用索引的numpy数组索引numpy数组 [英] Indexing a numpy array with a numpy array of indexes

问题描述

我有一个3D numpy数组data和另一个索引数组pos(索引本身就是一个numpy数组，这使后者成为2D数组):

I have a 3D numpy array data and another array pos of indexes (an index is a numpy array on its own, which makes the latter array a 2D array):

import numpy as np
data = np.arange(8).reshape(2, 2, -1)
#array([[[0, 1],
#    [2, 3]],
#
#  [[4, 5],
#    [6, 7]]])

pos = np.array([[1, 1, 0], [0, 1, 0], [1, 0, 0]])
#array([[1, 1, 0],
#       [0, 1, 0],
#       [1, 0, 0]])

我想使用pos的索引选择和/或变异data中的元素.我可以使用for循环或列表理解来进行选择:

I want to select and/or mutate the elements from data using the indexes from pos. I can do the selection using a for loop or a list comprehension:

[data[tuple(i)] for i in pos]
#[6, 2, 4]
data[[i for i in pos.T]]
#array([6, 2, 4])

但这似乎不是一种麻木的方式.是否有针对此问题的矢量化numpy解决方案?

But this does not seem to be a numpy way. Is there a vectorized numpy solution to this problem?

推荐答案

您可以将pos拆分为3个单独的数组和索引，就像这样—

You can split pos into 3 separate arrays and index, like so—

>>> i, j, k = pos.T
>>> data[i, j, k]
array([6, 2, 4])

在这里，pos中的列数与data的深度相对应.只要您要处理3D矩阵，就很好地获得i，j和k的复杂性.

Here, the number of columns in pos correspond to the depth of data. As long as you're dealing with 3D matrices, getting i, j, and k well never get more complicated than this.

在python-3.6 +上，您可以将其缩短为-

On python-3.6+, you can shorten this to—

>>> data[[*pos.T]]
array([6, 2, 4])

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