Hotelling在Python中的T ^ 2分数
我在python中使用matplotlib在数据集上应用了pca。 但是，matplotlib不提供像Matlab一样的t平方得分。 有没有办法像Matlab一样计算Hotelling的T ^ 2分数？
谢谢。
matplotlib的PCA类不包含Hotelling T2计算，但只需几行代码即可完成。 以下代码包含一个计算每个点的T2值的函数。 __main__脚本将PCA应用于Matlab的pca文档中使用的相同示例，因此您可以验证函数是否生成与Matlab相同的值。
from __future__ import print_function, division
import numpy as np
from matplotlib.mlab import PCA
def hotelling_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-1
print("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)
输出：
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: 517.7969 67.4964 12.4054 0.2372
tsquared:
[ 5.6803 3.0758 6.0002 2.6198 3.3681 0.5668 3.4818 3.9794 2.6086
7.4818 4.183 2.2327 2.7216]