在numpy数组中索引/切片以修改它 [英] indexing/slicing in numpy array to modify it

问题描述

我有一个numpy数组A，我想使用索引列表B修改它中的值。但事情是在我的切片中我可以有多次数组元素...
这个例子将更好地解释我的意思：

I have a numpy array A and I want to modify values in it using a indexing list B. But the thing is in my slicing I can have an element of the array multiple times... This example will explain better what I mean by that :

import numpy as np
A = np.arange(5) + 0.5
B = np.array([0, 1, 0, 2, 0, 3, 0, 4])
print A[B]

按预期返回 [0.5 1.5 0.5 2.5 0.5 3.5 0.5 4.5] 。
但是，如果我这样做：

returns as expected [ 0.5 1.5 0.5 2.5 0.5 3.5 0.5 4.5]. However if I do that :

A[B] += 1.
print A

我原本期望获得 [4.5 2.5 3.5 4.5 5.5] 因为第一个元素在索引向量B中重复了4次，但它返回 [1.5 2.5 3.5 4.5 5.5] 。
那么我该怎样做我真正想做的事情呢？ （不使用任何循环，因为我在非常大的数组上使用它）

I was expecting to obtain [ 4.5 2.5 3.5 4.5 5.5] as the first element is repeated 4 times in the indexing vector B, but it returns [ 1.5 2.5 3.5 4.5 5.5]. So how can I do what I actually wanted to do? (without using any loop as I'm using that on very large arrays)

推荐答案

为什么会发生这种情况的解释有点牵扯，但基本上，缓慢吃你的作业。这个numpy ufuncs问题有几种解决方法。适用于任何操作的合适的一个是使用相应的ufunc的 at 方法：

The explanation why this happens is a little involved, but basically, "buffering ate your homework." There are a couple of ways around this issue of numpy ufuncs. The proper one, that will work with any operation is to use the corresponding ufunc's at method:

>>> A = np.arange(5) + 0.5
>>> B = np.array([0, 1, 0, 2, 0, 3, 0, 4])
>>> np.add.at(A, B, 1)
>>> A
array([ 4.5,  2.5,  3.5,  4.5,  5.5])

这个趋势有点慢，所以为了尽可能快的性能，只有添加，你可以使用 np.bincount ：

This tends to be kind of slow, so for the fastest performance possible, and only for addition, you can use np.bincount:

>>> A = np.arange(5) + 0.5
>>> A += np.bincount(B) * 1  # replace the 1 with the number you want to add
>>> A
array([ 4.5,  2.5,  3.5,  4.5,  5.5])

编辑

如果要添加的是与 B 相同长度的数组，那么使用 bincount ，以下内容可能会比第一种方法运行得更快：

If what you want to add is an array of the same length as B, then the following, using bincount, is probably going to run faster than the first method:

>>> A = np.arange(5) + 0.5
>>> C = np.ones_like(B)  # They are all ones, but could be anything
>>> A += np.bincount(B, weights=C)
>>> A
array([ 4.5,  2.5,  3.5,  4.5,  5.5])

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