Python通过索引选择列表中的元素 [英] Python selecting elements in a list by indices

问题描述

我在python中有一个列表，想要从中获取一组索引并保存为原始列表的子集:

I have a list in python that I want to get the a set of indexes out of and save as a subset of the original list:

templist = [[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]]

我想要这个:

   sublist=[[1,    4,    7,                      16,      19,20]]

作为示例.

我无法提前知道list元素的内容.我所拥有的就是永远不变的索引.

I have no way of knowing ahead of time what the contents of the list elements will be . All I have is the indices that will always be the same.

有没有一种单行的方法可以做到这一点?

Is there a single line way of doing this?

推荐答案

假设您知道要选择的索引是什么，它的工作原理如下:

Assuming you know what the indices to be selected are, it would work something like this:

indices = [1, 4, 7, 16, 19, 20]
templist = [[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21]]
sublist = []

for i in indices:
    sublist.append(templist[i])

这也可以以列表理解的形式表达-

This can also be expressed in the form of a list comprehension -

sublist = [templist[0][i] for i in indices]

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