结合切片和广播的索引多维numpy的阵列 [英] Combining slicing and broadcasted indexing for multi-dimensional numpy arrays

问题描述

我有一个ND numpy的阵列（让说，例如3x3x3的）从至极，我想提取子阵列，结合片和数组索引。例如：

 导入numpy的是NP
A = np.arange（3 * 3 * 3）.reshape（（3,3,3））
I0，I1，I2 =（[0,1]，[0,1,2]，[0,2]）
IND1 = J0，J1，J2 = np.ix_（I0，I1，I2）
IND2 =（J0，切片（无），J2）
B1 = A [IND1]
B2 = A [IND2]
 

我预计B1 B2 ==，但实际上，形状不同

 ＆GT;＆GT;＆GT; B1.shape
（2，3，2）
＆GT;＆GT;＆GT; B2.shape
（2，1，2，3）
＆GT;＆GT;＆GT; B1
阵列（[[[0,2]，
        [3,5]
        [6,8]，       [9，11]，
        [12，14]，
        [15，17]]]）
＆GT;＆GT;＆GT; B2
阵列（[[[[0,3，6]
         [2，5,8]]]，       [[[9，12，15]，
         [11，14，17]]]]）
 

有人明白为什么呢？任何想法我怎么可能通过操纵只'A'和'IND2对象得到'B1'？我们的目标是，这将适用于任何次阵列，而且我也不会找我的尺寸要完全保持的形状（希望我不够清楚:)）。谢谢！结果
---编辑---结果
为了更清楚，我想有一个函数'有趣'，使得

  A [乐趣（IND2）] == B1
 

解决方案

在受限制的索引情况下，像这样使用九_ ，这是可以做到的索引在连续的步骤。

  A [IND1]
 

是相同的

  A [I1] [:, 12] [：，：，I3]
 

和自 I2 是品种齐全，

  A [I1] [...，I3]
 

如果你只有 IND2 可用

  A [IND2 [0] .flatten（）] [IND2 [2] .flatten（）]
 

在更广泛的背景下，你必须知道如何 J0，J1，J2 与对方，但播出时，它们是由生成IX _ ，关系很简单。

我可以想象它会是方便的分配情况 A1 = A [I1] ，其次是各种涉及行动 A1 ，包括但不限于 A1 [...，I3] 。你要知道，当 A1 是一个观点，当它是一个副本。

另一个索引工具是取：

  A.take（I0，轴= 0）。取（I2，轴= 2）
 

I have a ND numpy array (let say for instance 3x3x3) from wich I'd like to extract a sub-array, combining slices and index arrays. For instance:

import numpy as np  
A = np.arange(3*3*3).reshape((3,3,3))
i0, i1, i2 = ([0,1], [0,1,2], [0,2])
ind1 = j0, j1, j2 = np.ix_(i0, i1, i2)
ind2 = (j0, slice(None), j2)
B1 = A[ind1]
B2 = A[ind2]

I would expect that B1 == B2, but actually, the shapes are different

>>> B1.shape
(2, 3, 2)
>>> B2.shape
(2, 1, 2, 3)
>>> B1
array([[[ 0,  2],
        [ 3,  5],
        [ 6,  8]],

       [[ 9, 11],
        [12, 14],
        [15, 17]]])
>>> B2
array([[[[ 0,  3,  6],
         [ 2,  5,  8]]],

       [[[ 9, 12, 15],
         [11, 14, 17]]]])

Someone understands why? Any idea of how I could get 'B1' by manipulating only 'A' and 'ind2' objects? The goal is that it would work for any nD arrays, and that I would not have to look for the shape of dimensions I want to keep entirely (hope I'm clear enough:)). Thanks!!
---EDIT---
To be clearer, I would like to have a function 'fun' such that

A[fun(ind2)] == B1

解决方案

In restricted indexing cases like this using ix_, it is possible to do the indexing in successive steps.

A[ind1]

is the same as

A[i1][:,i2][:,:,i3]

and since i2 is the full range,

A[i1][...,i3]

If you only have ind2 available

A[ind2[0].flatten()][[ind2[2].flatten()]

In more general contexts you have to know how j0,j1,j2 broadcast with each other, but when they are generated by ix_, the relationship is simple.

I can imagine circumstances in which it would be convenient to assign A1 = A[i1], followed by a variety of actions involving A1, including, but not limited to A1[...,i3]. You have to be aware of when A1 is a view, and when it is a copy.

Another indexing tool is take:

A.take(i0,axis=0).take(i2,axis=2)

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