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Given the following text description, write Python code to implement the functionality described below step by step
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Description:
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Step1:  Simple preprocessing examples, breast cancer data and classification, Support Vector Machines
We show here how we can use a simple regression case on the breast
cancer data using support vector machines (SVM) as algorithm for
classification.
Step2:  More on Cancer Data, now with Logistic Regression
Step3:  Why should we think of reducing the dimensionality
In addition to the plot of the features, we study now also the covariance (and the correlation matrix).
We use also Pandas to compute the correlation matrix.
Step4:  In the above example we note two things. In the first plot we display
the overlap of benign and malignant tumors as functions of the various
features in the Wisconsing breast cancer data set. We see that for
some of the features we can distinguish clearly the benign and
malignant cases while for other features we cannot. This can point to
us which features may be of greater interest when we wish to classify
a benign or not benign tumour.
In the second figure we have computed the so-called correlation
matrix, which in our case with thirty features becomes a $30\times 30$
matrix.
We constructed this matrix using pandas via the statements
Step5:  and then
Step6:  Diagonalizing this matrix we can in turn say something about which
features are of relevance and which are not. But before we proceed we
need to define covariance and correlation matrices. This leads us to
the classical Principal Component Analysis (PCA) theorem with
applications.
Basic ideas of the Principal Component Analysis (PCA)
The principal component analysis deals with the problem of fitting a
low-dimensional affine subspace $S$ of dimension $d$ much smaller than
the totaldimension $D$ of the problem at hand (our data
set). Mathematically it can be formulated as a statistical problem or
a geometric problem.  In our discussion of the theorem for the
classical PCA, we will stay with a statistical approach. This is also
what set the scene historically which for the PCA.
We have a data set defined by a design/feature matrix $\boldsymbol{X}$ (see below for its definition) 
* Each data point is determined by $p$ extrinsic (measurement) variables
We may want to ask the following question
Step7:  Correlation Matrix
The previous example can be converted into the correlation matrix by
simply scaling the matrix elements with the variances.  We should also
subtract the mean values for each column. This leads to the following
code which sets up the correlations matrix for the previous example in
a more brute force way. Here we scale the mean values for each column of the design matrix, calculate the relevant mean values and variances and then finally set up the $2\times 2$ correlation matrix (since we have only two vectors).
Step8:  We see that the matrix elements along the diagonal are one as they
should be and that the matrix is symmetric. Furthermore, diagonalizing
this matrix we easily see that it is a positive definite matrix.
The above procedure with numpy can be made more compact if we use pandas.
Correlation Matrix with Pandas
We whow here how we can set up the correlation matrix using pandas, as done in this simple code
Step9:  We expand this model to the Franke function discussed above.
Correlation Matrix with Pandas and the Franke function
Step10:  We note here that the covariance is zero for the first rows and
columns since all matrix elements in the design matrix were set to one
(we are fitting the function in terms of a polynomial of degree $n$).
This means that the variance for these elements will be zero and will
cause problems when we set up the correlation matrix.  We can simply
drop these elements and construct a correlation
matrix without these elements. 
Rewriting the Covariance and/or Correlation Matrix
We can rewrite the covariance matrix in a more compact form in terms of the design/feature matrix $\boldsymbol{X}$ as
$$
\boldsymbol{C}[\boldsymbol{x}] = \frac{1}{n}\boldsymbol{X}\boldsymbol{X}^T= \mathbb{E}[\boldsymbol{X}\boldsymbol{X}^T].
$$
To see this let us simply look at a design matrix $\boldsymbol{X}\in {\mathbb{R}}^{2\times 2}$
$$
\boldsymbol{X}=\begin{bmatrix}
x_{00} & x_{01}\
x_{10} & x_{11}\
\end{bmatrix}=\begin{bmatrix}
\boldsymbol{x}{0} & \boldsymbol{x}{1}\
\end{bmatrix}.
$$
If we then compute the expectation value
$$
\mathbb{E}[\boldsymbol{X}\boldsymbol{X}^T] = \frac{1}{n}\boldsymbol{X}\boldsymbol{X}^T=\begin{bmatrix}
x_{00}^2+x_{01}^2 & x_{00}x_{10}+x_{01}x_{11}\
x_{10}x_{00}+x_{11}x_{01} & x_{10}^2+x_{11}^2\
\end{bmatrix},
$$
which is just
$$
\boldsymbol{C}[\boldsymbol{x}_0,\boldsymbol{x}_1] = \boldsymbol{C}[\boldsymbol{x}]=\begin{bmatrix} \mathrm{var}[\boldsymbol{x}_0] & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_1] \
                              \mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_0] & \mathrm{var}[\boldsymbol{x}_1] \
             \end{bmatrix},
$$
where we wrote $$\boldsymbol{C}[\boldsymbol{x}_0,\boldsymbol{x}_1] = \boldsymbol{C}[\boldsymbol{x}]$$ to indicate that this the covariance of the vectors $\boldsymbol{x}$ of the design/feature matrix $\boldsymbol{X}$.
It is easy to generalize this to a matrix $\boldsymbol{X}\in {\mathbb{R}}^{n\times p}$.
Towards the PCA theorem
We have that the covariance matrix (the correlation matrix involves a simple rescaling) is given as
$$
\boldsymbol{C}[\boldsymbol{x}] = \frac{1}{n}\boldsymbol{X}\boldsymbol{X}^T= \mathbb{E}[\boldsymbol{X}\boldsymbol{X}^T].
$$
Let us now assume that we can perform a series of orthogonal transformations where we employ some orthogonal matrices $\boldsymbol{S}$.
These matrices are defined as $\boldsymbol{S}\in {\mathbb{R}}^{p\times p}$ and obey the orthogonality requirements $\boldsymbol{S}\boldsymbol{S}^T=\boldsymbol{S}^T\boldsymbol{S}=\boldsymbol{I}$. The matrix can be written out in terms of the column vectors $\boldsymbol{s}i$ as $\boldsymbol{S}=[\boldsymbol{s}_0,\boldsymbol{s}_1,\dots,\boldsymbol{s}{p-1}]$ and $\boldsymbol{s}_i \in {\mathbb{R}}^{p}$.
Assume also that there is a transformation $\boldsymbol{S}\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}^T=\boldsymbol{C}[\boldsymbol{y}]$ such that the new matrix $\boldsymbol{C}[\boldsymbol{y}]$ is diagonal with elements $[\lambda_0,\lambda_1,\lambda_2,\dots,\lambda_{p-1}]$.  
That is we have
$$
\boldsymbol{C}[\boldsymbol{y}] = \mathbb{E}[\boldsymbol{S}\boldsymbol{X}\boldsymbol{X}^T\boldsymbol{S}^T]=\boldsymbol{S}\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}^T,
$$
since the matrix $\boldsymbol{S}$ is not a data dependent matrix.   Multiplying with $\boldsymbol{S}^T$ from the left we have
$$
\boldsymbol{S}^T\boldsymbol{C}[\boldsymbol{y}] = \boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}^T,
$$
and since $\boldsymbol{C}[\boldsymbol{y}]$ is diagonal we have for a given eigenvalue $i$ of the covariance matrix that
$$
\boldsymbol{S}^T_i\lambda_i = \boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}^T_i.
$$
In the derivation of the PCA theorem we will assume that the eigenvalues are ordered in descending order, that is
$\lambda_0 > \lambda_1 > \dots > \lambda_{p-1}$. 
The eigenvalues tell us then how much we need to stretch the
corresponding eigenvectors. Dimensions with large eigenvalues have
thus large variations (large variance) and define therefore useful
dimensions. The data points are more spread out in the direction of
these eigenvectors.  Smaller eigenvalues mean on the other hand that
the corresponding eigenvectors are shrunk accordingly and the data
points are tightly bunched together and there is not much variation in
these specific directions. Hopefully then we could leave it out
dimensions where the eigenvalues are very small. If $p$ is very large,
we could then aim at reducing $p$ to $l << p$ and handle only $l$
features/predictors.
The Algorithm before theorem
Here's how we would proceed in setting up the algorithm for the PCA, see also discussion below here. 
* Set up the datapoints for the design/feature matrix $\boldsymbol{X}$ with $\boldsymbol{X}\in {\mathbb{R}}^{n\times p}$, with the predictors/features $p$  referring to the column numbers and the entries $n$ being the row elements.
$$
\boldsymbol{X}=\begin{bmatrix}
x_{0,0} & x_{0,1} & x_{0,2}& \dots & \dots x_{0,p-1}\
x_{1,0} & x_{1,1} & x_{1,2}& \dots & \dots x_{1,p-1}\
x_{2,0} & x_{2,1} & x_{2,2}& \dots & \dots x_{2,p-1}\
\dots & \dots & \dots & \dots \dots & \dots \
x_{n-2,0} & x_{n-2,1} & x_{n-2,2}& \dots & \dots x_{n-2,p-1}\
x_{n-1,0} & x_{n-1,1} & x_{n-1,2}& \dots & \dots x_{n-1,p-1}\
\end{bmatrix},
$$
Center the data by subtracting the mean value for each column. This leads to a new matrix $\boldsymbol{X}\rightarrow \overline{\boldsymbol{X}}$.
Compute then the covariance/correlation matrix $\mathbb{E}[\overline{\boldsymbol{X}}\overline{\boldsymbol{X}}^T]$.
Find the eigenpairs of $\boldsymbol{C}$ with eigenvalues $[\lambda_0,\lambda_1,\dots,\lambda_{p-1}]$ and eigenvectors $[\boldsymbol{s}0,\boldsymbol{s}_1,\dots,\boldsymbol{s}{p-1}]$.
Order the eigenvalue (and the eigenvectors accordingly) in order of decreasing eigenvalues.
Keep only those $l$ eigenvalues larger than a selected threshold value, discarding thus $p-l$ features since we expect small variations in the data here.
Writing our own PCA code
We will use a simple example first with two-dimensional data
drawn from a multivariate normal distribution with the following mean and covariance matrix
Step11:  Now we are going to implement the PCA algorithm. We will break it down into various substeps.
Compute the sample mean and center the data
The first step of PCA is to compute the sample mean of the data and use it to center the data. Recall that the sample mean is
$$
\mu_n = \frac{1}{n} \sum_{i=1}^n x_i
$$
and the mean-centered data $\bar{X} = { \bar{x}_1, \ldots, \bar{x}_n }$ takes the form
$$
\bar{x}_i = x_i - \mu_n.
$$
When you are done with these steps, print out $\mu_n$ to verify it is
close to $\mu$ and plot your mean centered data to verify it is
centered at the origin! Compare your code with the functionality from Scikit-Learn discussed above.
The following code elements perform these operations using pandas or using our own functionality for doing so. The latter, using numpy is rather simple through the mean() function.
Step12:  Alternatively, we could use the functions we discussed
earlier for scaling the data set.  That is, we could have used the
StandardScaler function in Scikit-Learn, a function which ensures
that for each feature/predictor we study the mean value is zero and
the variance is one (every column in the design/feature matrix).  You
would then not get the same results, since we divide by the
variance. The diagonal covariance matrix elements will then be one,
while the non-diagonal ones need to be divided by $2\sqrt{2}$ for our
specific case.
Compute the sample covariance
Now we are going to use the mean centered data to compute the sample covariance of the data by using the following equation
$$
\Sigma_n = \frac{1}{n-1} \sum_{i=1}^n \bar{x}i^T \bar{x}_i = \frac{1}{n-1} \sum{i=1}^n (x_i - \mu_n)^T (x_i - \mu_n)
$$
where the data points $x_i \in \mathbb{R}^p$ (here in this example $p = 2$) are column vectors and $x^T$ is the transpose of $x$.
We can write our own code or simply use either the functionaly of numpy or that of pandas, as follows
Step13:  Note that the way we define the covariance matrix here has a factor $n-1$ instead of $n$. This is included in the cov() function by numpy and pandas. 
Our own code here is not very elegant and asks for obvious improvements. It is tailored to this specific $2\times 2$ covariance matrix.
Step14:  Depending on the number of points $n$, we will get results that are close to the covariance values defined above.
The plot shows how the data are clustered around a line with slope close to one. Is this expected?
Diagonalize the sample covariance matrix to obtain the principal components
Now we are ready to solve for the principal components! To do so we
diagonalize the sample covariance matrix $\Sigma$. We can use the
function np.linalg.eig to do so. It will return the eigenvalues and
eigenvectors of $\Sigma$. Once we have these we can perform the 
following tasks
Step15:  This code does not contain all the above elements, but it shows how we can use Scikit-Learn to extract the eigenvector which corresponds to the largest eigenvalue. Try to address the questions we pose before the above code.  Try also to change the values of the covariance matrix by making one of the diagonal elements much larger than the other. What do you observe then? 
Classical PCA Theorem
We assume now that we have a design matrix $\boldsymbol{X}$ which has been
centered as discussed above. For the sake of simplicity we skip the
overline symbol. The matrix is defined in terms of the various column
vectors $[\boldsymbol{x}0,\boldsymbol{x}_1,\dots, \boldsymbol{x}{p-1}]$ each with dimension
$\boldsymbol{x}\in {\mathbb{R}}^{n}$.
We assume also that we have an orthogonal transformation $\boldsymbol{W}\in {\mathbb{R}}^{p\times p}$.  We define the reconstruction error (which is similar to the mean squared error we have seen before) as
$$
J(\boldsymbol{W},\boldsymbol{Z}) = \frac{1}{n}\sum_i (\boldsymbol{x}_i - \overline{\boldsymbol{x}}_i)^2,
$$
with $\overline{\boldsymbol{x}}_i = \boldsymbol{W}\boldsymbol{z}_i$, where $\boldsymbol{z}_i$ is a row vector with dimension ${\mathbb{R}}^{n}$ of the matrix
$\boldsymbol{Z}\in{\mathbb{R}}^{p\times n}$.  When doing PCA we want to reduce this dimensionality. 
The PCA theorem states that minimizing the above reconstruction error
corresponds to setting $\boldsymbol{W}=\boldsymbol{S}$, the orthogonal matrix which
diagonalizes the empirical covariance(correlation) matrix. The optimal
low-dimensional encoding of the data is then given by a set of vectors
$\boldsymbol{z}_i$ with at most $l$ vectors, with $l << p$, defined by the
orthogonal projection of the data onto the columns spanned by the
eigenvectors of the covariance(correlations matrix).
The proof which follows will be updated by mid January 2020.
Proof of the PCA Theorem
To show the PCA theorem let us start with the assumption that there is one vector $\boldsymbol{w}_0$ which corresponds to a solution which minimized the reconstruction error $J$. This is an orthogonal vector. It means that we now approximate the reconstruction error in terms of $\boldsymbol{w}_0$ and $\boldsymbol{z}_0$ as
$$
J(\boldsymbol{w}0,\boldsymbol{z}_0)= \frac{1}{n}\sum_i (\boldsymbol{x}_i - z{i0}\boldsymbol{w}0)^2=\frac{1}{n}\sum_i (\boldsymbol{x}_i^T\boldsymbol{x}_i - 2z{i0}\boldsymbol{w}0^T\boldsymbol{x}_i+z{i0}^2\boldsymbol{w}_0^T\boldsymbol{w}_0),
$$
which we can rewrite due to the orthogonality of $\boldsymbol{w}_i$ as
$$
J(\boldsymbol{w}0,\boldsymbol{z}_0)=\frac{1}{n}\sum_i (\boldsymbol{x}_i^T\boldsymbol{x}_i - 2z{i0}\boldsymbol{w}0^T\boldsymbol{x}_i+z{i0}^2).
$$
Minimizing $J$ with respect to the unknown parameters $z_{0i}$ we obtain that
$$
z_{i0}=\boldsymbol{w}_0^T\boldsymbol{x}_i,
$$
where the vectors on the rhs are known. 
PCA Proof continued
We have now found the unknown parameters $z_{i0}$. These correspond to the projected coordinates and we can write
$$
J(\boldsymbol{w}0)= \frac{1}{p}\sum_i (\boldsymbol{x}_i^T\boldsymbol{x}_i - z{i0}^2)=\mathrm{const}-\frac{1}{n}\sum_i z_{i0}^2.
$$
We can show that the variance of the projected coordinates defined by $\boldsymbol{w}_0^T\boldsymbol{x}_i$ are given by
$$
\mathrm{var}[\boldsymbol{w}0^T\boldsymbol{x}_i] = \frac{1}{n}\sum_i z{i0}^2,
$$
since the expectation value of
$$
\mathbb{E}[\boldsymbol{w}0^T\boldsymbol{x}_i] = \mathbb{E}[z{i0}]= \boldsymbol{w}_0^T\mathbb{E}[\boldsymbol{x}_i]=0,
$$
where we have used the fact that our data are centered.
Recalling our definition of the covariance as
$$
\boldsymbol{C}[\boldsymbol{x}] = \frac{1}{n}\boldsymbol{X}\boldsymbol{X}^T=\mathbb{E}[\boldsymbol{X}\boldsymbol{X}^T],
$$
we have thus that
$$
\mathrm{var}[\boldsymbol{w}0^T\boldsymbol{x}_i] = \frac{1}{n}\sum_i z{i0}^2=\boldsymbol{w}_0^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0.
$$
We are almost there, we have obtained a relation between minimizing
the reconstruction error and the variance and the covariance
matrix. Minimizing the error is equivalent to maximizing the variance
of the projected data.
The final step
We could trivially maximize the variance of the projection (and
thereby minimize the error in the reconstruction function) by letting
the norm-2 of $\boldsymbol{w}_0$ go to infinity. However, this norm since we
want the matrix $\boldsymbol{W}$ to be an orthogonal matrix, is constrained by
$\vert\vert \boldsymbol{w}_0 \vert\vert_2^2=1$. Imposing this condition via a
Lagrange multiplier we can then in turn maximize
$$
J(\boldsymbol{w}_0)= \boldsymbol{w}_0^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0+\lambda_0(1-\boldsymbol{w}_0^T\boldsymbol{w}_0).
$$
Taking the derivative with respect to $\boldsymbol{w}_0$ we obtain
$$
\frac{\partial J(\boldsymbol{w}_0)}{\partial \boldsymbol{w}_0}= 2\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0-2\lambda_0\boldsymbol{w}_0=0,
$$
meaning that
$$
\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0=\lambda_0\boldsymbol{w}_0.
$$
The direction that maximizes the variance (or minimizes the construction error) is an eigenvector of the covariance matrix! If we left multiply with $\boldsymbol{w}_0^T$ we have the variance of the projected data is
$$
\boldsymbol{w}_0^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0=\lambda_0.
$$
If we want to maximize the variance (minimize the construction error)
we simply pick the eigenvector of the covariance matrix with the
largest eigenvalue. This establishes the link between the minimization
of the reconstruction function $J$ in terms of an orthogonal matrix
and the maximization of the variance and thereby the covariance of our
observations encoded in the design/feature matrix $\boldsymbol{X}$.
The proof
for the other eigenvectors $\boldsymbol{w}_1,\boldsymbol{w}_2,\dots$ can be
established by applying the above arguments and using the fact that
our basis of eigenvectors is orthogonal, see Murphy chapter
12.2.  The
discussion in chapter 12.2 of Murphy's text has also a nice link with
the Singular Value Decomposition theorem. For categorical data, see
chapter 12.4 and discussion therein.
Additional part of the proof for the other eigenvectors will be added by mid January 2020.
Geometric Interpretation and link with Singular Value Decomposition
This material will be added by mid January 2020.
Principal Component Analysis
Principal Component Analysis (PCA) is by far the most popular dimensionality reduction algorithm.
First it identifies the hyperplane that lies closest to the data, and then it projects the data onto it.
The following Python code uses NumPy’s svd() function to obtain all the principal components of the
training set, then extracts the first two principal components. First we center the data using either pandas or our own code
Step16:  PCA assumes that the dataset is centered around the origin. Scikit-Learn’s PCA classes take care of centering
the data for you. However, if you implement PCA yourself (as in the preceding example), or if you use other libraries, don’t
forget to center the data first.
Once you have identified all the principal components, you can reduce the dimensionality of the dataset
down to $d$ dimensions by projecting it onto the hyperplane defined by the first $d$ principal components.
Selecting this hyperplane ensures that the projection will preserve as much variance as possible.
Step17:  <!-- !split  -->
PCA and scikit-learn
Scikit-Learn’s PCA class implements PCA using SVD decomposition just like we did before. The
following code applies PCA to reduce the dimensionality of the dataset down to two dimensions (note
that it automatically takes care of centering the data)
Step18:  After fitting the PCA transformer to the dataset, you can access the principal components using the
components variable (note that it contains the PCs as horizontal vectors, so, for example, the first
principal component is equal to
Step19:  Another very useful piece of information is the explained variance ratio of each principal component,
available via the $explained_variance_ratio$ variable. It indicates the proportion of the dataset’s
variance that lies along the axis of each principal component. 
Back to the Cancer Data
We can now repeat the above but applied to real data, in this case our breast cancer data.
Here we compute performance scores on the training data using logistic regression.
Step20:  We see that our training data after the PCA decomposition has a performance similar to the non-scaled data. 
More on the PCA
Instead of arbitrarily choosing the number of dimensions to reduce down to, it is generally preferable to
choose the number of dimensions that add up to a sufficiently large portion of the variance (e.g., 95%).
Unless, of course, you are reducing dimensionality for data visualization — in that case you will
generally want to reduce the dimensionality down to 2 or 3.
The following code computes PCA without reducing dimensionality, then computes the minimum number
of dimensions required to preserve 95% of the training set’s variance
Step21:  You could then set $n_components=d$ and run PCA again. However, there is a much better option
Step22:  Incremental PCA
One problem with the preceding implementation of PCA is that it requires the whole training set to fit in
memory in order for the SVD algorithm to run. Fortunately, Incremental PCA (IPCA) algorithms have
been developed
<ASSISTANT_TASK:>
Python Code:
%matplotlib inline
# Common imports
import os
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import sklearn.linear_model as skl
from sklearn.metrics import mean_squared_error
from sklearn.model_selection import  train_test_split
from sklearn.preprocessing import MinMaxScaler, StandardScaler, Normalizer
from sklearn.svm import SVR
# Where to save the figures and data files
PROJECT_ROOT_DIR = "Results"
FIGURE_ID = "Results/FigureFiles"
DATA_ID = "DataFiles/"
if not os.path.exists(PROJECT_ROOT_DIR):
    os.mkdir(PROJECT_ROOT_DIR)
if not os.path.exists(FIGURE_ID):
    os.makedirs(FIGURE_ID)
if not os.path.exists(DATA_ID):
    os.makedirs(DATA_ID)
def image_path(fig_id):
    return os.path.join(FIGURE_ID, fig_id)
def data_path(dat_id):
    return os.path.join(DATA_ID, dat_id)
def save_fig(fig_id):
    plt.savefig(image_path(fig_id) + ".png", format='png')
def FrankeFunction(x,y):
	term1 = 0.75*np.exp(-(0.25*(9*x-2)**2) - 0.25*((9*y-2)**2))
	term2 = 0.75*np.exp(-((9*x+1)**2)/49.0 - 0.1*(9*y+1))
	term3 = 0.5*np.exp(-(9*x-7)**2/4.0 - 0.25*((9*y-3)**2))
	term4 = -0.2*np.exp(-(9*x-4)**2 - (9*y-7)**2)
	return term1 + term2 + term3 + term4
def create_X(x, y, n ):
	if len(x.shape) > 1:
		x = np.ravel(x)
		y = np.ravel(y)
	N = len(x)
	l = int((n+1)*(n+2)/2)		# Number of elements in beta
	X = np.ones((N,l))
	for i in range(1,n+1):
		q = int((i)*(i+1)/2)
		for k in range(i+1):
			X[:,q+k] = (x**(i-k))*(y**k)
	return X
# Making meshgrid of datapoints and compute Franke's function
n = 5
N = 1000
x = np.sort(np.random.uniform(0, 1, N))
y = np.sort(np.random.uniform(0, 1, N))
z = FrankeFunction(x, y)
X = create_X(x, y, n=n)    
# split in training and test data
X_train, X_test, y_train, y_test = train_test_split(X,z,test_size=0.2)
svm = SVR(gamma='auto',C=10.0)
svm.fit(X_train, y_train)
# The mean squared error and R2 score
print("MSE before scaling: {:.2f}".format(mean_squared_error(svm.predict(X_test), y_test)))
print("R2 score before scaling {:.2f}".format(svm.score(X_test,y_test)))
scaler = StandardScaler()
scaler.fit(X_train)
X_train_scaled = scaler.transform(X_train)
X_test_scaled = scaler.transform(X_test)
print("Feature min values before scaling:\n {}".format(X_train.min(axis=0)))
print("Feature max values before scaling:\n {}".format(X_train.max(axis=0)))
print("Feature min values after scaling:\n {}".format(X_train_scaled.min(axis=0)))
print("Feature max values after scaling:\n {}".format(X_train_scaled.max(axis=0)))
svm = SVR(gamma='auto',C=10.0)
svm.fit(X_train_scaled, y_train)
print("MSE after  scaling: {:.2f}".format(mean_squared_error(svm.predict(X_test_scaled), y_test)))
print("R2 score for  scaled data: {:.2f}".format(svm.score(X_test_scaled,y_test)))
Explanation: <!-- dom:TITLE: Data Analysis and Machine Learning: Preprocessing and Dimensionality Reduction -->
Data Analysis and Machine Learning: Preprocessing and Dimensionality Reduction
<!-- dom:AUTHOR: Morten Hjorth-Jensen at Department of Physics, University of Oslo & Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University -->
<!-- Author: -->
Morten Hjorth-Jensen, Department of Physics, University of Oslo and Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University
Date: Jan 4, 2020
Copyright 1999-2020, Morten Hjorth-Jensen. Released under CC Attribution-NonCommercial 4.0 license
Reducing the number of degrees of freedom, overarching view
Many Machine Learning problems involve thousands or even millions of
features for each training instance. Not only does this make training
extremely slow, it can also make it much harder to find a good
solution, as we will see. This problem is often referred to as the
curse of dimensionality.  Fortunately, in real-world problems, it is
often possible to reduce the number of features considerably, turning
an intractable problem into a tractable one.
Here we will discuss some of the most popular dimensionality reduction
techniques: the principal component analysis (PCA), Kernel PCA, and
Locally Linear Embedding (LLE).  Furthermore, we will start by looking
at some simple preprocessing of the data which allow us to rescale the
data.
Principal component analysis and its various variants deal with the
problem of fitting a low-dimensional affine
subspace to a set of of
data points in a high-dimensional space. With its family of methods it
is one of the most used tools in data modeling, compression and
visualization.
Preprocessing our data
Before we proceed however, we will discuss how to preprocess our
data. Till now and in connection with our previous examples we have
not met so many cases where we are too sensitive to the scaling of our
data. Normally the data may need a rescaling and/or may be sensitive
to extreme values. Scaling the data renders our inputs much more
suitable for the algorithms we want to employ.
Scikit-Learn has several functions which allow us to rescale the
data, normally resulting in much better results in terms of various
accuracy scores.  The StandardScaler function in Scikit-Learn
ensures that for each feature/predictor we study the mean value is
zero and the variance is one (every column in the design/feature
matrix).  This scaling has the drawback that it does not ensure that
we have a particular maximum or minimum in our data set. Another
function included in Scikit-Learn is the MinMaxScaler which
ensures that all features are exactly between $0$ and $1$. The
More preprocessing
The Normalizer scales each data
point such that the feature vector has a euclidean length of one. In other words, it
projects a data point on the circle (or sphere in the case of higher dimensions) with a
radius of 1. This means every data point is scaled by a different number (by the
inverse of it’s length).
This normalization is often used when only the direction (or angle) of the data matters,
not the length of the feature vector.
The RobustScaler works similarly to the StandardScaler in that it
ensures statistical properties for each feature that guarantee that
they are on the same scale. However, the RobustScaler uses the median
and quartiles, instead of mean and variance. This makes the
RobustScaler ignore data points that are very different from the rest
(like measurement errors). These odd data points are also called
outliers, and might often lead to trouble for other scaling
techniques.
Simple preprocessing examples, Franke function and regression
End of explanation
import matplotlib.pyplot as plt
import numpy as np
from sklearn.model_selection import  train_test_split 
from sklearn.datasets import load_breast_cancer
from sklearn.svm import SVC
cancer = load_breast_cancer()
X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)
print(X_train.shape)
print(X_test.shape)
svm = SVC(C=100)
svm.fit(X_train, y_train)
print("Test set accuracy: {:.2f}".format(svm.score(X_test,y_test)))
from sklearn.preprocessing import MinMaxScaler, StandardScaler
scaler = MinMaxScaler()
scaler.fit(X_train)
X_train_scaled = scaler.transform(X_train)
X_test_scaled = scaler.transform(X_test)
print("Feature min values before scaling:\n {}".format(X_train.min(axis=0)))
print("Feature max values before scaling:\n {}".format(X_train.max(axis=0)))
print("Feature min values before scaling:\n {}".format(X_train_scaled.min(axis=0)))
print("Feature max values before scaling:\n {}".format(X_train_scaled.max(axis=0)))
svm.fit(X_train_scaled, y_train)
print("Test set accuracy scaled data with Min-Max scaling: {:.2f}".format(svm.score(X_test_scaled,y_test)))
scaler = StandardScaler()
scaler.fit(X_train)
X_train_scaled = scaler.transform(X_train)
X_test_scaled = scaler.transform(X_test)
svm.fit(X_train_scaled, y_train)
print("Test set accuracy scaled data with Standar Scaler: {:.2f}".format(svm.score(X_test_scaled,y_test)))
Explanation: Simple preprocessing examples, breast cancer data and classification, Support Vector Machines
We show here how we can use a simple regression case on the breast
cancer data using support vector machines (SVM) as algorithm for
classification.
End of explanation
import matplotlib.pyplot as plt
import numpy as np
from sklearn.model_selection import  train_test_split 
from sklearn.datasets import load_breast_cancer
from sklearn.linear_model import LogisticRegression
cancer = load_breast_cancer()
# Set up training data
X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)
logreg = LogisticRegression()
logreg.fit(X_train, y_train)
print("Test set accuracy: {:.2f}".format(logreg.score(X_test,y_test)))
# Scale data
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
scaler.fit(X_train)
X_train_scaled = scaler.transform(X_train)
X_test_scaled = scaler.transform(X_test)
logreg.fit(X_train_scaled, y_train)
print("Test set accuracy scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
Explanation: More on Cancer Data, now with Logistic Regression
End of explanation
import matplotlib.pyplot as plt
import numpy as np
from sklearn.model_selection import  train_test_split 
from sklearn.datasets import load_breast_cancer
from sklearn.linear_model import LogisticRegression
cancer = load_breast_cancer()
import pandas as pd
# Making a data frame
cancerpd = pd.DataFrame(cancer.data, columns=cancer.feature_names)
fig, axes = plt.subplots(15,2,figsize=(10,20))
malignant = cancer.data[cancer.target == 0]
benign = cancer.data[cancer.target == 1]
ax = axes.ravel()
for i in range(30):
    _, bins = np.histogram(cancer.data[:,i], bins =50)
    ax[i].hist(malignant[:,i], bins = bins, alpha = 0.5)
    ax[i].hist(benign[:,i], bins = bins, alpha = 0.5)
    ax[i].set_title(cancer.feature_names[i])
    ax[i].set_yticks(())
ax[0].set_xlabel("Feature magnitude")
ax[0].set_ylabel("Frequency")
ax[0].legend(["Malignant", "Benign"], loc ="best")
fig.tight_layout()
plt.show()
import seaborn as sns
correlation_matrix = cancerpd.corr().round(1)
# use the heatmap function from seaborn to plot the correlation matrix
# annot = True to print the values inside the square
sns.heatmap(data=correlation_matrix, annot=True)
plt.show()
#print eigvalues of correlation matrix
EigValues, EigVectors = np.linalg.eig(correlation_matrix)
print(EigValues)
Explanation: Why should we think of reducing the dimensionality
In addition to the plot of the features, we study now also the covariance (and the correlation matrix).
We use also Pandas to compute the correlation matrix.
End of explanation
cancerpd = pd.DataFrame(cancer.data, columns=cancer.feature_names)
Explanation: In the above example we note two things. In the first plot we display
the overlap of benign and malignant tumors as functions of the various
features in the Wisconsing breast cancer data set. We see that for
some of the features we can distinguish clearly the benign and
malignant cases while for other features we cannot. This can point to
us which features may be of greater interest when we wish to classify
a benign or not benign tumour.
In the second figure we have computed the so-called correlation
matrix, which in our case with thirty features becomes a $30\times 30$
matrix.
We constructed this matrix using pandas via the statements
End of explanation
correlation_matrix = cancerpd.corr().round(1)
Explanation: and then
End of explanation
# Importing various packages
import numpy as np
n = 100
x = np.random.normal(size=n)
print(np.mean(x))
y = 4+3*x+np.random.normal(size=n)
print(np.mean(y))
W = np.vstack((x, y))
C = np.cov(W)
print(C)
Explanation: Diagonalizing this matrix we can in turn say something about which
features are of relevance and which are not. But before we proceed we
need to define covariance and correlation matrices. This leads us to
the classical Principal Component Analysis (PCA) theorem with
applications.
Basic ideas of the Principal Component Analysis (PCA)
The principal component analysis deals with the problem of fitting a
low-dimensional affine subspace $S$ of dimension $d$ much smaller than
the totaldimension $D$ of the problem at hand (our data
set). Mathematically it can be formulated as a statistical problem or
a geometric problem.  In our discussion of the theorem for the
classical PCA, we will stay with a statistical approach. This is also
what set the scene historically which for the PCA.
We have a data set defined by a design/feature matrix $\boldsymbol{X}$ (see below for its definition) 
* Each data point is determined by $p$ extrinsic (measurement) variables
We may want to ask the following question: Are there fewer intrinsic variables (say $d << p$) that still approximately describe the data?
If so, these intrinsic variables may tell us something important and finding these intrinsic variables is what dimension reduction methods do. 
Introducing the Covariance and Correlation functions
Before we discuss the PCA theorem, we need to remind ourselves about
the definition of the covariance and the correlation function. These are quantities 
Suppose we have defined two vectors
$\hat{x}$ and $\hat{y}$ with $n$ elements each. The covariance matrix $\boldsymbol{C}$ is defined as
$$
\boldsymbol{C}[\boldsymbol{x},\boldsymbol{y}] = \begin{bmatrix} \mathrm{cov}[\boldsymbol{x},\boldsymbol{x}] & \mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] \
                              \mathrm{cov}[\boldsymbol{y},\boldsymbol{x}] & \mathrm{cov}[\boldsymbol{y},\boldsymbol{y}] \
             \end{bmatrix},
$$
where for example
$$
\mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] =\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})(y_i- \overline{y}).
$$
With this definition and recalling that the variance is defined as
$$
\mathrm{var}[\boldsymbol{x}]=\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})^2,
$$
we can rewrite the covariance matrix as
$$
\boldsymbol{C}[\boldsymbol{x},\boldsymbol{y}] = \begin{bmatrix} \mathrm{var}[\boldsymbol{x}] & \mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] \
                              \mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] & \mathrm{var}[\boldsymbol{y}] \
             \end{bmatrix}.
$$
The covariance takes values between zero and infinity and may thus
lead to problems with loss of numerical precision for particularly
large values. It is common to scale the covariance matrix by
introducing instead the correlation matrix defined via the so-called
correlation function
$$
\mathrm{corr}[\boldsymbol{x},\boldsymbol{y}]=\frac{\mathrm{cov}[\boldsymbol{x},\boldsymbol{y}]}{\sqrt{\mathrm{var}[\boldsymbol{x}] \mathrm{var}[\boldsymbol{y}]}}.
$$
The correlation function is then given by values $\mathrm{corr}[\boldsymbol{x},\boldsymbol{y}]
\in [-1,1]$. This avoids eventual problems with too large values. We
can then define the correlation matrix for the two vectors $\boldsymbol{x}$
and $\boldsymbol{y}$ as
$$
\boldsymbol{K}[\boldsymbol{x},\boldsymbol{y}] = \begin{bmatrix} 1 & \mathrm{corr}[\boldsymbol{x},\boldsymbol{y}] \
                              \mathrm{corr}[\boldsymbol{y},\boldsymbol{x}] & 1 \
             \end{bmatrix},
$$
In the above example this is the function we constructed using pandas.
Correlation Function and Design/Feature Matrix
In our derivation of the various regression algorithms like Ordinary Least Squares or Ridge regression
we defined the design/feature matrix $\boldsymbol{X}$ as
$$
\boldsymbol{X}=\begin{bmatrix}
x_{0,0} & x_{0,1} & x_{0,2}& \dots & \dots x_{0,p-1}\
x_{1,0} & x_{1,1} & x_{1,2}& \dots & \dots x_{1,p-1}\
x_{2,0} & x_{2,1} & x_{2,2}& \dots & \dots x_{2,p-1}\
\dots & \dots & \dots & \dots \dots & \dots \
x_{n-2,0} & x_{n-2,1} & x_{n-2,2}& \dots & \dots x_{n-2,p-1}\
x_{n-1,0} & x_{n-1,1} & x_{n-1,2}& \dots & \dots x_{n-1,p-1}\
\end{bmatrix},
$$
with $\boldsymbol{X}\in {\mathbb{R}}^{n\times p}$, with the predictors/features $p$  refering to the column numbers and the
entries $n$ being the row elements.
We can rewrite the design/feature matrix in terms of its column vectors as
$$
\boldsymbol{X}=\begin{bmatrix} \boldsymbol{x}0 & \boldsymbol{x}_1 & \boldsymbol{x}_2 & \dots & \dots & \boldsymbol{x}{p-1}\end{bmatrix},
$$
with a given vector
$$
\boldsymbol{x}i^T = \begin{bmatrix}x{0,i} & x_{1,i} & x_{2,i}& \dots & \dots x_{n-1,i}\end{bmatrix}.
$$
With these definitions, we can now rewrite our $2\times 2$
correaltion/covariance matrix in terms of a moe general design/feature
matrix $\boldsymbol{X}\in {\mathbb{R}}^{n\times p}$. This leads to a $p\times p$
covariance matrix for the vectors $\boldsymbol{x}_i$ with $i=0,1,\dots,p-1$
$$
\boldsymbol{C}[\boldsymbol{x}] = \begin{bmatrix}
\mathrm{var}[\boldsymbol{x}0] & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_1]  & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_2] & \dots & \dots & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}{p-1}]\
\mathrm{cov}[\boldsymbol{x}1,\boldsymbol{x}_0] & \mathrm{var}[\boldsymbol{x}_1]  & \mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_2] & \dots & \dots & \mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}{p-1}]\
\mathrm{cov}[\boldsymbol{x}2,\boldsymbol{x}_0]   & \mathrm{cov}[\boldsymbol{x}_2,\boldsymbol{x}_1] & \mathrm{var}[\boldsymbol{x}_2] & \dots & \dots & \mathrm{cov}[\boldsymbol{x}_2,\boldsymbol{x}{p-1}]\
\dots & \dots & \dots & \dots & \dots & \dots \
\dots & \dots & \dots & \dots & \dots & \dots \
\mathrm{cov}[\boldsymbol{x}{p-1},\boldsymbol{x}_0]   & \mathrm{cov}[\boldsymbol{x}{p-1},\boldsymbol{x}1] & \mathrm{cov}[\boldsymbol{x}{p-1},\boldsymbol{x}{2}]  & \dots & \dots  & \mathrm{var}[\boldsymbol{x}{p-1}]\
\end{bmatrix},
$$
and the correlation matrix
$$
\boldsymbol{K}[\boldsymbol{x}] = \begin{bmatrix}
1 & \mathrm{corr}[\boldsymbol{x}0,\boldsymbol{x}_1]  & \mathrm{corr}[\boldsymbol{x}_0,\boldsymbol{x}_2] & \dots & \dots & \mathrm{corr}[\boldsymbol{x}_0,\boldsymbol{x}{p-1}]\
\mathrm{corr}[\boldsymbol{x}1,\boldsymbol{x}_0] & 1  & \mathrm{corr}[\boldsymbol{x}_1,\boldsymbol{x}_2] & \dots & \dots & \mathrm{corr}[\boldsymbol{x}_1,\boldsymbol{x}{p-1}]\
\mathrm{corr}[\boldsymbol{x}2,\boldsymbol{x}_0]   & \mathrm{corr}[\boldsymbol{x}_2,\boldsymbol{x}_1] & 1 & \dots & \dots & \mathrm{corr}[\boldsymbol{x}_2,\boldsymbol{x}{p-1}]\
\dots & \dots & \dots & \dots & \dots & \dots \
\dots & \dots & \dots & \dots & \dots & \dots \
\mathrm{corr}[\boldsymbol{x}{p-1},\boldsymbol{x}_0]   & \mathrm{corr}[\boldsymbol{x}{p-1},\boldsymbol{x}1] & \mathrm{corr}[\boldsymbol{x}{p-1},\boldsymbol{x}_{2}]  & \dots & \dots  & 1\
\end{bmatrix},
$$
Covariance Matrix Examples
The Numpy function np.cov calculates the covariance elements using
the factor $1/(n-1)$ instead of $1/n$ since it assumes we do not have
the exact mean values.  The following simple function uses the
np.vstack function which takes each vector of dimension $1\times n$
and produces a $2\times n$ matrix $\boldsymbol{W}$
$$
\boldsymbol{W} = \begin{bmatrix} x_0 & y_0 \
                          x_1 & y_1 \
                          x_2 & y_2\
                          \dots & \dots \
                          x_{n-2} & y_{n-2}\
                          x_{n-1} & y_{n-1} & 
             \end{bmatrix},
$$
which in turn is converted into into the $2\times 2$ covariance matrix
$\boldsymbol{C}$ via the Numpy function np.cov(). We note that we can also calculate
the mean value of each set of samples $\boldsymbol{x}$ etc using the Numpy
function np.mean(x). We can also extract the eigenvalues of the
covariance matrix through the np.linalg.eig() function.
End of explanation
import numpy as np
n = 100
# define two vectors                                                                                           
x = np.random.random(size=n)
y = 4+3*x+np.random.normal(size=n)
#scaling the x and y vectors                                                                                   
x = x - np.mean(x)
y = y - np.mean(y)
variance_x = np.sum(x@x)/n
variance_y = np.sum(y@y)/n
print(variance_x)
print(variance_y)
cov_xy = np.sum(x@y)/n
cov_xx = np.sum(x@x)/n
cov_yy = np.sum(y@y)/n
C = np.zeros((2,2))
C[0,0]= cov_xx/variance_x
C[1,1]= cov_yy/variance_y
C[0,1]= cov_xy/np.sqrt(variance_y*variance_x)
C[1,0]= C[0,1]
print(C)
Explanation: Correlation Matrix
The previous example can be converted into the correlation matrix by
simply scaling the matrix elements with the variances.  We should also
subtract the mean values for each column. This leads to the following
code which sets up the correlations matrix for the previous example in
a more brute force way. Here we scale the mean values for each column of the design matrix, calculate the relevant mean values and variances and then finally set up the $2\times 2$ correlation matrix (since we have only two vectors).
End of explanation
import numpy as np
import pandas as pd
n = 10
x = np.random.normal(size=n)
x = x - np.mean(x)
y = 4+3*x+np.random.normal(size=n)
y = y - np.mean(y)
X = (np.vstack((x, y))).T
print(X)
Xpd = pd.DataFrame(X)
print(Xpd)
correlation_matrix = Xpd.corr()
print(correlation_matrix)
Explanation: We see that the matrix elements along the diagonal are one as they
should be and that the matrix is symmetric. Furthermore, diagonalizing
this matrix we easily see that it is a positive definite matrix.
The above procedure with numpy can be made more compact if we use pandas.
Correlation Matrix with Pandas
We whow here how we can set up the correlation matrix using pandas, as done in this simple code
End of explanation
# Common imports
import numpy as np
import pandas as pd
def FrankeFunction(x,y):
	term1 = 0.75*np.exp(-(0.25*(9*x-2)**2) - 0.25*((9*y-2)**2))
	term2 = 0.75*np.exp(-((9*x+1)**2)/49.0 - 0.1*(9*y+1))
	term3 = 0.5*np.exp(-(9*x-7)**2/4.0 - 0.25*((9*y-3)**2))
	term4 = -0.2*np.exp(-(9*x-4)**2 - (9*y-7)**2)
	return term1 + term2 + term3 + term4
def create_X(x, y, n ):
	if len(x.shape) > 1:
		x = np.ravel(x)
		y = np.ravel(y)
	N = len(x)
	l = int((n+1)*(n+2)/2)		# Number of elements in beta
	X = np.ones((N,l))
	for i in range(1,n+1):
		q = int((i)*(i+1)/2)
		for k in range(i+1):
			X[:,q+k] = (x**(i-k))*(y**k)
	return X
# Making meshgrid of datapoints and compute Franke's function
n = 4
N = 100
x = np.sort(np.random.uniform(0, 1, N))
y = np.sort(np.random.uniform(0, 1, N))
z = FrankeFunction(x, y)
X = create_X(x, y, n=n)    
Xpd = pd.DataFrame(X)
# subtract the mean values and set up the covariance matrix
Xpd = Xpd - Xpd.mean()
covariance_matrix = Xpd.cov()
print(covariance_matrix)
Explanation: We expand this model to the Franke function discussed above.
Correlation Matrix with Pandas and the Franke function
End of explanation
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from IPython.display import display
n = 10000
mean = (-1, 2)
cov = [[4, 2], [2, 2]]
X = np.random.multivariate_normal(mean, cov, n)
Explanation: We note here that the covariance is zero for the first rows and
columns since all matrix elements in the design matrix were set to one
(we are fitting the function in terms of a polynomial of degree $n$).
This means that the variance for these elements will be zero and will
cause problems when we set up the correlation matrix.  We can simply
drop these elements and construct a correlation
matrix without these elements. 
Rewriting the Covariance and/or Correlation Matrix
We can rewrite the covariance matrix in a more compact form in terms of the design/feature matrix $\boldsymbol{X}$ as
$$
\boldsymbol{C}[\boldsymbol{x}] = \frac{1}{n}\boldsymbol{X}\boldsymbol{X}^T= \mathbb{E}[\boldsymbol{X}\boldsymbol{X}^T].
$$
To see this let us simply look at a design matrix $\boldsymbol{X}\in {\mathbb{R}}^{2\times 2}$
$$
\boldsymbol{X}=\begin{bmatrix}
x_{00} & x_{01}\
x_{10} & x_{11}\
\end{bmatrix}=\begin{bmatrix}
\boldsymbol{x}{0} & \boldsymbol{x}{1}\
\end{bmatrix}.
$$
If we then compute the expectation value
$$
\mathbb{E}[\boldsymbol{X}\boldsymbol{X}^T] = \frac{1}{n}\boldsymbol{X}\boldsymbol{X}^T=\begin{bmatrix}
x_{00}^2+x_{01}^2 & x_{00}x_{10}+x_{01}x_{11}\
x_{10}x_{00}+x_{11}x_{01} & x_{10}^2+x_{11}^2\
\end{bmatrix},
$$
which is just
$$
\boldsymbol{C}[\boldsymbol{x}_0,\boldsymbol{x}_1] = \boldsymbol{C}[\boldsymbol{x}]=\begin{bmatrix} \mathrm{var}[\boldsymbol{x}_0] & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_1] \
                              \mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_0] & \mathrm{var}[\boldsymbol{x}_1] \
             \end{bmatrix},
$$
where we wrote $$\boldsymbol{C}[\boldsymbol{x}_0,\boldsymbol{x}_1] = \boldsymbol{C}[\boldsymbol{x}]$$ to indicate that this the covariance of the vectors $\boldsymbol{x}$ of the design/feature matrix $\boldsymbol{X}$.
It is easy to generalize this to a matrix $\boldsymbol{X}\in {\mathbb{R}}^{n\times p}$.
Towards the PCA theorem
We have that the covariance matrix (the correlation matrix involves a simple rescaling) is given as
$$
\boldsymbol{C}[\boldsymbol{x}] = \frac{1}{n}\boldsymbol{X}\boldsymbol{X}^T= \mathbb{E}[\boldsymbol{X}\boldsymbol{X}^T].
$$
Let us now assume that we can perform a series of orthogonal transformations where we employ some orthogonal matrices $\boldsymbol{S}$.
These matrices are defined as $\boldsymbol{S}\in {\mathbb{R}}^{p\times p}$ and obey the orthogonality requirements $\boldsymbol{S}\boldsymbol{S}^T=\boldsymbol{S}^T\boldsymbol{S}=\boldsymbol{I}$. The matrix can be written out in terms of the column vectors $\boldsymbol{s}i$ as $\boldsymbol{S}=[\boldsymbol{s}_0,\boldsymbol{s}_1,\dots,\boldsymbol{s}{p-1}]$ and $\boldsymbol{s}_i \in {\mathbb{R}}^{p}$.
Assume also that there is a transformation $\boldsymbol{S}\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}^T=\boldsymbol{C}[\boldsymbol{y}]$ such that the new matrix $\boldsymbol{C}[\boldsymbol{y}]$ is diagonal with elements $[\lambda_0,\lambda_1,\lambda_2,\dots,\lambda_{p-1}]$.  
That is we have
$$
\boldsymbol{C}[\boldsymbol{y}] = \mathbb{E}[\boldsymbol{S}\boldsymbol{X}\boldsymbol{X}^T\boldsymbol{S}^T]=\boldsymbol{S}\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}^T,
$$
since the matrix $\boldsymbol{S}$ is not a data dependent matrix.   Multiplying with $\boldsymbol{S}^T$ from the left we have
$$
\boldsymbol{S}^T\boldsymbol{C}[\boldsymbol{y}] = \boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}^T,
$$
and since $\boldsymbol{C}[\boldsymbol{y}]$ is diagonal we have for a given eigenvalue $i$ of the covariance matrix that
$$
\boldsymbol{S}^T_i\lambda_i = \boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}^T_i.
$$
In the derivation of the PCA theorem we will assume that the eigenvalues are ordered in descending order, that is
$\lambda_0 > \lambda_1 > \dots > \lambda_{p-1}$. 
The eigenvalues tell us then how much we need to stretch the
corresponding eigenvectors. Dimensions with large eigenvalues have
thus large variations (large variance) and define therefore useful
dimensions. The data points are more spread out in the direction of
these eigenvectors.  Smaller eigenvalues mean on the other hand that
the corresponding eigenvectors are shrunk accordingly and the data
points are tightly bunched together and there is not much variation in
these specific directions. Hopefully then we could leave it out
dimensions where the eigenvalues are very small. If $p$ is very large,
we could then aim at reducing $p$ to $l << p$ and handle only $l$
features/predictors.
The Algorithm before theorem
Here's how we would proceed in setting up the algorithm for the PCA, see also discussion below here. 
* Set up the datapoints for the design/feature matrix $\boldsymbol{X}$ with $\boldsymbol{X}\in {\mathbb{R}}^{n\times p}$, with the predictors/features $p$  referring to the column numbers and the entries $n$ being the row elements.
$$
\boldsymbol{X}=\begin{bmatrix}
x_{0,0} & x_{0,1} & x_{0,2}& \dots & \dots x_{0,p-1}\
x_{1,0} & x_{1,1} & x_{1,2}& \dots & \dots x_{1,p-1}\
x_{2,0} & x_{2,1} & x_{2,2}& \dots & \dots x_{2,p-1}\
\dots & \dots & \dots & \dots \dots & \dots \
x_{n-2,0} & x_{n-2,1} & x_{n-2,2}& \dots & \dots x_{n-2,p-1}\
x_{n-1,0} & x_{n-1,1} & x_{n-1,2}& \dots & \dots x_{n-1,p-1}\
\end{bmatrix},
$$
Center the data by subtracting the mean value for each column. This leads to a new matrix $\boldsymbol{X}\rightarrow \overline{\boldsymbol{X}}$.
Compute then the covariance/correlation matrix $\mathbb{E}[\overline{\boldsymbol{X}}\overline{\boldsymbol{X}}^T]$.
Find the eigenpairs of $\boldsymbol{C}$ with eigenvalues $[\lambda_0,\lambda_1,\dots,\lambda_{p-1}]$ and eigenvectors $[\boldsymbol{s}0,\boldsymbol{s}_1,\dots,\boldsymbol{s}{p-1}]$.
Order the eigenvalue (and the eigenvectors accordingly) in order of decreasing eigenvalues.
Keep only those $l$ eigenvalues larger than a selected threshold value, discarding thus $p-l$ features since we expect small variations in the data here.
Writing our own PCA code
We will use a simple example first with two-dimensional data
drawn from a multivariate normal distribution with the following mean and covariance matrix:
$$
\mu = (-1,2) \qquad \Sigma = \begin{bmatrix} 4 & 2 \
2 & 2
\end{bmatrix}
$$
Note that the mean refers to each column of data. 
We will generate $n = 1000$ points $X = { x_1, \ldots, x_N }$ from
this distribution, and store them in the $1000 \times 2$ matrix $\boldsymbol{X}$.
The following Python code aids in setting up the data and writing out the design matrix.
Note that the function multivariate returns also the covariance discussed above and that it is defined by dividing by $n-1$ instead of $n$.
End of explanation
df = pd.DataFrame(X)
# Pandas does the centering for us
df = df -df.mean()
# we center it ourselves
X_centered = X - X.mean(axis=0)
Explanation: Now we are going to implement the PCA algorithm. We will break it down into various substeps.
Compute the sample mean and center the data
The first step of PCA is to compute the sample mean of the data and use it to center the data. Recall that the sample mean is
$$
\mu_n = \frac{1}{n} \sum_{i=1}^n x_i
$$
and the mean-centered data $\bar{X} = { \bar{x}_1, \ldots, \bar{x}_n }$ takes the form
$$
\bar{x}_i = x_i - \mu_n.
$$
When you are done with these steps, print out $\mu_n$ to verify it is
close to $\mu$ and plot your mean centered data to verify it is
centered at the origin! Compare your code with the functionality from Scikit-Learn discussed above.
The following code elements perform these operations using pandas or using our own functionality for doing so. The latter, using numpy is rather simple through the mean() function.
End of explanation
print(df.cov())
print(np.cov(X_centered.T))
Explanation: Alternatively, we could use the functions we discussed
earlier for scaling the data set.  That is, we could have used the
StandardScaler function in Scikit-Learn, a function which ensures
that for each feature/predictor we study the mean value is zero and
the variance is one (every column in the design/feature matrix).  You
would then not get the same results, since we divide by the
variance. The diagonal covariance matrix elements will then be one,
while the non-diagonal ones need to be divided by $2\sqrt{2}$ for our
specific case.
Compute the sample covariance
Now we are going to use the mean centered data to compute the sample covariance of the data by using the following equation
$$
\Sigma_n = \frac{1}{n-1} \sum_{i=1}^n \bar{x}i^T \bar{x}_i = \frac{1}{n-1} \sum{i=1}^n (x_i - \mu_n)^T (x_i - \mu_n)
$$
where the data points $x_i \in \mathbb{R}^p$ (here in this example $p = 2$) are column vectors and $x^T$ is the transpose of $x$.
We can write our own code or simply use either the functionaly of numpy or that of pandas, as follows
End of explanation
# extract the relevant columns from the centered design matrix of dim n x 2
x = X_centered[:,0]
y = X_centered[:,1]
Cov = np.zeros((2,2))
Cov[0,1] = np.sum(x.T@y)/(n-1.0)
Cov[0,0] = np.sum(x.T@x)/(n-1.0)
Cov[1,1] = np.sum(y.T@y)/(n-1.0)
Cov[1,0]= Cov[0,1]
print("Centered covariance using own code")
print(Cov)
plt.plot(x, y, 'x')
plt.axis('equal')
plt.show()
Explanation: Note that the way we define the covariance matrix here has a factor $n-1$ instead of $n$. This is included in the cov() function by numpy and pandas. 
Our own code here is not very elegant and asks for obvious improvements. It is tailored to this specific $2\times 2$ covariance matrix.
End of explanation
# diagonalize and obtain eigenvalues, not necessarily sorted
EigValues, EigVectors = np.linalg.eig(Cov)
# sort eigenvectors and eigenvalues
#permute = EigValues.argsort()
#EigValues = EigValues[permute]
#EigVectors = EigVectors[:,permute]
print("Eigenvalues of Covariance matrix")
for i in range(2):
    print(EigValues[i])
FirstEigvector = EigVectors[:,0]
SecondEigvector = EigVectors[:,1]
print("First eigenvector")
print(FirstEigvector)
print("Second eigenvector")
print(SecondEigvector)
#thereafter we do a PCA with Scikit-learn
from sklearn.decomposition import PCA
pca = PCA(n_components = 2)
X2Dsl = pca.fit_transform(X)
print("Eigenvector of largest eigenvalue")
print(pca.components_.T[:, 0])
Explanation: Depending on the number of points $n$, we will get results that are close to the covariance values defined above.
The plot shows how the data are clustered around a line with slope close to one. Is this expected?
Diagonalize the sample covariance matrix to obtain the principal components
Now we are ready to solve for the principal components! To do so we
diagonalize the sample covariance matrix $\Sigma$. We can use the
function np.linalg.eig to do so. It will return the eigenvalues and
eigenvectors of $\Sigma$. Once we have these we can perform the 
following tasks:
We compute the percentage of the total variance captured by the first principal component
We plot the mean centered data and lines along the first and second principal components
Then we project the mean centered data onto the first and second principal components, and plot the projected data. 
Finally, we approximate the data as
$$
x_i \approx \tilde{x}_i = \mu_n + \langle x_i, v_0 \rangle v_0
$$
where $v_0$ is the first principal component. 
Collecting all these steps we can write our own PCA function and
compare this with the functionality included in Scikit-Learn.  
The code here outlines some of the elements we could include in the
analysis. Feel free to extend upon this in order to address the above
questions.
End of explanation
import numpy as np
import pandas as pd
from IPython.display import display
np.random.seed(100)
# setting up a 10 x 5 vanilla matrix 
rows = 10
cols = 5
X = np.random.randn(rows,cols)
df = pd.DataFrame(X)
# Pandas does the centering for us
df = df -df.mean()
display(df)
# we center it ourselves
X_centered = X - X.mean(axis=0)
# Then check the difference between pandas and our own set up
print(X_centered-df)
#Now we do an SVD
U, s, V = np.linalg.svd(X_centered)
c1 = V.T[:, 0]
c2 = V.T[:, 1]
W2 = V.T[:, :2]
X2D = X_centered.dot(W2)
print(X2D)
Explanation: This code does not contain all the above elements, but it shows how we can use Scikit-Learn to extract the eigenvector which corresponds to the largest eigenvalue. Try to address the questions we pose before the above code.  Try also to change the values of the covariance matrix by making one of the diagonal elements much larger than the other. What do you observe then? 
Classical PCA Theorem
We assume now that we have a design matrix $\boldsymbol{X}$ which has been
centered as discussed above. For the sake of simplicity we skip the
overline symbol. The matrix is defined in terms of the various column
vectors $[\boldsymbol{x}0,\boldsymbol{x}_1,\dots, \boldsymbol{x}{p-1}]$ each with dimension
$\boldsymbol{x}\in {\mathbb{R}}^{n}$.
We assume also that we have an orthogonal transformation $\boldsymbol{W}\in {\mathbb{R}}^{p\times p}$.  We define the reconstruction error (which is similar to the mean squared error we have seen before) as
$$
J(\boldsymbol{W},\boldsymbol{Z}) = \frac{1}{n}\sum_i (\boldsymbol{x}_i - \overline{\boldsymbol{x}}_i)^2,
$$
with $\overline{\boldsymbol{x}}_i = \boldsymbol{W}\boldsymbol{z}_i$, where $\boldsymbol{z}_i$ is a row vector with dimension ${\mathbb{R}}^{n}$ of the matrix
$\boldsymbol{Z}\in{\mathbb{R}}^{p\times n}$.  When doing PCA we want to reduce this dimensionality. 
The PCA theorem states that minimizing the above reconstruction error
corresponds to setting $\boldsymbol{W}=\boldsymbol{S}$, the orthogonal matrix which
diagonalizes the empirical covariance(correlation) matrix. The optimal
low-dimensional encoding of the data is then given by a set of vectors
$\boldsymbol{z}_i$ with at most $l$ vectors, with $l << p$, defined by the
orthogonal projection of the data onto the columns spanned by the
eigenvectors of the covariance(correlations matrix).
The proof which follows will be updated by mid January 2020.
Proof of the PCA Theorem
To show the PCA theorem let us start with the assumption that there is one vector $\boldsymbol{w}_0$ which corresponds to a solution which minimized the reconstruction error $J$. This is an orthogonal vector. It means that we now approximate the reconstruction error in terms of $\boldsymbol{w}_0$ and $\boldsymbol{z}_0$ as
$$
J(\boldsymbol{w}0,\boldsymbol{z}_0)= \frac{1}{n}\sum_i (\boldsymbol{x}_i - z{i0}\boldsymbol{w}0)^2=\frac{1}{n}\sum_i (\boldsymbol{x}_i^T\boldsymbol{x}_i - 2z{i0}\boldsymbol{w}0^T\boldsymbol{x}_i+z{i0}^2\boldsymbol{w}_0^T\boldsymbol{w}_0),
$$
which we can rewrite due to the orthogonality of $\boldsymbol{w}_i$ as
$$
J(\boldsymbol{w}0,\boldsymbol{z}_0)=\frac{1}{n}\sum_i (\boldsymbol{x}_i^T\boldsymbol{x}_i - 2z{i0}\boldsymbol{w}0^T\boldsymbol{x}_i+z{i0}^2).
$$
Minimizing $J$ with respect to the unknown parameters $z_{0i}$ we obtain that
$$
z_{i0}=\boldsymbol{w}_0^T\boldsymbol{x}_i,
$$
where the vectors on the rhs are known. 
PCA Proof continued
We have now found the unknown parameters $z_{i0}$. These correspond to the projected coordinates and we can write
$$
J(\boldsymbol{w}0)= \frac{1}{p}\sum_i (\boldsymbol{x}_i^T\boldsymbol{x}_i - z{i0}^2)=\mathrm{const}-\frac{1}{n}\sum_i z_{i0}^2.
$$
We can show that the variance of the projected coordinates defined by $\boldsymbol{w}_0^T\boldsymbol{x}_i$ are given by
$$
\mathrm{var}[\boldsymbol{w}0^T\boldsymbol{x}_i] = \frac{1}{n}\sum_i z{i0}^2,
$$
since the expectation value of
$$
\mathbb{E}[\boldsymbol{w}0^T\boldsymbol{x}_i] = \mathbb{E}[z{i0}]= \boldsymbol{w}_0^T\mathbb{E}[\boldsymbol{x}_i]=0,
$$
where we have used the fact that our data are centered.
Recalling our definition of the covariance as
$$
\boldsymbol{C}[\boldsymbol{x}] = \frac{1}{n}\boldsymbol{X}\boldsymbol{X}^T=\mathbb{E}[\boldsymbol{X}\boldsymbol{X}^T],
$$
we have thus that
$$
\mathrm{var}[\boldsymbol{w}0^T\boldsymbol{x}_i] = \frac{1}{n}\sum_i z{i0}^2=\boldsymbol{w}_0^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0.
$$
We are almost there, we have obtained a relation between minimizing
the reconstruction error and the variance and the covariance
matrix. Minimizing the error is equivalent to maximizing the variance
of the projected data.
The final step
We could trivially maximize the variance of the projection (and
thereby minimize the error in the reconstruction function) by letting
the norm-2 of $\boldsymbol{w}_0$ go to infinity. However, this norm since we
want the matrix $\boldsymbol{W}$ to be an orthogonal matrix, is constrained by
$\vert\vert \boldsymbol{w}_0 \vert\vert_2^2=1$. Imposing this condition via a
Lagrange multiplier we can then in turn maximize
$$
J(\boldsymbol{w}_0)= \boldsymbol{w}_0^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0+\lambda_0(1-\boldsymbol{w}_0^T\boldsymbol{w}_0).
$$
Taking the derivative with respect to $\boldsymbol{w}_0$ we obtain
$$
\frac{\partial J(\boldsymbol{w}_0)}{\partial \boldsymbol{w}_0}= 2\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0-2\lambda_0\boldsymbol{w}_0=0,
$$
meaning that
$$
\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0=\lambda_0\boldsymbol{w}_0.
$$
The direction that maximizes the variance (or minimizes the construction error) is an eigenvector of the covariance matrix! If we left multiply with $\boldsymbol{w}_0^T$ we have the variance of the projected data is
$$
\boldsymbol{w}_0^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0=\lambda_0.
$$
If we want to maximize the variance (minimize the construction error)
we simply pick the eigenvector of the covariance matrix with the
largest eigenvalue. This establishes the link between the minimization
of the reconstruction function $J$ in terms of an orthogonal matrix
and the maximization of the variance and thereby the covariance of our
observations encoded in the design/feature matrix $\boldsymbol{X}$.
The proof
for the other eigenvectors $\boldsymbol{w}_1,\boldsymbol{w}_2,\dots$ can be
established by applying the above arguments and using the fact that
our basis of eigenvectors is orthogonal, see Murphy chapter
12.2.  The
discussion in chapter 12.2 of Murphy's text has also a nice link with
the Singular Value Decomposition theorem. For categorical data, see
chapter 12.4 and discussion therein.
Additional part of the proof for the other eigenvectors will be added by mid January 2020.
Geometric Interpretation and link with Singular Value Decomposition
This material will be added by mid January 2020.
Principal Component Analysis
Principal Component Analysis (PCA) is by far the most popular dimensionality reduction algorithm.
First it identifies the hyperplane that lies closest to the data, and then it projects the data onto it.
The following Python code uses NumPy’s svd() function to obtain all the principal components of the
training set, then extracts the first two principal components. First we center the data using either pandas or our own code
End of explanation
W2 = V.T[:, :2]
X2D = X_centered.dot(W2)
Explanation: PCA assumes that the dataset is centered around the origin. Scikit-Learn’s PCA classes take care of centering
the data for you. However, if you implement PCA yourself (as in the preceding example), or if you use other libraries, don’t
forget to center the data first.
Once you have identified all the principal components, you can reduce the dimensionality of the dataset
down to $d$ dimensions by projecting it onto the hyperplane defined by the first $d$ principal components.
Selecting this hyperplane ensures that the projection will preserve as much variance as possible.
End of explanation
#thereafter we do a PCA with Scikit-learn
from sklearn.decomposition import PCA
pca = PCA(n_components = 2)
X2D = pca.fit_transform(X)
print(X2D)
Explanation: <!-- !split  -->
PCA and scikit-learn
Scikit-Learn’s PCA class implements PCA using SVD decomposition just like we did before. The
following code applies PCA to reduce the dimensionality of the dataset down to two dimensions (note
that it automatically takes care of centering the data):
End of explanation
pca.components_.T[:, 0].
Explanation: After fitting the PCA transformer to the dataset, you can access the principal components using the
components variable (note that it contains the PCs as horizontal vectors, so, for example, the first
principal component is equal to
End of explanation
import matplotlib.pyplot as plt
import numpy as np
from sklearn.model_selection import  train_test_split 
from sklearn.datasets import load_breast_cancer
from sklearn.linear_model import LogisticRegression
cancer = load_breast_cancer()
X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)
logreg = LogisticRegression()
logreg.fit(X_train, y_train)
print("Train set accuracy from Logistic Regression: {:.2f}".format(logreg.score(X_train,y_train)))
# We scale the data
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
scaler.fit(X_train)
X_train_scaled = scaler.transform(X_train)
X_test_scaled = scaler.transform(X_test)
# Then perform again a log reg fit
logreg.fit(X_train_scaled, y_train)
print("Train set accuracy scaled data: {:.2f}".format(logreg.score(X_train_scaled,y_train)))
#thereafter we do a PCA with Scikit-learn
from sklearn.decomposition import PCA
pca = PCA(n_components = 2)
X2D_train = pca.fit_transform(X_train_scaled)
# and finally compute the log reg fit and the score on the training data	
logreg.fit(X2D_train,y_train)
print("Train set accuracy scaled and PCA data: {:.2f}".format(logreg.score(X2D_train,y_train)))
Explanation: Another very useful piece of information is the explained variance ratio of each principal component,
available via the $explained_variance_ratio$ variable. It indicates the proportion of the dataset’s
variance that lies along the axis of each principal component. 
Back to the Cancer Data
We can now repeat the above but applied to real data, in this case our breast cancer data.
Here we compute performance scores on the training data using logistic regression.
End of explanation
pca = PCA()
pca.fit(X)
cumsum = np.cumsum(pca.explained_variance_ratio_)
d = np.argmax(cumsum >= 0.95) + 1
Explanation: We see that our training data after the PCA decomposition has a performance similar to the non-scaled data. 
More on the PCA
Instead of arbitrarily choosing the number of dimensions to reduce down to, it is generally preferable to
choose the number of dimensions that add up to a sufficiently large portion of the variance (e.g., 95%).
Unless, of course, you are reducing dimensionality for data visualization — in that case you will
generally want to reduce the dimensionality down to 2 or 3.
The following code computes PCA without reducing dimensionality, then computes the minimum number
of dimensions required to preserve 95% of the training set’s variance:
End of explanation
pca = PCA(n_components=0.95)
X_reduced = pca.fit_transform(X)
Explanation: You could then set $n_components=d$ and run PCA again. However, there is a much better option: instead
of specifying the number of principal components you want to preserve, you can set $n_components$ to be
a float between 0.0 and 1.0, indicating the ratio of variance you wish to preserve:
End of explanation
from sklearn.decomposition import KernelPCA
rbf_pca = KernelPCA(n_components = 2, kernel="rbf", gamma=0.04)
X_reduced = rbf_pca.fit_transform(X)
Explanation: Incremental PCA
One problem with the preceding implementation of PCA is that it requires the whole training set to fit in
memory in order for the SVD algorithm to run. Fortunately, Incremental PCA (IPCA) algorithms have
been developed: you can split the training set into mini-batches and feed an IPCA algorithm one minibatch
at a time. This is useful for large training sets, and also to apply PCA online (i.e., on the fly, as new
instances arrive).
Randomized PCA
Scikit-Learn offers yet another option to perform PCA, called Randomized PCA. This is a stochastic
algorithm that quickly finds an approximation of the first d principal components. Its computational
complexity is $O(m \times d^2)+O(d^3)$, instead of $O(m \times n^2) + O(n^3)$, so it is dramatically faster than the
previous algorithms when $d$ is much smaller than $n$.
Kernel PCA
The kernel trick is a mathematical technique that implicitly maps instances into a
very high-dimensional space (called the feature space), enabling nonlinear classification and regression
with Support Vector Machines. Recall that a linear decision boundary in the high-dimensional feature
space corresponds to a complex nonlinear decision boundary in the original space.
It turns out that the same trick can be applied to PCA, making it possible to perform complex nonlinear
projections for dimensionality reduction. This is called Kernel PCA (kPCA). It is often good at
preserving clusters of instances after projection, or sometimes even unrolling datasets that lie close to a
twisted manifold.
For example, the following code uses Scikit-Learn’s KernelPCA class to perform kPCA with an
End of explanation
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