按任意Lambda排序 [英] Sorting by arbitrary lambda

问题描述

如何通过任意函数描述的键对列表进行排序?例如，如果我有:

How can I sort a list by a key described by an arbitrary function? For example, if I have:

mylist = [["quux", 1, "a"], ["bar", 0, "b"]]

我想按每个成员的第二个元素对"mylist"进行排序，例如

I'd like to sort "mylist" by the second element of each member, e.g.

sort(mylist, key=lambda x: x[1])

我该怎么做?

推荐答案

您基本上已经拥有它:

>>> mylist = [["quux", 1, "a"], ["bar", 0, "b"]]
>>> mylist.sort(key=lambda x: x[1])
>>> print mylist

给予:

[['bar', 0, 'b'], ['quux', 1, 'a']]

这将对mylist进行排序.

That will sort mylist in place.

[由于@Daniel的更正，此段进行了编辑.] sorted将返回已排序的新列表，而不是实际更改输入，如

[this para edited thanks to @Daniel's correction.] sorted will return a new list that is sorted rather than actually changing the input, as described in http://wiki.python.org/moin/HowTo/Sorting/.

这篇关于按任意Lambda排序的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！