使用高级索引复制数组数据分配吗？ [英] Does assignment with advanced indexing copy array data?

问题描述

我慢慢想了解查看 s和 copy在numpy中的区别，以及可变

如果我使用高级索引，它应该返回一个副本。这似乎是真的：

 在[1]：import numpy as np 
 In [2]：a = np.zeros（（3,3））
 In [3]：b = np.array（np.identity（3），dtype = bool）
 
 In [4]：c = a [b] 
 
在[5]：c [：] = 9 
 
在[6]：a 
 Out [6]：
 array（[[0.,0,0,0]，
 [0.，0.，0]，
 [0.，0.，0 。]]）
  

由于 c 一个副本，它不共享数据，更改不会改变 a 。但是，这让我困惑：

 在[7]：a [b] = 1 
 
 In [8]：a 
 Out [8]：
 array（[[1,0,0,0]，
 [ 1.，0.]，
 [0.，0.，1.]]）
  

所以，看来，即使我使用高级索引，分配仍然把左边的东西视为一个视图。显然，第2行中的 a 与第6行中的 a 是相同的对象/数据，因为突变 c 对此没有影响。

所以我的问题是： a 在第8行与以前相同的对象/数据（不计算对角线当然）还是它是一个副本？换句话说， a 的数据是否复制到新的 a ，或者是其数据是否在原位？ / p>



例如，是否如下：

  x = 2,3] 
x + = [4] 
  

或类似：

  y =（1,2,3）
y + =（4，）
  

我不知道如何检查这个，因为在这两种情况下， a.flags.owndata 是 True 。

解决方案

如果你这么做的话，请随时解释或回答一个不同的问题。 c = a [b] ， a .__ get_item __ 使用 b 作为其唯一的参数，并且返回的任何值都分配给 c 。

使用 b a [b] = c ， a .__ setitem __

因此，尽管有相同的<$ c $> c> a [b] 语法，两个表达式都在做不同的事情。你可以子类化 ndarray ，重载这两个函数，并使它们的行为不同。默认情况下在numpy中，前者返回一个副本（如果 b 是数组），但后者修改 a

I am slowly trying to understand the difference between views and copys in numpy, as well as mutable vs. immutable types.

If I access part of an array with 'advanced indexing' it is supposed to return a copy. This seems to be true:

In [1]: import numpy as np
In [2]: a = np.zeros((3,3))
In [3]: b = np.array(np.identity(3), dtype=bool)

In [4]: c = a[b]

In [5]: c[:] = 9

In [6]: a
Out[6]: 
array([[ 0.,  0.,  0.],
       [ 0.,  0.,  0.],
       [ 0.,  0.,  0.]])

Since c is just a copy, it does not share data and changing it does not mutate a. However, this is what confuses me:

In [7]: a[b] = 1

In [8]: a
Out[8]: 
array([[ 1.,  0.,  0.],
       [ 0.,  1.,  0.],
       [ 0.,  0.,  1.]])

So, it seems, even if I use advanced indexing, assignment still treats the thing on the left as a view. Clearly the a in line 2 is the same object/data as the a in line 6, since mutating c has no effect on it.

So my question: is the a in line 8 the same object/data as before (not counting the diagonal of course) or is it a copy? In other words, was a's data copied to the new a, or was its data mutated in place?

For example, is it like:

x = [1,2,3]
x += [4]

or like:

y = (1,2,3)
y += (4,)

I don't know how to check for this because in either case, a.flags.owndata is True. Please feel free to elaborate or answer a different question if I'm thinking about this in a confusing way.

解决方案

When you do c = a[b], a.__get_item__ is called with b as its only argument, and whatever gets returned is assigned to c.

When you doa[b] = c, a.__setitem__ is called with b and c as arguments and whatever gets returned is silently discarded.

So despite having the same a[b] syntax, both expressions are doing different things. You could subclass ndarray, overload this two functions, and have them behave differently. As is by default in numpy, the former returns a copy (if b is an array) but the latter modifies a in place.

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