在稀疏矩阵中采用对数函数的有效方法 [英] Efficient way of taking Logarithm function in a sparse matrix

问题描述

我有一个大的稀疏矩阵.我想对该稀疏矩阵中的所有元素采用log4.

I have a big sparse matrix. I want to take log4 for all element in that sparse matrix.

我尝试使用numpy.log()，但不适用于矩阵.

I try to use numpy.log() but it doesn't work with matrices.

我也可以逐行取对数.然后，我用新的一行粉碎旧的一行.

I can also take logarithm row by row. Then I crush old row with a new one.

# Assume A is a sparse matrix (Linked List Format) with float values as data
# It is only for one row

import numpy as np
c = np.log(A.getrow(0)) / numpy.log(4)
A[0, :] = c

这没有我预期的那么快.有没有更快的方法可以做到这一点?

This was not as quick as I'd expected. Is there a faster way to do this?

推荐答案

您可以直接修改data属性:

>>> a = np.array([[5,0,0,0,0,0,0],[0,0,0,0,2,0,0]])
>>> coo = coo_matrix(a)
>>> coo.data
array([5, 2])
>>> coo.data = np.log(coo.data)
>>> coo.data
array([ 1.60943791,  0.69314718])
>>> coo.todense()
matrix([[ 1.60943791,  0.        ,  0.        ,  0.        ,  0.        ,
          0.        ,  0.        ],
        [ 0.        ,  0.        ,  0.        ,  0.        ,  0.69314718,
          0.        ,  0.        ]])

请注意，如果稀疏格式包含重复的元素(在COO格式中有效)，则此方法将无法正常工作；它将分别获取日志和log(a) + log(b) != log(a + b).您可能要先转换为CSR或CSC(速度很快)，以避免出现此问题.

Note that this doesn't work properly if the sparse format has repeated elements (which is valid in the COO format); it'll take the logs individually, and log(a) + log(b) != log(a + b). You probably want to convert to CSR or CSC first (which is fast) to avoid this problem.

当然，您还必须添加检查，如果稀疏矩阵的格式不同.而且，如果您不想就地修改矩阵，只需像在答案中一样构造一个新的稀疏矩阵，但无需添加3，因为这里完全没有必要.

You'll also have to add checks if the sparse matrix is in a different format, of course. And if you don't want to modify the matrix in-place, just construct a new sparse matrix as you did in your answer, but without adding 3 because that's completely unnecessary here.

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