更改满足条件的NumPy数组的第n个条目 [英] Change n-th entry of NumPy array that fulfills condition

问题描述

我有一个NumPy数组 arr 和一个（反向）掩码掩码。为简单起见，我们假设它们都是1d。我想更改 arr 中的 n 非屏蔽值。

I have a NumPy array arr and an (inverse) mask mask. For simplicity let us assume they are both 1d. I want to change the nth non-masked value in arr.

一个例子：

import numpy as np
arr = np.arange(5)
mask = np.array((True, False, True, True, False))

不幸的是，

arr[mask][-1] = 100

我预计会返回

array([0, 1, 2, 100, 4])

无效关于非连续项目的NumPy数组视图中列出的原因。

解决方法是将允许的值存储在新变量中，更改相应的值，并将所有值复制回原始数组：

A workaround would be to store the allowed values in a new variable, change the respective value, and copy all values back into the original array:

tmp = arr[mask]
tmp[-1] = 100
arr[mask] = tmp

然而，这个解决方案很丑陋且效率低下，因为我必须复制许多我根本不想改变的值。

However, this solution is ugly and inefficient, since I have to copy many values that I do not want to change at all.

有没有人有一种优雅的方式来处理这类问题？我会对最通用的解决方案感兴趣，这样我就可以用 tmp 完成所有经典赋值操作。但是，如果有一种有效的方法仅适用于具体的案例，我仍然会对它感兴趣！

Does anyone have an elegant way to deal with this kind of problem? I would be interested in a maximally general solution, so that I could do all classic assignment operations with tmp. However, if there is an efficient way that works only for the concrete dscribed case, I would still be interested in it!

推荐答案

一种选择是使用 np.where 获取掩码条件为 True 。然后，您可以使用这些索引的子集索引到 arr 并进行分配：

One option would be to use np.where to obtain the set of indices where your mask condition is True. You can then index into arr using a subset of these indices and make your assignment:

# np.where returns a tuple of index arrays, one per dimension
arr[np.where(mask)[0][-1]] = 100

print(repr(arr))
# array([  0,   1,   2, 100,   4])

您可以将此方法与切片索引，布尔索引等结合使用。例如：

You could combine this approach with slice indexing, boolean indexing etc. For example:

arr[np.where(mask)[0][::-1]] = 100, 200, 300
print(repr(arr))
# array([300,   1, 200, 100,   4])

这篇关于更改满足条件的NumPy数组的第n个条目的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！