From f5b9eb32a26d34cf37cd92f11df41a01da17a3df Mon Sep 17 00:00:00 2001 From: mhjensen Date: Thu, 6 Sep 2018 11:28:45 +0200 Subject: [PATCH] update on regression analysis --- .../Regression/html/._Regression-bs000.html | 155 +- .../Regression/html/._Regression-bs001.html | 153 +- .../Regression/html/._Regression-bs002.html | 179 +- .../Regression/html/._Regression-bs003.html | 172 +- .../Regression/html/._Regression-bs004.html | 192 +- .../Regression/html/._Regression-bs005.html | 195 +- .../Regression/html/._Regression-bs006.html | 180 +- .../Regression/html/._Regression-bs007.html | 181 +- .../Regression/html/._Regression-bs008.html | 177 +- .../Regression/html/._Regression-bs009.html | 182 +- .../Regression/html/._Regression-bs010.html | 178 +- .../Regression/html/._Regression-bs011.html | 167 +- .../Regression/html/._Regression-bs012.html | 176 +- .../Regression/html/._Regression-bs013.html | 172 +- .../Regression/html/._Regression-bs014.html | 167 +- .../Regression/html/._Regression-bs015.html | 170 +- .../Regression/html/._Regression-bs016.html | 172 +- .../Regression/html/._Regression-bs017.html | 187 +- .../Regression/html/._Regression-bs018.html | 224 +- .../Regression/html/._Regression-bs019.html | 178 +- .../Regression/html/._Regression-bs020.html | 206 +- .../Regression/html/._Regression-bs021.html | 221 +- .../Regression/html/._Regression-bs022.html | 176 +- .../Regression/html/._Regression-bs023.html | 176 +- .../Regression/html/._Regression-bs024.html | 184 +- .../Regression/html/._Regression-bs025.html | 202 +- .../Regression/html/._Regression-bs026.html | 204 +- .../Regression/html/._Regression-bs027.html | 199 +- .../Regression/html/._Regression-bs028.html | 178 +- .../Regression/html/._Regression-bs029.html | 187 +- .../Regression/html/._Regression-bs030.html | 212 +- .../Regression/html/._Regression-bs031.html | 226 +- .../Regression/html/._Regression-bs032.html | 185 +- .../Regression/html/._Regression-bs033.html | 204 +- .../Regression/html/._Regression-bs034.html | 258 +- .../Regression/html/._Regression-bs035.html | 180 +- .../Regression/html/._Regression-bs036.html | 236 +- .../Regression/html/._Regression-bs037.html | 207 +- .../Regression/html/._Regression-bs038.html | 307 +++ .../Regression/html/._Regression-bs039.html | 392 +++ .../Regression/html/._Regression-bs040.html | 304 +++ .../Regression/html/._Regression-bs041.html | 291 +++ .../Regression/html/._Regression-bs042.html | 365 +++ .../Regression/html/._Regression-bs043.html | 327 +++ .../Regression/html/._Regression-bs044.html | 278 +++ .../Regression/html/._Regression-bs045.html | 275 +++ doc/pub/Regression/html/Regression-bs.html | 155 +- .../Regression/html/Regression-reveal.html | 656 ++++- .../Regression/html/Regression-solarized.html | 725 ++++-- doc/pub/Regression/html/Regression.html | 725 ++++-- doc/pub/Regression/ipynb/Regression.ipynb | 916 ++++--- .../ipynb/ipynb-Regression-src.tar.gz | Bin 211 -> 211 bytes .../pdf/Regression-beamer-handouts2x3.pdf | Bin 345650 -> 389595 bytes doc/pub/Regression/pdf/Regression-beamer.pdf | Bin 324702 -> 351781 bytes doc/pub/Regression/pdf/Regression-minted.pdf | Bin 341147 -> 391306 bytes doc/src/Regression/Regression.do.txt | 2135 ++--------------- 56 files changed, 9374 insertions(+), 5675 deletions(-) create mode 100644 doc/pub/Regression/html/._Regression-bs038.html create mode 100644 doc/pub/Regression/html/._Regression-bs039.html create mode 100644 doc/pub/Regression/html/._Regression-bs040.html create mode 100644 doc/pub/Regression/html/._Regression-bs041.html create mode 100644 doc/pub/Regression/html/._Regression-bs042.html create mode 100644 doc/pub/Regression/html/._Regression-bs043.html create mode 100644 doc/pub/Regression/html/._Regression-bs044.html create mode 100644 doc/pub/Regression/html/._Regression-bs045.html diff --git a/doc/pub/Regression/html/._Regression-bs000.html b/doc/pub/Regression/html/._Regression-bs000.html index 9560a3323..c92c12f8a 100644 --- a/doc/pub/Regression/html/._Regression-bs000.html +++ b/doc/pub/Regression/html/._Regression-bs000.html @@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source @@ -152,42 +173,50 @@ MathJax.Hub.Config({ Contents @@ -222,7 +251,7 @@ MathJax.Hub.Config({
[2] Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University

-

Aug 24, 2018

+

Sep 6, 2018


@@ -246,7 +275,7 @@ MathJax.Hub.Config({

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    General linear models

    +

    Regression analysis, overarching aims II

    -Before we proceed let us study a case from linear algebra where we aim at fitting a set of data \( \hat{y}=[y_0,y_1,\dots,y_{n-1}] \). We could think of these data as a result of an experiment or a complicated numerical experiment. These data are functions of a series of variables \( \hat{x}=[x_0,x_1,\dots,x_{n-1}] \), that is \( y_i = y(x_i) \) with \( i=0,1,2,\dots,n-1 \). The variables \( x_i \) could represent physical quantities like time, temperature, position etc. We assume that \( y(x) \) is a smooth function.

    -Since obtaining these data points may not be trivial, we want to use these data to fit a function which can allow us to make predictions for values of \( y \) which are not in the present set. The perhaps simplest approach is to assume we can parametrize our function in terms of a polynomial of degree \( n-1 \) with \( n \) points, that is -$$ -y=y(x) \rightarrow y(x_i)=\tilde{y}_i+\epsilon_i=\sum_{j=0}^{n-1} \beta_i x_i^j+\epsilon_i, -$$ - -where \( \epsilon_i \) is the error in our approximation. +Consider an experiment in which \( p \) characteristics of \( n \) samples are +measured. The data from this experiment are denoted \( \mathbf{X} \), with +\( \mathbf{X} \) as above. The matrix \( \mathbf{X} \) is called the design +matrix. Additional information of the samples is available in the +form of \( \mathbf{Y} \) (also as above). The variable \( \mathbf{Y} \) is +generally referred to as the response variable. The aim of +regression analysis is to explain \( \mathbf{Y} \) in terms of +\( \mathbf{X} \) through a functional relationship like \( Y_i = +f(\mathbf{X}_{i,\ast}) \). When no prior knowledge on the form of +\( f(\cdot) \) is available, it is common to assume a linear relationship +between \( \mathbf{X} \) and \( \mathbf{Y} \). This assumption gives rise to +the linear regression model where \( \beta = (\beta_1, \ldots, +\beta_p)^{\top} \) is the regression parameter. The parameter +\( \beta_j \), \( j=1, \ldots, p \), represents the effect size of covariate +\( j \) on the response. That is, for each unit change in covariate \( j \) +(while keeping the other covariates fixed) the observed change in the +response is equal to \( \beta_j \).

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    Rewriting the fitting procedure as a linear algebra problem

    +

    General linear models

    -For every set of values \( y_i,x_i \) we have thus the corresponding set of equations +Before we proceed let us study a case from linear algebra where we aim at fitting a set of data \( \hat{y}=[y_0,y_1,\dots,y_{n-1}] \). We could think of these data as a result of an experiment or a complicated numerical experiment. These data are functions of a series of variables \( \hat{x}=[x_0,x_1,\dots,x_{n-1}] \), that is \( y_i = y(x_i) \) with \( i=0,1,2,\dots,n-1 \). The variables \( x_i \) could represent physical quantities like time, temperature, position etc. We assume that \( y(x) \) is a smooth function. + +

    +Since obtaining these data points may not be trivial, we want to use these data to fit a function which can allow us to make predictions for values of \( y \) which are not in the present set. The perhaps simplest approach is to assume we can parametrize our function in terms of a polynomial of degree \( n-1 \) with \( n \) points, that is $$ -\begin{align*} -y_0&=\beta_0+\beta_1x_0^1+\beta_2x_0^2+\dots+\beta_{n-1}x_0^{n-1}+\epsilon_0\\ -y_1&=\beta_0+\beta_1x_1^1+\beta_2x_1^2+\dots+\beta_{n-1}x_1^{n-1}+\epsilon_1\\ -y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2\\ -\dots & \dots \\ -y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_1x_{n-1}^{n-1}+\epsilon_{n-1}.\\ -\end{align*} +y=y(x) \rightarrow y(x_i)=\tilde{y}_i+\epsilon_i=\sum_{j=0}^{n-1} \beta_i x_i^j+\epsilon_i, $$ + +where \( \epsilon_i \) is the error in our approximation. + +

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    Rewriting the fitting procedure as a linear algebra problem, follows

    +

    Rewriting the fitting procedure as a linear algebra problem

    -Defining the vectors +For every set of values \( y_i,x_i \) we have thus the corresponding set of equations $$ -\hat{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T, -$$ - -and -$$ -\hat{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T, -$$ - -and -$$ -\hat{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T, -$$ - -and the matrix -$$ -\hat{X}= -\begin{bmatrix} -1& x_{0}^1 &x_{0}^2& \dots & \dots &x_{0}^{n-1}\\ -1& x_{1}^1 &x_{1}^2& \dots & \dots &x_{1}^{n-1}\\ -1& x_{2}^1 &x_{2}^2& \dots & \dots &x_{2}^{n-1}\\ -\dots& \dots &\dots& \dots & \dots &\dots\\ -1& x_{n-1}^1 &x_{n-1}^2& \dots & \dots &x_{n-1}^{n-1}\\ -\end{bmatrix} -$$ - -we can rewrite our equations as -$$ -\hat{y} = \hat{X}\hat{\beta}+\hat{\epsilon}. +\begin{align*} +y_0&=\beta_0+\beta_1x_0^1+\beta_2x_0^2+\dots+\beta_{n-1}x_0^{n-1}+\epsilon_0\\ +y_1&=\beta_0+\beta_1x_1^1+\beta_2x_1^2+\dots+\beta_{n-1}x_1^{n-1}+\epsilon_1\\ +y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2\\ +\dots & \dots \\ +y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_1x_{n-1}^{n-1}+\epsilon_{n-1}.\\ +\end{align*} $$

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    Generalizing the fitting procedure as a linear algebra problem

    +

    Rewriting the fitting procedure as a linear algebra problem, follows

    -We are obviously not limited to the above polynomial. We could replace the various powers of \( x \) with elements of Fourier series, that is, instead of \( x_i^j \) we could have \( \cos{(j x_i)} \) or \( \sin{(j x_i)} \), or time series or other orthogonal functions. -For every set of values \( y_i,x_i \) we can then generalize the equations to +Defining the vectors $$ -\begin{align*} -y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ -y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ -y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_2\\ -\dots & \dots \\ -y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_i\\ -\dots & \dots \\ -y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1,n-1}+\epsilon_{n-1}.\\ -\end{align*} +\hat{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T, +$$ + +and +$$ +\hat{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T, +$$ + +and +$$ +\hat{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T, +$$ + +and the matrix +$$ +\hat{X}= +\begin{bmatrix} +1& x_{0}^1 &x_{0}^2& \dots & \dots &x_{0}^{n-1}\\ +1& x_{1}^1 &x_{1}^2& \dots & \dots &x_{1}^{n-1}\\ +1& x_{2}^1 &x_{2}^2& \dots & \dots &x_{2}^{n-1}\\ +\dots& \dots &\dots& \dots & \dots &\dots\\ +1& x_{n-1}^1 &x_{n-1}^2& \dots & \dots &x_{n-1}^{n-1}\\ +\end{bmatrix} +$$ + +we can rewrite our equations as +$$ +\hat{y} = \hat{X}\hat{\beta}+\hat{\epsilon}. $$

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    -We redefine in turn the matrix \( \hat{X} \) as +We are obviously not limited to the above polynomial. We could replace the various powers of \( x \) with elements of Fourier series, that is, instead of \( x_i^j \) we could have \( \cos{(j x_i)} \) or \( \sin{(j x_i)} \), or time series or other orthogonal functions. +For every set of values \( y_i,x_i \) we can then generalize the equations to $$ -\hat{X}= -\begin{bmatrix} -x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\ -x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\ -x_{20}& x_{21} &x_{22}& \dots & \dots &x_{2,n-1}\\ -\dots& \dots &\dots& \dots & \dots &\dots\\ -x_{n-1,0}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\ -\end{bmatrix} +\begin{align*} +y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ +y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ +y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_2\\ +\dots & \dots \\ +y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_i\\ +\dots & \dots \\ +y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1,n-1}+\epsilon_{n-1}.\\ +\end{align*} $$ - -and without loss of generality we rewrite again our equations as -$$ -\hat{y} = \hat{X}\hat{\beta}+\hat{\epsilon}. -$$ - -The left-hand side of this equation forms know. Our error vector \( \hat{\epsilon} \) and the parameter vector \( \hat{\beta} \) are our unknow quantities. How can we obtain the optimal set of \( \beta_i \) values?

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    Optimizing our parameters

    +

    Generalizing the fitting procedure as a linear algebra problem

    -We have defined the matrix \( \hat{X} \) +We redefine in turn the matrix \( \hat{X} \) as $$ -\begin{align*} -y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ -y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ -y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1\\ -\dots & \dots \\ -y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1\\ -\dots & \dots \\ -y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1,n-1}+\epsilon_{n-1}.\\ -\end{align*} +\hat{X}= +\begin{bmatrix} +x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\ +x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\ +x_{20}& x_{21} &x_{22}& \dots & \dots &x_{2,n-1}\\ +\dots& \dots &\dots& \dots & \dots &\dots\\ +x_{n-1,0}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\ +\end{bmatrix} $$ + +and without loss of generality we rewrite again our equations as +$$ +\hat{y} = \hat{X}\hat{\beta}+\hat{\epsilon}. +$$ + +The left-hand side of this equation forms know. Our error vector \( \hat{\epsilon} \) and the parameter vector \( \hat{\beta} \) are our unknow quantities. How can we obtain the optimal set of \( \beta_i \) values?

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    Optimizing our parameters, more details

    +

    Optimizing our parameters

    -We well use this matrix to define the approximation \( \hat{\tilde{y}} \) via the unknown quantity \( \hat{\beta} \) as +We have defined the matrix \( \hat{X} \) $$ -\hat{\tilde{y}}= \hat{X}\hat{\beta}, -$$ - -and in order to find the optimal parameters \( \beta_i \) instead of solving the above linear algebra problem, we define a function which gives a measure of the spread between the values \( y_i \) (which represent hopefully the exact values) and the parametrized values \( \tilde{y}_i \), namely -$$ -Q(\hat{\beta})=\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\left(\hat{y}-\hat{\tilde{y}}\right)^T\left(\hat{y}-\hat{\tilde{y}}\right), -$$ - -or using the matrix \( \hat{X} \) as -$$ -Q(\hat{\beta})=\left(\hat{y}-\hat{X}\hat{\beta}\right)^T\left(\hat{y}-\hat{X}\hat{\beta}\right). +\begin{align*} +y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ +y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ +y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1\\ +\dots & \dots \\ +y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1\\ +\dots & \dots \\ +y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1,n-1}+\epsilon_{n-1}.\\ +\end{align*} $$

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    Interpretations and optimizing our parameters

    +

    Optimizing our parameters, more details

    -The function +We well use this matrix to define the approximation \( \hat{\tilde{y}} \) via the unknown quantity \( \hat{\beta} \) as $$ -Q(\hat{\beta})=\left(\hat{y}-\hat{X}\hat{\beta}\right)^T\left(\hat{y}-\hat{X}\hat{\beta}\right), +\hat{\tilde{y}}= \hat{X}\hat{\beta}, $$ -can be linked to the variance of the quantity \( y_i \) if we interpret the latter as the mean value of for example a numerical experiment. When linking below with the maximum likelihood approach below, we will indeed interpret \( y_i \) as a mean value +and in order to find the optimal parameters \( \beta_i \) instead of solving the above linear algebra problem, we define a function which gives a measure of the spread between the values \( y_i \) (which represent hopefully the exact values) and the parametrized values \( \tilde{y}_i \), namely $$ -y_{i}=\langle y_i \rangle = \beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}+\epsilon_i, +Q(\hat{\beta})=\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\left(\hat{y}-\hat{\tilde{y}}\right)^T\left(\hat{y}-\hat{\tilde{y}}\right), $$ -where \( \langle y_i \rangle \) is the mean value. Keep in mind also that till now we have treated \( y_i \) as the exact value. Normally, the response (dependent or outcome) variable \( y_i \) the outcome of a numerical experiment or another type of experiment and is thus only an approximation to the true value. It is then always accompanied by an error estimate, often limited to a statistical error estimate given by the standard deviation discussed earlier. In the discussion here we will treat \( y_i \) as our exact value for the response variable. - -

    -In order to find the parameters \( \beta_i \) we will then minimize the spread of \( Q(\hat{\beta}) \) by requiring +or using the matrix \( \hat{X} \) as $$ -\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0, +Q(\hat{\beta})=\left(\hat{y}-\hat{X}\hat{\beta}\right)^T\left(\hat{y}-\hat{X}\hat{\beta}\right). $$ - -which results in -$$ -\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0, -$$ - -or in a matrix-vector form as -$$ -\frac{\partial Q(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right). -$$ - -

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    -We can rewrite +The function $$ -\frac{\partial Q(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right), +Q(\hat{\beta})=\left(\hat{y}-\hat{X}\hat{\beta}\right)^T\left(\hat{y}-\hat{X}\hat{\beta}\right), $$ -as +can be linked to the variance of the quantity \( y_i \) if we interpret the latter as the mean value of for example a numerical experiment. When linking below with the maximum likelihood approach below, we will indeed interpret \( y_i \) as a mean value $$ -\hat{X}^T\hat{y} = \hat{X}^T\hat{X}\hat{\beta}, +y_{i}=\langle y_i \rangle = \beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}+\epsilon_i, $$ -and if the matrix \( \hat{X}^T\hat{X} \) is invertible we have the solution +where \( \langle y_i \rangle \) is the mean value. Keep in mind also that till now we have treated \( y_i \) as the exact value. Normally, the response (dependent or outcome) variable \( y_i \) the outcome of a numerical experiment or another type of experiment and is thus only an approximation to the true value. It is then always accompanied by an error estimate, often limited to a statistical error estimate given by the standard deviation discussed earlier. In the discussion here we will treat \( y_i \) as our exact value for the response variable. + +

    +In order to find the parameters \( \beta_i \) we will then minimize the spread of \( Q(\hat{\beta}) \) by requiring $$ -\hat{\beta} =\left(\hat{X}^T\hat{X}\right)^{-1}\hat{X}^T\hat{y}. +\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0, +$$ + +which results in +$$ +\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0, +$$ + +or in a matrix-vector form as +$$ +\frac{\partial Q(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right). $$

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    -The residuals \( \hat{\epsilon} \) are in turn given by +We can rewrite $$ -\hat{\epsilon} = \hat{y}-\hat{\tilde{y}} = \hat{y}-\hat{X}\hat{\beta}, +\frac{\partial Q(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right), $$ -and with +as $$ -\hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right)= 0, +\hat{X}^T\hat{y} = \hat{X}^T\hat{X}\hat{\beta}, $$ -we have +and if the matrix \( \hat{X}^T\hat{X} \) is invertible we have the solution $$ -\hat{X}^T\hat{\epsilon}=\hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right)= 0, +\hat{\beta} =\left(\hat{X}^T\hat{X}\right)^{-1}\hat{X}^T\hat{y}. $$ -meaning that the solution for \( \hat{\beta} \) is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach. -

    @@ -255,7 +282,7 @@ meaning that the solution for \( \hat{\beta} \) is the one which minimizes the r
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    The \( \chi^2 \) function

    +

    Interpretations and optimizing our parameters

    - -

    -Normally, the response (dependent or outcome) variable \( y_i \) the outcome of a numerical experiment or another type of experiment and is thus only an approximation to the true value. It is then always accompanied by an error estimate, often limited to a statistical error estimate given by the standard deviation discussed earlier. In the discussion here we will treat \( y_i \) as our exact value for the response variable. - -

    -Introducing the standard deviation \( \sigma_i \) for each measurement \( y_i \), we define now the \( \chi^2 \) function as +The residuals \( \hat{\epsilon} \) are in turn given by $$ -\chi^2(\hat{\beta})=\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\left(\hat{y}-\hat{\tilde{y}}\right)^T\frac{1}{\hat{\Sigma^2}}\left(\hat{y}-\hat{\tilde{y}}\right), +\hat{\epsilon} = \hat{y}-\hat{\tilde{y}} = \hat{y}-\hat{X}\hat{\beta}, $$ -where the matrix \( \hat{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as matrix elements. +and with +$$ +\hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right)= 0, +$$ + +we have +$$ +\hat{X}^T\hat{\epsilon}=\hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right)= 0, +$$ + +meaning that the solution for \( \hat{\beta} \) is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach.

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    -In order to find the parameters \( \beta_i \) we will then minimize the spread of \( \chi^2(\hat{\beta}) \) by requiring +Normally, the response (dependent or outcome) variable \( y_i \) the outcome of a numerical experiment or another type of experiment and is thus only an approximation to the true value. It is then always accompanied by an error estimate, often limited to a statistical error estimate given by the standard deviation discussed earlier. In the discussion here we will treat \( y_i \) as our exact value for the response variable. + +

    +Introducing the standard deviation \( \sigma_i \) for each measurement \( y_i \), we define now the \( \chi^2 \) function as $$ -\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0, +\chi^2(\hat{\beta})=\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\left(\hat{y}-\hat{\tilde{y}}\right)^T\frac{1}{\hat{\Sigma^2}}\left(\hat{y}-\hat{\tilde{y}}\right), $$ -which results in -$$ -\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0, -$$ +where the matrix \( \hat{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as matrix elements. -or in a matrix-vector form as -$$ -\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right). -$$ - -where we have defined the matrix \( \hat{A} =\hat{X}/\hat{\Sigma} \) with matrix elements \( a_{ij} = x_{ij}/\sigma_i \) and the vector \( \hat{b} \) with elements \( b_i = y_i/\sigma_i \). +

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    -We can rewrite +In order to find the parameters \( \beta_i \) we will then minimize the spread of \( \chi^2(\hat{\beta}) \) by requiring $$ -\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right), +\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0, $$ -as +which results in $$ -\hat{A}^T\hat{b} = \hat{A}^T\hat{A}\hat{\beta}, +\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0, $$ -and if the matrix \( \hat{A}^T\hat{A} \) is invertible we have the solution +or in a matrix-vector form as $$ -\hat{\beta} =\left(\hat{A}^T\hat{A}\right)^{-1}\hat{A}^T\hat{b}. +\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right). $$ + +where we have defined the matrix \( \hat{A} =\hat{X}/\hat{\Sigma} \) with matrix elements \( a_{ij} = x_{ij}/\sigma_i \) and the vector \( \hat{b} \) with elements \( b_i = y_i/\sigma_i \).

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    -If we then introduce the matrix +We can rewrite $$ -\hat{H} = \left(\hat{A}^T\hat{A}\right)^{-1}, +\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right), $$ -we have then the following expression for the parameters \( \beta_j \) (the matrix elements of \( \hat{H} \) are \( h_{ij} \)) +as $$ -\beta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik} +\hat{A}^T\hat{b} = \hat{A}^T\hat{A}\hat{\beta}, $$ -We state without proof the expression for the uncertainty in the parameters \( \beta_j \) as (we leave this as an exercise) +and if the matrix \( \hat{A}^T\hat{A} \) is invertible we have the solution $$ -\sigma^2(\beta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \beta_j}{\partial y_i}\right)^2, -$$ - -resulting in -$$ -\sigma^2(\beta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}! +\hat{\beta} =\left(\hat{A}^T\hat{A}\right)^{-1}\hat{A}^T\hat{b}. $$ @@ -258,7 +282,7 @@ $$

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    -The first step here is to approximate the function \( y \) with a first-order polynomial, that is we write + +

    +If we then introduce the matrix $$ -y=y(x) \rightarrow y(x_i) \approx \beta_0+\beta_1 x_i. +\hat{H} = \left(\hat{A}^T\hat{A}\right)^{-1}, $$ -By computing the derivatives of \( \chi^2 \) with respect to \( \beta_0 \) and \( \beta_1 \) show that these are given by +we have then the following expression for the parameters \( \beta_j \) (the matrix elements of \( \hat{H} \) are \( h_{ij} \)) $$ -\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0, +\beta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik} $$ -and +We state without proof the expression for the uncertainty in the parameters \( \beta_j \) as (we leave this as an exercise) $$ -\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{n-1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0. +\sigma^2(\beta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \beta_j}{\partial y_i}\right)^2, +$$ + +resulting in +$$ +\sigma^2(\beta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}! $$

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    - -

    -For a linear fit we don't need to invert a matrix!! -Defining +The first step here is to approximate the function \( y \) with a first-order polynomial, that is we write $$ -\gamma = \sum_{i=0}^{n-1}\frac{1}{\sigma_i^2}, +y=y(x) \rightarrow y(x_i) \approx \beta_0+\beta_1 x_i. $$ - +By computing the derivatives of \( \chi^2 \) with respect to \( \beta_0 \) and \( \beta_1 \) show that these are given by $$ -\gamma_x = \sum_{i=0}^{n-1}\frac{x_{i}}{\sigma_i^2}, +\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0, $$ +and $$ -\gamma_y = \sum_{i=0}^{n-1}\left(\frac{y_i}{\sigma_i^2}\right), +\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{n-1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0. $$ - -$$ -\gamma_{xx} = \sum_{i=0}^{n-1}\frac{x_ix_{i}}{\sigma_i^2}, -$$ - -$$ -\gamma_{xy} = \sum_{i=0}^{n-1}\frac{y_ix_{i}}{\sigma_i^2}, -$$ - -we obtain -$$ -\beta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}, -$$ - -$$ -\beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}. -$$ - -

    -This approach (different linear and non-linear regression) suffers often from both being underdetermined and overdetermined in the unknown coefficients \( \beta_i \). A better approach is to use the Singular Value Decomposition (SVD) method discussed below. Or using Lasso and Ridge regression. See below.

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    Simple regression model

    -We are now ready to write our first program which aims at solving the above linear regression equations. We start with data we have produced ourselves, in this case normally distributed random numbers along the \( x \)-axis. These numbers define then the value of a function \( y(x)=4+3x+N(0,1) \). Thereafter we order the \( x \) values and employ our linear regression algorithm to set up the best fit. Here we find it useful to use the numpy function \( c\_ \) arrays where arrays are stacked along their last axis after being upgraded to at least two dimensions with ones post-pended to the shape. The following examples help in understanding what happens +

    The \( \chi^2 \) function

    +
    +
    +

    +

    +For a linear fit we don't need to invert a matrix!! +Defining +$$ +\gamma = \sum_{i=0}^{n-1}\frac{1}{\sigma_i^2}, +$$ + + +$$ +\gamma_x = \sum_{i=0}^{n-1}\frac{x_{i}}{\sigma_i^2}, +$$ + +$$ +\gamma_y = \sum_{i=0}^{n-1}\left(\frac{y_i}{\sigma_i^2}\right), +$$ + +$$ +\gamma_{xx} = \sum_{i=0}^{n-1}\frac{x_ix_{i}}{\sigma_i^2}, +$$ + +$$ +\gamma_{xy} = \sum_{i=0}^{n-1}\frac{y_ix_{i}}{\sigma_i^2}, +$$ + +we obtain +$$ +\beta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}, +$$ + +$$ +\beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}. +$$ - -

    import numpy as np
-print(np.c_[np.array([1,2,3]), np.array([4,5,6])])
-print(np.c_[np.array([[1,2,3]]), 0, 0, np.array([[4,5,6]])])
-

    +This approach (different linear and non-linear regression) suffers often from both being underdetermined and overdetermined in the unknown coefficients \( \beta_i \). A better approach is to use the Singular Value Decomposition (SVD) method discussed below. Or using Lasso and Ridge regression. See below. +

    +
    - -
    # Importing various packages
-from random import random, seed
-import numpy as np
-import matplotlib.pyplot as plt
-
-x = 2*np.random.rand(100,1)
-y = 4+3*x+np.random.randn(100,1)
-
-xb = np.c_[np.ones((100,1)), x]
-beta = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-xnew = np.array([[0],[2]])
-xbnew = np.c_[np.ones((2,1)), xnew]
-ypredict = xbnew.dot(beta)
-
-plt.plot(xnew, ypredict, "r-")
-plt.plot(x, y ,'ro')
-plt.axis([0,2.0,0, 15.0])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$y$')
-plt.title(r'Linear Regression')
-plt.show()
-
    -

    -We see that, as expected, a linear fit gives a seemingly (from the graph) good representation of the data.

    @@ -266,7 +302,7 @@ We see that, as expected, a linear fit gives a seemingly (from the graph) good r

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    Simple regression model, now using scikit-learn

    - +

    Simple regression model

    +We are now ready to write our first program which aims at solving the above linear regression equations. We start with data we have produced ourselves, in this case normally distributed random numbers along the \( x \)-axis. These numbers define then the value of a function \( y(x)=4+3x+N(0,1) \). Thereafter we order the \( x \) values and employ our linear regression algorithm to set up the best fit. Here we find it useful to use the numpy function \( c\_ \) arrays where arrays are stacked along their last axis after being upgraded to at least two dimensions with ones post-pended to the shape. The following examples help in understanding what happens

    -We can repeat the above algorithm using scikit-learn as follows + + +

    import numpy as np
+print(np.c_[np.array([1,2,3]), np.array([4,5,6])])
+print(np.c_[np.array([[1,2,3]]), 0, 0, np.array([[4,5,6]])])
+

    @@ -214,23 +248,27 @@ We can repeat the above algorithm using scikit-learn as follows from random import random, seed import numpy as np import matplotlib.pyplot as plt -from sklearn.linear_model import LinearRegression x = 2*np.random.rand(100,1) y = 4+3*x+np.random.randn(100,1) -linreg = LinearRegression() -linreg.fit(x,y) + +xb = np.c_[np.ones((100,1)), x] +beta = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y) xnew = np.array([[0],[2]]) -ypredict = linreg.predict(xnew) +xbnew = np.c_[np.ones((2,1)), xnew] +ypredict = xbnew.dot(beta) plt.plot(xnew, ypredict, "r-") plt.plot(x, y ,'ro') plt.axis([0,2.0,0, 15.0]) plt.xlabel(r'$x$') plt.ylabel(r'$y$') -plt.title(r'Random numbers ') +plt.title(r'Linear Regression') plt.show() +

    +We see that, as expected, a linear fit gives a seemingly (from the graph) good representation of the data. +

    @@ -257,7 +295,7 @@ plt.show()

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    Simple linear regression model using scikit-learn

    +

    Simple regression model, now using scikit-learn

    -We start with perhaps our simplest possible example, using scikit-learn to perform linear regression analysis on a data set produced by us. -What follows is a simple Python code where we have defined function \( y \) in terms of the variable \( x \). Both are defined as vectors of dimension \( 1\times 100 \). The entries to the vector \( \hat{x} \) are given by random numbers generated with a uniform distribution with entries \( x_i \in [0,1] \) (more about probability distribution functions later). These values are then used to define a function \( y(x) \) (tabulated again as a vector) with a linear dependence on \( x \) plus a random noise added via the normal distribution. - -

    -The Numpy functions are imported used the import numpy as np -statement and the random number generator for the uniform distribution -is called using the function np.random.rand(), where we specificy -that we want \( 100 \) random variables. Using Numpy we define -automatically an array with the specified number of elements, \( 100 \) in -our case. With the Numpy function randn() we can compute random -numbers with the normal distribution (mean value \( \mu \) equal to zero and -variance \( \sigma^2 \) set to one) and produce the values of \( y \) assuming a linear -dependence as function of \( x \) - -$$ -y = 2x+N(0,1), -$$ - -

    -where \( N(0,1) \) represents random numbers generated by the normal -distribution. From scikit-learn we import then the -LinearRegression functionality and make a prediction \( \tilde{y} = -\alpha + \beta x \) using the function fit(x,y). We call the set of -data \( (\hat{x},\hat{y}) \) for our training data. The Python package -scikit-learn has also a functionality which extracts the above -fitting parameters \( \alpha \) and \( \beta \) (see below). Later we will -distinguish between training data and test data. - -

    -For plotting we use the Python package -matplotlib which produces publication -quality figures. Feel free to explore the extensive -gallery of examples. In -this example we plot our original values of \( x \) and \( y \) as well as the -prediction ypredict (\( \tilde{y} \)), which attempts at fitting our -data with a straight line. - -

    -The Python code follows here. +We can repeat the above algorithm using scikit-learn as follows 

    # Importing various packages
+from random import random, seed
 import numpy as np
 import matplotlib.pyplot as plt
 from sklearn.linear_model import LinearRegression
 
-x = np.random.rand(100,1)
-y = 2*x+np.random.randn(100,1)
+x = 2*np.random.rand(100,1)
+y = 4+3*x+np.random.randn(100,1)
 linreg = LinearRegression()
 linreg.fit(x,y)
-xnew = np.array([[0],[1]])
+xnew = np.array([[0],[2]])
 ypredict = linreg.predict(xnew)
 
 plt.plot(xnew, ypredict, "r-")
 plt.plot(x, y ,'ro')
-plt.axis([0,1.0,0, 5.0])
+plt.axis([0,2.0,0, 15.0])
 plt.xlabel(r'$x$')
 plt.ylabel(r'$y$')
-plt.title(r'Simple Linear Regression')
+plt.title(r'Random numbers ')
 plt.show()

    @@ -294,7 +286,7 @@ plt.show()

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    • diff --git a/doc/pub/Regression/html/._Regression-bs021.html b/doc/pub/Regression/html/._Regression-bs021.html index c1f975f55..a768c9fd4 100644 --- a/doc/pub/Regression/html/._Regression-bs021.html +++ b/doc/pub/Regression/html/._Regression-bs021.html @@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source @@ -152,42 +173,50 @@ MathJax.Hub.Config({ Contents @@ -203,25 +232,71 @@ MathJax.Hub.Config({ -

    Simple linear regression model

    +

    Simple linear regression model using scikit-learn

    -This example serves several aims. It allows us to demonstrate several -aspects of data analysis and later machine learning algorithms. The -immediate visualization shows that our linear fit is not -impressive. It goes through the data points, but there are many -outliers which are not reproduced by our linear regression. We could -now play around with this small program and change for example the -factor in front of \( x \) and the normal distribution. Try to change the -function \( y \) to +We start with perhaps our simplest possible example, using scikit-learn to perform linear regression analysis on a data set produced by us. +What follows is a simple Python code where we have defined function \( y \) in terms of the variable \( x \). Both are defined as vectors of dimension \( 1\times 100 \). The entries to the vector \( \hat{x} \) are given by random numbers generated with a uniform distribution with entries \( x_i \in [0,1] \) (more about probability distribution functions later). These values are then used to define a function \( y(x) \) (tabulated again as a vector) with a linear dependence on \( x \) plus a random noise added via the normal distribution. + +

    +The Numpy functions are imported used the import numpy as np +statement and the random number generator for the uniform distribution +is called using the function np.random.rand(), where we specificy +that we want \( 100 \) random variables. Using Numpy we define +automatically an array with the specified number of elements, \( 100 \) in +our case. With the Numpy function randn() we can compute random +numbers with the normal distribution (mean value \( \mu \) equal to zero and +variance \( \sigma^2 \) set to one) and produce the values of \( y \) assuming a linear +dependence as function of \( x \) $$ -y = 10x+0.01 \times N(0,1), +y = 2x+N(0,1), $$

    -where \( x \) is defined as before. +where \( N(0,1) \) represents random numbers generated by the normal +distribution. From scikit-learn we import then the +LinearRegression functionality and make a prediction \( \tilde{y} = +\alpha + \beta x \) using the function fit(x,y). We call the set of +data \( (\hat{x},\hat{y}) \) for our training data. The Python package +scikit-learn has also a functionality which extracts the above +fitting parameters \( \alpha \) and \( \beta \) (see below). Later we will +distinguish between training data and test data. +

    +For plotting we use the Python package +matplotlib which produces publication +quality figures. Feel free to explore the extensive +gallery of examples. In +this example we plot our original values of \( x \) and \( y \) as well as the +prediction ypredict (\( \tilde{y} \)), which attempts at fitting our +data with a straight line. + +

    +The Python code follows here. +

    + + +

    # Importing various packages
+import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import LinearRegression
+
+x = np.random.rand(100,1)
+y = 2*x+np.random.randn(100,1)
+linreg = LinearRegression()
+linreg.fit(x,y)
+xnew = np.array([[0],[1]])
+ypredict = linreg.predict(xnew)
+
+plt.plot(xnew, ypredict, "r-")
+plt.plot(x, y ,'ro')
+plt.axis([0,1.0,0, 5.0])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$y$')
+plt.title(r'Simple Linear Regression')
+plt.show()
+

    @@ -248,7 +323,7 @@ where \( x \) is defined as before.

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    Less noise

    +

    Simple linear regression model

    -Does the fit look better? Indeed, by -reducing the role of the normal distribution we see immediately that -our linear prediction seemingly reproduces better the training -set. However, this testing 'by the eye' is obviouly not satisfactory in the -long run. Here we have only defined the training data and our model, and -have not discussed a more rigorous approach to the cost function. +This example serves several aims. It allows us to demonstrate several +aspects of data analysis and later machine learning algorithms. The +immediate visualization shows that our linear fit is not +impressive. It goes through the data points, but there are many +outliers which are not reproduced by our linear regression. We could +now play around with this small program and change for example the +factor in front of \( x \) and the normal distribution. Try to change the +function \( y \) to + +$$ +y = 10x+0.01 \times N(0,1), +$$ + +

    +where \( x \) is defined as before.

    @@ -239,7 +277,7 @@ have not discussed a more rigorous approach to the cost function.

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    • diff --git a/doc/pub/Regression/html/._Regression-bs023.html b/doc/pub/Regression/html/._Regression-bs023.html index bd1b9f6a5..193801b1c 100644 --- a/doc/pub/Regression/html/._Regression-bs023.html +++ b/doc/pub/Regression/html/._Regression-bs023.html @@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source @@ -152,42 +173,50 @@ MathJax.Hub.Config({ Contents @@ -203,24 +232,15 @@ MathJax.Hub.Config({ -

    How to study our fits

    +

    Less noise

    -We need more rigorous criteria in defining whether we have succeeded or -not in modeling our training data. You will be surprised to see that -many scientists seldomly venture beyond this 'by the eye' approach. A -standard approach for the cost function is the so-called \( \chi^2 \) -function - -$$ \chi^2 = \frac{1}{n} -\sum_{i=0}^{n-1}\frac{(y_i-\tilde{y}_i)^2}{\sigma_i^2}, -$$ - -

    -where \( \sigma_i^2 \) is the variance (to be defined later) of the entry -\( y_i \). We may not know the explicit value of \( \sigma_i^2 \), it serves -however the aim of scaling the equations and make the cost function -dimensionless. +Does the fit look better? Indeed, by +reducing the role of the normal distribution we see immediately that +our linear prediction seemingly reproduces better the training +set. However, this testing 'by the eye' is obviouly not satisfactory in the +long run. Here we have only defined the training data and our model, and +have not discussed a more rigorous approach to the cost function.

    @@ -248,7 +268,7 @@ dimensionless.

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    Minimizing the cost function

    +

    How to study our fits

    -Minimizing the cost function is a central aspect of -our discussions to come. Finding its minima as function of the model -parameters (\( \alpha \) and \( \beta \) in our case) will be a recurring -theme in these series of lectures. Essentially all machine learning -algorithms we will discuss center around the minimization of the -chosen cost function. This depends in turn on our specific -model for describing the data, a typical situation in supervised -learning. Automatizing the search for the minima of the cost function is a -central ingredient in all algorithms. Typical methods which are -employed are various variants of gradient methods. These will be -discussed in more detail later. Again, you'll be surprised to hear that -many practitioners minimize the above function ''by the eye', popularly dubbed as -'chi by the eye'. That is, change a parameter and see (visually and numerically) that -the \( \chi^2 \) function becomes smaller. +We need more rigorous criteria in defining whether we have succeeded or +not in modeling our training data. You will be surprised to see that +many scientists seldomly venture beyond this 'by the eye' approach. A +standard approach for the cost function is the so-called \( \chi^2 \) +function + +$$ \chi^2 = \frac{1}{n} +\sum_{i=0}^{n-1}\frac{(y_i-\tilde{y}_i)^2}{\sigma_i^2}, +$$ + +

    +where \( \sigma_i^2 \) is the variance (to be defined later) of the entry +\( y_i \). We may not know the explicit value of \( \sigma_i^2 \), it serves +however the aim of scaling the equations and make the cost function +dimensionless.

    @@ -247,7 +277,7 @@ the \( \chi^2 \) function becomes smaller.

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    Relative error

    +

    Minimizing the cost function

    -There are many ways to define the cost function. A simpler approach is to look at the relative difference between the training data and the predicted data, that is we define -the relative error as - -$$ -\epsilon_{\mathrm{relative}}= \frac{\vert \hat{y} -\hat{\tilde{y}}\vert}{\vert \hat{y}\vert}. -$$ - -We can modify easily the above Python code and plot the relative error instead -

    - - -

    import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.linear_model import LinearRegression
-
-x = np.random.rand(100,1)
-y = 5*x+0.01*np.random.randn(100,1)
-linreg = LinearRegression()
-linreg.fit(x,y)
-ypredict = linreg.predict(x)
-
-plt.plot(x, np.abs(ypredict-y)/abs(y), "ro")
-plt.axis([0,1.0,0.0, 0.5])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$\epsilon_{\mathrm{relative}}$')
-plt.title(r'Relative error')
-plt.show()
-
    -

    -Depending on the parameter in front of the normal distribution, we may -have a small or larger relative error. Try to play around with -different training data sets and study (graphically) the value of the -relative error. +Minimizing the cost function is a central aspect of +our discussions to come. Finding its minima as function of the model +parameters (\( \alpha \) and \( \beta \) in our case) will be a recurring +theme in these series of lectures. Essentially all machine learning +algorithms we will discuss center around the minimization of the +chosen cost function. This depends in turn on our specific +model for describing the data, a typical situation in supervised +learning. Automatizing the search for the minima of the cost function is a +central ingredient in all algorithms. Typical methods which are +employed are various variants of gradient methods. These will be +discussed in more detail later. Again, you'll be surprised to hear that +many practitioners minimize the above function ''by the eye', popularly dubbed as +'chi by the eye'. That is, change a parameter and see (visually and numerically) that +the \( \chi^2 \) function becomes smaller.

    @@ -266,7 +276,7 @@ relative error.

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    The richness of scikit-learn

    +

    Relative error

    -As mentioned above, scikit-learn has an impressive functionality. -We can for example extract the values of \( \alpha \) and \( \beta \) and -their error estimates, or the variance and standard deviation and many -other properties from the statistical data analysis. +There are many ways to define the cost function. A simpler approach is to look at the relative difference between the training data and the predicted data, that is we define +the relative error as -

    -Here we show an -example of the functionality of scikit-learn. +$$ +\epsilon_{\mathrm{relative}}= \frac{\vert \hat{y} -\hat{\tilde{y}}\vert}{\vert \hat{y}\vert}. +$$ + +We can modify easily the above Python code and plot the relative error instead

    -

    import numpy as np 
-import matplotlib.pyplot as plt 
-from sklearn.linear_model import LinearRegression 
-from sklearn.metrics import mean_squared_error, r2_score, mean_squared_log_error, mean_absolute_error
+
    import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import LinearRegression
 
 x = np.random.rand(100,1)
-y = 2.0+ 5*x+0.5*np.random.randn(100,1)
+y = 5*x+0.01*np.random.randn(100,1)
 linreg = LinearRegression()
 linreg.fit(x,y)
 ypredict = linreg.predict(x)
-print('The intercept alpha: \n', linreg.intercept_)
-print('Coefficient beta : \n', linreg.coef_)
-# The mean squared error                               
-print("Mean squared error: %.2f" % mean_squared_error(y, ypredict))
-# Explained variance score: 1 is perfect prediction                                 
-print('Variance score: %.2f' % r2_score(y, ypredict))
-# Mean squared log error                                                        
-print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) )
-# Mean absolute error                                                           
-print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict))
-plt.plot(x, ypredict, "r-")
-plt.plot(x, y ,'ro')
-plt.axis([0.0,1.0,1.5, 7.0])
+
+plt.plot(x, np.abs(ypredict-y)/abs(y), "ro")
+plt.axis([0,1.0,0.0, 0.5])
 plt.xlabel(r'$x$')
-plt.ylabel(r'$y$')
-plt.title(r'Linear Regression fit ')
+plt.ylabel(r'$\epsilon_{\mathrm{relative}}$')
+plt.title(r'Relative error')
 plt.show()
 
    
+
    
+Depending on the parameter in front of the normal distribution, we may
+have a small or larger relative error. Try to play around with
+different training data sets and study (graphically) the value of the
+relative error.
+
 
    
 
    
 
@@ -271,7 +295,7 @@ plt.show()
   
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diff --git a/doc/pub/Regression/html/._Regression-bs027.html b/doc/pub/Regression/html/._Regression-bs027.html
index d68ae908b..79b422d69 100644
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@@ -152,42 +173,50 @@ MathJax.Hub.Config({
         Contents 
         
       
@@ -203,20 +232,48 @@ MathJax.Hub.Config({
 
 
 
-
    Functions in scikit-learn 
    
+
    The richness of scikit-learn 
    
 
 
    
-The function coef gives us the parameter \( \beta \) of our fit while intercept yields 
-\( \alpha \). Depending on the constant in front of the normal distribution, we get values near or far from \( alpha =2 \) and \( \beta =5 \). Try to play around with different parameters in front of the normal distribution. The function meansquarederror gives us the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error or loss defined as
-$$ MSE(\hat{y},\hat{\tilde{y}}) = \frac{1}{n}
-\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, 
-$$
+As mentioned above, scikit-learn has an impressive functionality.
+We can for example extract the values of \( \alpha \) and \( \beta \) and
+their error estimates, or the variance and standard deviation and many
+other properties from the statistical data analysis.
 
 
    
-The smaller the value, the better the fit. Ideally we would like to
-have an MSE equal zero.  The attentive reader has probably recognized
-this function as being similar to the \( \chi^2 \) function defined above.
+Here we show an
+example of the functionality of scikit-learn.
+
    
 
+
+
    import numpy as np 
+import matplotlib.pyplot as plt 
+from sklearn.linear_model import LinearRegression 
+from sklearn.metrics import mean_squared_error, r2_score, mean_squared_log_error, mean_absolute_error
+
+x = np.random.rand(100,1)
+y = 2.0+ 5*x+0.5*np.random.randn(100,1)
+linreg = LinearRegression()
+linreg.fit(x,y)
+ypredict = linreg.predict(x)
+print('The intercept alpha: \n', linreg.intercept_)
+print('Coefficient beta : \n', linreg.coef_)
+# The mean squared error                               
+print("Mean squared error: %.2f" % mean_squared_error(y, ypredict))
+# Explained variance score: 1 is perfect prediction                                 
+print('Variance score: %.2f' % r2_score(y, ypredict))
+# Mean squared log error                                                        
+print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) )
+# Mean absolute error                                                           
+print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict))
+plt.plot(x, ypredict, "r-")
+plt.plot(x, y ,'ro')
+plt.axis([0.0,1.0,1.5, 7.0])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$y$')
+plt.title(r'Linear Regression fit ')
+plt.show()
+
    
 
    
 
    
 
@@ -243,7 +300,7 @@ this function as being similar to the \( \chi^2 \) function defined above.
   
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@@ -152,42 +173,50 @@ MathJax.Hub.Config({
         Contents 
         
       
@@ -203,26 +232,19 @@ MathJax.Hub.Config({
 
 
 
-
    Other functions in  scikit-learn 
    
+
    Functions in scikit-learn 
    
 
 
    
-The r2score function computes \( R^2 \), the coefficient of
-determination. It provides a measure of how well future samples are
-likely to be predicted by the model. Best possible score is 1.0 and it
-can be negative (because the model can be arbitrarily worse). A
-constant model that always predicts the expected value of \( \hat{y} \),
-disregarding the input features, would get a \( R^2 \) score of \( 0.0 \).
+The function coef gives us the parameter \( \beta \) of our fit while intercept yields 
+\( \alpha \). Depending on the constant in front of the normal distribution, we get values near or far from \( alpha =2 \) and \( \beta =5 \). Try to play around with different parameters in front of the normal distribution. The function meansquarederror gives us the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error or loss defined as
+$$ MSE(\hat{y},\hat{\tilde{y}}) = \frac{1}{n}
+\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, 
+$$
 
 
    
-If \( \tilde{\hat{y}}_i \) is the predicted value of the \( i-th \) sample and \( y_i \) is the corresponding true value, then the score \( R^2 \) is defined as
-$$
-R^2(\hat{y}, \tilde{\hat{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2},
-$$
-
-where we have defined the mean value  of \( \hat{y} \) as
-$$
-\bar{y} =  \frac{1}{n} \sum_{i=0}^{n - 1} y_i.
-$$
+The smaller the value, the better the fit. Ideally we would like to
+have an MSE equal zero.  The attentive reader has probably recognized
+this function as being similar to the \( \chi^2 \) function defined above.
 
 
    
 
    
@@ -249,6 +271,8 @@ $$
   
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diff --git a/doc/pub/Regression/html/._Regression-bs029.html b/doc/pub/Regression/html/._Regression-bs029.html
index 6a182d377..33cd41ef8 100644
--- a/doc/pub/Regression/html/._Regression-bs029.html
+++ b/doc/pub/Regression/html/._Regression-bs029.html
@@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source
 
 
 
@@ -152,42 +173,50 @@ MathJax.Hub.Config({
         Contents 
         
       
@@ -203,27 +232,26 @@ MathJax.Hub.Config({
 
 
 
-
    The mean absolute error and other functions in  scikit-learn 
    
+
    Other functions in  scikit-learn 
    
 
 
    
-Another quantity will meet again in our discussions of regression analysis is 
- mean absolute error (MAE), a risk metric corresponding to the expected value of the absolute error loss or what we call the \( l1 \)-norm loss. In our discussion above we presented the relative error.
-The MAE is defined as follows
-$$
-\text{MAE}(\hat{y}, \hat{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n-1} \left| y_i - \tilde{y}_i \right|.
-$$
-
-Finally we present the 
-squared logarithmic (quadratic) error
-$$
-\text{MSLE}(\hat{y}, \hat{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n - 1} (\log_e (1 + y_i) - \log_e (1 + \tilde{y}_i) )^2,
-$$
+The r2score function computes \( R^2 \), the coefficient of
+determination. It provides a measure of how well future samples are
+likely to be predicted by the model. Best possible score is 1.0 and it
+can be negative (because the model can be arbitrarily worse). A
+constant model that always predicts the expected value of \( \hat{y} \),
+disregarding the input features, would get a \( R^2 \) score of \( 0.0 \).
 
 
    
-where \( \log_e (x) \) stands for the natural logarithm of \( x \). This error
-estimate is best to use when targets having exponential growth, such
-as population counts, average sales of a commodity over a span of
-years etc.
+If \( \tilde{\hat{y}}_i \) is the predicted value of the \( i-th \) sample and \( y_i \) is the corresponding true value, then the score \( R^2 \) is defined as
+$$
+R^2(\hat{y}, \tilde{\hat{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2},
+$$
+
+where we have defined the mean value  of \( \hat{y} \) as
+$$
+\bar{y} =  \frac{1}{n} \sum_{i=0}^{n - 1} y_i.
+$$
 
 
    
 
    
@@ -249,6 +277,9 @@ years etc.
   
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diff --git a/doc/pub/Regression/html/._Regression-bs030.html b/doc/pub/Regression/html/._Regression-bs030.html
index a0e4dfa92..69b76d122 100644
--- a/doc/pub/Regression/html/._Regression-bs030.html
+++ b/doc/pub/Regression/html/._Regression-bs030.html
@@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source
 
 
 
@@ -152,42 +173,50 @@ MathJax.Hub.Config({
         Contents 
         
       
@@ -203,50 +232,27 @@ MathJax.Hub.Config({
 
 
 
-
    Cubic polynomial in  scikit-learn 
    
+
    The mean absolute error and other functions in  scikit-learn 
    
 
 
    
-We will discuss in more
-detail these and other functions in the various lectures.  We conclude this part with another example. Instead of 
-a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn.
+Another quantity will meet again in our discussions of regression analysis is 
+ mean absolute error (MAE), a risk metric corresponding to the expected value of the absolute error loss or what we call the \( l1 \)-norm loss. In our discussion above we presented the relative error.
+The MAE is defined as follows
+$$
+\text{MAE}(\hat{y}, \hat{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n-1} \left| y_i - \tilde{y}_i \right|.
+$$
+
+Finally we present the 
+squared logarithmic (quadratic) error
+$$
+\text{MSLE}(\hat{y}, \hat{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n - 1} (\log_e (1 + y_i) - \log_e (1 + \tilde{y}_i) )^2,
+$$
 
 
    
-
-
-
    import matplotlib.pyplot as plt
-import numpy as np
-import random
-from sklearn.linear_model import Ridge
-from sklearn.preprocessing import PolynomialFeatures
-from sklearn.pipeline import make_pipeline
-from sklearn.linear_model import LinearRegression
-
-x=np.linspace(0.02,0.98,200)
-noise = np.asarray(random.sample((range(200)),200))
-y=x**3*noise
-yn=x**3*100
-poly3 = PolynomialFeatures(degree=3)
-X = poly3.fit_transform(x[:,np.newaxis])
-clf3 = LinearRegression()
-clf3.fit(X,y)
-
-Xplot=poly3.fit_transform(x[:,np.newaxis])
-poly3_plot=plt.plot(x, clf3.predict(Xplot), label='Cubic Fit')
-plt.plot(x,yn, color='red', label="True Cubic")
-plt.scatter(x, y, label='Data', color='orange', s=15)
-plt.legend()
-plt.show()
-
-def error(a):
-    for i in y:
-        err=(y-yn)/yn
-    return abs(np.sum(err))/len(err)
-
-print (error(y))
-
    
-
    
-Similarly, using R, we can perform similar studies. 
-(more details on R will be inserted later).
+where \( \log_e (x) \) stands for the natural logarithm of \( x \). This error
+estimate is best to use when targets having exponential growth, such
+as population counts, average sales of a commodity over a span of
+years etc.
 
 
    
 
    
@@ -271,6 +277,10 @@ Similarly, using R, we can perform similar studies.
   
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diff --git a/doc/pub/Regression/html/._Regression-bs031.html b/doc/pub/Regression/html/._Regression-bs031.html
index 3ad7cabb6..3033d8063 100644
--- a/doc/pub/Regression/html/._Regression-bs031.html
+++ b/doc/pub/Regression/html/._Regression-bs031.html
@@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source
 
 
 
@@ -152,42 +173,50 @@ MathJax.Hub.Config({
         Contents 
         
       
@@ -203,47 +232,51 @@ MathJax.Hub.Config({
 
 
 
-
    Simple regression model with gradient descent  
    
-Add info about the equations, play around with different learning rates
+
    Cubic polynomial in  scikit-learn 
    
+
+
    
+We will discuss in more
+detail these and other functions in the various lectures.  We conclude this part with another example. Instead of 
+a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn. 
+Add description of the various python commands.
+
 
    
 
 
-
    # Importing various packages
-from math import exp, sqrt
-from random import random, seed
+
    import matplotlib.pyplot as plt
 import numpy as np
-import matplotlib.pyplot as plt
+import random
+from sklearn.linear_model import Ridge
+from sklearn.preprocessing import PolynomialFeatures
+from sklearn.pipeline import make_pipeline
+from sklearn.linear_model import LinearRegression
 
-x = 2*np.random.rand(100,1)
-y = 4+3*x+np.random.randn(100,1)
+x=np.linspace(0.02,0.98,200)
+noise = np.asarray(random.sample((range(200)),200))
+y=x**3*noise
+yn=x**3*100
+poly3 = PolynomialFeatures(degree=3)
+X = poly3.fit_transform(x[:,np.newaxis])
+clf3 = LinearRegression()
+clf3.fit(X,y)
 
-xb = np.c_[np.ones((100,1)), x]
-theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-print(theta_linreg)
-theta = np.random.randn(2,1)
-
-eta = 0.1
-Niterations = 1000
-m = 100
-
-for iter in range(Niterations):
-    gradients = 2.0/m*xb.T.dot(xb.dot(theta)-y)
-    theta -= eta*gradients
-
-print(theta)
-xnew = np.array([[0],[2]])
-xbnew = np.c_[np.ones((2,1)), xnew]
-ypredict = xbnew.dot(theta)
-ypredict2 = xbnew.dot(theta_linreg)
-plt.plot(xnew, ypredict, "r-")
-plt.plot(xnew, ypredict2, "b-")
-plt.plot(x, y ,'ro')
-plt.axis([0,2.0,0, 15.0])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$y$')
-plt.title(r'Random numbers ')
+Xplot=poly3.fit_transform(x[:,np.newaxis])
+poly3_plot=plt.plot(x, clf3.predict(Xplot), label='Cubic Fit')
+plt.plot(x,yn, color='red', label="True Cubic")
+plt.scatter(x, y, label='Data', color='orange', s=15)
+plt.legend()
 plt.show()
+
+def error(a):
+    for i in y:
+        err=(y-yn)/yn
+    return abs(np.sum(err))/len(err)
+
+print (error(y))
 
    
+
    
+Using R, we can perform similar studies.
+
 
    
 
    
 
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diff --git a/doc/pub/Regression/html/._Regression-bs032.html b/doc/pub/Regression/html/._Regression-bs032.html
index cae569fdf..0f9b89474 100644
--- a/doc/pub/Regression/html/._Regression-bs032.html
+++ b/doc/pub/Regression/html/._Regression-bs032.html
@@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source
 
 
 
@@ -152,42 +173,50 @@ MathJax.Hub.Config({
         Contents 
         
       
@@ -203,8 +232,7 @@ MathJax.Hub.Config({
 
 
 
-
    Simple regression model with stochastic gradient descent  
    
-Add info about the equations, play around with different learning rates
+
    Polynomial Regression 
    
 
    
 
 
@@ -213,17 +241,24 @@ Add info about the equations, play around with different learning rates
 from random import random, seed
 import numpy as np
 import matplotlib.pyplot as plt
-from sklearn.linear_model import SGDRegressor
 
-x = 2*np.random.rand(100,1)
-y = 4+3*x+np.random.randn(100,1)
+m = 100
+x = 2*np.random.rand(m,1)+4.
+y = 4+3*x*x+ +x-np.random.randn(m,1)
 
-xb = np.c_[np.ones((100,1)), x]
-theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-print(theta_linreg)
-sgdreg = SGDRegressor(n_iter = 50, penalty=None, eta0=0.1)
-sgdreg.fit(x,y.ravel())
-print(sgdreg.intercept_, sgdreg.coef_)
+xb = np.c_[np.ones((m,1)), x]
+theta = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
+xnew = np.array([[0],[2]])
+xbnew = np.c_[np.ones((2,1)), xnew]
+ypredict = xbnew.dot(theta)
+
+plt.plot(xnew, ypredict, "r-")
+plt.plot(x, y ,'ro')
+plt.axis([0,2.0,0, 15.0])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$y$')
+plt.title(r'Random numbers ')
+plt.show()
 
    
 
    
 
    
@@ -246,6 +281,12 @@ sgdreg.fit(x,y.
   
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diff --git a/doc/pub/Regression/html/._Regression-bs033.html b/doc/pub/Regression/html/._Regression-bs033.html
index 21186180b..d55114e2c 100644
--- a/doc/pub/Regression/html/._Regression-bs033.html
+++ b/doc/pub/Regression/html/._Regression-bs033.html
@@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source
 
 
 
@@ -152,42 +173,50 @@ MathJax.Hub.Config({
         Contents 
         
       
@@ -201,36 +230,32 @@ MathJax.Hub.Config({
 
     
     
     
     
 
 
-
+
+
+
    Linking the regression analysis with a statistical interpretation 
    
 
-
    Polynomial Regression 
    
 
    
+Before we proceed, and to link with our discussions of Bayesian statistics to come, it is useful the derive the standard regression analysis equations using a statistical interpretation. This allows us also to derive quantities like the variance and other expectation values in a rather straightforward way.
 
-
-
    # Importing various packages
-from math import exp, sqrt
-from random import random, seed
-import numpy as np
-import matplotlib.pyplot as plt
+
    
+It is assumed that \( \varepsilon_i
+\sim \mathcal{N}(0, \sigma^2) \) and the \( \varepsilon_{i} \) are
+independent, i.e.: 
+$$
+\begin{align*} 
+\mbox{Cov}(\varepsilon_{i_1},
+\varepsilon_{i_2}) & = \left\{ \begin{array}{lcc} \sigma^2 & \mbox{if}
+& i_1 = i_2, \\ 0 & \mbox{if} & i_1 \not= i_2.  \end{array} \right.
+\end{align*} 
+$$
 
-m = 100
-x = 2*np.random.rand(m,1)+4.
-y = 4+3*x*x+ +x-np.random.randn(m,1)
+The randomness of \( \varepsilon_i \) implies that
+\( \mathbf{Y}_i \) is also a random variable. In particular,
+\( \mathbf{Y}_i \) is normally distributed, because \( \varepsilon_i \sim
+\mathcal{N}(0, \sigma^2) \) and \( \mathbf{X}_{i,\ast} \, \beta \) is a
+non-random scalar. To specify the parameters of the distribution of
+\( \mathbf{Y}_i \) we need to calculate its first two moments.
 
-xb = np.c_[np.ones((m,1)), x]
-theta = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-xnew = np.array([[0],[2]])
-xbnew = np.c_[np.ones((2,1)), xnew]
-ypredict = xbnew.dot(theta)
-
-plt.plot(xnew, ypredict, "r-")
-plt.plot(x, y ,'ro')
-plt.axis([0,2.0,0, 15.0])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$y$')
-plt.title(r'Random numbers ')
-plt.show()
-
    
 
    
 
    
 
@@ -251,6 +276,13 @@ plt.show()
   
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diff --git a/doc/pub/Regression/html/._Regression-bs034.html b/doc/pub/Regression/html/._Regression-bs034.html
index 364fa6719..a5c44e236 100644
--- a/doc/pub/Regression/html/._Regression-bs034.html
+++ b/doc/pub/Regression/html/._Regression-bs034.html
@@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source
 
 
 
@@ -152,42 +173,50 @@ MathJax.Hub.Config({
         Contents 
         
       
@@ -203,84 +232,37 @@ MathJax.Hub.Config({
 
 
 
-
    Ridge and Lasso Regression 
    
+
    Expectation value and variance 
    
 
 
    
+Its expectation equals: 
+$$
+\begin{align*} 
+\mathbb{E}(Y_i) & =
+\mathbb{E}(\mathbf{X}_{i, \ast} \, \beta) + \mathbb{E}(\varepsilon_i)
+\, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \beta, 
+\end{align*} 
+$$
 
-
-
    import matplotlib.pyplot as plt
-import numpy as np
-from sklearn import linear_model
-from sklearn.linear_model import LinearRegression
-from sklearn.metrics import mean_squared_error, r2_score
+while
+its variance is 
+$$
+\begin{align*} \mbox{Var}(Y_i) & = \mathbb{E} \{ [Y_i
+- \mathbb{E}(Y_i)]^2 \} \, \, \, = \, \, \, \mathbb{E} ( Y_i^2 ) -
+[\mathbb{E}(Y_i)]^2  \\  & = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \,
+\beta + \varepsilon_i )^2] - ( \mathbf{X}_{i, \ast} \, \beta)^2 \\ &
+= \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2 \varepsilon_i
+\mathbf{X}_{i, \ast} \, \beta + \varepsilon_i^2 ] - ( \mathbf{X}_{i,
+\ast} \, \beta)^2 \\  & = ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2
+\mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \beta +
+\mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \beta)^2 
+\\ & = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \,
+\mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2.  
+\end{align*}
+$$
 
-#creating data with random noise
-x=np.arange(50)
+Hence, \( Y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \beta, \sigma^2) \).
 
-delta=np.random.uniform(-2.5,2.5, size=(50))
-np.random.shuffle(delta)
-y =0.5*x+5+delta
-
-#arranging data into 2x50 matrix
-a=np.array(x) #inputs
-b=np.array(y) #outputs
-
-#Split into training and test
-X_train=a[:37, np.newaxis]
-X_test=a[37:, np.newaxis]
-y_train=b[:37]
-y_test=b[37:]
-
-print ("X_train: ", X_train.shape)
-print ("y_train: ", y_train.shape)
-print ("X_test: ", X_test.shape)
-print ("y_test: ", y_test.shape)
-
-print ("------------------------------------")
-
-print ("Ordinary Least Squares")
-#Add Ordinary Least Squares fit
-reg=LinearRegression()
-reg.fit(X_train, y_train)
-pred=reg.predict(X_test)
-print ("Prediction Shape: ", pred.shape)
-
-print('Coefficients: \n', reg.coef_)
-# The mean squared error
-print("Mean squared error: %.2f"
-      % mean_squared_error(y_test, pred))
-# Explained variance score: 1 is perfect prediction
-print('Variance score: %.2f' % r2_score(y_test, pred))
-
-#plot
-plt.scatter(X_test,y_test,color='green', label="Training Data")
-plt.plot(X_test, pred, color='black', label="Fit Line")
-plt.legend()
-plt.show()
-
-print ("------------------------------------")
-
-print ("Ridge Regression")
-
-ridge=linear_model.RidgeCV(alphas=[0.1,1.0,10.0])
-ridge.fit(X_train,y_train)
-print ("Ridge Coefficient: ",ridge.coef_)
-print ("Ridge Intercept: ", ridge.intercept_)
-#Look into graphing with Ridge fit
-
-print ("------------------------------------")
-
-print ("Lasso")
-lasso=linear_model.Lasso(alpha=0.1)
-lasso.fit(X_train,y_train)
-predl=lasso.predict(X_test)
-print("Lasso Coefficient: ", lasso.coef_)
-print("Lasso Intercept: ", lasso.intercept_)
-plt.scatter(X_test,y_test,color='green', label="Training Data")
-plt.plot(X_test, predl, color='blue', label="Lasso")
-plt.legend()
-plt.show()
-
    
 
    
 
    
 
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diff --git a/doc/pub/Regression/html/._Regression-bs035.html b/doc/pub/Regression/html/._Regression-bs035.html
index 476608042..3902c8b08 100644
--- a/doc/pub/Regression/html/._Regression-bs035.html
+++ b/doc/pub/Regression/html/._Regression-bs035.html
@@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source
 
 
 
@@ -152,42 +173,50 @@ MathJax.Hub.Config({
         Contents 
         
       
@@ -207,16 +236,20 @@ MathJax.Hub.Config({
 
    
 
    
 
     
-How can we use the singular value decomposition to find the parameters \( \beta_j \)? More details will come. We first note that a general \( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal matrix \( \hat{\Sigma} \) of dimensionality \( n\times n \) and two orthognal matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality \( m \times n \) and the last dimensionality \( n\times n \). We have then
-$$
-\hat{A} = \hat{U}\hat{\Sigma}\hat{V}
-$$
-
    
-
    
-
 
 
    
-Add codes and discuss this in connection with lasso and ridge, show example where the standard inversion of a matrix fails and where SVD comes to rescue
+A general
+\( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal
+matrix \( \hat{D} \) of dimensionality \( n\times n \) and two orthognal
+matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality
+\( m \times m \) and the last dimensionality \( n\times n \). 
+We have then 
+$$ 
+\hat{A} = \hat{U}\hat{D}\hat{V}^T 
+$$
+
    + +

    @@ -236,6 +269,15 @@ Add codes and discuss this in connection with lasso and ridge, show example wher

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    • diff --git a/doc/pub/Regression/html/._Regression-bs036.html b/doc/pub/Regression/html/._Regression-bs036.html index b5551cea1..a2a01b043 100644 --- a/doc/pub/Regression/html/._Regression-bs036.html +++ b/doc/pub/Regression/html/._Regression-bs036.html @@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source @@ -152,42 +173,50 @@ MathJax.Hub.Config({ Contents @@ -203,11 +232,84 @@ MathJax.Hub.Config({ -

    Lasso and Ridge regression

    +

    Code examples for Ridge and Lasso Regression

    -Discuss the mathematics here + +

    import matplotlib.pyplot as plt
+import numpy as np
+from sklearn import linear_model
+from sklearn.linear_model import LinearRegression
+from sklearn.metrics import mean_squared_error, r2_score
+
+#creating data with random noise
+x=np.arange(50)
+
+delta=np.random.uniform(-2.5,2.5, size=(50))
+np.random.shuffle(delta)
+y =0.5*x+5+delta
+
+#arranging data into 2x50 matrix
+a=np.array(x) #inputs
+b=np.array(y) #outputs
+
+#Split into training and test
+X_train=a[:37, np.newaxis]
+X_test=a[37:, np.newaxis]
+y_train=b[:37]
+y_test=b[37:]
+
+print ("X_train: ", X_train.shape)
+print ("y_train: ", y_train.shape)
+print ("X_test: ", X_test.shape)
+print ("y_test: ", y_test.shape)
+
+print ("------------------------------------")
+
+print ("Ordinary Least Squares")
+#Add Ordinary Least Squares fit
+reg=LinearRegression()
+reg.fit(X_train, y_train)
+pred=reg.predict(X_test)
+print ("Prediction Shape: ", pred.shape)
+
+print('Coefficients: \n', reg.coef_)
+# The mean squared error
+print("Mean squared error: %.2f"
+      % mean_squared_error(y_test, pred))
+# Explained variance score: 1 is perfect prediction
+print('Variance score: %.2f' % r2_score(y_test, pred))
+
+#plot
+plt.scatter(X_test,y_test,color='green', label="Training Data")
+plt.plot(X_test, pred, color='black', label="Fit Line")
+plt.legend()
+plt.show()
+
+print ("------------------------------------")
+
+print ("Ridge Regression")
+
+ridge=linear_model.RidgeCV(alphas=[0.1,1.0,10.0])
+ridge.fit(X_train,y_train)
+print ("Ridge Coefficient: ",ridge.coef_)
+print ("Ridge Intercept: ", ridge.intercept_)
+#Look into graphing with Ridge fit
+
+print ("------------------------------------")
+
+print ("Lasso")
+lasso=linear_model.Lasso(alpha=0.1)
+lasso.fit(X_train,y_train)
+predl=lasso.predict(X_test)
+print("Lasso Coefficient: ", lasso.coef_)
+print("Lasso Intercept: ", lasso.intercept_)
+plt.scatter(X_test,y_test,color='green', label="Training Data")
+plt.plot(X_test, predl, color='blue', label="Lasso")
+plt.legend()
+plt.show()
+

    @@ -225,6 +327,14 @@ Discuss the mathematics here

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    • diff --git a/doc/pub/Regression/html/._Regression-bs037.html b/doc/pub/Regression/html/._Regression-bs037.html index 1a3204fc2..d7790e7d6 100644 --- a/doc/pub/Regression/html/._Regression-bs037.html +++ b/doc/pub/Regression/html/._Regression-bs037.html @@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source @@ -152,42 +173,50 @@ MathJax.Hub.Config({ Contents @@ -203,14 +232,55 @@ MathJax.Hub.Config({ -

    Logistic regression

    -Add discussion about classification versus regression, show examples of more than two cases and why regression is not the best approach. Motivate for k-nearest neighbors +

    From standard regression to Ridge regressions

    -Add examples on classification problems +One of the typical problems we encounter with linear regression, in particular +when the matrix \( \hat{X} \) (our so-called design matrix) is high-dimensional, +are problems with near singular or singular matrices. The column vectors of \( \hat{X} \) +may be linearly dependent, normally referred to as super-collinearity. +This means that the matrix may be rank deficient and it is basically impossible to +to model the data using linear regression. As an example, consider the matrix +$$ +\begin{align*} +\mathbf{X} & = \left[ +\begin{array}{rrr} +1 & -1 & 2 +\\ +1 & 0 & 1 +\\ +1 & 2 & -1 +\\ +1 & 1 & 0 +\end{array} \right] +\end{align*} +$$

    +The columns of \( \hat{X} \) are linearly dependent. We se this easily since the +the first column is the row-wise sum of the other two columns. The rank (more correct, +the column rank) of a matrix is the dimension of the space spanned by the +column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number +of linearly independent columns. In this particular case the matrix has rank 2. +

    +Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies +that the inverse of the matrix \( \hat{X}^T\hat{x} \) (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this +$$ +\begin{align*} +\hat{X} & = \left[ +\begin{array}{rr} +1 & -1 +\\ +1 & -1 +\end{array} \right]. +\end{align*} +$$ + +We see easily that \( \mbox{det}(\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0 \). Hence, \( \mathbf{X} \) is singular and its inverse is undefined. +This is equivalent to saying that the matrix \( \hat{X} \) has at least an eigenvalue which is zero. + +

    diff --git a/doc/pub/Regression/html/._Regression-bs038.html b/doc/pub/Regression/html/._Regression-bs038.html new file mode 100644 index 000000000..cca079168 --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs038.html @@ -0,0 +1,307 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    + + +
    + +

     

     

     

    + + + + +

    Fixing the singularity

    + +

    +If our design matrix \( \hat{X} \) which enters the linear regression problem +$$ +\begin{align} +\hat{\beta} & = (\hat{X}^{T} \hat{X})^{-1} \hat{X}^{T} \hat{y}, +\tag{1} +\end{align} +$$ + +has linearly dependent column vectors, we will not be able to compute the inverse +of \( \hat{X}^T\hat{X} \) and we cannot find the parameters (estimators) \( \beta_i \). +The estimators are only well-defined if \( (\hat{X}^{T}\hat{X})^{-1} \) exits. +This is more likely to happen when the matrix \( \hat{X} \) is high-dimensional. In this case it is likely to encounter a situation where +the regression parameters \( \beta_i \) cannot be estimated. + +

    +The ad hoc approach which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change +$$ +\hat{X}^{T} \hat{X} \rightarrow \hat{X}^{T} \hat{X}+\lambda \hat{I}, +$$ + +where \( \hat{I} \) is the identity matrix. + +

    +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/._Regression-bs039.html b/doc/pub/Regression/html/._Regression-bs039.html new file mode 100644 index 000000000..aa8c96dc3 --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs039.html @@ -0,0 +1,392 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    + + +
    + +

     

     

     

    + + + + +

    A second-order polynomial with Ridge and Lasso

    +

    + + +

    import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import Ridge
+from sklearn.metrics import r2_score
+
+np.random.seed(4155)
+
+n_samples = 100
+
+x = np.random.rand(n_samples,1)
+y = 5*x*x + 0.1*np.random.rand(n_samples,1)
+
+# Centering  x and y.
+x_ = x - np.mean(x)
+y_ = y - np.mean(y) # beta_0 = mean(y)
+
+X = np.c_[np.ones((n_samples,1)), x, x**2]
+X_ = np.c_[x_, x_**2]
+
+
+### 1.
+lmb_values = [1e-4, 1e-3, 1e-2, 10, 1e2, 1e4]
+num_values = len(lmb_values)
+
+## Ridge-regression of centered and not centered data
+beta_ridge = np.zeros((3,num_values))
+beta_ridge_centered = np.zeros((3,num_values))
+
+I3 = np.eye(3)
+I2 = np.eye(2)
+
+for i,lmb in enumerate(lmb_values):
+    beta_ridge[:,i] = (np.linalg.inv( X.T @ X + lmb*I3) @ X.T @ y).flatten()
+    beta_ridge_centered[1:,i] = (np.linalg.inv( X_.T @ X_ + lmb*I2) @ X_.T @ y_).flatten()
+
+# sett beta_0 = np.mean(y)
+beta_ridge_centered[0,:] = np.mean(y)
+
+## OLS (ordinary least squares) solution 
+beta_ls = np.linalg.inv( X.T @ X ) @ X.T @ y
+
+## Evaluate the models
+pred_ls = X @ beta_ls
+pred_ridge =  X @ beta_ridge
+pred_ridge_centered =  X_ @ beta_ridge_centered[1:] + beta_ridge_centered[0,:]
+
+## Plot the results
+
+# Sorting
+sort_ind = np.argsort(x[:,0])
+
+x_plot = x[sort_ind,0]
+x_centered_plot = x_[sort_ind,0]
+
+pred_ls_plot = pred_ls[sort_ind,0]
+pred_ridge_plot = pred_ridge[sort_ind,:]
+pred_ridge_centered_plot = pred_ridge_centered[sort_ind,:]
+
+# Plott not centered
+plt.plot(x_plot,pred_ls_plot,label='ls')
+
+for i in range(num_values):
+    plt.plot(x_plot,pred_ridge_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x,y,'ro')
+
+plt.title('linear regression on un-centered data')
+plt.legend()
+
+# Plott centered
+plt.figure()
+
+for i in range(num_values):
+    plt.plot(x_centered_plot,pred_ridge_centered_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x_,y,'ro')
+
+plt.title('linear regression on centered data')
+plt.legend()
+
+
+# 2.
+
+pred_ridge_scikit =  np.zeros((n_samples,num_values))
+for i,lmb in enumerate(lmb_values):
+    pred_ridge_scikit[:,i] = (Ridge(alpha=lmb,fit_intercept=False).fit(X,y).predict(X)).flatten() # fit_intercept=False fordi bias er allerede i X
+
+plt.figure()
+
+plt.plot(x_plot,pred_ls_plot,label='ls')
+
+for i in range(num_values):
+    plt.plot(x_plot,pred_ridge_scikit[sort_ind,i],label='scikit-ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x,y,'ro')
+plt.legend()
+plt.title('linear regression using scikit')
+
+plt.show()
+
+### R2-score of the results
+for i in range(num_values):
+    print('lambda = %g'%lmb_values[i])
+    print('r2 for scikit: %g'%r2_score(y,pred_ridge_scikit[:,i]))
+    print('r2 for own code, not centered: %g'%r2_score(y,pred_ridge[:,i]))
+    print('r2 for own, centered: %g\n'%r2_score(y,pred_ridge_centered[:,i]))
+
    +

    +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/._Regression-bs040.html b/doc/pub/Regression/html/._Regression-bs040.html new file mode 100644 index 000000000..e62c176d7 --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs040.html @@ -0,0 +1,304 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    + + +
    + +

     

     

     

    + + + + +

    Fitting vs. predicting when data is in the model class

    + +

    +We start by considering the case +\( f(x)=2x \). + +

    +Then the data is clearly generated by a model that is contained within +all three model classes we are using to make predictions (linear +models, third order polynomials, and tenth order polynomials). + +

    +Run the code for the following cases: + +

      +
    1. For \( f(x)=2x \) , \( Ntrain=10 \) and \( \sigma =0 \) (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when \( x \in [0,1] \) . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?
      2. +
    2. Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?
      3. +
    3. Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example \( x \in [0,1.2] \) ) using the code below. How well do you do on points in the range of x where you trained the model? How about points outside the original training data set?
      4. +
    4. Repeat the above for \( f(x)=2x \) , \( Ntrain=10 \) , and \( \sigma=1 \) . What changes?
      5. +
    + +Repeat the exercises above for \( f(x)=2x \) , \( Ntrain=100 \) , and \( \sigma=1 \) . What changes? +Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well. + +

    +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/._Regression-bs041.html b/doc/pub/Regression/html/._Regression-bs041.html new file mode 100644 index 000000000..0449005ac --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs041.html @@ -0,0 +1,291 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    + + +
    + +

     

     

     

    + + + + +

    Fitting versus predicting when data is not in the model class

    + +

    +Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider \( f(x)=2x-10x^5+15x^{10} \) . Notice that the for linear and third-order polynomial the true model \( f(x) \) is not contained in model class. + +

      +
    1. Do better fits lead to better predictions?
      2. +
    2. What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points \( Ntrain \) and \( \sigma \)?
      3. +
    + +Summarize what you think you learned about the relationship of knowing the true model class and predictive power. + +

    +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/._Regression-bs042.html b/doc/pub/Regression/html/._Regression-bs042.html new file mode 100644 index 000000000..ff3f3130a --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs042.html @@ -0,0 +1,365 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    + + +
    + +

     

     

     

    + + + + +

    The code

    + +

    + + +

    import numpy as np
+import sklearn as sk
+from sklearn import datasets, linear_model
+from sklearn.preprocessing import PolynomialFeatures
+
+import matplotlib as mpl
+from matplotlib import pyplot as plt
+
+%matplotlib notebook
+
+# The Training Data
+
+N_train=100
+
+sigma_train=1;
+
+# Train on integers
+x=np.linspace(0.05,0.95,N_train)
+# Draw random noise
+s = sigma_train*np.random.randn(N_train)
+
+#linear
+y=2*x+s
+
+#Tenth Order
+#y=2*x-10*x**5+15*x**10+s
+
+p1=plt.plot(x,y, "o",ms=15, label='Training')
+
+#Linear Regression
+# Create linear regression object
+clf = linear_model.LinearRegression()
+
+# Train the model using the training sets
+clf.fit(x[:, np.newaxis], y)
+# The coefficients
+
+xplot=np.linspace(0.02,0.98,200)
+linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')
+
+#Polynomial Regression
+
+
+poly3 = PolynomialFeatures(degree=3)
+X = poly3.fit_transform(x[:,np.newaxis])
+clf3 = linear_model.LinearRegression()
+clf3.fit(X,y)
+
+
+Xplot=poly3.fit_transform(xplot[:,np.newaxis])
+poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')
+
+
+
+#poly5 = PolynomialFeatures(degree=5)
+#X = poly5.fit_transform(x[:,np.newaxis])
+#clf5 = linear_model.LinearRegression()
+#clf5.fit(X,y)
+
+#Xplot=poly5.fit_transform(xplot[:,np.newaxis])
+#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)
+
+poly10 = PolynomialFeatures(degree=10)
+X = poly10.fit_transform(x[:,np.newaxis])
+clf10 = linear_model.LinearRegression()
+clf10.fit(X,y)
+
+Xplot=poly10.fit_transform(xplot[:,np.newaxis])
+poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')
+
+axes = plt.gca()
+axes.set_ylim([-7,7])
+
+handles, labels=axes.get_legend_handles_labels()
+plt.legend(handles,labels, loc='lower center')
+plt.xlabel("$x$")
+plt.ylabel("$y$")
+Title="$N=$"+str(N_train)+", $\sigma=$"+str(sigma_train)
+plt.title(Title+" (train)")
+plt.tight_layout()
+plt.show()
+
    +

    +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/._Regression-bs043.html b/doc/pub/Regression/html/._Regression-bs043.html new file mode 100644 index 000000000..b45774529 --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs043.html @@ -0,0 +1,327 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    + + +
    + +

     

     

     

    + + + + +

    Generating test data

    +

    + + +

    # Generate Test Data
+
+#Number of test data
+N_test=20
+
+sigma_test=sigma_train
+
+max_x=1.2
+x_test=max_x*np.random.random(N_test)
+# Draw random noise
+s_test = sigma_test*np.random.randn(N_test)
+
+#Linear
+y_test=2*x_test+s_test
+#Tenth order
+#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test
+
+#Make design matrices for prediction
+x_plot=np.linspace(0,max_x, 200)
+X3 = poly3.fit_transform(x_plot[:,np.newaxis])
+X10 = poly10.fit_transform(x_plot[:,np.newaxis])
+
+%matplotlib notebook
+
+fig = plt.figure() 
+p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')
+p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')
+p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')
+p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')
+
+
+plt.legend(loc=2)
+plt.xlabel('$x$')
+plt.ylabel('$y$')
+plt.legend(loc='best')
+plt.title(Title+" (pred.)")
+plt.tight_layout()
+plt.show()
+
+#Linear Filename
+#filename_test=Title+"pred-linear.pdf"
+#Tenth Order Filename
+#filename_test=Title+"pred-o10.pdf"
+#plt.savefig(filename_test)
+#plt.ylim((-6,12))
+
    +

    +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/._Regression-bs044.html b/doc/pub/Regression/html/._Regression-bs044.html new file mode 100644 index 000000000..1ad5ddece --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs044.html @@ -0,0 +1,278 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    + + +
    + +

     

     

     

    + + + + +

    Lasso regression

    + +

    +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/._Regression-bs045.html b/doc/pub/Regression/html/._Regression-bs045.html new file mode 100644 index 000000000..59fd87d05 --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs045.html @@ -0,0 +1,275 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    + + +
    + +

     

     

     

    + + + + +

    Logistic regression

    + +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/Regression-bs.html b/doc/pub/Regression/html/Regression-bs.html index 9560a3323..c92c12f8a 100644 --- a/doc/pub/Regression/html/Regression-bs.html +++ b/doc/pub/Regression/html/Regression-bs.html @@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source @@ -152,42 +173,50 @@ MathJax.Hub.Config({ Contents @@ -222,7 +251,7 @@ MathJax.Hub.Config({
    [2] Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University

    -

    Aug 24, 2018

    +

    Sep 6, 2018


    @@ -246,7 +275,7 @@ MathJax.Hub.Config({

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    • diff --git a/doc/pub/Regression/html/Regression-reveal.html b/doc/pub/Regression/html/Regression-reveal.html index d5aaa438c..13e0c45e3 100644 --- a/doc/pub/Regression/html/Regression-reveal.html +++ b/doc/pub/Regression/html/Regression-reveal.html @@ -148,7 +148,7 @@ MathJax.Hub.Config({
    [2] Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University

     
    -

    Aug 24, 2018

    +

    Sep 6, 2018


    @@ -182,7 +182,35 @@ A regression model aims at finding a likelihood function \( p(y\vert \hat{x}) \)

    -

    General linear models

    +

    Regression analysis, overarching aims II

    +
    + +

    +Consider an experiment in which \( p \) characteristics of \( n \) samples are +measured. The data from this experiment are denoted \( \mathbf{X} \), with +\( \mathbf{X} \) as above. The matrix \( \mathbf{X} \) is called the design +matrix. Additional information of the samples is available in the +form of \( \mathbf{Y} \) (also as above). The variable \( \mathbf{Y} \) is +generally referred to as the response variable. The aim of +regression analysis is to explain \( \mathbf{Y} \) in terms of +\( \mathbf{X} \) through a functional relationship like \( Y_i = +f(\mathbf{X}_{i,\ast}) \). When no prior knowledge on the form of +\( f(\cdot) \) is available, it is common to assume a linear relationship +between \( \mathbf{X} \) and \( \mathbf{Y} \). This assumption gives rise to +the linear regression model where \( \beta = (\beta_1, \ldots, +\beta_p)^{\top} \) is the regression parameter. The parameter +\( \beta_j \), \( j=1, \ldots, p \), represents the effect size of covariate +\( j \) on the response. That is, for each unit change in covariate \( j \) +(while keeping the other covariates fixed) the observed change in the +response is equal to \( \beta_j \). + + +

    +
    + + +
    +

    General linear models

    @@ -204,7 +232,7 @@ where \( \epsilon_i \) is the error in our approximation.

    -

    Rewriting the fitting procedure as a linear algebra problem

    +

    Rewriting the fitting procedure as a linear algebra problem

    @@ -225,7 +253,7 @@ $$

    -

    Rewriting the fitting procedure as a linear algebra problem, follows

    +

    Rewriting the fitting procedure as a linear algebra problem, follows

    @@ -275,7 +303,7 @@ $$

    -

    Generalizing the fitting procedure as a linear algebra problem

    +

    Generalizing the fitting procedure as a linear algebra problem

    @@ -299,7 +327,7 @@ $$

    -

    Generalizing the fitting procedure as a linear algebra problem

    +

    Generalizing the fitting procedure as a linear algebra problem

    @@ -330,7 +358,7 @@ The left-hand side of this equation forms know. Our error vector \( \hat{\epsilo

    -

    Optimizing our parameters

    +

    Optimizing our parameters

    @@ -353,7 +381,7 @@ $$

    -

    Optimizing our parameters, more details

    +

    Optimizing our parameters, more details

    @@ -382,7 +410,7 @@ $$

    -

    Interpretations and optimizing our parameters

    +

    Interpretations and optimizing our parameters

    @@ -430,7 +458,7 @@ $$

    -

    Interpretations and optimizing our parameters

    +

    Interpretations and optimizing our parameters

    @@ -461,7 +489,7 @@ $$

    -

    Interpretations and optimizing our parameters

    +

    Interpretations and optimizing our parameters

    @@ -494,7 +522,7 @@ meaning that the solution for \( \hat{\beta} \) is the one which minimizes the r

    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -517,7 +545,7 @@ where the matrix \( \hat{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as

    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -548,7 +576,7 @@ where we have defined the matrix \( \hat{A} =\hat{X}/\hat{\Sigma} \) with matrix

    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -577,7 +605,7 @@ $$

    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -613,7 +641,7 @@ $$

    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -642,7 +670,7 @@ $$

    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -698,7 +726,7 @@ This approach (different linear and non-linear regression) suffers often from bo

    -

    Simple regression model

    +

    Simple regression model

    We are now ready to write our first program which aims at solving the above linear regression equations. We start with data we have produced ourselves, in this case normally distributed random numbers along the \( x \)-axis. These numbers define then the value of a function \( y(x)=4+3x+N(0,1) \). Thereafter we order the \( x \) values and employ our linear regression algorithm to set up the best fit. Here we find it useful to use the numpy function \( c\_ \) arrays where arrays are stacked along their last axis after being upgraded to at least two dimensions with ones post-pended to the shape. The following examples help in understanding what happens

    @@ -738,7 +766,7 @@ We see that, as expected, a linear fit gives a seemingly (from the graph) good r

    -

    Simple regression model, now using scikit-learn

    +

    Simple regression model, now using scikit-learn

    We can repeat the above algorithm using scikit-learn as follows @@ -770,7 +798,7 @@ plt.show()

    -

    Simple linear regression model using scikit-learn

    +

    Simple linear regression model using scikit-learn

    We start with perhaps our simplest possible example, using scikit-learn to perform linear regression analysis on a data set produced by us. @@ -841,7 +869,7 @@ plt.show()

    -

    Simple linear regression model

    +

    Simple linear regression model

    This example serves several aims. It allows us to demonstrate several @@ -865,7 +893,7 @@ where \( x \) is defined as before.

    -

    Less noise

    +

    Less noise

    Does the fit look better? Indeed, by @@ -878,7 +906,7 @@ have not discussed a more rigorous approach to the cost function.

    -

    How to study our fits

    +

    How to study our fits

    We need more rigorous criteria in defining whether we have succeeded or @@ -902,7 +930,7 @@ dimensionless.

    -

    Minimizing the cost function

    +

    Minimizing the cost function

    Minimizing the cost function is a central aspect of @@ -923,7 +951,7 @@ the \( \chi^2 \) function becomes smaller.

    -

    Relative error

    +

    Relative error

    There are many ways to define the cost function. A simpler approach is to look at the relative difference between the training data and the predicted data, that is we define @@ -965,7 +993,7 @@ relative error.

    -

    The richness of scikit-learn

    +

    The richness of scikit-learn

    As mentioned above, scikit-learn has an impressive functionality. @@ -1011,7 +1039,7 @@ plt.show()

    -

    Functions in scikit-learn

    +

    Functions in scikit-learn

    The function coef gives us the parameter \( \beta \) of our fit while intercept yields @@ -1030,7 +1058,7 @@ this function as being similar to the \( \chi^2 \) function defined above.

    -

    Other functions in scikit-learn

    +

    Other functions in scikit-learn

    The r2score function computes \( R^2 \), the coefficient of @@ -1058,7 +1086,7 @@ $$

    -

    The mean absolute error and other functions in scikit-learn

    +

    The mean absolute error and other functions in scikit-learn

    Another quantity will meet again in our discussions of regression analysis is @@ -1087,12 +1115,13 @@ years etc.

    -

    Cubic polynomial in scikit-learn

    +

    Cubic polynomial in scikit-learn

    We will discuss in more detail these and other functions in the various lectures. We conclude this part with another example. Instead of -a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn. +a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn. +Add description of the various python commands.

    @@ -1129,84 +1158,12 @@ plt.show() print (error(y))

    -Similarly, using R, we can perform similar studies. -(more details on R will be inserted later). +Using R, we can perform similar studies.

    -

    Simple regression model with gradient descent

    -Add info about the equations, play around with different learning rates -

    - - -

    # Importing various packages
-from math import exp, sqrt
-from random import random, seed
-import numpy as np
-import matplotlib.pyplot as plt
-
-x = 2*np.random.rand(100,1)
-y = 4+3*x+np.random.randn(100,1)
-
-xb = np.c_[np.ones((100,1)), x]
-theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-print(theta_linreg)
-theta = np.random.randn(2,1)
-
-eta = 0.1
-Niterations = 1000
-m = 100
-
-for iter in range(Niterations):
-    gradients = 2.0/m*xb.T.dot(xb.dot(theta)-y)
-    theta -= eta*gradients
-
-print(theta)
-xnew = np.array([[0],[2]])
-xbnew = np.c_[np.ones((2,1)), xnew]
-ypredict = xbnew.dot(theta)
-ypredict2 = xbnew.dot(theta_linreg)
-plt.plot(xnew, ypredict, "r-")
-plt.plot(xnew, ypredict2, "b-")
-plt.plot(x, y ,'ro')
-plt.axis([0,2.0,0, 15.0])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$y$')
-plt.title(r'Random numbers ')
-plt.show()
-
    -
    - - -
    -

    Simple regression model with stochastic gradient descent

    -Add info about the equations, play around with different learning rates -

    - - -

    # Importing various packages
-from math import exp, sqrt
-from random import random, seed
-import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.linear_model import SGDRegressor
-
-x = 2*np.random.rand(100,1)
-y = 4+3*x+np.random.randn(100,1)
-
-xb = np.c_[np.ones((100,1)), x]
-theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-print(theta_linreg)
-sgdreg = SGDRegressor(n_iter = 50, penalty=None, eta0=0.1)
-sgdreg.fit(x,y.ravel())
-print(sgdreg.intercept_, sgdreg.coef_)
-
    -
    - - -
    -

    Polynomial Regression

    +

    Polynomial Regression

    @@ -1238,7 +1195,94 @@ plt.show()

    -

    Ridge and Lasso Regression

    +

    Linking the regression analysis with a statistical interpretation

    + +

    +Before we proceed, and to link with our discussions of Bayesian statistics to come, it is useful the derive the standard regression analysis equations using a statistical interpretation. This allows us also to derive quantities like the variance and other expectation values in a rather straightforward way. + +

    +It is assumed that \( \varepsilon_i +\sim \mathcal{N}(0, \sigma^2) \) and the \( \varepsilon_{i} \) are +independent, i.e.: +

     
    +$$ +\begin{align*} +\mbox{Cov}(\varepsilon_{i_1}, +\varepsilon_{i_2}) & = \left\{ \begin{array}{lcc} \sigma^2 & \mbox{if} +& i_1 = i_2, \\ 0 & \mbox{if} & i_1 \not= i_2. \end{array} \right. +\end{align*} +$$ +

     
    + +The randomness of \( \varepsilon_i \) implies that +\( \mathbf{Y}_i \) is also a random variable. In particular, +\( \mathbf{Y}_i \) is normally distributed, because \( \varepsilon_i \sim +\mathcal{N}(0, \sigma^2) \) and \( \mathbf{X}_{i,\ast} \, \beta \) is a +non-random scalar. To specify the parameters of the distribution of +\( \mathbf{Y}_i \) we need to calculate its first two moments. +

    + + +
    +

    Expectation value and variance

    + +

    +Its expectation equals: +

     
    +$$ +\begin{align*} +\mathbb{E}(Y_i) & = +\mathbb{E}(\mathbf{X}_{i, \ast} \, \beta) + \mathbb{E}(\varepsilon_i) +\, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \beta, +\end{align*} +$$ +

     
    + +while +its variance is +

     
    +$$ +\begin{align*} \mbox{Var}(Y_i) & = \mathbb{E} \{ [Y_i +- \mathbb{E}(Y_i)]^2 \} \, \, \, = \, \, \, \mathbb{E} ( Y_i^2 ) - +[\mathbb{E}(Y_i)]^2 \\ & = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, +\beta + \varepsilon_i )^2] - ( \mathbf{X}_{i, \ast} \, \beta)^2 \\ & += \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2 \varepsilon_i +\mathbf{X}_{i, \ast} \, \beta + \varepsilon_i^2 ] - ( \mathbf{X}_{i, +\ast} \, \beta)^2 \\ & = ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2 +\mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \beta + +\mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \beta)^2 +\\ & = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \, +\mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2. +\end{align*} +$$ +

     
    + +Hence, \( Y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \beta, \sigma^2) \). +

    + + +
    +

    The singular value decompostion

    +
    + +

    +A general +\( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal +matrix \( \hat{D} \) of dimensionality \( n\times n \) and two orthognal +matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality +\( m \times m \) and the last dimensionality \( n\times n \). +We have then +

     
    +$$ +\hat{A} = \hat{U}\hat{D}\hat{V}^T +$$ +

     
    +

    +
    + + +
    +

    Code examples for Ridge and Lasso Regression

    @@ -1320,37 +1364,403 @@ plt.show()

    -

    The singular value decompostion

    -
    - -

    -How can we use the singular value decomposition to find the parameters \( \beta_j \)? More details will come. We first note that a general \( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal matrix \( \hat{\Sigma} \) of dimensionality \( n\times n \) and two orthognal matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality \( m \times n \) and the last dimensionality \( n\times n \). We have then -

     
    -$$ -\hat{A} = \hat{U}\hat{\Sigma}\hat{V} -$$ -

     
    -

    +

    From standard regression to Ridge regressions

    -Add codes and discuss this in connection with lasso and ridge, show example where the standard inversion of a matrix fails and where SVD comes to rescue +One of the typical problems we encounter with linear regression, in particular +when the matrix \( \hat{X} \) (our so-called design matrix) is high-dimensional, +are problems with near singular or singular matrices. The column vectors of \( \hat{X} \) +may be linearly dependent, normally referred to as super-collinearity. +This means that the matrix may be rank deficient and it is basically impossible to +to model the data using linear regression. As an example, consider the matrix +

     
    +$$ +\begin{align*} +\mathbf{X} & = \left[ +\begin{array}{rrr} +1 & -1 & 2 +\\ +1 & 0 & 1 +\\ +1 & 2 & -1 +\\ +1 & 1 & 0 +\end{array} \right] +\end{align*} +$$ +

     
    + +

    +The columns of \( \hat{X} \) are linearly dependent. We se this easily since the +the first column is the row-wise sum of the other two columns. The rank (more correct, +the column rank) of a matrix is the dimension of the space spanned by the +column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number +of linearly independent columns. In this particular case the matrix has rank 2. + +

    +Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies +that the inverse of the matrix \( \hat{X}^T\hat{x} \) (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this +

     
    +$$ +\begin{align*} +\hat{X} & = \left[ +\begin{array}{rr} +1 & -1 +\\ +1 & -1 +\end{array} \right]. +\end{align*} +$$ +

     
    + +We see easily that \( \mbox{det}(\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0 \). Hence, \( \mathbf{X} \) is singular and its inverse is undefined. +This is equivalent to saying that the matrix \( \hat{X} \) has at least an eigenvalue which is zero.

    -

    Lasso and Ridge regression

    +

    Fixing the singularity

    -Discuss the mathematics here +If our design matrix \( \hat{X} \) which enters the linear regression problem +

     
    +$$ +\begin{align} +\hat{\beta} & = (\hat{X}^{T} \hat{X})^{-1} \hat{X}^{T} \hat{y}, +\tag{1} +\end{align} +$$ +

     
    + +has linearly dependent column vectors, we will not be able to compute the inverse +of \( \hat{X}^T\hat{X} \) and we cannot find the parameters (estimators) \( \beta_i \). +The estimators are only well-defined if \( (\hat{X}^{T}\hat{X})^{-1} \) exits. +This is more likely to happen when the matrix \( \hat{X} \) is high-dimensional. In this case it is likely to encounter a situation where +the regression parameters \( \beta_i \) cannot be estimated. + +

    +The ad hoc approach which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change +

     
    +$$ +\hat{X}^{T} \hat{X} \rightarrow \hat{X}^{T} \hat{X}+\lambda \hat{I}, +$$ +

     
    + +where \( \hat{I} \) is the identity matrix.

    -

    Logistic regression

    -Add discussion about classification versus regression, show examples of more than two cases and why regression is not the best approach. Motivate for k-nearest neighbors +

    A second-order polynomial with Ridge and Lasso

    +

    + + +

    import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import Ridge
+from sklearn.metrics import r2_score
+
+np.random.seed(4155)
+
+n_samples = 100
+
+x = np.random.rand(n_samples,1)
+y = 5*x*x + 0.1*np.random.rand(n_samples,1)
+
+# Centering  x and y.
+x_ = x - np.mean(x)
+y_ = y - np.mean(y) # beta_0 = mean(y)
+
+X = np.c_[np.ones((n_samples,1)), x, x**2]
+X_ = np.c_[x_, x_**2]
+
+
+### 1.
+lmb_values = [1e-4, 1e-3, 1e-2, 10, 1e2, 1e4]
+num_values = len(lmb_values)
+
+## Ridge-regression of centered and not centered data
+beta_ridge = np.zeros((3,num_values))
+beta_ridge_centered = np.zeros((3,num_values))
+
+I3 = np.eye(3)
+I2 = np.eye(2)
+
+for i,lmb in enumerate(lmb_values):
+    beta_ridge[:,i] = (np.linalg.inv( X.T @ X + lmb*I3) @ X.T @ y).flatten()
+    beta_ridge_centered[1:,i] = (np.linalg.inv( X_.T @ X_ + lmb*I2) @ X_.T @ y_).flatten()
+
+# sett beta_0 = np.mean(y)
+beta_ridge_centered[0,:] = np.mean(y)
+
+## OLS (ordinary least squares) solution 
+beta_ls = np.linalg.inv( X.T @ X ) @ X.T @ y
+
+## Evaluate the models
+pred_ls = X @ beta_ls
+pred_ridge =  X @ beta_ridge
+pred_ridge_centered =  X_ @ beta_ridge_centered[1:] + beta_ridge_centered[0,:]
+
+## Plot the results
+
+# Sorting
+sort_ind = np.argsort(x[:,0])
+
+x_plot = x[sort_ind,0]
+x_centered_plot = x_[sort_ind,0]
+
+pred_ls_plot = pred_ls[sort_ind,0]
+pred_ridge_plot = pred_ridge[sort_ind,:]
+pred_ridge_centered_plot = pred_ridge_centered[sort_ind,:]
+
+# Plott not centered
+plt.plot(x_plot,pred_ls_plot,label='ls')
+
+for i in range(num_values):
+    plt.plot(x_plot,pred_ridge_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x,y,'ro')
+
+plt.title('linear regression on un-centered data')
+plt.legend()
+
+# Plott centered
+plt.figure()
+
+for i in range(num_values):
+    plt.plot(x_centered_plot,pred_ridge_centered_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x_,y,'ro')
+
+plt.title('linear regression on centered data')
+plt.legend()
+
+
+# 2.
+
+pred_ridge_scikit =  np.zeros((n_samples,num_values))
+for i,lmb in enumerate(lmb_values):
+    pred_ridge_scikit[:,i] = (Ridge(alpha=lmb,fit_intercept=False).fit(X,y).predict(X)).flatten() # fit_intercept=False fordi bias er allerede i X
+
+plt.figure()
+
+plt.plot(x_plot,pred_ls_plot,label='ls')
+
+for i in range(num_values):
+    plt.plot(x_plot,pred_ridge_scikit[sort_ind,i],label='scikit-ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x,y,'ro')
+plt.legend()
+plt.title('linear regression using scikit')
+
+plt.show()
+
+### R2-score of the results
+for i in range(num_values):
+    print('lambda = %g'%lmb_values[i])
+    print('r2 for scikit: %g'%r2_score(y,pred_ridge_scikit[:,i]))
+    print('r2 for own code, not centered: %g'%r2_score(y,pred_ridge[:,i]))
+    print('r2 for own, centered: %g\n'%r2_score(y,pred_ridge_centered[:,i]))
+
    +
    + + +
    +

    Fitting vs. predicting when data is in the model class

    -Add examples on classification problems +We start by considering the case +\( f(x)=2x \). + +

    +Then the data is clearly generated by a model that is contained within +all three model classes we are using to make predictions (linear +models, third order polynomials, and tenth order polynomials). + +

    +Run the code for the following cases: + +

      +

    1. For \( f(x)=2x \) , \( Ntrain=10 \) and \( \sigma =0 \) (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when \( x \in [0,1] \) . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?
      2. +

    2. Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?
      3. +

    3. Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example \( x \in [0,1.2] \) ) using the code below. How well do you do on points in the range of x where you trained the model? How about points outside the original training data set?
      4. +

    4. Repeat the above for \( f(x)=2x \) , \( Ntrain=10 \) , and \( \sigma=1 \) . What changes?
      5. +
    +

    + +Repeat the exercises above for \( f(x)=2x \) , \( Ntrain=100 \) , and \( \sigma=1 \) . What changes? +Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well. +

    + + +
    +

    Fitting versus predicting when data is not in the model class

    + +

    +Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider \( f(x)=2x-10x^5+15x^{10} \) . Notice that the for linear and third-order polynomial the true model \( f(x) \) is not contained in model class. + +

      +

    1. Do better fits lead to better predictions?
      2. +

    2. What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points \( Ntrain \) and \( \sigma \)?
      3. +
    +

    + +Summarize what you think you learned about the relationship of knowing the true model class and predictive power. +

    + + +
    +

    The code

    + +

    + + +

    import numpy as np
+import sklearn as sk
+from sklearn import datasets, linear_model
+from sklearn.preprocessing import PolynomialFeatures
+
+import matplotlib as mpl
+from matplotlib import pyplot as plt
+
+%matplotlib notebook
+
+# The Training Data
+
+N_train=100
+
+sigma_train=1;
+
+# Train on integers
+x=np.linspace(0.05,0.95,N_train)
+# Draw random noise
+s = sigma_train*np.random.randn(N_train)
+
+#linear
+y=2*x+s
+
+#Tenth Order
+#y=2*x-10*x**5+15*x**10+s
+
+p1=plt.plot(x,y, "o",ms=15, label='Training')
+
+#Linear Regression
+# Create linear regression object
+clf = linear_model.LinearRegression()
+
+# Train the model using the training sets
+clf.fit(x[:, np.newaxis], y)
+# The coefficients
+
+xplot=np.linspace(0.02,0.98,200)
+linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')
+
+#Polynomial Regression
+
+
+poly3 = PolynomialFeatures(degree=3)
+X = poly3.fit_transform(x[:,np.newaxis])
+clf3 = linear_model.LinearRegression()
+clf3.fit(X,y)
+
+
+Xplot=poly3.fit_transform(xplot[:,np.newaxis])
+poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')
+
+
+
+#poly5 = PolynomialFeatures(degree=5)
+#X = poly5.fit_transform(x[:,np.newaxis])
+#clf5 = linear_model.LinearRegression()
+#clf5.fit(X,y)
+
+#Xplot=poly5.fit_transform(xplot[:,np.newaxis])
+#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)
+
+poly10 = PolynomialFeatures(degree=10)
+X = poly10.fit_transform(x[:,np.newaxis])
+clf10 = linear_model.LinearRegression()
+clf10.fit(X,y)
+
+Xplot=poly10.fit_transform(xplot[:,np.newaxis])
+poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')
+
+axes = plt.gca()
+axes.set_ylim([-7,7])
+
+handles, labels=axes.get_legend_handles_labels()
+plt.legend(handles,labels, loc='lower center')
+plt.xlabel("$x$")
+plt.ylabel("$y$")
+Title="$N=$"+str(N_train)+", $\sigma=$"+str(sigma_train)
+plt.title(Title+" (train)")
+plt.tight_layout()
+plt.show()
+
    +
    + + +
    +

    Generating test data

    +

    + + +

    # Generate Test Data
+
+#Number of test data
+N_test=20
+
+sigma_test=sigma_train
+
+max_x=1.2
+x_test=max_x*np.random.random(N_test)
+# Draw random noise
+s_test = sigma_test*np.random.randn(N_test)
+
+#Linear
+y_test=2*x_test+s_test
+#Tenth order
+#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test
+
+#Make design matrices for prediction
+x_plot=np.linspace(0,max_x, 200)
+X3 = poly3.fit_transform(x_plot[:,np.newaxis])
+X10 = poly10.fit_transform(x_plot[:,np.newaxis])
+
+%matplotlib notebook
+
+fig = plt.figure() 
+p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')
+p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')
+p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')
+p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')
+
+
+plt.legend(loc=2)
+plt.xlabel('$x$')
+plt.ylabel('$y$')
+plt.legend(loc='best')
+plt.title(Title+" (pred.)")
+plt.tight_layout()
+plt.show()
+
+#Linear Filename
+#filename_test=Title+"pred-linear.pdf"
+#Tenth Order Filename
+#filename_test=Title+"pred-o10.pdf"
+#plt.savefig(filename_test)
+#plt.ylim((-6,12))
+
    +
    + + +
    +

    Lasso regression

    +
    + + +
    +

    Logistic regression

    diff --git a/doc/pub/Regression/html/Regression-solarized.html b/doc/pub/Regression/html/Regression-solarized.html index 9b2167152..9b2031654 100644 --- a/doc/pub/Regression/html/Regression-solarized.html +++ b/doc/pub/Regression/html/Regression-solarized.html @@ -61,17 +61,14 @@ div { text-align: justify; text-justify: inter-word; } @@ -175,7 +196,7 @@ MathJax.Hub.Config({
    [2] Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University

    -

    Aug 24, 2018

    +

    Sep 6, 2018












    @@ -205,7 +226,38 @@ A regression model aims at finding a likelihood function \( p(y\vert \hat{x}) \)











    -

    General linear models

    +

    Regression analysis, overarching aims II

    +
    + +

    + +

    +Consider an experiment in which \( p \) characteristics of \( n \) samples are +measured. The data from this experiment are denoted \( \mathbf{X} \), with +\( \mathbf{X} \) as above. The matrix \( \mathbf{X} \) is called the design +matrix. Additional information of the samples is available in the +form of \( \mathbf{Y} \) (also as above). The variable \( \mathbf{Y} \) is +generally referred to as the response variable. The aim of +regression analysis is to explain \( \mathbf{Y} \) in terms of +\( \mathbf{X} \) through a functional relationship like \( Y_i = +f(\mathbf{X}_{i,\ast}) \). When no prior knowledge on the form of +\( f(\cdot) \) is available, it is common to assume a linear relationship +between \( \mathbf{X} \) and \( \mathbf{Y} \). This assumption gives rise to +the linear regression model where \( \beta = (\beta_1, \ldots, +\beta_p)^{\top} \) is the regression parameter. The parameter +\( \beta_j \), \( j=1, \ldots, p \), represents the effect size of covariate +\( j \) on the response. That is, for each unit change in covariate \( j \) +(while keeping the other covariates fixed) the observed change in the +response is equal to \( \beta_j \). + + +

    + + +

    +









    + +

    General linear models

    @@ -226,7 +278,7 @@ where \( \epsilon_i \) is the error in our approximation.











    -

    Rewriting the fitting procedure as a linear algebra problem

    +

    Rewriting the fitting procedure as a linear algebra problem

    @@ -246,7 +298,7 @@ $$











    -

    Rewriting the fitting procedure as a linear algebra problem, follows

    +

    Rewriting the fitting procedure as a linear algebra problem, follows

    @@ -287,7 +339,7 @@ $$











    -

    Generalizing the fitting procedure as a linear algebra problem

    +

    Generalizing the fitting procedure as a linear algebra problem

    @@ -310,7 +362,7 @@ $$











    -

    Generalizing the fitting procedure as a linear algebra problem

    +

    Generalizing the fitting procedure as a linear algebra problem

    @@ -338,7 +390,7 @@ The left-hand side of this equation forms know. Our error vector \( \hat{\epsilo











    -

    Optimizing our parameters

    +

    Optimizing our parameters

    @@ -360,7 +412,7 @@ $$











    -

    Optimizing our parameters, more details

    +

    Optimizing our parameters, more details

    @@ -384,7 +436,7 @@ $$











    -

    Interpretations and optimizing our parameters

    +

    Interpretations and optimizing our parameters

    @@ -423,7 +475,7 @@ $$











    -

    Interpretations and optimizing our parameters

    +

    Interpretations and optimizing our parameters

    @@ -449,7 +501,7 @@ $$











    -

    Interpretations and optimizing our parameters

    +

    Interpretations and optimizing our parameters

    @@ -477,7 +529,7 @@ meaning that the solution for \( \hat{\beta} \) is the one which minimizes the r











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -500,7 +552,7 @@ where the matrix \( \hat{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -528,7 +580,7 @@ where we have defined the matrix \( \hat{A} =\hat{X}/\hat{\Sigma} \) with matrix











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -554,7 +606,7 @@ $$











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -585,7 +637,7 @@ $$











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -609,7 +661,7 @@ $$











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -655,7 +707,7 @@ This approach (different linear and non-linear regression) suffers often from bo











    -

    Simple regression model

    +

    Simple regression model

    We are now ready to write our first program which aims at solving the above linear regression equations. We start with data we have produced ourselves, in this case normally distributed random numbers along the \( x \)-axis. These numbers define then the value of a function \( y(x)=4+3x+N(0,1) \). Thereafter we order the \( x \) values and employ our linear regression algorithm to set up the best fit. Here we find it useful to use the numpy function \( c\_ \) arrays where arrays are stacked along their last axis after being upgraded to at least two dimensions with ones post-pended to the shape. The following examples help in understanding what happens

    @@ -695,7 +747,7 @@ We see that, as expected, a linear fit gives a seemingly (from the graph) good r











    -

    Simple regression model, now using scikit-learn

    +

    Simple regression model, now using scikit-learn

    We can repeat the above algorithm using scikit-learn as follows @@ -726,7 +778,7 @@ plt.show()











    -

    Simple linear regression model using scikit-learn

    +

    Simple linear regression model using scikit-learn

    We start with perhaps our simplest possible example, using scikit-learn to perform linear regression analysis on a data set produced by us. @@ -794,7 +846,7 @@ plt.show()











    -

    Simple linear regression model

    +

    Simple linear regression model

    This example serves several aims. It allows us to demonstrate several @@ -816,7 +868,7 @@ where \( x \) is defined as before.











    -

    Less noise

    +

    Less noise

    Does the fit look better? Indeed, by @@ -829,7 +881,7 @@ have not discussed a more rigorous approach to the cost function.











    -

    How to study our fits

    +

    How to study our fits

    We need more rigorous criteria in defining whether we have succeeded or @@ -851,7 +903,7 @@ dimensionless.











    -

    Minimizing the cost function

    +

    Minimizing the cost function

    Minimizing the cost function is a central aspect of @@ -872,7 +924,7 @@ the \( \chi^2 \) function becomes smaller.











    -

    Relative error

    +

    Relative error

    There are many ways to define the cost function. A simpler approach is to look at the relative difference between the training data and the predicted data, that is we define @@ -912,7 +964,7 @@ relative error.











    -

    The richness of scikit-learn

    +

    The richness of scikit-learn

    As mentioned above, scikit-learn has an impressive functionality. @@ -957,7 +1009,7 @@ plt.show()











    -

    Functions in scikit-learn

    +

    Functions in scikit-learn

    The function coef gives us the parameter \( \beta \) of our fit while intercept yields @@ -974,7 +1026,7 @@ this function as being similar to the \( \chi^2 \) function defined above.











    -

    Other functions in scikit-learn

    +

    Other functions in scikit-learn

    The r2score function computes \( R^2 \), the coefficient of @@ -998,7 +1050,7 @@ $$











    -

    The mean absolute error and other functions in scikit-learn

    +

    The mean absolute error and other functions in scikit-learn

    Another quantity will meet again in our discussions of regression analysis is @@ -1023,12 +1075,13 @@ years etc.











    -

    Cubic polynomial in scikit-learn

    +

    Cubic polynomial in scikit-learn

    We will discuss in more detail these and other functions in the various lectures. We conclude this part with another example. Instead of -a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn. +a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn. +Add description of the various python commands.

    @@ -1065,82 +1118,12 @@ plt.show() print (error(y))

    -Similarly, using R, we can perform similar studies. -(more details on R will be inserted later). +Using R, we can perform similar studies.











    -

    Simple regression model with gradient descent

    -Add info about the equations, play around with different learning rates -

    - - -

    # Importing various packages
-from math import exp, sqrt
-from random import random, seed
-import numpy as np
-import matplotlib.pyplot as plt
-
-x = 2*np.random.rand(100,1)
-y = 4+3*x+np.random.randn(100,1)
-
-xb = np.c_[np.ones((100,1)), x]
-theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-print(theta_linreg)
-theta = np.random.randn(2,1)
-
-eta = 0.1
-Niterations = 1000
-m = 100
-
-for iter in range(Niterations):
-    gradients = 2.0/m*xb.T.dot(xb.dot(theta)-y)
-    theta -= eta*gradients
-
-print(theta)
-xnew = np.array([[0],[2]])
-xbnew = np.c_[np.ones((2,1)), xnew]
-ypredict = xbnew.dot(theta)
-ypredict2 = xbnew.dot(theta_linreg)
-plt.plot(xnew, ypredict, "r-")
-plt.plot(xnew, ypredict2, "b-")
-plt.plot(x, y ,'ro')
-plt.axis([0,2.0,0, 15.0])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$y$')
-plt.title(r'Random numbers ')
-plt.show()
-
    -

    -









    - -

    Simple regression model with stochastic gradient descent

    -Add info about the equations, play around with different learning rates -

    - - -

    # Importing various packages
-from math import exp, sqrt
-from random import random, seed
-import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.linear_model import SGDRegressor
-
-x = 2*np.random.rand(100,1)
-y = 4+3*x+np.random.randn(100,1)
-
-xb = np.c_[np.ones((100,1)), x]
-theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-print(theta_linreg)
-sgdreg = SGDRegressor(n_iter = 50, penalty=None, eta0=0.1)
-sgdreg.fit(x,y.ravel())
-print(sgdreg.intercept_, sgdreg.coef_)
-
    -

    -









    - -

    Polynomial Regression

    +

    Polynomial Regression

    @@ -1168,10 +1151,92 @@ plt.ylabel(r'$y$') plt.title(r'Random numbers ') plt.show()

    +

    + + +

    Linking the regression analysis with a statistical interpretation

    + +

    +Before we proceed, and to link with our discussions of Bayesian statistics to come, it is useful the derive the standard regression analysis equations using a statistical interpretation. This allows us also to derive quantities like the variance and other expectation values in a rather straightforward way. + +

    +It is assumed that \( \varepsilon_i +\sim \mathcal{N}(0, \sigma^2) \) and the \( \varepsilon_{i} \) are +independent, i.e.: +$$ +\begin{align*} +\mbox{Cov}(\varepsilon_{i_1}, +\varepsilon_{i_2}) & = \left\{ \begin{array}{lcc} \sigma^2 & \mbox{if} +& i_1 = i_2, \\ 0 & \mbox{if} & i_1 \not= i_2. \end{array} \right. +\end{align*} +$$ + +The randomness of \( \varepsilon_i \) implies that +\( \mathbf{Y}_i \) is also a random variable. In particular, +\( \mathbf{Y}_i \) is normally distributed, because \( \varepsilon_i \sim +\mathcal{N}(0, \sigma^2) \) and \( \mathbf{X}_{i,\ast} \, \beta \) is a +non-random scalar. To specify the parameters of the distribution of +\( \mathbf{Y}_i \) we need to calculate its first two moments. +











    -

    Ridge and Lasso Regression

    +

    Expectation value and variance

    + +

    +Its expectation equals: +$$ +\begin{align*} +\mathbb{E}(Y_i) & = +\mathbb{E}(\mathbf{X}_{i, \ast} \, \beta) + \mathbb{E}(\varepsilon_i) +\, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \beta, +\end{align*} +$$ + +while +its variance is +$$ +\begin{align*} \mbox{Var}(Y_i) & = \mathbb{E} \{ [Y_i +- \mathbb{E}(Y_i)]^2 \} \, \, \, = \, \, \, \mathbb{E} ( Y_i^2 ) - +[\mathbb{E}(Y_i)]^2 \\ & = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, +\beta + \varepsilon_i )^2] - ( \mathbf{X}_{i, \ast} \, \beta)^2 \\ & += \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2 \varepsilon_i +\mathbf{X}_{i, \ast} \, \beta + \varepsilon_i^2 ] - ( \mathbf{X}_{i, +\ast} \, \beta)^2 \\ & = ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2 +\mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \beta + +\mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \beta)^2 +\\ & = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \, +\mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2. +\end{align*} +$$ + +Hence, \( Y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \beta, \sigma^2) \). + +

    +









    + +

    The singular value decompostion

    +
    + +

    + +

    +A general +\( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal +matrix \( \hat{D} \) of dimensionality \( n\times n \) and two orthognal +matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality +\( m \times m \) and the last dimensionality \( n\times n \). +We have then +$$ +\hat{A} = \hat{U}\hat{D}\hat{V}^T +$$ +

    + + +

    +









    + +

    Code examples for Ridge and Lasso Regression

    @@ -1252,38 +1317,390 @@ plt.show()











    -

    The singular value decompostion

    -
    - -

    -How can we use the singular value decomposition to find the parameters \( \beta_j \)? More details will come. We first note that a general \( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal matrix \( \hat{\Sigma} \) of dimensionality \( n\times n \) and two orthognal matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality \( m \times n \) and the last dimensionality \( n\times n \). We have then -$$ -\hat{A} = \hat{U}\hat{\Sigma}\hat{V} -$$ -

    - +

    From standard regression to Ridge regressions

    -Add codes and discuss this in connection with lasso and ridge, show example where the standard inversion of a matrix fails and where SVD comes to rescue +One of the typical problems we encounter with linear regression, in particular +when the matrix \( \hat{X} \) (our so-called design matrix) is high-dimensional, +are problems with near singular or singular matrices. The column vectors of \( \hat{X} \) +may be linearly dependent, normally referred to as super-collinearity. +This means that the matrix may be rank deficient and it is basically impossible to +to model the data using linear regression. As an example, consider the matrix +$$ +\begin{align*} +\mathbf{X} & = \left[ +\begin{array}{rrr} +1 & -1 & 2 +\\ +1 & 0 & 1 +\\ +1 & 2 & -1 +\\ +1 & 1 & 0 +\end{array} \right] +\end{align*} +$$ + +

    +The columns of \( \hat{X} \) are linearly dependent. We se this easily since the +the first column is the row-wise sum of the other two columns. The rank (more correct, +the column rank) of a matrix is the dimension of the space spanned by the +column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number +of linearly independent columns. In this particular case the matrix has rank 2. + +

    +Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies +that the inverse of the matrix \( \hat{X}^T\hat{x} \) (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this +$$ +\begin{align*} +\hat{X} & = \left[ +\begin{array}{rr} +1 & -1 +\\ +1 & -1 +\end{array} \right]. +\end{align*} +$$ + +We see easily that \( \mbox{det}(\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0 \). Hence, \( \mathbf{X} \) is singular and its inverse is undefined. +This is equivalent to saying that the matrix \( \hat{X} \) has at least an eigenvalue which is zero.











    -

    Lasso and Ridge regression

    +

    Fixing the singularity

    -Discuss the mathematics here +If our design matrix \( \hat{X} \) which enters the linear regression problem +$$ +\begin{align} +\hat{\beta} & = (\hat{X}^{T} \hat{X})^{-1} \hat{X}^{T} \hat{y}, +\label{_auto1} +\end{align} +$$ + +has linearly dependent column vectors, we will not be able to compute the inverse +of \( \hat{X}^T\hat{X} \) and we cannot find the parameters (estimators) \( \beta_i \). +The estimators are only well-defined if \( (\hat{X}^{T}\hat{X})^{-1} \) exits. +This is more likely to happen when the matrix \( \hat{X} \) is high-dimensional. In this case it is likely to encounter a situation where +the regression parameters \( \beta_i \) cannot be estimated. + +

    +The ad hoc approach which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change +$$ +\hat{X}^{T} \hat{X} \rightarrow \hat{X}^{T} \hat{X}+\lambda \hat{I}, +$$ + +where \( \hat{I} \) is the identity matrix.











    -

    Logistic regression

    -Add discussion about classification versus regression, show examples of more than two cases and why regression is not the best approach. Motivate for k-nearest neighbors +

    A second-order polynomial with Ridge and Lasso

    +

    + + +

    import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import Ridge
+from sklearn.metrics import r2_score
+
+np.random.seed(4155)
+
+n_samples = 100
+
+x = np.random.rand(n_samples,1)
+y = 5*x*x + 0.1*np.random.rand(n_samples,1)
+
+# Centering  x and y.
+x_ = x - np.mean(x)
+y_ = y - np.mean(y) # beta_0 = mean(y)
+
+X = np.c_[np.ones((n_samples,1)), x, x**2]
+X_ = np.c_[x_, x_**2]
+
+
+### 1.
+lmb_values = [1e-4, 1e-3, 1e-2, 10, 1e2, 1e4]
+num_values = len(lmb_values)
+
+## Ridge-regression of centered and not centered data
+beta_ridge = np.zeros((3,num_values))
+beta_ridge_centered = np.zeros((3,num_values))
+
+I3 = np.eye(3)
+I2 = np.eye(2)
+
+for i,lmb in enumerate(lmb_values):
+    beta_ridge[:,i] = (np.linalg.inv( X.T @ X + lmb*I3) @ X.T @ y).flatten()
+    beta_ridge_centered[1:,i] = (np.linalg.inv( X_.T @ X_ + lmb*I2) @ X_.T @ y_).flatten()
+
+# sett beta_0 = np.mean(y)
+beta_ridge_centered[0,:] = np.mean(y)
+
+## OLS (ordinary least squares) solution 
+beta_ls = np.linalg.inv( X.T @ X ) @ X.T @ y
+
+## Evaluate the models
+pred_ls = X @ beta_ls
+pred_ridge =  X @ beta_ridge
+pred_ridge_centered =  X_ @ beta_ridge_centered[1:] + beta_ridge_centered[0,:]
+
+## Plot the results
+
+# Sorting
+sort_ind = np.argsort(x[:,0])
+
+x_plot = x[sort_ind,0]
+x_centered_plot = x_[sort_ind,0]
+
+pred_ls_plot = pred_ls[sort_ind,0]
+pred_ridge_plot = pred_ridge[sort_ind,:]
+pred_ridge_centered_plot = pred_ridge_centered[sort_ind,:]
+
+# Plott not centered
+plt.plot(x_plot,pred_ls_plot,label='ls')
+
+for i in range(num_values):
+    plt.plot(x_plot,pred_ridge_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x,y,'ro')
+
+plt.title('linear regression on un-centered data')
+plt.legend()
+
+# Plott centered
+plt.figure()
+
+for i in range(num_values):
+    plt.plot(x_centered_plot,pred_ridge_centered_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x_,y,'ro')
+
+plt.title('linear regression on centered data')
+plt.legend()
+
+
+# 2.
+
+pred_ridge_scikit =  np.zeros((n_samples,num_values))
+for i,lmb in enumerate(lmb_values):
+    pred_ridge_scikit[:,i] = (Ridge(alpha=lmb,fit_intercept=False).fit(X,y).predict(X)).flatten() # fit_intercept=False fordi bias er allerede i X
+
+plt.figure()
+
+plt.plot(x_plot,pred_ls_plot,label='ls')
+
+for i in range(num_values):
+    plt.plot(x_plot,pred_ridge_scikit[sort_ind,i],label='scikit-ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x,y,'ro')
+plt.legend()
+plt.title('linear regression using scikit')
+
+plt.show()
+
+### R2-score of the results
+for i in range(num_values):
+    print('lambda = %g'%lmb_values[i])
+    print('r2 for scikit: %g'%r2_score(y,pred_ridge_scikit[:,i]))
+    print('r2 for own code, not centered: %g'%r2_score(y,pred_ridge[:,i]))
+    print('r2 for own, centered: %g\n'%r2_score(y,pred_ridge_centered[:,i]))
+
    +

    +









    + +

    Fitting vs. predicting when data is in the model class

    -Add examples on classification problems +We start by considering the case +\( f(x)=2x \).

    +Then the data is clearly generated by a model that is contained within +all three model classes we are using to make predictions (linear +models, third order polynomials, and tenth order polynomials). + +

    +Run the code for the following cases: + +

      +
    1. For \( f(x)=2x \) , \( Ntrain=10 \) and \( \sigma =0 \) (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when \( x \in [0,1] \) . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?
      2. +
    2. Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?
      3. +
    3. Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example \( x \in [0,1.2] \) ) using the code below. How well do you do on points in the range of x where you trained the model? How about points outside the original training data set?
      4. +
    4. Repeat the above for \( f(x)=2x \) , \( Ntrain=10 \) , and \( \sigma=1 \) . What changes?
      5. +
    + +Repeat the exercises above for \( f(x)=2x \) , \( Ntrain=100 \) , and \( \sigma=1 \) . What changes? +Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well. + +

    +









    + +

    Fitting versus predicting when data is not in the model class

    + +

    +Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider \( f(x)=2x-10x^5+15x^{10} \) . Notice that the for linear and third-order polynomial the true model \( f(x) \) is not contained in model class. + +

      +
    1. Do better fits lead to better predictions?
      2. +
    2. What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points \( Ntrain \) and \( \sigma \)?
      3. +
    + +Summarize what you think you learned about the relationship of knowing the true model class and predictive power. + +

    +









    + +

    The code

    + +

    + + +

    import numpy as np
+import sklearn as sk
+from sklearn import datasets, linear_model
+from sklearn.preprocessing import PolynomialFeatures
+
+import matplotlib as mpl
+from matplotlib import pyplot as plt
+
+%matplotlib notebook
+
+# The Training Data
+
+N_train=100
+
+sigma_train=1;
+
+# Train on integers
+x=np.linspace(0.05,0.95,N_train)
+# Draw random noise
+s = sigma_train*np.random.randn(N_train)
+
+#linear
+y=2*x+s
+
+#Tenth Order
+#y=2*x-10*x**5+15*x**10+s
+
+p1=plt.plot(x,y, "o",ms=15, label='Training')
+
+#Linear Regression
+# Create linear regression object
+clf = linear_model.LinearRegression()
+
+# Train the model using the training sets
+clf.fit(x[:, np.newaxis], y)
+# The coefficients
+
+xplot=np.linspace(0.02,0.98,200)
+linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')
+
+#Polynomial Regression
+
+
+poly3 = PolynomialFeatures(degree=3)
+X = poly3.fit_transform(x[:,np.newaxis])
+clf3 = linear_model.LinearRegression()
+clf3.fit(X,y)
+
+
+Xplot=poly3.fit_transform(xplot[:,np.newaxis])
+poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')
+
+
+
+#poly5 = PolynomialFeatures(degree=5)
+#X = poly5.fit_transform(x[:,np.newaxis])
+#clf5 = linear_model.LinearRegression()
+#clf5.fit(X,y)
+
+#Xplot=poly5.fit_transform(xplot[:,np.newaxis])
+#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)
+
+poly10 = PolynomialFeatures(degree=10)
+X = poly10.fit_transform(x[:,np.newaxis])
+clf10 = linear_model.LinearRegression()
+clf10.fit(X,y)
+
+Xplot=poly10.fit_transform(xplot[:,np.newaxis])
+poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')
+
+axes = plt.gca()
+axes.set_ylim([-7,7])
+
+handles, labels=axes.get_legend_handles_labels()
+plt.legend(handles,labels, loc='lower center')
+plt.xlabel("$x$")
+plt.ylabel("$y$")
+Title="$N=$"+str(N_train)+", $\sigma=$"+str(sigma_train)
+plt.title(Title+" (train)")
+plt.tight_layout()
+plt.show()
+
    +

    + + +

    Generating test data

    +

    + + +

    # Generate Test Data
+
+#Number of test data
+N_test=20
+
+sigma_test=sigma_train
+
+max_x=1.2
+x_test=max_x*np.random.random(N_test)
+# Draw random noise
+s_test = sigma_test*np.random.randn(N_test)
+
+#Linear
+y_test=2*x_test+s_test
+#Tenth order
+#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test
+
+#Make design matrices for prediction
+x_plot=np.linspace(0,max_x, 200)
+X3 = poly3.fit_transform(x_plot[:,np.newaxis])
+X10 = poly10.fit_transform(x_plot[:,np.newaxis])
+
+%matplotlib notebook
+
+fig = plt.figure() 
+p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')
+p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')
+p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')
+p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')
+
+
+plt.legend(loc=2)
+plt.xlabel('$x$')
+plt.ylabel('$y$')
+plt.legend(loc='best')
+plt.title(Title+" (pred.)")
+plt.tight_layout()
+plt.show()
+
+#Linear Filename
+#filename_test=Title+"pred-linear.pdf"
+#Tenth Order Filename
+#filename_test=Title+"pred-o10.pdf"
+#plt.savefig(filename_test)
+#plt.ylim((-6,12))
+
    +

    +









    + +

    Lasso regression

    + +

    +









    + +

    Logistic regression

    diff --git a/doc/pub/Regression/html/Regression.html b/doc/pub/Regression/html/Regression.html index 7e1a6bbe5..fd86d2474 100644 --- a/doc/pub/Regression/html/Regression.html +++ b/doc/pub/Regression/html/Regression.html @@ -66,17 +66,14 @@ div { text-align: justify; text-justify: inter-word; } @@ -180,7 +201,7 @@ MathJax.Hub.Config({
    [2] Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University

    -

    Aug 24, 2018

    +

    Sep 6, 2018












    @@ -210,7 +231,38 @@ A regression model aims at finding a likelihood function \( p(y\vert \hat{x}) \)











    -

    General linear models

    +

    Regression analysis, overarching aims II

    +
    + +

    + +

    +Consider an experiment in which \( p \) characteristics of \( n \) samples are +measured. The data from this experiment are denoted \( \mathbf{X} \), with +\( \mathbf{X} \) as above. The matrix \( \mathbf{X} \) is called the design +matrix. Additional information of the samples is available in the +form of \( \mathbf{Y} \) (also as above). The variable \( \mathbf{Y} \) is +generally referred to as the response variable. The aim of +regression analysis is to explain \( \mathbf{Y} \) in terms of +\( \mathbf{X} \) through a functional relationship like \( Y_i = +f(\mathbf{X}_{i,\ast}) \). When no prior knowledge on the form of +\( f(\cdot) \) is available, it is common to assume a linear relationship +between \( \mathbf{X} \) and \( \mathbf{Y} \). This assumption gives rise to +the linear regression model where \( \beta = (\beta_1, \ldots, +\beta_p)^{\top} \) is the regression parameter. The parameter +\( \beta_j \), \( j=1, \ldots, p \), represents the effect size of covariate +\( j \) on the response. That is, for each unit change in covariate \( j \) +(while keeping the other covariates fixed) the observed change in the +response is equal to \( \beta_j \). + + +

    + + +

    +









    + +

    General linear models

    @@ -231,7 +283,7 @@ where \( \epsilon_i \) is the error in our approximation.











    -

    Rewriting the fitting procedure as a linear algebra problem

    +

    Rewriting the fitting procedure as a linear algebra problem

    @@ -251,7 +303,7 @@ $$











    -

    Rewriting the fitting procedure as a linear algebra problem, follows

    +

    Rewriting the fitting procedure as a linear algebra problem, follows

    @@ -292,7 +344,7 @@ $$











    -

    Generalizing the fitting procedure as a linear algebra problem

    +

    Generalizing the fitting procedure as a linear algebra problem

    @@ -315,7 +367,7 @@ $$











    -

    Generalizing the fitting procedure as a linear algebra problem

    +

    Generalizing the fitting procedure as a linear algebra problem

    @@ -343,7 +395,7 @@ The left-hand side of this equation forms know. Our error vector \( \hat{\epsilo











    -

    Optimizing our parameters

    +

    Optimizing our parameters

    @@ -365,7 +417,7 @@ $$











    -

    Optimizing our parameters, more details

    +

    Optimizing our parameters, more details

    @@ -389,7 +441,7 @@ $$











    -

    Interpretations and optimizing our parameters

    +

    Interpretations and optimizing our parameters

    @@ -428,7 +480,7 @@ $$











    -

    Interpretations and optimizing our parameters

    +

    Interpretations and optimizing our parameters

    @@ -454,7 +506,7 @@ $$











    -

    Interpretations and optimizing our parameters

    +

    Interpretations and optimizing our parameters

    @@ -482,7 +534,7 @@ meaning that the solution for \( \hat{\beta} \) is the one which minimizes the r











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -505,7 +557,7 @@ where the matrix \( \hat{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -533,7 +585,7 @@ where we have defined the matrix \( \hat{A} =\hat{X}/\hat{\Sigma} \) with matrix











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -559,7 +611,7 @@ $$











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -590,7 +642,7 @@ $$











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -614,7 +666,7 @@ $$











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -660,7 +712,7 @@ This approach (different linear and non-linear regression) suffers often from bo











    -

    Simple regression model

    +

    Simple regression model

    We are now ready to write our first program which aims at solving the above linear regression equations. We start with data we have produced ourselves, in this case normally distributed random numbers along the \( x \)-axis. These numbers define then the value of a function \( y(x)=4+3x+N(0,1) \). Thereafter we order the \( x \) values and employ our linear regression algorithm to set up the best fit. Here we find it useful to use the numpy function \( c\_ \) arrays where arrays are stacked along their last axis after being upgraded to at least two dimensions with ones post-pended to the shape. The following examples help in understanding what happens

    @@ -700,7 +752,7 @@ We see that, as expected, a linear fit gives a seemingly (from the graph) good r











    -

    Simple regression model, now using scikit-learn

    +

    Simple regression model, now using scikit-learn

    We can repeat the above algorithm using scikit-learn as follows @@ -731,7 +783,7 @@ plt.show()











    -

    Simple linear regression model using scikit-learn

    +

    Simple linear regression model using scikit-learn

    We start with perhaps our simplest possible example, using scikit-learn to perform linear regression analysis on a data set produced by us. @@ -799,7 +851,7 @@ plt.show()











    -

    Simple linear regression model

    +

    Simple linear regression model

    This example serves several aims. It allows us to demonstrate several @@ -821,7 +873,7 @@ where \( x \) is defined as before.











    -

    Less noise

    +

    Less noise

    Does the fit look better? Indeed, by @@ -834,7 +886,7 @@ have not discussed a more rigorous approach to the cost function.











    -

    How to study our fits

    +

    How to study our fits

    We need more rigorous criteria in defining whether we have succeeded or @@ -856,7 +908,7 @@ dimensionless.











    -

    Minimizing the cost function

    +

    Minimizing the cost function

    Minimizing the cost function is a central aspect of @@ -877,7 +929,7 @@ the \( \chi^2 \) function becomes smaller.











    -

    Relative error

    +

    Relative error

    There are many ways to define the cost function. A simpler approach is to look at the relative difference between the training data and the predicted data, that is we define @@ -917,7 +969,7 @@ relative error.











    -

    The richness of scikit-learn

    +

    The richness of scikit-learn

    As mentioned above, scikit-learn has an impressive functionality. @@ -962,7 +1014,7 @@ plt.show()











    -

    Functions in scikit-learn

    +

    Functions in scikit-learn

    The function coef gives us the parameter \( \beta \) of our fit while intercept yields @@ -979,7 +1031,7 @@ this function as being similar to the \( \chi^2 \) function defined above.











    -

    Other functions in scikit-learn

    +

    Other functions in scikit-learn

    The r2score function computes \( R^2 \), the coefficient of @@ -1003,7 +1055,7 @@ $$











    -

    The mean absolute error and other functions in scikit-learn

    +

    The mean absolute error and other functions in scikit-learn

    Another quantity will meet again in our discussions of regression analysis is @@ -1028,12 +1080,13 @@ years etc.











    -

    Cubic polynomial in scikit-learn

    +

    Cubic polynomial in scikit-learn

    We will discuss in more detail these and other functions in the various lectures. We conclude this part with another example. Instead of -a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn. +a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn. +Add description of the various python commands.

    @@ -1070,82 +1123,12 @@ plt.show() print (error(y))

    -Similarly, using R, we can perform similar studies. -(more details on R will be inserted later). +Using R, we can perform similar studies.











    -

    Simple regression model with gradient descent

    -Add info about the equations, play around with different learning rates -

    - - -

    # Importing various packages
-from math import exp, sqrt
-from random import random, seed
-import numpy as np
-import matplotlib.pyplot as plt
-
-x = 2*np.random.rand(100,1)
-y = 4+3*x+np.random.randn(100,1)
-
-xb = np.c_[np.ones((100,1)), x]
-theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-print(theta_linreg)
-theta = np.random.randn(2,1)
-
-eta = 0.1
-Niterations = 1000
-m = 100
-
-for iter in range(Niterations):
-    gradients = 2.0/m*xb.T.dot(xb.dot(theta)-y)
-    theta -= eta*gradients
-
-print(theta)
-xnew = np.array([[0],[2]])
-xbnew = np.c_[np.ones((2,1)), xnew]
-ypredict = xbnew.dot(theta)
-ypredict2 = xbnew.dot(theta_linreg)
-plt.plot(xnew, ypredict, "r-")
-plt.plot(xnew, ypredict2, "b-")
-plt.plot(x, y ,'ro')
-plt.axis([0,2.0,0, 15.0])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$y$')
-plt.title(r'Random numbers ')
-plt.show()
-
    -

    -









    - -

    Simple regression model with stochastic gradient descent

    -Add info about the equations, play around with different learning rates -

    - - -

    # Importing various packages
-from math import exp, sqrt
-from random import random, seed
-import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.linear_model import SGDRegressor
-
-x = 2*np.random.rand(100,1)
-y = 4+3*x+np.random.randn(100,1)
-
-xb = np.c_[np.ones((100,1)), x]
-theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-print(theta_linreg)
-sgdreg = SGDRegressor(n_iter = 50, penalty=None, eta0=0.1)
-sgdreg.fit(x,y.ravel())
-print(sgdreg.intercept_, sgdreg.coef_)
-
    -

    -









    - -

    Polynomial Regression

    +

    Polynomial Regression

    @@ -1173,10 +1156,92 @@ plt.ylabel(r plt.title(r'Random numbers ') plt.show()

    +

    + + +

    Linking the regression analysis with a statistical interpretation

    + +

    +Before we proceed, and to link with our discussions of Bayesian statistics to come, it is useful the derive the standard regression analysis equations using a statistical interpretation. This allows us also to derive quantities like the variance and other expectation values in a rather straightforward way. + +

    +It is assumed that \( \varepsilon_i +\sim \mathcal{N}(0, \sigma^2) \) and the \( \varepsilon_{i} \) are +independent, i.e.: +$$ +\begin{align*} +\mbox{Cov}(\varepsilon_{i_1}, +\varepsilon_{i_2}) & = \left\{ \begin{array}{lcc} \sigma^2 & \mbox{if} +& i_1 = i_2, \\ 0 & \mbox{if} & i_1 \not= i_2. \end{array} \right. +\end{align*} +$$ + +The randomness of \( \varepsilon_i \) implies that +\( \mathbf{Y}_i \) is also a random variable. In particular, +\( \mathbf{Y}_i \) is normally distributed, because \( \varepsilon_i \sim +\mathcal{N}(0, \sigma^2) \) and \( \mathbf{X}_{i,\ast} \, \beta \) is a +non-random scalar. To specify the parameters of the distribution of +\( \mathbf{Y}_i \) we need to calculate its first two moments. +











    -

    Ridge and Lasso Regression

    +

    Expectation value and variance

    + +

    +Its expectation equals: +$$ +\begin{align*} +\mathbb{E}(Y_i) & = +\mathbb{E}(\mathbf{X}_{i, \ast} \, \beta) + \mathbb{E}(\varepsilon_i) +\, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \beta, +\end{align*} +$$ + +while +its variance is +$$ +\begin{align*} \mbox{Var}(Y_i) & = \mathbb{E} \{ [Y_i +- \mathbb{E}(Y_i)]^2 \} \, \, \, = \, \, \, \mathbb{E} ( Y_i^2 ) - +[\mathbb{E}(Y_i)]^2 \\ & = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, +\beta + \varepsilon_i )^2] - ( \mathbf{X}_{i, \ast} \, \beta)^2 \\ & += \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2 \varepsilon_i +\mathbf{X}_{i, \ast} \, \beta + \varepsilon_i^2 ] - ( \mathbf{X}_{i, +\ast} \, \beta)^2 \\ & = ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2 +\mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \beta + +\mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \beta)^2 +\\ & = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \, +\mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2. +\end{align*} +$$ + +Hence, \( Y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \beta, \sigma^2) \). + +

    +









    + +

    The singular value decompostion

    +
    + +

    + +

    +A general +\( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal +matrix \( \hat{D} \) of dimensionality \( n\times n \) and two orthognal +matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality +\( m \times m \) and the last dimensionality \( n\times n \). +We have then +$$ +\hat{A} = \hat{U}\hat{D}\hat{V}^T +$$ +

    + + +

    +









    + +

    Code examples for Ridge and Lasso Regression

    @@ -1257,38 +1322,390 @@ plt.show()











    -

    The singular value decompostion

    -
    - -

    -How can we use the singular value decomposition to find the parameters \( \beta_j \)? More details will come. We first note that a general \( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal matrix \( \hat{\Sigma} \) of dimensionality \( n\times n \) and two orthognal matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality \( m \times n \) and the last dimensionality \( n\times n \). We have then -$$ -\hat{A} = \hat{U}\hat{\Sigma}\hat{V} -$$ -

    - +

    From standard regression to Ridge regressions

    -Add codes and discuss this in connection with lasso and ridge, show example where the standard inversion of a matrix fails and where SVD comes to rescue +One of the typical problems we encounter with linear regression, in particular +when the matrix \( \hat{X} \) (our so-called design matrix) is high-dimensional, +are problems with near singular or singular matrices. The column vectors of \( \hat{X} \) +may be linearly dependent, normally referred to as super-collinearity. +This means that the matrix may be rank deficient and it is basically impossible to +to model the data using linear regression. As an example, consider the matrix +$$ +\begin{align*} +\mathbf{X} & = \left[ +\begin{array}{rrr} +1 & -1 & 2 +\\ +1 & 0 & 1 +\\ +1 & 2 & -1 +\\ +1 & 1 & 0 +\end{array} \right] +\end{align*} +$$ + +

    +The columns of \( \hat{X} \) are linearly dependent. We se this easily since the +the first column is the row-wise sum of the other two columns. The rank (more correct, +the column rank) of a matrix is the dimension of the space spanned by the +column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number +of linearly independent columns. In this particular case the matrix has rank 2. + +

    +Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies +that the inverse of the matrix \( \hat{X}^T\hat{x} \) (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this +$$ +\begin{align*} +\hat{X} & = \left[ +\begin{array}{rr} +1 & -1 +\\ +1 & -1 +\end{array} \right]. +\end{align*} +$$ + +We see easily that \( \mbox{det}(\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0 \). Hence, \( \mathbf{X} \) is singular and its inverse is undefined. +This is equivalent to saying that the matrix \( \hat{X} \) has at least an eigenvalue which is zero.











    -

    Lasso and Ridge regression

    +

    Fixing the singularity

    -Discuss the mathematics here +If our design matrix \( \hat{X} \) which enters the linear regression problem +$$ +\begin{align} +\hat{\beta} & = (\hat{X}^{T} \hat{X})^{-1} \hat{X}^{T} \hat{y}, +\label{_auto1} +\end{align} +$$ + +has linearly dependent column vectors, we will not be able to compute the inverse +of \( \hat{X}^T\hat{X} \) and we cannot find the parameters (estimators) \( \beta_i \). +The estimators are only well-defined if \( (\hat{X}^{T}\hat{X})^{-1} \) exits. +This is more likely to happen when the matrix \( \hat{X} \) is high-dimensional. In this case it is likely to encounter a situation where +the regression parameters \( \beta_i \) cannot be estimated. + +

    +The ad hoc approach which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change +$$ +\hat{X}^{T} \hat{X} \rightarrow \hat{X}^{T} \hat{X}+\lambda \hat{I}, +$$ + +where \( \hat{I} \) is the identity matrix.











    -

    Logistic regression

    -Add discussion about classification versus regression, show examples of more than two cases and why regression is not the best approach. Motivate for k-nearest neighbors +

    A second-order polynomial with Ridge and Lasso

    +

    + + +

    import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import Ridge
+from sklearn.metrics import r2_score
+
+np.random.seed(4155)
+
+n_samples = 100
+
+x = np.random.rand(n_samples,1)
+y = 5*x*x + 0.1*np.random.rand(n_samples,1)
+
+# Centering  x and y.
+x_ = x - np.mean(x)
+y_ = y - np.mean(y) # beta_0 = mean(y)
+
+X = np.c_[np.ones((n_samples,1)), x, x**2]
+X_ = np.c_[x_, x_**2]
+
+
+### 1.
+lmb_values = [1e-4, 1e-3, 1e-2, 10, 1e2, 1e4]
+num_values = len(lmb_values)
+
+## Ridge-regression of centered and not centered data
+beta_ridge = np.zeros((3,num_values))
+beta_ridge_centered = np.zeros((3,num_values))
+
+I3 = np.eye(3)
+I2 = np.eye(2)
+
+for i,lmb in enumerate(lmb_values):
+    beta_ridge[:,i] = (np.linalg.inv( X.T @ X + lmb*I3) @ X.T @ y).flatten()
+    beta_ridge_centered[1:,i] = (np.linalg.inv( X_.T @ X_ + lmb*I2) @ X_.T @ y_).flatten()
+
+# sett beta_0 = np.mean(y)
+beta_ridge_centered[0,:] = np.mean(y)
+
+## OLS (ordinary least squares) solution 
+beta_ls = np.linalg.inv( X.T @ X ) @ X.T @ y
+
+## Evaluate the models
+pred_ls = X @ beta_ls
+pred_ridge =  X @ beta_ridge
+pred_ridge_centered =  X_ @ beta_ridge_centered[1:] + beta_ridge_centered[0,:]
+
+## Plot the results
+
+# Sorting
+sort_ind = np.argsort(x[:,0])
+
+x_plot = x[sort_ind,0]
+x_centered_plot = x_[sort_ind,0]
+
+pred_ls_plot = pred_ls[sort_ind,0]
+pred_ridge_plot = pred_ridge[sort_ind,:]
+pred_ridge_centered_plot = pred_ridge_centered[sort_ind,:]
+
+# Plott not centered
+plt.plot(x_plot,pred_ls_plot,label='ls')
+
+for i in range(num_values):
+    plt.plot(x_plot,pred_ridge_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x,y,'ro')
+
+plt.title('linear regression on un-centered data')
+plt.legend()
+
+# Plott centered
+plt.figure()
+
+for i in range(num_values):
+    plt.plot(x_centered_plot,pred_ridge_centered_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x_,y,'ro')
+
+plt.title('linear regression on centered data')
+plt.legend()
+
+
+# 2.
+
+pred_ridge_scikit =  np.zeros((n_samples,num_values))
+for i,lmb in enumerate(lmb_values):
+    pred_ridge_scikit[:,i] = (Ridge(alpha=lmb,fit_intercept=False).fit(X,y).predict(X)).flatten() # fit_intercept=False fordi bias er allerede i X
+
+plt.figure()
+
+plt.plot(x_plot,pred_ls_plot,label='ls')
+
+for i in range(num_values):
+    plt.plot(x_plot,pred_ridge_scikit[sort_ind,i],label='scikit-ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x,y,'ro')
+plt.legend()
+plt.title('linear regression using scikit')
+
+plt.show()
+
+### R2-score of the results
+for i in range(num_values):
+    print('lambda = %g'%lmb_values[i])
+    print('r2 for scikit: %g'%r2_score(y,pred_ridge_scikit[:,i]))
+    print('r2 for own code, not centered: %g'%r2_score(y,pred_ridge[:,i]))
+    print('r2 for own, centered: %g\n'%r2_score(y,pred_ridge_centered[:,i]))
+
    +

    +









    + +

    Fitting vs. predicting when data is in the model class

    -Add examples on classification problems +We start by considering the case +\( f(x)=2x \).

    +Then the data is clearly generated by a model that is contained within +all three model classes we are using to make predictions (linear +models, third order polynomials, and tenth order polynomials). + +

    +Run the code for the following cases: + +

      +
    1. For \( f(x)=2x \) , \( Ntrain=10 \) and \( \sigma =0 \) (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when \( x \in [0,1] \) . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?
      2. +
    2. Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?
      3. +
    3. Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example \( x \in [0,1.2] \) ) using the code below. How well do you do on points in the range of x where you trained the model? How about points outside the original training data set?
      4. +
    4. Repeat the above for \( f(x)=2x \) , \( Ntrain=10 \) , and \( \sigma=1 \) . What changes?
      5. +
    + +Repeat the exercises above for \( f(x)=2x \) , \( Ntrain=100 \) , and \( \sigma=1 \) . What changes? +Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well. + +

    +









    + +

    Fitting versus predicting when data is not in the model class

    + +

    +Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider \( f(x)=2x-10x^5+15x^{10} \) . Notice that the for linear and third-order polynomial the true model \( f(x) \) is not contained in model class. + +

      +
    1. Do better fits lead to better predictions?
      2. +
    2. What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points \( Ntrain \) and \( \sigma \)?
      3. +
    + +Summarize what you think you learned about the relationship of knowing the true model class and predictive power. + +

    +









    + +

    The code

    + +

    + + +

    import numpy as np
+import sklearn as sk
+from sklearn import datasets, linear_model
+from sklearn.preprocessing import PolynomialFeatures
+
+import matplotlib as mpl
+from matplotlib import pyplot as plt
+
+%matplotlib notebook
+
+# The Training Data
+
+N_train=100
+
+sigma_train=1;
+
+# Train on integers
+x=np.linspace(0.05,0.95,N_train)
+# Draw random noise
+s = sigma_train*np.random.randn(N_train)
+
+#linear
+y=2*x+s
+
+#Tenth Order
+#y=2*x-10*x**5+15*x**10+s
+
+p1=plt.plot(x,y, "o",ms=15, label='Training')
+
+#Linear Regression
+# Create linear regression object
+clf = linear_model.LinearRegression()
+
+# Train the model using the training sets
+clf.fit(x[:, np.newaxis], y)
+# The coefficients
+
+xplot=np.linspace(0.02,0.98,200)
+linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')
+
+#Polynomial Regression
+
+
+poly3 = PolynomialFeatures(degree=3)
+X = poly3.fit_transform(x[:,np.newaxis])
+clf3 = linear_model.LinearRegression()
+clf3.fit(X,y)
+
+
+Xplot=poly3.fit_transform(xplot[:,np.newaxis])
+poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')
+
+
+
+#poly5 = PolynomialFeatures(degree=5)
+#X = poly5.fit_transform(x[:,np.newaxis])
+#clf5 = linear_model.LinearRegression()
+#clf5.fit(X,y)
+
+#Xplot=poly5.fit_transform(xplot[:,np.newaxis])
+#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)
+
+poly10 = PolynomialFeatures(degree=10)
+X = poly10.fit_transform(x[:,np.newaxis])
+clf10 = linear_model.LinearRegression()
+clf10.fit(X,y)
+
+Xplot=poly10.fit_transform(xplot[:,np.newaxis])
+poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')
+
+axes = plt.gca()
+axes.set_ylim([-7,7])
+
+handles, labels=axes.get_legend_handles_labels()
+plt.legend(handles,labels, loc='lower center')
+plt.xlabel("$x$")
+plt.ylabel("$y$")
+Title="$N=$"+str(N_train)+", $\sigma=$"+str(sigma_train)
+plt.title(Title+" (train)")
+plt.tight_layout()
+plt.show()
+
    +

    + + +

    Generating test data

    +

    + + +

    # Generate Test Data
+
+#Number of test data
+N_test=20
+
+sigma_test=sigma_train
+
+max_x=1.2
+x_test=max_x*np.random.random(N_test)
+# Draw random noise
+s_test = sigma_test*np.random.randn(N_test)
+
+#Linear
+y_test=2*x_test+s_test
+#Tenth order
+#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test
+
+#Make design matrices for prediction
+x_plot=np.linspace(0,max_x, 200)
+X3 = poly3.fit_transform(x_plot[:,np.newaxis])
+X10 = poly10.fit_transform(x_plot[:,np.newaxis])
+
+%matplotlib notebook
+
+fig = plt.figure() 
+p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')
+p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')
+p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')
+p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')
+
+
+plt.legend(loc=2)
+plt.xlabel('$x$')
+plt.ylabel('$y$')
+plt.legend(loc='best')
+plt.title(Title+" (pred.)")
+plt.tight_layout()
+plt.show()
+
+#Linear Filename
+#filename_test=Title+"pred-linear.pdf"
+#Tenth Order Filename
+#filename_test=Title+"pred-o10.pdf"
+#plt.savefig(filename_test)
+#plt.ylim((-6,12))
+
    +

    +









    + +

    Lasso regression

    + +

    +









    + +

    Logistic regression

    diff --git a/doc/pub/Regression/ipynb/Regression.ipynb b/doc/pub/Regression/ipynb/Regression.ipynb index a4a85095f..af7e36ea4 100644 --- a/doc/pub/Regression/ipynb/Regression.ipynb +++ b/doc/pub/Regression/ipynb/Regression.ipynb @@ -10,7 +10,7 @@ " \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: **Aug 24, 2018**\n", + "Date: **Sep 6, 2018**\n", "\n", "Copyright 1999-2018, Morten Hjorth-Jensen. Released under CC Attribution-NonCommercial 4.0 license\n", "\n", @@ -33,6 +33,28 @@ "\n", "\n", "\n", + "## Regression analysis, overarching aims II\n", + "\n", + "\n", + "Consider an experiment in which $p$ characteristics of $n$ samples are\n", + "measured. The data from this experiment are denoted $\\mathbf{X}$, with\n", + "$\\mathbf{X}$ as above. The matrix $\\mathbf{X}$ is called the *design\n", + "matrix*. Additional information of the samples is available in the\n", + "form of $\\mathbf{Y}$ (also as above). The variable $\\mathbf{Y}$ is\n", + "generally referred to as the *response variable*. The aim of\n", + "regression analysis is to explain $\\mathbf{Y}$ in terms of\n", + "$\\mathbf{X}$ through a functional relationship like $Y_i =\n", + "f(\\mathbf{X}_{i,\\ast})$. When no prior knowledge on the form of\n", + "$f(\\cdot)$ is available, it is common to assume a linear relationship\n", + "between $\\mathbf{X}$ and $\\mathbf{Y}$. This assumption gives rise to\n", + "the *linear regression model* where $\\beta = (\\beta_1, \\ldots,\n", + "\\beta_p)^{\\top}$ is the *regression parameter*. The parameter\n", + "$\\beta_j$, $j=1, \\ldots, p$, represents the effect size of covariate\n", + "$j$ on the response. That is, for each unit change in covariate $j$\n", + "(while keeping the other covariates fixed) the observed change in the\n", + "response is equal to $\\beta_j$.\n", + "\n", + "\n", "\n", "## General linear models\n", "Before we proceed let us study a case from linear algebra where we aim at fitting a set of data $\\hat{y}=[y_0,y_1,\\dots,y_{n-1}]$. We could think of these data as a result of an experiment or a complicated numerical experiment. These data are functions of a series of variables $\\hat{x}=[x_0,x_1,\\dots,x_{n-1}]$, that is $y_i = y(x_i)$ with $i=0,1,2,\\dots,n-1$. The variables $x_i$ could represent physical quantities like time, temperature, position etc. We assume that $y(x)$ is a smooth function. \n", @@ -897,19 +919,10 @@ { "cell_type": "code", "execution_count": 1, - "metadata": {}, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "[[1 4]\n", - " [2 5]\n", - " [3 6]]\n", - "[[1 2 3 0 0 4 5 6]]\n" - ] - } - ], + "metadata": { + "collapsed": false + }, + "outputs": [], "source": [ "import numpy as np\n", "print(np.c_[np.array([1,2,3]), np.array([4,5,6])])\n", @@ -918,20 +931,11 @@ }, { "cell_type": "code", - "execution_count": 15, - "metadata": {}, - "outputs": [ - { - "data": { - "image/png": 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\n", - "text/plain": [ - "
    " - ] - }, - "metadata": {}, - "output_type": "display_data" - } - ], + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [], "source": [ "%matplotlib inline\n", "\n", @@ -941,7 +945,7 @@ "import matplotlib.pyplot as plt\n", "\n", "x = 2*np.random.rand(100,1)\n", - "y = 4+3*x+0.01*np.random.randn(100,1)\n", + "y = 4+3*x+np.random.randn(100,1)\n", "\n", "xb = np.c_[np.ones((100,1)), x]\n", "beta = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)\n", @@ -975,20 +979,11 @@ }, { "cell_type": "code", - "execution_count": 17, - "metadata": {}, - "outputs": [ - { - "data": { - "image/png": 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SJWkOpuo5mZYsGV6sNmaU2ksqIs4BzikzBrOOVmmLqF67YbiN07UsWDBwn3s1WQ2DljAkLZf0xnYEY2Y5+UFykJIFNC9ZTJiQqplccrCCilRJfQZYLOkSSXu0OiCzcaWnB6ZO7RtRPWFCX1fXZkzaV61SvVRZPMjJwoZg0IQREbdGxHuAHwM/k3SOpJ1aH5rZGJSfh2nqVDj2WHjuub7jvb2pwbm7O3WJHa7q2V0rScLzMtkIFGr0liTgXuBC4FTgPknHtTIwszEjP1ju2GP7urvmE0W1detSQmnkFa+ovb8yiG7TJk/eZ01VpA3jN8DDwNeAmcAJwEHAWyS5V5NZIwsXphJDpf1hKC66aOCkfRVTpsDFF9eej8kJwlqkSAljATAzIg6NiP8VET+OiPsj4lTgP7Y4PrPO16gdYukw/6aaMKH/kqOVfeDBclaaEU0NImmfiFjdxHga8tQg1jHy3V3rmTYtTbkxHO69ZE3UEVODtDNZmLVdrYWCKvvz3V3r2bJl6IsHSU4W1rG84p5ZLdXzMa1Zkx5ffjl86lPFu7uefHLj45XG8EpPpt5eJwvrWE4YNr5Vry5XWVnutNMGru2wfTt85CPw2GPFrj1hQvryP+WUgdNvABxySLqmezLZKOGEYeNXrdXlKivLNZrddfr0YtevTLmxZEkaJFfdm+mXvxx26GZlcMKwsa9eW0S91eXqrSpXsXhx/e6ukF7H7RA2BpW9RKtZa+R7MeWn+16zBj76UbjlluHNydTV1Vd1VFmgqLu7bxlSszGstBX3hsPdaq2ufILYYYfULjCYRutGdHXBM8/0L4FMnuwlR21U6ohutWalqlQ1SXDccX0lhqLJ4tJLUxKoNmlSWllu2bL+a0I4Wdg45yopG30qvZjyDdNDLSl3d6ceTxMm9L9WV1dKFl5ZzmwAlzCss1U3WC9cmHofNerFNJgpU1KbA6SEsGlTX8+lTZucJMzqcAnDOldlRHVlkNyaNbXXmi6i0l4xe7YbqM2GySUMK0+97q4Ajz4Kp5/enAWEurrSCG3P5Go2Ik4Y1h71qpbyU2/Mnw/vex8ccADsuWeqHhqKyrxN+Vldr7jC1UxmTVJqwpD0KklXSbpH0t2S/rLMeKyJ8j2Ydtih/8JBa9aktR6qSw8vvphGP++1F3zlK7D77rWvXWtCv3wpojKq2qUJs6Yquw3jPOBnEfFfJE0GGgyftVFj4cKUECo9l2r1YGrUq+lnP0u/Z87s34YBqcH6+OPhJz/xoDmzNistYUjaFXgXaQU/ImIbUGOeBhtVenr6J4uh6u7u2/aIarOOUmYJY29gI3CJpDcCK4HTIqLfQseSFpBW/aM7/2VinSUC7rkHTj21eLKoHmmd7+5aMW+eE4RZhyizDWMi8Gbgwog4EHgOOLP6pIhYGhFzI2LujBkz2h2j1Zv+G1JbxLJlqX1i5kzYf3944oli150yJa0VkR9J7WVHzTpamSWM9cD6iLg5e3wVNRKGtVFlPqZK9c/hh8M3v9l/XYjNm9MI6U98AjZuTPte/Wo4+OC0vsM558AjjzR+nerR1GY2KpSWMCLiMUnrJO0XEfcChwC/Lyueca3WVBuNBsn19sKzz6Yv/YMPTt1gKz2XdtppYEN1hQfNmY1qZfeSOhXoyXpIrQZOLDme8ad6NHVRW7fCxz8+cL8bqs3GLE9vPp5t357GPDz++NCfO3t2GudgZh3P05tbffWm3OjthVWr4J/+CY44AqZNG16ymDx5YG8mMxvznDBGu/yI6spP9ajqk06Ct741NU4feCB86lPwwANpDYki61PvkPuYdHV5XQizcarsNgwbrloN1fVs2wYrV6ZEcsgh8J73wKxZ6dg73tG4DWPKFHd3NTPACWN0Gk5DdW9vWmGuWnUj9bRp6fGWLW6wNrN+XCXViRpN+/3002kMxFB7NTUaJT9vXmrA7u1NM7tu2pS2PXmfmeW4hNFpai0aNH8+fO97qYH6d7+Dl18e2jVrTblhZjZELmF0mrPPrj3t9w9/mLbPPDM1XhfV1eU2CDNrCieMVmtUvQSp6uf222HxYjjyyNSOUIsEN94IX/winHtuKjU0svPOXjzIzJrKVVKtUJmTac2a/jOyrlmTqpsefzx9oV93HVx/fd+cTK95DUydmqbdqOZpv82sZB7p3SxD6eZaseeeqZvrwQenn+7u2j2g3LXVzEagWSO9XcIYrvzMrtOmwZNPDq0x+p574LWvHbjcqEsPZtahnDCKqk4QzzyTBsTB0EoVkOZh2m+/+se9aJCZdSAnjCKqq4mGmiDy3MXVzEYp95IqolZX16GoVDt5VTkzG8XGd8Ko1+U1Au68E84/H446qn5X1yK6uuDyy9M1PXLazEax8VMlVb386Gtek7q15ru8nnQSXHABrF4NGzak/fvsU7+ra96kSWna7+eeS4+9DKmZjTFjv4TR05Om8K6e8vvaa/uSRcW2bXDLLXDooXDxxfDgg2ka8IsuGjhQbtKklBSkVNV0ySUpqUSkHw+YM7MxZmyWMOoNnCsiIo2QznNXVzOzUZww6iWFqVPT3Evbt6fHQx2YWG9WV3d1NbNxbnQmjOpurvmkMFhbQyOSu7yamdVRehuGpAmS/k3Sjws/adGikXVzrR0InHyySxFmZnWUnjCA04C7C525dSt84xupGmokurrglFNSY3Wl0fryy2HJkpFd18xsDCu1SkrSLOAI4EvA3w36hLvugo99DCZMGNq8TZU2jtmz3VhtZjZMZZcwFgNnAL31TpC0QNIKSSuenTYtdXP91rcarwcxcWL/Lq8eOGdmNmKllTAkfRDYEBErJR1U77yIWAoshTS9OfvskwbTQe1eUh4wZ2bWEmVWSb0D+JCkw4EdgVdKuiIiji30bHdzNTNrq9KqpCLirIiYFRFzgKOB6wonCzMza7uy2zDMzGyU6IiBexFxA3BDyWGYmVkDLmGYmVkhThhmZlaIE4aZmRXihGFmZoU4YZiZWSFOGGZmVogThpmZFeKEYWZmhThhmJlZIU4YZmZWiBOGmZkV4oRhZmaFOGGYmVkhThhmZlaIE4aZmRXihGFmZoU4YZiZWSFOGGZmVogThpmZFeKEYWZmhZSWMCTtJel6Sb+XdJek08qKxczMBjexxNd+CfhkRNwqaRdgpaTlEfH7EmMyM7M6SithRMSjEXFrtv0McDcws6x4zMyssY5ow5A0BzgQuLnGsQWSVkhasXHjxnaHZmZmmdIThqSpwPeB0yPi6erjEbE0IuZGxNwZM2a0P0AzMwNKThiSJpGSRU9EXF1mLGZm1liZvaQEXAzcHRHnlhWHmZkVU2YJ4x3AccDBklZlP4eXGI+ZmTVQWrfaiPg1oLJe38zMhqb0Rm8zMxsdnDDMzKwQJwwzMyvECcPMzApxwjAzs0KcMMzMrBAnDDMzK8QJw8zMCnHCMDOzQpwwzMysECcMMzMrxAnDzMwKccIwM7NCnDDMzKwQJwwzMyvECcPMzApxwjAzs0KcMMzMrBAnDDMzK8QJw8zMCik1YUg6TNK9ku6XdGaZsZiZWWOlJQxJE4BvAB8A9geOkbR/WfGYmVljZZYw3gLcHxGrI2Ib8G3gqBLjMTOzBiaW+NozgXW5x+uBt1afJGkBsCB7+KKkO9sQ20hNBzaVHUQBjrN5RkOM4DibbbTEuV8zLlJmwigkIpYCSwEkrYiIuSWHNCjH2VyjIc7RECM4zmYbTXE24zplVkk9DOyVezwr22dmZh2ozITxO2BfSXtLmgwcDVxTYjxmZtZAaVVSEfGSpI8BPwcmAMsi4q5Bnra09ZE1heNsrtEQ52iIERxns42rOBURzbiOmZmNcR7pbWZmhThhmJlZIR2TMAabJkTSKyR9Jzt+s6Q5uWNnZfvvlfT+EmP8O0m/l3S7pGslzc4de1nSquynpY37BeI8QdLGXDz/I3fseEn3ZT/Hlxzn13Ix/kHSk7ljbbmfkpZJ2lBv/I+S87P3cLukN+eOtfNeDhbnvCy+OyT9VtIbc8ceyvavalb3yxHEeZCkp3L/tp/NHWvbVEIF4vx0LsY7s8/jtOxYW+6npL0kXZ9959wl6bQa5zT38xkRpf+QGr0fAPYBJgO3AftXnbMQuCjbPhr4Tra9f3b+K4C9s+tMKCnG9wBTsu1TKjFmj5/toHt5AnBBjedOA1Znv3fLtncrK86q808ldYxo9/18F/Bm4M46xw8HfgoIeBtwc7vvZcE43155fdJ0PDfnjj0ETO+Q+3kQ8OORfl5aHWfVuUcC17X7fgJ7AG/OtncB/lDj/3pTP5+dUsIoMk3IUcC3su2rgEMkKdv/7Yh4MSIeBO7Prtf2GCPi+oh4Pnt4E2lsSbuNZMqV9wPLI2JLRDwBLAcO65A4jwGubFEsdUXEr4AtDU45CrgskpuAV0nag/bey0HjjIjfZnFAeZ/NIveznrZOJTTEOMv6bD4aEbdm288Ad5Nm0Mhr6uezUxJGrWlCqt/4H8+JiJeAp4Cugs9tV4x580mZvWJHSSsk3STpwy2Ir6JonP85K6JeJakygLJd93JIr5VV7e0NXJfb3a77OZh676Od93Koqj+bAfxC0kqlqXjK9peSbpP0U0kHZPs68n5KmkL6ov1+bnfb76dSFf2BwM1Vh5r6+ez4qUFGI0nHAnOBd+d2z46IhyXtA1wn6Y6IeKCcCPkRcGVEvCjpb0glt4NLiqWIo4GrIuLl3L5Oup+jhqT3kBLGO3O735ndy1cDyyXdk/2FXYZbSf+2z0o6HPgXYN+SYiniSOA3EZEvjbT1fkqaSkpYp0fE0616HeicEkaRaUL+eI6kicCuwOaCz21XjEh6L7AI+FBEvFjZHxEPZ79XAzeQ/hpohUHjjIjNudi+Cfx50ee2M86co6kq8rfxfg6m3vvouKlvJL2B9O99VERsruzP3csNwA9oTZVuIRHxdEQ8m23/BJgkaTodeD8zjT6bLb+fkiaRkkVPRFxd45Tmfj5b3TBTsPFmIqnRZW/6GrQOqDrnb+nf6P3dbPsA+jd6r6Y1jd5FYjyQ1DC3b9X+3YBXZNvTgftoUYNdwTj3yG3/FXBT9DWEPZjFu1u2Pa2sOLPzXkdqRFQZ9zN7jTnUb6Q9gv6Nire0+14WjLOb1L739qr9OwO75LZ/CxxWYpy7V/6tSV+0a7N7W+jz0q44s+O7kto5di7jfmb35TJgcYNzmvr5bNnNHsabP5zUyv8AsCjb93nSX+oAOwLfyz70twD75J67KHvevcAHSozxl8DjwKrs55ps/9uBO7IP+R3A/JLv5T8Ad2XxXA+8Lvfck7J7fD9wYplxZo8/B3y56nltu5+kvx4fBbaT6nnnAycDJ2fHRVoI7IEslrkl3cvB4vwm8ETus7ki279Pdh9vyz4Ti0qO82O5z+ZN5BJcrc9LWXFm55xA6nCTf17b7iepWjGA23P/roe38vPpqUHMzKyQTmnDMDOzDueEYWZmhThhmJlZIU4YZmZWiBOGmZkV4oRhZmaFOGGYmVkhThhmI5CtR3Botv1FSV8vOyazVvHkg2Yjcw7w+WyiuQOBD5Ucj1nLeKS32QhJ+ldgKnBQpHUJzMYkV0mZjYCk/0Ba+Wybk4WNdU4YZsOUrVzWQ1rV7FlJLVtRz6wTOGGYDUO20trVwCcj4m7gC6T2DLMxy20YZmZWiEsYZmZWiBOGmZkV4oRhZmaFOGGYmVkhThhmZlaIE4aZmRXihGFmZoX8OwP7A67UJe89AAAAAElFTkSuQmCC\n", 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    " - ] - }, - "metadata": {}, - "output_type": "display_data" - } - ], + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [], "source": [ "# Importing various packages\n", "from random import random, seed\n", @@ -997,7 +992,7 @@ "from sklearn.linear_model import LinearRegression\n", "\n", "x = 2*np.random.rand(100,1)\n", - "y = 4+3*x+0.01*np.random.randn(100,1)\n", + "y = 4+3*x+np.random.randn(100,1)\n", "linreg = LinearRegression()\n", "linreg.fit(x,y)\n", "xnew = np.array([[0],[2]])\n", @@ -1068,20 +1063,11 @@ }, { "cell_type": "code", - "execution_count": 11, - "metadata": {}, - "outputs": [ - { - "data": { - "image/png": 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\n", 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    " - ] - }, - "metadata": {}, - "output_type": "display_data" - } - ], + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [], "source": [ "# Importing various packages\n", "import numpy as np\n", @@ -1216,27 +1202,18 @@ }, { "cell_type": "code", - "execution_count": 19, - "metadata": {}, - "outputs": [ - { - "data": { - "image/png": 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\n", - "text/plain": [ - "
    " - ] - }, - "metadata": {}, - "output_type": "display_data" - } - ], + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [], "source": [ "import numpy as np\n", "import matplotlib.pyplot as plt\n", "from sklearn.linear_model import LinearRegression\n", "\n", "x = np.random.rand(100,1)\n", - "y = 5*x+np.random.randn(100,1)\n", + "y = 5*x+0.01*np.random.randn(100,1)\n", "linreg = LinearRegression()\n", "linreg.fit(x,y)\n", "ypredict = linreg.predict(x)\n", @@ -1272,34 +1249,11 @@ }, { "cell_type": "code", - "execution_count": 21, - "metadata": {}, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The intercept alpha: \n", - " [2.00137415]\n", - "Coefficient beta : \n", - " [[4.99698001]]\n", - "Mean squared error: 0.00\n", - "Variance score: 1.00\n", - "Mean squared log error: 0.00\n", - "Mean absolute error: 0.01\n" - ] - }, - { - "data": { - "image/png": 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- "text/plain": [ - "
    " - ] - }, - "metadata": {}, - "output_type": "display_data" - } - ], + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [], "source": [ "import numpy as np \n", "import matplotlib.pyplot as plt \n", @@ -1307,7 +1261,7 @@ "from sklearn.metrics import mean_squared_error, r2_score, mean_squared_log_error, mean_absolute_error\n", "\n", "x = np.random.rand(100,1)\n", - "y = 2.0+ 5*x+0.01*np.random.randn(100,1)\n", + "y = 2.0+ 5*x+0.5*np.random.randn(100,1)\n", "linreg = LinearRegression()\n", "linreg.fit(x,y)\n", "ypredict = linreg.predict(x)\n", @@ -1446,32 +1400,17 @@ "\n", "We will discuss in more\n", "detail these and other functions in the various lectures. We conclude this part with another example. Instead of \n", - "a linear $x$-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn." + "a linear $x$-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn. \n", + "Add description of the various python commands." ] }, { "cell_type": "code", - "execution_count": 22, - "metadata": {}, - "outputs": [ - { - "data": { - "image/png": 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\n", 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    " - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "name": "stdout", - "output_type": "stream", - "text": [ - "0.004999999999999991\n" - ] - } - ], + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [], "source": [ "import matplotlib.pyplot as plt\n", "import numpy as np\n", @@ -1509,110 +1448,22 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "Similarly, using **R**, we can perform similar studies. \n", - "(more details on **R** will be inserted later).\n", + "Using **R**, we can perform similar studies. \n", "\n", "\n", "\n", "\n", "\n", "\n", - "\n", - "\n", - "\n", - "\n", - "\n", - "\n", - "## Simple regression model with gradient descent\n", - "Add info about the equations, play around with different learning rates" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": {}, - "outputs": [], - "source": [ - "# Importing various packages\n", - "from math import exp, sqrt\n", - "from random import random, seed\n", - "import numpy as np\n", - "import matplotlib.pyplot as plt\n", - "\n", - "x = 2*np.random.rand(100,1)\n", - "y = 4+3*x+np.random.randn(100,1)\n", - "\n", - "xb = np.c_[np.ones((100,1)), x]\n", - "theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)\n", - "print(theta_linreg)\n", - "theta = np.random.randn(2,1)\n", - "\n", - "eta = 0.1\n", - "Niterations = 1000\n", - "m = 100\n", - "\n", - "for iter in range(Niterations):\n", - " gradients = 2.0/m*xb.T.dot(xb.dot(theta)-y)\n", - " theta -= eta*gradients\n", - "\n", - "print(theta)\n", - "xnew = np.array([[0],[2]])\n", - "xbnew = np.c_[np.ones((2,1)), xnew]\n", - "ypredict = xbnew.dot(theta)\n", - "ypredict2 = xbnew.dot(theta_linreg)\n", - "plt.plot(xnew, ypredict, \"r-\")\n", - "plt.plot(xnew, ypredict2, \"b-\")\n", - "plt.plot(x, y ,'ro')\n", - "plt.axis([0,2.0,0, 15.0])\n", - "plt.xlabel(r'$x$')\n", - "plt.ylabel(r'$y$')\n", - "plt.title(r'Random numbers ')\n", - "plt.show()" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Simple regression model with stochastic gradient descent\n", - "Add info about the equations, play around with different learning rates" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": {}, - "outputs": [], - "source": [ - "# Importing various packages\n", - "from math import exp, sqrt\n", - "from random import random, seed\n", - "import numpy as np\n", - "import matplotlib.pyplot as plt\n", - "from sklearn.linear_model import SGDRegressor\n", - "\n", - "x = 2*np.random.rand(100,1)\n", - "y = 4+3*x+np.random.randn(100,1)\n", - "\n", - "xb = np.c_[np.ones((100,1)), x]\n", - "theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)\n", - "print(theta_linreg)\n", - "sgdreg = SGDRegressor(n_iter = 50, penalty=None, eta0=0.1)\n", - "sgdreg.fit(x,y.ravel())\n", - "print(sgdreg.intercept_, sgdreg.coef_)" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ "## Polynomial Regression" ] }, { "cell_type": "code", - "execution_count": 10, - "metadata": {}, + "execution_count": 8, + "metadata": { + "collapsed": false + }, "outputs": [], "source": [ "# Importing various packages\n", @@ -1644,72 +1495,131 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "## Ridge and Lasso Regression" + "\n", + "## Linking the regression analysis with a statistical interpretation\n", + "\n", + "Before we proceed, and to link with our discussions of Bayesian statistics to come, it is useful the derive the standard regression analysis equations using a statistical interpretation. This allows us also to derive quantities like the variance and other expectation values in a rather straightforward way. \n", + "\n", + "It is assumed that $\\varepsilon_i\n", + "\\sim \\mathcal{N}(0, \\sigma^2)$ and the $\\varepsilon_{i}$ are\n", + "independent, i.e.:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*} \n", + "\\mbox{Cov}(\\varepsilon_{i_1},\n", + "\\varepsilon_{i_2}) & = \\left\\{ \\begin{array}{lcc} \\sigma^2 & \\mbox{if}\n", + "& i_1 = i_2, \\\\ 0 & \\mbox{if} & i_1 \\not= i_2. \\end{array} \\right.\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The randomness of $\\varepsilon_i$ implies that\n", + "$\\mathbf{Y}_i$ is also a random variable. In particular,\n", + "$\\mathbf{Y}_i$ is normally distributed, because $\\varepsilon_i \\sim\n", + "\\mathcal{N}(0, \\sigma^2)$ and $\\mathbf{X}_{i,\\ast} \\, \\beta$ is a\n", + "non-random scalar. To specify the parameters of the distribution of\n", + "$\\mathbf{Y}_i$ we need to calculate its first two moments. \n", + "\n", + "## Expectation value and variance\n", + "\n", + "Its expectation equals:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*} \n", + "\\mathbb{E}(Y_i) & =\n", + "\\mathbb{E}(\\mathbf{X}_{i, \\ast} \\, \\beta) + \\mathbb{E}(\\varepsilon_i)\n", + "\\, \\, \\, = \\, \\, \\, \\mathbf{X}_{i, \\ast} \\, \\beta, \n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "while\n", + "its variance is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*} \\mbox{Var}(Y_i) & = \\mathbb{E} \\{ [Y_i\n", + "- \\mathbb{E}(Y_i)]^2 \\} \\, \\, \\, = \\, \\, \\, \\mathbb{E} ( Y_i^2 ) -\n", + "[\\mathbb{E}(Y_i)]^2 \\\\ & = \\mathbb{E} [ ( \\mathbf{X}_{i, \\ast} \\,\n", + "\\beta + \\varepsilon_i )^2] - ( \\mathbf{X}_{i, \\ast} \\, \\beta)^2 \\\\ &\n", + "= \\mathbb{E} [ ( \\mathbf{X}_{i, \\ast} \\, \\beta)^2 + 2 \\varepsilon_i\n", + "\\mathbf{X}_{i, \\ast} \\, \\beta + \\varepsilon_i^2 ] - ( \\mathbf{X}_{i,\n", + "\\ast} \\, \\beta)^2 \\\\ & = ( \\mathbf{X}_{i, \\ast} \\, \\beta)^2 + 2\n", + "\\mathbb{E}(\\varepsilon_i) \\mathbf{X}_{i, \\ast} \\, \\beta +\n", + "\\mathbb{E}(\\varepsilon_i^2 ) - ( \\mathbf{X}_{i, \\ast} \\, \\beta)^2 \n", + "\\\\ & = \\mathbb{E}(\\varepsilon_i^2 ) \\, \\, \\, = \\, \\, \\,\n", + "\\mbox{Var}(\\varepsilon_i) \\, \\, \\, = \\, \\, \\, \\sigma^2. \n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Hence, $Y_i \\sim \\mathcal{N}( \\mathbf{X}_{i, \\ast} \\, \\beta, \\sigma^2)$. \n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "## The singular value decompostion\n", + "\n", + "\n", + "A general\n", + "$m\\times n$ matrix $\\hat{A}$ can be written in terms of a diagonal\n", + "matrix $\\hat{D}$ of dimensionality $n\\times n$ and two orthognal\n", + "matrices $\\hat{U}$ and $\\hat{V}$, where the first has dimensionality\n", + "$m \\times m$ and the last dimensionality $n\\times n$. \n", + "We have then" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\hat{A} = \\hat{U}\\hat{D}\\hat{V}^T\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Code examples for Ridge and Lasso Regression" ] }, { "cell_type": "code", - "execution_count": 1, - "metadata": {}, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "X_train: (37, 1)\n", - "y_train: (37,)\n", - "X_test: (13, 1)\n", - "y_test: (13,)\n", - "------------------------------------\n", - "Ordinary Least Squares\n", - "Prediction Shape: (13,)\n", - "Coefficients: \n", - " [0.54090544]\n", - "Mean squared error: 4.92\n", - "Variance score: 0.11\n" - ] - }, - { - "name": "stderr", - "output_type": "stream", - "text": [ - "/usr/local/lib/python3.7/site-packages/sklearn/linear_model/base.py:509: RuntimeWarning: internal gelsd driver lwork query error, required iwork dimension not returned. This is likely the result of LAPACK bug 0038, fixed in LAPACK 3.2.2 (released July 21, 2010). Falling back to 'gelss' driver.\n", - " linalg.lstsq(X, y)\n" - ] - }, - { - "data": { - "text/plain": [ - "
    " - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "name": "stdout", - "output_type": "stream", - "text": [ - "------------------------------------\n", - "Ridge Regression\n", - "Ridge Coefficient: [0.54089262]\n", - "Ridge Intercept: 4.6076494432587065\n", - "------------------------------------\n", - "Lasso\n", - "Lasso Coefficient: [0.54002825]\n", - "Lasso Intercept: 4.623208095009552\n" - ] - }, - { - "data": { - "text/plain": [ - "
    " - ] - }, - "metadata": {}, - "output_type": "display_data" - } - ], + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [], "source": [ "import matplotlib.pyplot as plt\n", "import numpy as np\n", @@ -1789,8 +1699,14 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "## The singular value decompostion\n", - "How can we use the singular value decomposition to find the parameters $\\beta_j$? More details will come. We first note that a general $m\\times n$ matrix $\\hat{A}$ can be written in terms of a diagonal matrix $\\hat{\\Sigma}$ of dimensionality $n\\times n$ and two orthognal matrices $\\hat{U}$ and $\\hat{V}$, where the first has dimensionality $m \\times n$ and the last dimensionality $n\\times n$. We have then" + "## From standard regression to Ridge regressions\n", + "\n", + "One of the typical problems we encounter with linear regression, in particular \n", + "when the matrix $\\hat{X}$ (our so-called design matrix) is high-dimensional, \n", + "are problems with near singular or singular matrices. The column vectors of $\\hat{X}$ \n", + "may be linearly dependent, normally referred to as super-collinearity. \n", + "This means that the matrix may be rank deficient and it is basically impossible to \n", + "to model the data using linear regression. As an example, consider the matrix" ] }, { @@ -1798,7 +1714,18 @@ "metadata": {}, "source": [ "$$\n", - "\\hat{A} = \\hat{U}\\hat{\\Sigma}\\hat{V}\n", + "\\begin{align*}\n", + "\\mathbf{X} & = \\left[\n", + "\\begin{array}{rrr}\n", + "1 & -1 & 2\n", + "\\\\\n", + "1 & 0 & 1\n", + "\\\\\n", + "1 & 2 & -1\n", + "\\\\\n", + "1 & 1 & 0\n", + "\\end{array} \\right]\n", + "\\end{align*}\n", "$$" ] }, @@ -1806,39 +1733,414 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "Add codes and discuss this in connection with lasso and ridge, show example where the standard inversion of a matrix fails and where SVD comes to rescue\n", + "The columns of $\\hat{X}$ are linearly dependent. We se this easily since the \n", + "the first column is the row-wise sum of the other two columns. The rank (more correct,\n", + "the column rank) of a matrix is the dimension of the space spanned by the\n", + "column vectors. Hence, the rank of $\\mathbf{X}$ is equal to the number\n", + "of linearly independent columns. In this particular case the matrix has rank 2.\n", + "\n", + "Super-collinearity of an $(n \\times p)$-dimensional design matrix $\\mathbf{X}$ implies\n", + "that the inverse of the matrix $\\hat{X}^T\\hat{x}$ (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + "\\hat{X} & = \\left[\n", + "\\begin{array}{rr}\n", + "1 & -1\n", + "\\\\\n", + "1 & -1\n", + "\\end{array} \\right].\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We see easily that $\\mbox{det}(\\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \\times (-1) - 1 \\times (-1) = 0$. Hence, $\\mathbf{X}$ is singular and its inverse is undefined.\n", + "This is equivalent to saying that the matrix $\\hat{X}$ has at least an eigenvalue which is zero.\n", + "\n", + "## Fixing the singularity\n", + "\n", + "If our design matrix $\\hat{X}$ which enters the linear regression problem" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
    \n", + "\n", + "$$\n", + "\\begin{equation}\n", + "\\hat{\\beta} = (\\hat{X}^{T} \\hat{X})^{-1} \\hat{X}^{T} \\hat{y},\n", + "\\label{_auto1} \\tag{1}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "has linearly dependent column vectors, we will not be able to compute the inverse\n", + "of $\\hat{X}^T\\hat{X}$ and we cannot find the parameters (estimators) $\\beta_i$. \n", + "The estimators are only well-defined if $(\\hat{X}^{T}\\hat{X})^{-1}$ exits. \n", + "This is more likely to happen when the matrix $\\hat{X}$ is high-dimensional. In this case it is likely to encounter a situation where \n", + "the regression parameters $\\beta_i$ cannot be estimated.\n", + "\n", + "The *ad hoc* approach which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\hat{X}^{T} \\hat{X} \\rightarrow \\hat{X}^{T} \\hat{X}+\\lambda \\hat{I},\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $\\hat{I}$ is the identity matrix.\n", "\n", "\n", - "## Lasso and Ridge regression\n", "\n", - "Discuss the mathematics here\n", + "## A second-order polynomial with Ridge and Lasso" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + "import numpy as np\n", + "import matplotlib.pyplot as plt\n", + "from sklearn.linear_model import Ridge\n", + "from sklearn.metrics import r2_score\n", "\n", - "## Logistic regression\n", - "Add discussion about classification versus regression, show examples of more than two cases and why regression is not the best approach. Motivate for k-nearest neighbors\n", + "np.random.seed(4155)\n", "\n", - "Add examples on classification problems" + "n_samples = 100\n", + "\n", + "x = np.random.rand(n_samples,1)\n", + "y = 5*x*x + 0.1*np.random.rand(n_samples,1)\n", + "\n", + "# Centering x and y.\n", + "x_ = x - np.mean(x)\n", + "y_ = y - np.mean(y) # beta_0 = mean(y)\n", + "\n", + "X = np.c_[np.ones((n_samples,1)), x, x**2]\n", + "X_ = np.c_[x_, x_**2]\n", + "\n", + "\n", + "### 1.\n", + "lmb_values = [1e-4, 1e-3, 1e-2, 10, 1e2, 1e4]\n", + "num_values = len(lmb_values)\n", + "\n", + "## Ridge-regression of centered and not centered data\n", + "beta_ridge = np.zeros((3,num_values))\n", + "beta_ridge_centered = np.zeros((3,num_values))\n", + "\n", + "I3 = np.eye(3)\n", + "I2 = np.eye(2)\n", + "\n", + "for i,lmb in enumerate(lmb_values):\n", + " beta_ridge[:,i] = (np.linalg.inv( X.T @ X + lmb*I3) @ X.T @ y).flatten()\n", + " beta_ridge_centered[1:,i] = (np.linalg.inv( X_.T @ X_ + lmb*I2) @ X_.T @ y_).flatten()\n", + "\n", + "# sett beta_0 = np.mean(y)\n", + "beta_ridge_centered[0,:] = np.mean(y)\n", + "\n", + "## OLS (ordinary least squares) solution \n", + "beta_ls = np.linalg.inv( X.T @ X ) @ X.T @ y\n", + "\n", + "## Evaluate the models\n", + "pred_ls = X @ beta_ls\n", + "pred_ridge = X @ beta_ridge\n", + "pred_ridge_centered = X_ @ beta_ridge_centered[1:] + beta_ridge_centered[0,:]\n", + "\n", + "## Plot the results\n", + "\n", + "# Sorting\n", + "sort_ind = np.argsort(x[:,0])\n", + "\n", + "x_plot = x[sort_ind,0]\n", + "x_centered_plot = x_[sort_ind,0]\n", + "\n", + "pred_ls_plot = pred_ls[sort_ind,0]\n", + "pred_ridge_plot = pred_ridge[sort_ind,:]\n", + "pred_ridge_centered_plot = pred_ridge_centered[sort_ind,:]\n", + "\n", + "# Plott not centered\n", + "plt.plot(x_plot,pred_ls_plot,label='ls')\n", + "\n", + "for i in range(num_values):\n", + " plt.plot(x_plot,pred_ridge_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])\n", + "\n", + "plt.plot(x,y,'ro')\n", + "\n", + "plt.title('linear regression on un-centered data')\n", + "plt.legend()\n", + "\n", + "# Plott centered\n", + "plt.figure()\n", + "\n", + "for i in range(num_values):\n", + " plt.plot(x_centered_plot,pred_ridge_centered_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])\n", + "\n", + "plt.plot(x_,y,'ro')\n", + "\n", + "plt.title('linear regression on centered data')\n", + "plt.legend()\n", + "\n", + "\n", + "# 2.\n", + "\n", + "pred_ridge_scikit = np.zeros((n_samples,num_values))\n", + "for i,lmb in enumerate(lmb_values):\n", + " pred_ridge_scikit[:,i] = (Ridge(alpha=lmb,fit_intercept=False).fit(X,y).predict(X)).flatten() # fit_intercept=False fordi bias er allerede i X\n", + "\n", + "plt.figure()\n", + "\n", + "plt.plot(x_plot,pred_ls_plot,label='ls')\n", + "\n", + "for i in range(num_values):\n", + " plt.plot(x_plot,pred_ridge_scikit[sort_ind,i],label='scikit-ridge, lmb=%g'%lmb_values[i])\n", + "\n", + "plt.plot(x,y,'ro')\n", + "plt.legend()\n", + "plt.title('linear regression using scikit')\n", + "\n", + "plt.show()\n", + "\n", + "### R2-score of the results\n", + "for i in range(num_values):\n", + " print('lambda = %g'%lmb_values[i])\n", + " print('r2 for scikit: %g'%r2_score(y,pred_ridge_scikit[:,i]))\n", + " print('r2 for own code, not centered: %g'%r2_score(y,pred_ridge[:,i]))\n", + " print('r2 for own, centered: %g\\n'%r2_score(y,pred_ridge_centered[:,i]))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Fitting vs. predicting when data is in the model class\n", + "\n", + "We start by considering the case\n", + "$f(x)=2x$.\n", + "\n", + "Then the data is clearly generated by a model that is contained within\n", + "all three model classes we are using to make predictions (linear\n", + "models, third order polynomials, and tenth order polynomials).\n", + "\n", + "Run the code for the following cases:\n", + "\n", + "1. For $f(x)=2x$ , $Ntrain=10$ and $\\sigma =0$ (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when $x \\in [0,1]$ . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?\n", + "\n", + "2. Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?\n", + "\n", + "3. Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example $x \\in [0,1.2]$ ) using the code below. How well do you do on points in the range of x where you trained the model? How about points outside the original training data set?\n", + "\n", + "4. Repeat the above for $f(x)=2x$ , $Ntrain=10$ , and $\\sigma=1$ . What changes?\n", + "\n", + "Repeat the exercises above for $f(x)=2x$ , $Ntrain=100$ , and $\\sigma=1$ . What changes?\n", + "Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well.\n", + "\n", + "\n", + "## Fitting versus predicting when data is not in the model class\n", + "\n", + "Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider $f(x)=2x-10x^5+15x^{10}$ . Notice that the for linear and third-order polynomial the true model $f(x)$ is not contained in model class.\n", + "\n", + "1. Do better fits lead to better predictions?\n", + "\n", + "2. What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points $Ntrain$ and $\\sigma$?\n", + "\n", + "Summarize what you think you learned about the relationship of knowing the true model class and predictive power.\n", + "\n", + "## The code" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + "import numpy as np\n", + "import sklearn as sk\n", + "from sklearn import datasets, linear_model\n", + "from sklearn.preprocessing import PolynomialFeatures\n", + "\n", + "import matplotlib as mpl\n", + "from matplotlib import pyplot as plt\n", + "\n", + "%matplotlib notebook\n", + "\n", + "# The Training Data\n", + "\n", + "N_train=100\n", + "\n", + "sigma_train=1;\n", + "\n", + "# Train on integers\n", + "x=np.linspace(0.05,0.95,N_train)\n", + "# Draw random noise\n", + "s = sigma_train*np.random.randn(N_train)\n", + "\n", + "#linear\n", + "y=2*x+s\n", + "\n", + "#Tenth Order\n", + "#y=2*x-10*x**5+15*x**10+s\n", + "\n", + "p1=plt.plot(x,y, \"o\",ms=15, label='Training')\n", + "\n", + "#Linear Regression\n", + "# Create linear regression object\n", + "clf = linear_model.LinearRegression()\n", + "\n", + "# Train the model using the training sets\n", + "clf.fit(x[:, np.newaxis], y)\n", + "# The coefficients\n", + "\n", + "xplot=np.linspace(0.02,0.98,200)\n", + "linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')\n", + "\n", + "#Polynomial Regression\n", + "\n", + "\n", + "poly3 = PolynomialFeatures(degree=3)\n", + "X = poly3.fit_transform(x[:,np.newaxis])\n", + "clf3 = linear_model.LinearRegression()\n", + "clf3.fit(X,y)\n", + "\n", + "\n", + "Xplot=poly3.fit_transform(xplot[:,np.newaxis])\n", + "poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')\n", + "\n", + "\n", + "\n", + "#poly5 = PolynomialFeatures(degree=5)\n", + "#X = poly5.fit_transform(x[:,np.newaxis])\n", + "#clf5 = linear_model.LinearRegression()\n", + "#clf5.fit(X,y)\n", + "\n", + "#Xplot=poly5.fit_transform(xplot[:,np.newaxis])\n", + "#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)\n", + "\n", + "poly10 = PolynomialFeatures(degree=10)\n", + "X = poly10.fit_transform(x[:,np.newaxis])\n", + "clf10 = linear_model.LinearRegression()\n", + "clf10.fit(X,y)\n", + "\n", + "Xplot=poly10.fit_transform(xplot[:,np.newaxis])\n", + "poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')\n", + "\n", + "axes = plt.gca()\n", + "axes.set_ylim([-7,7])\n", + "\n", + "handles, labels=axes.get_legend_handles_labels()\n", + "plt.legend(handles,labels, loc='lower center')\n", + "plt.xlabel(\"$x$\")\n", + "plt.ylabel(\"$y$\")\n", + "Title=\"$N=$\"+str(N_train)+\", $\\sigma=$\"+str(sigma_train)\n", + "plt.title(Title+\" (train)\")\n", + "plt.tight_layout()\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "## Generating test data" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + "# Generate Test Data\n", + "\n", + "#Number of test data\n", + "N_test=20\n", + "\n", + "sigma_test=sigma_train\n", + "\n", + "max_x=1.2\n", + "x_test=max_x*np.random.random(N_test)\n", + "# Draw random noise\n", + "s_test = sigma_test*np.random.randn(N_test)\n", + "\n", + "#Linear\n", + "y_test=2*x_test+s_test\n", + "#Tenth order\n", + "#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test\n", + "\n", + "#Make design matrices for prediction\n", + "x_plot=np.linspace(0,max_x, 200)\n", + "X3 = poly3.fit_transform(x_plot[:,np.newaxis])\n", + "X10 = poly10.fit_transform(x_plot[:,np.newaxis])\n", + "\n", + "%matplotlib notebook\n", + "\n", + "fig = plt.figure() \n", + "p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')\n", + "p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')\n", + "p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')\n", + "p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')\n", + "\n", + "\n", + "plt.legend(loc=2)\n", + "plt.xlabel('$x$')\n", + "plt.ylabel('$y$')\n", + "plt.legend(loc='best')\n", + "plt.title(Title+\" (pred.)\")\n", + "plt.tight_layout()\n", + "plt.show()\n", + "\n", + "#Linear Filename\n", + "#filename_test=Title+\"pred-linear.pdf\"\n", + "#Tenth Order Filename\n", + "#filename_test=Title+\"pred-o10.pdf\"\n", + "#plt.savefig(filename_test)\n", + "#plt.ylim((-6,12))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Lasso regression\n", + "\n", + "\n", + "## Logistic regression" ] } ], - "metadata": { - "kernelspec": { - "display_name": "Python 3", - "language": "python", - "name": "python3" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 3 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython3", - "version": "3.7.0" - } - }, + "metadata": {}, "nbformat": 4, "nbformat_minor": 2 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