Hotelling's T^2 Scores In Python

I applied pca on a data set using matplotlib in python. However, matplotlib does not provide a t-squared scores like Matlab. Is there a way to compute Hotelling's T^2 score like Ma

Solution 1:

matplotlib's PCA class doesn't include the Hotelling T calculation, but it can be done with just a couple lines of code. The following code includes a function to compute the T values for each point. The __main__ script applies PCA to the same example as used in Matlab's pca documentation, so you can verify that the function generates the same values as Matlab.

from __future__ import print_function, division

import numpy as np
from matplotlib.mlab import PCA


defhotelling_tsquared(pc):
    """`pc` should be the object returned by matplotlib.mlab.PCA()."""
    x = pc.a.T
    cov = pc.Wt.T.dot(np.diag(pc.s)).dot(pc.Wt) / (x.shape[1] - 1)
    w = np.linalg.solve(cov, x)
    t2 = (x * w).sum(axis=0)
    return t2


if __name__ == "__main__":

    hald_text = """Y       X1      X2      X3      X4
    78.5    7       26      6       60
    74.3    1       29      15      52
    104.3   11      56      8       20
    87.6    11      31      8       47
    95.9    7       52      6       33
    109.2   11      55      9       22
    102.7   3       71      17      6
    72.5    1       31      22      44
    93.1    2       54      18      22
    115.9   21      47      4       26
    83.8    1       40      23      34
    113.3   11      66      9       12
    109.4   10      68      8       12
    """
    hald = np.loadtxt(hald_text.splitlines(), skiprows=1)
    ingredients = hald[:, 1:]

    pc = PCA(ingredients, standardize=False)
    coeff = pc.Wt

    np.set_printoptions(precision=4)

    # For coeff and latent, compare to#     http://www.mathworks.com/help/stats/pca.html#btjpztu-1print("coeff:")
    print(coeff)
    print()

    latent = pc.s / (ingredients.shape[0] - 1)
    print("latent:" + (" %9.4f"*len(latent)) % tuple(latent))
    print()

    # For tsquared, compare to#     http://www.mathworks.com/help/stats/pca.html#bti6r0c-1
    tsquared = hotelling_tsquared(pc)
    print("tsquared:")
    print(tsquared)

Output:

coeff:
[[ 0.0678  0.6785 -0.029  -0.7309]
 [ 0.646   0.02   -0.7553  0.1085]
 [-0.5673  0.544  -0.4036  0.4684]
 [ 0.5062  0.4933  0.5156  0.4844]]

latent:  517.796967.496412.40540.2372

tsquared:
[ 5.68033.07586.00022.61983.36810.56683.48183.97942.60867.48184.1832.23272.7216]

Solution 2:

Even though this is an old question, I am posting the code as it may help someone. Here is the code, as a bonus this does multiple hotelling tests at once

import numpy as np
from scipy.stats import f as f_distrib


def hotelling_t2(X, Y):

# X and Y are 3D arrays# dim 0: number of features# dim 1: number of subjects# dim 2: number of mesh nodes or voxels (numer of tests)

nx = X.shape[1]
ny = Y.shape[1]
p = X.shape[0]
Xbar = X.mean(1)
Ybar = Y.mean(1)
Xbar = Xbar.reshape(Xbar.shape[0], 1, Xbar.shape[1])
Ybar = Ybar.reshape(Ybar.shape[0], 1, Ybar.shape[1])

X_Xbar = X - Xbar
Y_Ybar = Y - Ybar
Wx = np.einsum('ijk,ljk->ilk', X_Xbar, X_Xbar)
Wy = np.einsum('ijk,ljk->ilk', Y_Ybar, Y_Ybar)
W = (Wx + Wy) / float(nx + ny - 2)
Xbar_minus_Ybar = Xbar - Ybar
x = np.linalg.solve(W.transpose(2, 0, 1),
Xbar_minus_Ybar.transpose(2, 0, 1))
x = x.transpose(1, 2, 0)

t2 = np.sum(Xbar_minus_Ybar * x, 0)
t2 = t2 * float(nx * ny) / float(nx + ny)
stat = (t2 * float(nx + ny - 1 - p) / (float(nx + ny - 2) * p))

pval = 1 - np.squeeze(f_distrib.cdf(stat, p, nx + ny - 1 - p))
return pval, t2