Python scikit学习pca.explained_variance_ratio_截止 [英] Python scikit learn pca.explained_variance_ratio_ cutoff

问题描述

当选择主成分(k)的数目，我们选择k以可使得例如，方差为99％，被保持在最小值.

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.

但是，在Python Scikit学习中，我不是100％确定pca.explained_variance_ratio_ = 0.99等于保留了99％的方差"吗?有人可以开导吗?谢谢.

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.

• Python Scikit学习PCA手册在这里

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

推荐答案

是的，您几乎是正确的. pca.explained_variance_ratio_参数返回每个维度说明的方差矢量.因此，pca.explained_variance_ratio_[i]给出了仅由第i + 1维解释的方差.

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.

您可能想执行pca.explained_variance_ratio_.cumsum().这将返回向量x，以使x[i]返回由前i + 1个维度解释的累积方差.

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.        ]

所以在我的随机玩具数据中，如果我选择k=4，我将保留93.3％的差异.

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

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