添加两个列表然后 sort = None(?) [英] add two lists then sort = None(?)

问题描述

第二个列表将列表中的每个项目平方，xs.运行下面的代码，python 给了我 'None'

Second list squared each item on list, xs. Running the code below, python gives me 'None'

xs = [12, 10, 32, 3, 66, 17, 42, 99, 20]
a = [b**2 for b in xs]
c = (a + xs).sort()
print(c, end=', ')

相同的列表但不同的代码--

Same list but different code--

xs = [12, 10, 32, 3, 66, 17, 42, 99, 20]
a = [b**2 for b in xs]
c = a + xs
c.sort()
print(c, end=', ')

...python 给了我我的列表(c)，全部排序.我不明白.或者有没有更pythonic的方法来做到这一点?

...python gives me my list(c), all sorted. I don't get it. Or is there a more pythonic way to do this?

谢谢！

推荐答案

一般来说，按照约定，任何对就地进行操作的操作都将返回 None.(但是，不一定总是遵循此约定.) somelist.sort() 将就地对列表进行排序.

Generally speaking, anything that operates on something in-place will return None, by convention. (This convention is not necessarily always followed, however.) somelist.sort() will sort the list in-place.

如果您想要一个有序的副本，您可以调用c = sorted(a + xs).sorted 对原始副本进行操作，因此返回副本.

If you'd rather have a sorted copy, you can just call c = sorted(a + xs). sorted operates on a copy of the original, and therefore returns the copy.

这里有更多解释:http://wiki.python.org/moin/HowTo/排序/

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