用 2d 数组索引 3d numpy 数组 [英] Indexing of 3d numpy arrays with 2d arrays

问题描述

我正在尝试从 3d numpy 数组中提取值.目前我可以执行以下操作:

I am attempting to extract values from a 3d numpy array. At the moment I can perform the following operations:

newmesh.shape
(40,40,40)

newmesh[2,5,6]
6

然而，如果我尝试用数组索引它，结果并不如预期；

However, if I try to index it with an array, the result is not as expected;

newmesh[np.array([2,5,6])].shape
(3, 42, 42)

我曾尝试使用 np.take，但它会产生以下结果；

I have tried using np.take, however it produces the following;

np.take(newmesh,np.array([2,5,6]))
[-1 -1 -1]

任何想法为什么会发生这种情况?我的目标是输入一个 (n,3) 数组，其中每一行对应一个 newmesh 的值，即输入一个 (n,3) 数组将返回一个长度为 n 的一维数组.

Any ideas why this is happening? My goal is to input a (n,3) array, where each row corresponds to a value of newmesh, i.e. inputting a (n,3) array would give back a 1d array of length n.

推荐答案

使用 idx 作为 (n,3) 索引数组，一种使用 linear 的方法-indexing 将与 np.ravel_multi_index -

With idx as the (n,3) indexing array, one approach using linear-indexing would be with np.ravel_multi_index -

np.take(newmesh,np.ravel_multi_index(idx.T,newmesh.shape))

元组形成的方法看起来像这样 -

An approach with tuple formation would look like this -

newmesh[tuple(idx.T)]

如果只有三个维度，你甚至可以只使用柱状切片来索引每个维度，就像这样 -

If there are just three dimensions, you can even just use columnar slices for indexing into each dimension, like so -

newmesh[idx[:,0],idx[:,1],idx[:,2]]

运行时测试如果有人有兴趣查看与所列方法相关的性能数据，这里有一个快速运行时测试 -

Runtime test If anyone's interested in seeing the performance numbers associated with the listed approaches, here's a quick runtime test -

In [18]: newmesh = np.random.rand(40,40,40)

In [19]: idx = np.random.randint(0,40,(1000,3))

In [20]: %timeit np.take(newmesh,np.ravel_multi_index(idx.T,newmesh.shape))
10000 loops, best of 3: 22.5 µs per loop

In [21]: %timeit newmesh[tuple(idx.T)]
10000 loops, best of 3: 20.9 µs per loop

In [22]: %timeit newmesh[idx[:,0],idx[:,1],idx[:,2]]
100000 loops, best of 3: 17.2 µs per loop

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