用偶数和奇数索引将列表分成两半? [英] Split a list into half by even and odd indexes?

问题描述

可能的重复:
Python程序，用于将列表分为两部分具有交替元素的列表

我有一个这样的列表:

list1 = [blah, 3, haha, 2, pointer, 1, poop, fire]

我想要的输出是:

list = [3, 2, 1, fire]

所以我想要的是列出前一个列表中的偶数元素.我尝试使用 for 语句，并尝试在将第2n个元素追加到列表时删除它们，但是没有成功:

So what I want is to make a list of even elements of the former list. I tried using a for statement and tried to delete 2nth element while appending them to the list, but it didn't work out:

count = 0
for a in list1:
 list2.append(a)
 if count % 2 = = 1:
  list2.pop(count)

print list2

有什么建议吗?

推荐答案

这应该可以为您提供所需的信息-定期从偏移量0或1采样列表:

This should give you what you need - sampling a list at regular intervals from an offset 0 or 1:

>>> a = ['blah', 3,'haha', 2, 'pointer', 1, 'poop', 'fire']
>>> a[0:][::2] # even
['blah', 'haha', 'pointer', 'poop']
>>> a[1:][::2] # odd
[3, 2, 1, 'fire']

请注意，在上面的示例中，第一个切片操作( a [1:] )演示了从所需起始索引中选择所有元素的方法，而第二个切片操作( a [::2]) 演示了如何选择列表中的所有其他项目.

Note that in the examples above, the first slice operation (a[1:]) demonstrates the selection of all elements from desired start index, whereas the second slice operation (a[::2]) demonstrates how to select every other item in the list.

一种更加惯用和高效的切片操作将两者合并为一个，即 a [:: 2] (可以省略 0 )和 a [1:: 2] ，它避免了不必要的列表复制，应该在生产代码中使用，就像其他人在评论中指出的那样.

A more idiomatic and efficient slice operation combines the two into one, namely a[::2] (0 can be omitted) and a[1::2], which avoids the unnecessary list copy and should be used in production code, as others have pointed out in the comments.

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