== Simple example =====
|
|
|
|
The following example on data for coronary heart disease (CHD) as function of age may serve as an illustration. In the code here we read and plot whether a person has had CHD (output = 1) or not (output = 0). This ouput is plotted the person's against age. Clearly, the figure shows that attempting to make a standard linear regression fit may not be very meaningful.
|
|
|
|
!bc pycod
|
|
# Common imports
|
|
import os
|
|
import numpy as np
|
|
import pandas as pd
|
|
import matplotlib.pyplot as plt
|
|
from sklearn.linear_model import LinearRegression, Ridge, Lasso
|
|
from sklearn.model_selection import train_test_split
|
|
from sklearn.utils import resample
|
|
from sklearn.metrics import mean_squared_error
|
|
from IPython.display import display
|
|
from pylab import plt, mpl
|
|
plt.style.use('seaborn')
|
|
mpl.rcParams['font.family'] = 'serif'
|
|
|
|
# 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("chddata.csv"),'r')
|
|
|
|
# Read the chd data as csv file and organize the data into arrays with age group, age, and chd
|
|
chd = pd.read_csv(infile, names=('ID', 'Age', 'Agegroup', 'CHD'))
|
|
chd.columns = ['ID', 'Age', 'Agegroup', 'CHD']
|
|
output = chd['CHD']
|
|
age = chd['Age']
|
|
agegroup = chd['Agegroup']
|
|
numberID = chd['ID']
|
|
display(chd)
|
|
|
|
plt.scatter(age, output, marker='o')
|
|
plt.axis([18,70.0,-0.1, 1.2])
|
|
plt.xlabel(r'Age')
|
|
plt.ylabel(r'CHD')
|
|
plt.title(r'Age distribution and Coronary heart disease')
|
|
plt.show()
|
|
!ec
|
|
|
|
!split
|
|
===== Plotting the mean value for each group =====
|
|
|
|
What we could attempt however is to plot the mean value for each group.
|
|
|
|
!bc pycod
|
|
agegroupmean = np.array([0.1, 0.133, 0.250, 0.333, 0.462, 0.625, 0.765, 0.800])
|
|
group = np.array([1, 2, 3, 4, 5, 6, 7, 8])
|
|
plt.plot(group, agegroupmean, "r-")
|
|
plt.axis([0,9,0, 1.0])
|
|
plt.xlabel(r'Age group')
|
|
plt.ylabel(r'CHD mean values')
|
|
plt.title(r'Mean values for each age group')
|
|
plt.show()
|
|
!ec
|
|
|
|
We are now trying to find a function $f(y\vert x)$, that is a function which gives us an expected value for the output $y$ with a given input $x$.
|
|
In standard linear regression with a linear dependence on $x$, we would write this in terms of our model
|
|
!bt
|
|
\[
|
|
f(y_i\vert x_i)=\beta_0+\beta_1 x_i.
|
|
\]
|
|
!et
|
|
|
|
This expression implies however that $f(y_i\vert x_i)$ could take any
|
|
value from minus infinity to plus infinity. If we however let
|
|
$f(y\vert y)$ be represented by the mean value, the above example
|
|
shows us that we can constrain the function to take values between
|
|
zero and one, that is we have $0 \le f(y_i\vert x_i) \le 1$. Looking
|
|
at our last curve we see also that it has an S-shaped form. This leads
|
|
us to a very popular model for the function $f$, namely the so-called
|
|
Sigmoid function or logistic model. We will consider this function as
|
|
representing the probability for finding a value of $y_i$ with a given
|
|
$x_i$.
|
|
|
|
!split
|
|
===== The logistic function =====
|
|
|
|
Another widely studied model, is the so-called
|
|
perceptron model, which is an example of a ``hard classification'' model. We
|
|
will encounter this model when we discuss neural networks as
|
|
well. Each datapoint is deterministically assigned to a category (i.e
|
|
$y_i=0$ or $y_i=1$). In many cases, and the coronary heart disease data forms one of many such examples, it is favorable to have a ``soft''
|
|
classifier that outputs the probability of a given category rather
|
|
than a single value. For example, given $x_i$, the classifier
|
|
outputs the probability of being in a category $k$. Logistic regression
|
|
is the most common example of a so-called soft classifier. In logistic
|
|
regression, the probability that a data point $x_i$
|
|
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,
|
|
!bt
|
|
\[
|
|
p(t) = \frac{1}{1+\mathrm \exp{-t}}=\frac{\exp{t}}{1+\mathrm \exp{t}}.
|
|
\]
|
|
!et
|
|
Note that $1-p(t)= p(-t)$.
|
|
|
|
!split
|
|
===== Examples of likelihood functions used in logistic regression and nueral networks =====
|
|
|
|
|
|
The following code plots the logistic function, the step function and other functions we will encounter from here and on.
|
|
|
|
|
|
!bc pycod
|
|
"""The sigmoid function (or the logistic curve) is a
|
|
function that takes any real number, z, and outputs a number (0,1).
|
|
It is useful in neural networks for assigning weights on a relative scale.
|
|
The value z is the weighted sum of parameters involved in the learning algorithm."""
|
|
|
|
import numpy
|
|
import matplotlib.pyplot as plt
|
|
import math as mt
|
|
|
|
z = numpy.arange(-5, 5, .1)
|
|
sigma_fn = numpy.vectorize(lambda z: 1/(1+numpy.exp(-z)))
|
|
sigma = sigma_fn(z)
|
|
|
|
fig = plt.figure()
|
|
ax = fig.add_subplot(111)
|
|
ax.plot(z, sigma)
|
|
ax.set_ylim([-0.1, 1.1])
|
|
ax.set_xlim([-5,5])
|
|
ax.grid(True)
|
|
ax.set_xlabel('z')
|
|
ax.set_title('sigmoid function')
|
|
|
|
plt.show()
|
|
|
|
"""Step Function"""
|
|
z = numpy.arange(-5, 5, .02)
|
|
step_fn = numpy.vectorize(lambda z: 1.0 if z >= 0.0 else 0.0)
|
|
step = step_fn(z)
|
|
|
|
fig = plt.figure()
|
|
ax = fig.add_subplot(111)
|
|
ax.plot(z, step)
|
|
ax.set_ylim([-0.5, 1.5])
|
|
ax.set_xlim([-5,5])
|
|
ax.grid(True)
|
|
ax.set_xlabel('z')
|
|
ax.set_title('step function')
|
|
|
|
plt.show()
|
|
|
|
"""tanh Function"""
|
|
z = numpy.arange(-2*mt.pi, 2*mt.pi, 0.1)
|
|
t = numpy.tanh(z)
|
|
|
|
fig = plt.figure()
|
|
ax = fig.add_subplot(111)
|
|
ax.plot(z, t)
|
|
ax.set_ylim([-1.0, 1.0])
|
|
ax.set_xlim([-2*mt.pi,2*mt.pi])
|
|
ax.grid(True)
|
|
ax.set_xlabel('z')
|
|
ax.set_title('tanh function')
|
|
|
|
plt.show()
|
|
!ec
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
!split
|
|
===== Two parameters =====
|
|
|
|
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
|
|
!bt
|
|
\begin{align*}
|
|
p(y_i=1|x_i,\hat{\beta}) &= \frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}},\nonumber\\
|
|
p(y_i=0|x_i,\hat{\beta}) &= 1 - p(y_i=1|x_i,\hat{\beta}),
|
|
\end{align*}
|
|
!et
|
|
where $\hat{\beta}$ are the weights we wish to extract from data, in our case $\beta_0$ and $\beta_1$.
|
|
|
|
Note that we used
|
|
!bt
|
|
\[
|
|
p(y_i=0\vert x_i, \hat{\beta}) = 1-p(y_i=1\vert x_i, \hat{\beta}).
|
|
\]
|
|
!et
|
|
|
|
!split
|
|
===== Maximum likelihood =====
|
|
|
|
In order to define the total likelihood for all possible outcomes from a
|
|
dataset $\mathcal{D}=\{(y_i,x_i)\}$, with the binary labels
|
|
$y_i\in\{0,1\}$ 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.
|
|
We aim thus at maximizing
|
|
the probability of seeing the observed data. We can then approximate the
|
|
likelihood in terms of the product of the individual probabilities of a specific outcome $y_i$, that is
|
|
!bt
|
|
\begin{align*}
|
|
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 \\
|
|
\end{align*}
|
|
!et
|
|
from which we obtain the log-likelihood and our _cost/loss_ function
|
|
!bt
|
|
\[
|
|
\mathcal{C}(\hat{\beta}) = \sum_{i=1}^n \left( y_i\log{p(y_i=1|x_i,\hat{\beta})} + (1-y_i)\log\left[1-p(y_i=1|x_i,\hat{\beta}))\right]\right).
|
|
\]
|
|
!et
|
|
|
|
!split
|
|
===== The cost function rewritten =====
|
|
|
|
Reordering the logarithms, we can rewrite the _cost/loss_ function as
|
|
!bt
|
|
\[
|
|
\mathcal{C}(\hat{\beta}) = \sum_{i=1}^n \left(y_i(\beta_0+\beta_1x_i) -\log{(1+\exp{(\beta_0+\beta_1x_i)})}\right).
|
|
\]
|
|
!et
|
|
|
|
The maximum likelihood estimator is defined as the set of parameters that maximize the log-likelihood where we maximize with respect to $\beta$.
|
|
Since the cost (error) function is just the negative log-likelihood, for logistic regression we have that
|
|
!bt
|
|
\[
|
|
\mathcal{C}(\hat{\beta})=-\sum_{i=1}^n \left(y_i(\beta_0+\beta_1x_i) -\log{(1+\exp{(\beta_0+\beta_1x_i)})}\right).
|
|
\]
|
|
!et
|
|
This equation is known in statistics as the _cross entropy_. Finally, we note that just as in linear regression,
|
|
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.
|
|
|
|
!split
|
|
===== Minimizing the cross entropy =====
|
|
|
|
The cross entropy is a convex function of the weights $\hat{\beta}$ and,
|
|
therefore, any local minimizer is a global minimizer.
|
|
|
|
|
|
Minimizing this
|
|
cost function with respect to the two parameters $\beta_0$ and $\beta_1$ we obtain
|
|
|
|
!bt
|
|
\[
|
|
\frac{\partial \mathcal{C}(\hat{\beta})}{\partial \beta_0} = -\sum_{i=1}^n \left(y_i -\frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}}\right),
|
|
\]
|
|
!et
|
|
and
|
|
!bt
|
|
\[
|
|
\frac{\partial \mathcal{C}(\hat{\beta})}{\partial \beta_1} = -\sum_{i=1}^n \left(y_ix_i -x_i\frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}}\right).
|
|
\]
|
|
!et
|
|
|
|
!split
|
|
===== A more compact expression =====
|
|
|
|
Let us now define a vector $\hat{y}$ with $n$ elements $y_i$, an
|
|
$n\times p$ matrix $\hat{X}$ which contains the $x_i$ values and a
|
|
vector $\hat{p}$ of fitted probabilities $p(y_i\vert x_i,\hat{\beta})$. We can rewrite in a more compact form the first
|
|
derivative of cost function as
|
|
|
|
!bt
|
|
\[
|
|
\frac{\partial \mathcal{C}(\hat{\beta})}{\partial \hat{\beta}} = -\hat{X}^T\left(\hat{y}-\hat{p}\right).
|
|
\]
|
|
!et
|
|
|
|
If we in addition define a diagonal matrix $\hat{W}$ with elements
|
|
$p(y_i\vert x_i,\hat{\beta})(1-p(y_i\vert x_i,\hat{\beta})$, we can obtain a compact expression of the second derivative as
|
|
|
|
!bt
|
|
\[
|
|
\frac{\partial^2 \mathcal{C}(\hat{\beta})}{\partial \hat{\beta}\partial \hat{\beta}^T} = \hat{X}^T\hat{W}\hat{X}.
|
|
\]
|
|
!et
|
|
|
|
!split
|
|
===== Extending to more predictors =====
|
|
|
|
Within a binary classification problem, we can easily expand our model to include multiple predictors. Our ratio between likelihoods is then with $p$ predictors
|
|
!bt
|
|
\[
|
|
\log{ \frac{p(\hat{\beta}\hat{x})}{1-p(\hat{\beta}\hat{x})}} = \beta_0+\beta_1x_1+\beta_2x_2+\dots+\beta_px_p.
|
|
\]
|
|
!et
|
|
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
|
|
!bt
|
|
\[
|
|
p(\hat{\beta}\hat{x})=\frac{ \exp{(\beta_0+\beta_1x_1+\beta_2x_2+\dots+\beta_px_p)}}{1+\exp{(\beta_0+\beta_1x_1+\beta_2x_2+\dots+\beta_px_p)}}.
|
|
\]
|
|
!et
|
|
|
|
!split
|
|
===== Including more classes =====
|
|
|
|
Till now we have mainly focused on two classes, the so-called binary
|
|
system. Suppose we wish to extend to $K$ classes. Let us for the sake
|
|
of simplicity assume we have only two predictors. We have then following model
|
|
|
|
!bt
|
|
\[
|
|
\log{\frac{p(C=1\vert x)}{p(K\vert x)}} = \beta_{10}+\beta_{11}x_1,
|
|
\]
|
|
!et
|
|
and
|
|
!bt
|
|
\[
|
|
\log{\frac{p(C=2\vert x)}{p(K\vert x)}} = \beta_{20}+\beta_{21}x_1,
|
|
\]
|
|
!et
|
|
and so on till the class $C=K-1$ class
|
|
!bt
|
|
\[
|
|
\log{\frac{p(C=K-1\vert x)}{p(K\vert x)}} = \beta_{(K-1)0}+\beta_{(K-1)1}x_1,
|
|
\]
|
|
!et
|
|
|
|
and the model is specified in term of $K-1$ so-called log-odds or
|
|
_logit_ transformations.
|
|
|
|
|
|
!split
|
|
===== More classes =====
|
|
|
|
In our discussion of neural networks we will encounter the above again
|
|
in terms of a slightly modified function, the so-called _Softmax_ function.
|
|
|
|
The softmax function is used in various multiclass classification
|
|
methods, such as multinomial logistic regression (also known as
|
|
softmax regression), multiclass linear discriminant analysis, naive
|
|
Bayes classifiers, and artificial neural networks. Specifically, in
|
|
multinomial logistic regression and linear discriminant analysis, the
|
|
input to the function is the result of $K$ distinct linear functions,
|
|
and the predicted probability for the $k$-th class given a sample
|
|
vector $\hat{x}$ and a weighting vector $\hat{\beta}$ is (with two
|
|
predictors):
|
|
|
|
!bt
|
|
\[
|
|
p(C=k\vert \mathbf {x} )=\frac{\exp{(\beta_{k0}+\beta_{k1}x_1)}}{1+\sum_{l=1}^{K-1}\exp{(\beta_{l0}+\beta_{l1}x_1)}}.
|
|
\]
|
|
!et
|
|
It is easy to extend to more predictors. The final class is
|
|
!bt
|
|
\[
|
|
p(C=K\vert \mathbf {x} )=\frac{1}{1+\sum_{l=1}^{K-1}\exp{(\beta_{l0}+\beta_{l1}x_1)}},
|
|
\]
|
|
!et
|
|
|
|
and they sum to one. Our earlier discussions were all specialized to
|
|
the case with two classes only. It is easy to see from the above that
|
|
what we derived earlier is compatible with these equations.
|
|
|
|
To find the optimal parameters we would typically use a gradient
|
|
descent method. Newton's method and gradient descent methods are
|
|
discussed in the material on "optimization
|
|
methods":"https://compphysics.github.io/MachineLearning/doc/pub/Splines/html/Splines-bs.html".
|
|
|
|
This will be discussed next week. Before we develop our own codes for logistic regression, we end this lecture by studying the functionality that _Scikit-learn_ offers.
|
|
|
|
|
|
|
|
|
|
|
|
!split
|
|
===== Wisconsin Cancer Data =====
|
|
|
|
We show here how we can use a simple regression case on the breast
|
|
cancer data using Logistic regression as our algorithm for
|
|
classification.
|
|
|
|
|
|
!bc pycod
|
|
import matplotlib.pyplot as plt
|
|
import numpy as np
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from sklearn.model_selection import train_test_split
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from sklearn.datasets import load_breast_cancer
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from sklearn.linear_model import LogisticRegression
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# Load the data
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cancer = load_breast_cancer()
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X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)
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print(X_train.shape)
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print(X_test.shape)
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# Logistic Regression
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logreg = LogisticRegression(solver='lbfgs')
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logreg.fit(X_train, y_train)
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print("Test set accuracy with Logistic Regression: {:.2f}".format(logreg.score(X_test,y_test)))
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#now scale the data
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from sklearn.preprocessing import StandardScaler
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scaler = StandardScaler()
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scaler.fit(X_train)
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X_train_scaled = scaler.transform(X_train)
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X_test_scaled = scaler.transform(X_test)
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# Logistic Regression
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logreg.fit(X_train_scaled, y_train)
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print("Test set accuracy Logistic Regression with scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
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!ec
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!split
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===== Using the correlation matrix =====
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In addition to the above scores, we could also study the covariance (and the correlation matrix).
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We use _Pandas_ to compute the correlation matrix.
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!bc pycod
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import matplotlib.pyplot as plt
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import numpy as np
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from sklearn.model_selection import train_test_split
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from sklearn.datasets import load_breast_cancer
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from sklearn.linear_model import LogisticRegression
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cancer = load_breast_cancer()
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import pandas as pd
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# Making a data frame
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cancerpd = pd.DataFrame(cancer.data, columns=cancer.feature_names)
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fig, axes = plt.subplots(15,2,figsize=(10,20))
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malignant = cancer.data[cancer.target == 0]
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benign = cancer.data[cancer.target == 1]
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ax = axes.ravel()
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for i in range(30):
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_, bins = np.histogram(cancer.data[:,i], bins =50)
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ax[i].hist(malignant[:,i], bins = bins, alpha = 0.5)
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ax[i].hist(benign[:,i], bins = bins, alpha = 0.5)
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ax[i].set_title(cancer.feature_names[i])
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ax[i].set_yticks(())
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ax[0].set_xlabel("Feature magnitude")
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ax[0].set_ylabel("Frequency")
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ax[0].legend(["Malignant", "Benign"], loc ="best")
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fig.tight_layout()
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plt.show()
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import seaborn as sns
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correlation_matrix = cancerpd.corr().round(1)
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# use the heatmap function from seaborn to plot the correlation matrix
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# annot = True to print the values inside the square
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plt.figure(figsize=(15,8))
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sns.heatmap(data=correlation_matrix, annot=True)
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plt.show()
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!ec
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!split
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===== Discussing the correlation data =====
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In the above example we note two things. In the first plot we display
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the overlap of benign and malignant tumors as functions of the various
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features in the Wisconsing breast cancer data set. We see that for
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some of the features we can distinguish clearly the benign and
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malignant cases while for other features we cannot. This can point to
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us which features may be of greater interest when we wish to classify
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a benign or not benign tumour.
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In the second figure we have computed the so-called correlation
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matrix, which in our case with thirty features becomes a $30\times 30$
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matrix.
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We constructed this matrix using _pandas_ via the statements
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!bc pycod
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cancerpd = pd.DataFrame(cancer.data, columns=cancer.feature_names)
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!ec
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and then
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!bc pycod
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correlation_matrix = cancerpd.corr().round(1)
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!ec
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Diagonalizing this matrix we can in turn say something about which
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features are of relevance and which are not. This leads us to
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the classical Principal Component Analysis (PCA) theorem with
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applications. This will be discussed later this semester ("week 43":"https://compphysics.github.io/MachineLearning/doc/pub/week43/html/week43-bs.html").
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!split
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===== Other measures in classification studies: Cancer Data again =====
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!bc pycod
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import matplotlib.pyplot as plt
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import numpy as np
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from sklearn.model_selection import train_test_split
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from sklearn.datasets import load_breast_cancer
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from sklearn.linear_model import LogisticRegression
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# Load the data
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cancer = load_breast_cancer()
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X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)
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print(X_train.shape)
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print(X_test.shape)
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# Logistic Regression
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logreg = LogisticRegression(solver='lbfgs')
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logreg.fit(X_train, y_train)
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print("Test set accuracy with Logistic Regression: {:.2f}".format(logreg.score(X_test,y_test)))
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#now scale the data
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from sklearn.preprocessing import StandardScaler
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scaler = StandardScaler()
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scaler.fit(X_train)
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X_train_scaled = scaler.transform(X_train)
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X_test_scaled = scaler.transform(X_test)
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# Logistic Regression
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logreg.fit(X_train_scaled, y_train)
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print("Test set accuracy Logistic Regression with scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
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from sklearn.preprocessing import LabelEncoder
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from sklearn.model_selection import cross_validate
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#Cross validation
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accuracy = cross_validate(logreg,X_test_scaled,y_test,cv=10)['test_score']
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print(accuracy)
|
|
print("Test set accuracy with Logistic Regression and scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
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import scikitplot as skplt
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|
y_pred = logreg.predict(X_test_scaled)
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skplt.metrics.plot_confusion_matrix(y_test, y_pred, normalize=True)
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plt.show()
|
|
y_probas = logreg.predict_proba(X_test_scaled)
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skplt.metrics.plot_roc(y_test, y_probas)
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plt.show()
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skplt.metrics.plot_cumulative_gain(y_test, y_probas)
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plt.show()
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!ec
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