如何在Scipy中计算稀疏矩阵的一列的方差? [英] How do I compute the variance of a column of a sparse matrix in Scipy?

问题描述

我有一个大的scipy.sparse.csc_matrix，并想将其标准化.也就是说，从每个元素中减去列平均值，然后除以列标准偏差(std)i.

I have a large scipy.sparse.csc_matrix and would like to normalize it. That is subtract the column mean from each element and divide by the column standard deviation (std)i.

scipy.sparse.csc_matrix具有.mean()，但是有一种有效的方法来计算方差或标准差吗?

scipy.sparse.csc_matrix has a .mean() but is there an efficient way to compute the variance or std?

推荐答案

您可以使用均值来自己计算方差，并使用以下

You can calculate the variance yourself using the mean, with the following formula:

E[X^2] - (E[X])^2

E[X]代表平均值.因此，要计算E[X^2]，您必须将csc_matrix平方，然后使用mean函数.要获得(E[X])^2，您只需要对使用常规输入获得的mean函数的结果求平方即可.

E[X] stands for the mean. So to calculate E[X^2] you would have to square the csc_matrix and then use the mean function. To get (E[X])^2 you simply need to square the result of the mean function obtained using the normal input.

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