替换索引数组中与另一个数组对应的值 [英] Replace values in array of indexes corresponding to another array

问题描述

我有一个大小为[1, x]的值数组A和一个大小为[1, y](y> x)的索引数组B与数组A对应.因此，我想要一个大小为[1,y]的数组C，其中填充了A的值.

I have an array A of size [1, x] of values and an array B of size [1, y] (y > x) of indexes corresponding to array A. I want as result an array C of size [1,y] filled with values of A.

以下是输入和输出的示例:

Here is an example of inputs and outputs:

>>> A = [6, 7, 8]
>>> B = [0, 2, 0, 0, 1]
>>> C = #Some operations
>>> C
[6, 8, 6, 6, 7]

我当然可以这样解决:

>>> C = []
>>> for val in B:
>>>     C.append(A[val])

但是实际上我被认为是一种更好的方法.特别是因为我想将其用作另一个函数的参数.看起来像A[B]的表达式(但有效)是理想的.我不介意使用NumPy或Pandas解决方案.

But I was actually expected a nicer way to do it. Especially because I want to use it as an argument of another function. An expression looking like A[B] (but a working one) would be ideal. I don't mind solution using NumPy or pandas.

推荐答案

用于获取多个项目 operator.itemgetter 派上用场了:

For fetching multiple items operator.itemgetter comes in handy:

from operator import itemgetter
A = [6, 7, 8]
B = [0, 2, 0, 0, 1]

itemgetter(*B)(A)
# (6, 8, 6, 6, 7)

---

也正如您提到的numpy一样，可以直接按指定的索引数组来完成此操作，即A[B]:

---

Also as you've mentioned numpy, this could be done directly by indexing the array as you've specified, i.e. A[B]:

import numpy as np
A = np.array([6, 7, 8])
B = np.array([0, 2, 0, 0, 1])

A[B]
# array([6, 8, 6, 6, 7])

另一种选择是使用 :

np.take(A,B)
# array([6, 8, 6, 6, 7])

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