Python scikit learn pca.explained_variance_ratio_ cutoff

Chubaka picture Chubaka · Sep 30, 2015 · Viewed 47.6k times · Source

When choosing the number of principal components (k), we choose k to be the smallest value so that for example, 99% of variance, is retained.

However, in the Python Scikit learn, I am not 100% sure pca.explained_variance_ratio_ = 0.99 is equal to "99% of variance is retained"? Could anyone enlighten? Thanks.

  • The Python Scikit learn PCA manual is here

http://scikit-learn.org/stable/modules/generated/sklearn.decomposition.PCA.html#sklearn.decomposition.PCA

Answer

Curt F. picture Curt F. · Sep 30, 2015

Yes, you are nearly right. The pca.explained_variance_ratio_ parameter returns a vector of the variance explained by each dimension. Thus pca.explained_variance_ratio_[i] gives the variance explained solely by the i+1st dimension.

You probably want to do pca.explained_variance_ratio_.cumsum(). That will return a vector x such that x[i] returns the cumulative variance explained by the first i+1 dimensions.

import numpy as np
from sklearn.decomposition import PCA

np.random.seed(0)
my_matrix = np.random.randn(20, 5)

my_model = PCA(n_components=5)
my_model.fit_transform(my_matrix)

print my_model.explained_variance_
print my_model.explained_variance_ratio_
print my_model.explained_variance_ratio_.cumsum()

[ 1.50756565  1.29374452  0.97042041  0.61712667  0.31529082]
[ 0.32047581  0.27502207  0.20629036  0.13118776  0.067024  ]
[ 0.32047581  0.59549787  0.80178824  0.932976    1.        ]

So in my random toy data, if I picked k=4 I would retain 93.3% of the variance.