不区分大小写的列表排序，而不小写结果? [英] case-insensitive list sorting, without lowercasing the result?

问题描述

我有一个像这样的字符串列表:

['Aden', 'abel']

我想对项目进行排序，不区分大小写.所以我想得到:

['abel', 'Aden']

但我用 sorted() 或 list.sort() 得到相反的结果，因为大写出现在小写之前.

我怎样才能忽略这个案例?我见过涉及小写所有列表项的解决方案，但我不想更改列表项的大小写.

解决方案

在 Python 3.3+ 中有 str.casefold 专为无壳匹配设计的方法:

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

在 Python 2 中使用 lower():

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

它适用于普通和 unicode 字符串，因为它们都有一个 lower 方法.

在 Python 2 中，它适用于普通字符串和 unicode 字符串的混合，因为这两种类型的值可以相互比较.然而，Python 3 不是这样工作的:你不能比较字节字符串和 unicode 字符串，所以在 Python 3 中你应该做明智的事情，只对一种类型的字符串列表进行排序.

<预><代码>>>>lst = ['亚丁', u'abe1']>>>排序(lst)['亚丁'，你'abe1']>>>排序(lst，key = lambda s:s.lower())[u'abe1', '亚丁']

I have a list of strings like this:

['Aden', 'abel']

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

['abel', 'Aden']

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

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.

解决方案

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)

In Python 2 use lower():

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

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

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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