不区分大小写的列表排序，而不降低结果的大小? [英] case-insensitive list sorting, without lowercasing the result?

问题描述

我有一个这样的字符串列表:

I have a list of strings like this:

['Aden', 'abel']

我要对项目进行排序，不区分大小写. 所以我想得到:

I want to sort the items, case-insensitive. So I want to get:

['abel', 'Aden']

但是与sorted()或list.sort()相反，因为大写字母出现在小写字母之前.

But I get the opposite with sorted() or list.sort(), because uppercase appears before lowercase.

我如何忽略这种情况?我已经看到了涉及降低所有列表项的解决方案，但是我不想更改列表项的大小写.

How can I ignore the case? I've seen solutions which involves lowercasing all list items, but I don't want to change the case of the list items.

推荐答案

在Python 3.3+中，存在

In Python 3.3+ there is the str.casefold method that's specifically designed for caseless matching:

sorted_list = sorted(unsorted_list, key=str.casefold)

在Python 2中使用lower():

In Python 2 use lower():

sorted_list = sorted(unsorted_list, key=lambda s: s.lower())

它适用于普通字符串和unicode字符串，因为它们都具有lower方法.

It works for both normal and unicode strings, since they both have a lower method.

在Python 2中，它可以将普通字符串和unicode字符串混合使用，因为这两种类型的值可以相互比较.但是，Python 3不能那样工作:您不能比较字节字符串和unicode字符串，因此在Python 3中，您应该做明智的事情，并且只能对一种类型的字符串列表进行排序.

In Python 2 it works for a mix of normal and unicode strings, since values of the two types can be compared with each other. Python 3 doesn't work like that, though: you can't compare a byte string and a unicode string, so in Python 3 you should do the sane thing and only sort lists of one type of string.

>>> lst = ['Aden', u'abe1']
>>> sorted(lst)
['Aden', u'abe1']
>>> sorted(lst, key=lambda s: s.lower())
[u'abe1', 'Aden']

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