结合索引操作时，会查看和复制NumPy数组的混乱情况 [英] Views and copies confusion with NumPy arrays when combining index operations

问题描述

>>> a = np.arange(12).reshape(3,4)

>>> a[0:3:2, :][:, [0,2]] = 100   ### the first time
>>> a  
array([[100, 1, 100, 3],
[ 4, 5, 6, 7],
[100, 9, 100, 11]])

>>> a[:, [0, 2]][0:3:2, :] = 0    ### second time
>>> a   
array([[100, 1, 100, 3],
       [ 4, 5, 6, 7],
       [100, 9, 100, 11]])

我真的很困惑python中的视图和副本.上面的代码显示，数组a中的给定行和列第一次更改为100，这将更改原始数组a.

I am really confused about views and copies in python. The code above shows that the first time the given rows and columns in array a was changed to 100, which changed the original array a.

但是，第二次未更改原始数组.为什么会这样?

However, the second time the original array was not changed. Why is that?

推荐答案

具有[:, [0,2]]的查找将返回一个副本，因为它是

A lookup with [:, [0,2]] will return a copy because it's advanced indexing. However when you assign to a slice (e.g. array[whatever] = sth) it won't create the copy but assign to the specified items, even if it's advanced indexing.

第一个示例之所以可行，是因为第一个切片会返回一个视图，然后将分配用于该视图的切片.

So the first example works because the first slicing returns a view and then it uses assignment for the slice of the view.

但是第二个失败"是因为您分配给副本的一部分(因为高级索引是在常规索引之前完成的).

However the second one "fails" because you assign to a slice of a copy (because the advanced indexing is done before the regular indexing).

区别主要是因为另一种方法负责设置( __setitem__ )，而不是获取( __getitem__ )这些切片.要反汇编您的声明:

The difference is mainly because another method is responsible for setting (__setitem__) to slices than for getting (__getitem__) these slices. To disassemble your statements:

a[0:3:2, :][:, [0,2]] = 100 

a.__getitem__((slice(0, 3, 2), slice(None))).__setitem__((slice(None), [0, 2]), 100)
|-------------- returns a view ------------|

而第二个是:

a[:, [0,2]][0:3:2, :] = 0

a.__getitem__((slice(None), [0, 2])).__setitem__((slice(0, 3, 2), slice(None)), 0)
|--------- returns a copy ---------|

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