如何最有效地替换python中数组的每个第n个值? [英] how to replace every n-th value of an array in python most efficiently?

问题描述

我想知道是否有更pythonic(和有效)的方法来执行以下操作:

I was wondering whether there is a more pythonic (and efficient) way of doing the following:

MAX_SIZE = 100
nbr_elements = 10000
y = np.random.randint(1, MAX_SIZE, nbr_elements)


REPLACE_EVERY_Nth = 100
REPLACE_WITH = 120
c = 0

for index, item in enumerate(y):
    c += 1
    if (c % REPLACE_EVERY_Nth == 0):
        y[index] = REPLACE_WITH

所以基本上我生成了一堆从 1 到 MAX_SIZE-1 的数字，然后我想用 REPLACE_WITH 替换每个 REPLACE_EVERY_Nth 元素.这工作正常，但我想可以在不使用 enumerate 的情况下以某种方式完成?

So basically I generate a bunch of numbers from 1 to MAX_SIZE-1, and then I want to replace every REPLACE_EVERY_Nth element with REPLACE_WITH. This works fine but I guess it could be somehow done without using enumerate?

我在想这样的事情(我知道这是错误的，因为我用 y 的索引替换了原来的 y):

I was thinking something like this (which I know is wrong, because I replace the original y with the indices of y):

y = map(lambda x: REPLACE_WITH if not x%REPLACE_EVERY_Nth else x, range(len(y)))

有没有办法对索引进行取模但替换值?

is there a way to do modulo on the indices but replace the values?

推荐答案

使用以 REPLACE_EVERY_Nth 作为步长值的切片:

Use a slicing with REPLACE_EVERY_Nth as step value:

y[::REPLACE_EVERY_Nth] = REPLACE_WITH

这与您的代码略有不同，因为它将从第一个项目(即索引 0)开始.要准确了解您的代码的作用，请使用

This is slightly different from your code, since it will start with the very first item (i.e. index 0). To get exactly what your code does, use

y[REPLACE_EVERY_Nth - 1::REPLACE_EVERY_Nth] = REPLACE_WITH

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