如何通过给定索引从嵌套列表中提取元素 [英] How to extract elements from nested lists by given index

问题描述

我有一个看起来像这样的列表列表:

I have a list of lists which look like this:

[[[12, 15, 0], [13, 15, 25], [14, 15, 25], [16, 16, 66], [18, 15, 55]]]

提取出现在索引位置1的所有元素的最佳方法是什么?我知道我可以使用for循环；

What would be the best way to extract all elements occurring at index position 1. I know I can use a for loop like;

for i in list:
    for j in i:
        print j[2]

但是还有更多的"pythonic"方式(简短的/简单的/更少的代码/更有效的方式)吗?

But is there a more "pythonic" (short /easy /less code/ efficient) way to do this?

推荐答案

您可以使用列表理解:

>>> lst = [[[12, 15, 0], [13, 15, 25], [14, 15, 25], [16, 16, 66], [18, 15, 55]]]
>>> [x[1] for x in lst[0]]
[15, 15, 15, 16, 15]
>>>

以上等同于:

lst = [[[12, 15, 0], [13, 15, 25], [14, 15, 25], [16, 16, 66], [18, 15, 55]]]
final_list = []
for sub_list in lst[0]:
    final_list.append(sub_list[1])

除了更加简洁之外，还避免了对 list.append 的所有调用(这意味着效率更高).

except that it is a lot more concise and also avoids all those calls to list.append (which means that it is more efficient).

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