索引numpy多维数组取决于切片方法 [英] Indexing numpy multidimensional arrays depends on a slicing method

问题描述

我有一个3-D阵列.当我对它进行二维切片时，结果取决于它是用列表索引还是用切片索引.在第一种情况下，结果被转置.在手册中未找到此行为.. >

I have a 3-D array. When I take a 2-D slice of it the result depends on whether it is indexed with a list or with a slice. In the first case the result is transposed. Didn't find this behaviour in the manual.

>>> import numpy as np
>>> x = np.array([[[1,1,1],[2,2,2]], [[3,3,3],[4,4,4]]])
>>> x
array([[[1, 1, 1],
        [2, 2, 2]],
       [[3, 3, 3],
        [4, 4, 4]]])
>>> x[0,:,[0,1]]
array([[1, 2],
       [1, 2]])
>>> x[0,:,slice(2)]
array([[1, 1],
       [2, 2]])
>>> 

有人可以为此指出理由吗?

Could anyone point a rationale for this?

推荐答案

据我了解，当给定类似list/tuple的索引时，NumPy吐出结果时会遵循轴编号原理.

As I understand it, NumPy is following the axis numbering philosophy when it spits out the result when given a list/tuple-like index.

array([[[1, 1, 1],
        [2, 2, 2]],

       [[3, 3, 3],
        [4, 4, 4]]])

当您已经指定了前两个索引(x[0, :, ])时，接下来的问题是如何提取第三维.现在，当您指定一个元组(0,1)时，它首先会明智地提取0个切片轴，因此它会得到[1, 2]，因为它位于第0个轴上，接下来它将同样提取1 st个切片并堆叠在已存在的行[1, 2]下.

When you already specify the first two indices (x[0, :, ]), now the next question is how to extract the third dimension. Now, when you specify a tuple (0,1), it first extracts the 0th slice axis wise, so it gets [1, 2] since it lies in 0th axis, next it extracts 1st slice likewise and stacks below the already existing row [1, 2].

[[1, 1, 1],          array([[1, 2],
 [2, 2, 2]]  =====>         [1, 2]])

(由于第0轴向下生长，因此已可视化堆栈(如下所示(不在上方))

(visualize this stacking as below (not on top of) the already existing row since axis-0 grows downwards)

或者，当为索引指定slice(n)时，它遵循切片原理(start:stop:step).请注意，在您的示例中，使用slice(2)本质上等于0:2.因此，它首先提取[1, 1]，然后提取[2, 2].请注意，这里[1, 1]是如何出现在[2, 2]之上的，再次遵循这里相同的轴原理，因为我们还没有离开三维.这就是为什么这个结果是另一个的转置的原因.

Alternatively, it is following the slicing philosophy (start:stop:step) when slice(n) is given for the index. Note that using slice(2) is essentially equal to 0:2 in your example. So, it extracts [1, 1] first, then [2, 2]. Note, here to how [1, 1] comes on top of [2, 2], again following the same axis philosophy here since we didn't leave the third dimension yet. This is why this result is the transpose of the other.

array([[1, 1],
       [2, 2]])

---

此外，请注意，从3-D数组开始，将保持这种一致性.下面是4-D数组和切片结果的示例.

---

Also, note that starting from 3-D arrays this consistency is preserved. Below is an example from 4-D array and the slicing results.

In [327]: xa
Out[327]: 
array([[[[ 0,  1,  2],
         [ 3,  4,  5],
         [ 6,  7,  8]],

        [[ 9, 10, 11],
         [12, 13, 14],
         [15, 16, 17]]],


       [[[18, 19, 20],
         [21, 22, 23],
         [24, 25, 26]],

        [[27, 28, 29],
         [30, 31, 32],
         [33, 34, 35]]]])

In [328]: xa[0, 0, :, [0, 1]]
Out[328]: 
array([[0, 3, 6],
       [1, 4, 7]])

In [329]: xa[0, 0, :, 0:2]
Out[329]: 
array([[0, 1],
       [3, 4],
       [6, 7]])

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