Zero-center

It is often useful to normalise data. One type of normalisation is to 'zero center' the data by subtracting it's mean. This will yield data with a new mean of zero. Often we want to do this separately for each column or row of a matrix. Also it's often the case that we want to perform this operation on a square matrix. In that case, we can multiply by a 'centering matrix':

\[\begin{equation} C = I - \frac{1}{N} \end{equation}\] \[\begin{equation} C = % \begin{bmatrix} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \\ \end{bmatrix} % -% \begin{bmatrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\[5pt] \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\[5pt] \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{bmatrix} % =% \begin{bmatrix} \frac{2}{3} & -\frac{1}{3} & -\frac{1}{3} \\[5pt] -\frac{1}{3} & \frac{2}{3} & -\frac{1}{3} \\[5pt] -\frac{1}{3} & -\frac{1}{3} & \frac{2}{3} \end{bmatrix} \end{equation}\] \[\begin{equation} CA =% \begin{bmatrix} \frac{2}{3} & -\frac{1}{3} & -\frac{1}{3} \\[5pt] -\frac{1}{3} & \frac{2}{3} & -\frac{1}{3} \\[5pt] -\frac{1}{3} & -\frac{1}{3} & \frac{2}{3} \end{bmatrix} \begin{bmatrix} 1 & 1 & 1\\[5pt] 0 & 2 & 0\\[5pt] 0 & 3 & -4 \end{bmatrix} % =% \begin{bmatrix} \frac{2}{3} & -1 & 2 \\[5pt] -\frac{1}{3} & 0 & 1 \\[5pt] -\frac{1}{3} & 1 & -3 \end{bmatrix} \end{equation}\]

Here the matrix $C$ had the effect of subtracting $1/3$ from the first column, subtracting $2/3$ from the second column, and adding $1$ to the last column - ensuring that all columns sum to zero.

We create a function to build the centering matrix:

def gen_centering(size):
    if type(size) is int:
        size = [size, size]
    return la.eye(size).subtract(1/size[0])

Code implementation

We then define a function that can zero center each column, each row, or that matrix as a whole (if axis=2). If the matrix is square, then we use the centering matrix approach above. Otherwise, after taking the mean across the relevant axis (columns, rows, or all elements), we make a matrix of the same size but filled with the mean of each axis and subtract it off the original matrix:

def zero_center(A, axis=0):
    if axis == 2:
        global_mean = mean(mean(A)).data[0][0]
        return A.subtract(global_mean)
    elif axis == 1:
        A = A.tr()
    if A.is_square():
        A = gen_centering(la.size(A)).multiply(A)
    else:
        A_mean = mean(A)
        ones = la.gen_mat([la.size(A)[0], 1], values=[1])
        A_mean_mat = ones.multiply(A_mean)
        A = A.subtract(A_mean_mat)
    if axis == 1:
        A = A.tr()
    return A

Demo

We create a matrix, call the zero center method twice with a different axis as an argument and print the results (we only print 2 decimal places in one case to make it look pretty):

import linalg as la

A = la.Mat([[1, 2, 3],
            [-2, 1, 4],
            [0, 1, 2],
            [3, 6, 1]])

B = la.Mat([[1, 1, 1],
            [0, 2, 0],
            [0, 3, -4]])

result = la.stats.zero_center(A)
la.print_mat(result)

result = la.stats.zero_center(A, axis=1)
la.print_mat(result, 2)

result = la.stats.zero_center(B)
la.print_mat(result, 2)

Outputs:

>>> la.print_mat(result)
[0.5, -0.5, 0.5]
[-2.5, -1.5, 1.5]
[-0.5, -1.5, -0.5]
[2.5, 3.5, -1.5]

>>> la.print_mat(result, 2)
[-1.0, 0.0, 1.0]
[-3.0, 0.0, 3.0]
[-1.0, 0.0, 1.0]
[-0.33, 2.67, -2.33]

>>> la.print_mat(result, 2)
[0.67, -1.0, 2.0]
[-0.33, 0.0, 1.0]
[-0.33, 1.0, -3.0]

< Sum and mean

Variance, covariance, standard deviation and standard error >

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