使用布尔索引就地修改numpy数组节 [英] Modify numpy array section in-place using boolean indexing

问题描述

给出2D numpy数组，即；

Given a 2D numpy array, i.e.;

import numpy as np

data = np.array([
     [11,12,13],
     [21,22,23],
     [31,32,33],
     [41,42,43],         
     ])

我需要针对所需的行和列，基于两个掩蔽向量修改子阵列；

I need modify in place a sub-array based on two masking vectors for the desired rows and columns;

rows = np.array([False, False, True, True], dtype=bool)
cols = np.array([True, True, False], dtype=bool)

就是这样；

print data

 #[[11,12,13],
 # [21,22,23],
 # [0,0,33],
 # [0,0,43]]      

推荐答案

Now that you know how to access the rows/cols you want, just assigne the value you want to your subarray. It's a tad trickier, though:

mask = rows[:,None]*cols[None,:]
data[mask] = 0

原因是，当我们以data[rows][:,cols]的形式访问子数组时(如您的

The reason is that when we access the subarray as data[rows][:,cols] (as illustrated in your previous question, we're taking a view of a view, and some references to the original data get lost in the way.

相反，这里我们通过广播您的两个1D数组rows和cols彼此构建一个2D布尔数组.您的mask数组现在具有形状(len(rows),len(cols).我们可以使用mask直接访问data的原始项目，并将它们设置为新值.请注意，当您执行data[mask]时，将获得一维数组，这不是您在

Instead, here we construct a 2D boolean array by broadcasting your two 1D arrays rows and cols one with the other. Your mask array has now the shape (len(rows),len(cols). We can use mask to directly access the original items of data, and we set them to a new value. Note that when you do data[mask], you get a 1D array, which was not the answer you wanted in your previous question.

要构造掩码，可以使用&运算符代替*(因为我们正在处理布尔数组)，或更简单的

To construct the mask, we could have used the & operator instead of * (because we're dealing with boolean arrays), or the simpler np.outer function:

mask = np.outer(rows,cols)

为np.outer解决方案提供@Marcus Jones的道具.

props to @Marcus Jones for the np.outer solution.

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