python - 不区分大小写的列表排序，不小写结果？

Question

我有一个这样的字符串列表：

['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']

但我得到相反的sorted()or list.sort()，因为大写出现在小写之前。

我怎么能忽略这个案子？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。

Answer 1

在 Python 3.3+ 中有str.casefold专门为无大小写匹配设计的方法：

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())

它适用于普通字符串和 unicode 字符串，因为它们都有一个lower方法。

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']

Answer 2

>>> x = ['Aden', 'abel']
>>> sorted(x, key=str.lower) # Or unicode.lower if all items are unicode
['abel', 'Aden']

在 Python 3str中是 unicode，但在 Python 2 中，您可以使用这种更通用的方法，它适用于str和unicode：

>>> sorted(x, key=lambda s: s.lower())
['abel', 'Aden']

Answer 3

You can also try this to sort the list in-place:

>>> x = ['Aden', 'abel']
>>> x.sort(key=lambda y: y.lower())
>>> x
['abel', 'Aden']

Answer 4

This works in Python 3 and does not involves lowercasing the result (!).

values.sort(key=str.lower)

Answer 5

In python3 you can use

list1.sort(key=lambda x: x.lower()) #Case In-sensitive             
list1.sort() #Case Sensitive

Answer 6

I did it this way for Python 3.3:

 def sortCaseIns(lst):
    lst2 = [[x for x in range(0, 2)] for y in range(0, len(lst))]
    for i in range(0, len(lst)):
        lst2[i][0] = lst[i].lower()
        lst2[i][1] = lst[i]
    lst2.sort()
    for i in range(0, len(lst)):
        lst[i] = lst2[i][1]

Then you just can call this function:

sortCaseIns(yourListToSort)

Answer 7

Case-insensitive sort, sorting the string in place, in Python 2 OR 3 (tested in Python 2.7.17 and Python 3.6.9):

>>> x = ["aa", "A", "bb", "B", "cc", "C"]
>>> x.sort()
>>> x
['A', 'B', 'C', 'aa', 'bb', 'cc']
>>> x.sort(key=str.lower)           # <===== there it is!
>>> x
['A', 'aa', 'B', 'bb', 'C', 'cc']

The key is key=str.lower. Here's what those commands look like with just the commands, for easy copy-pasting so you can test them:

x = ["aa", "A", "bb", "B", "cc", "C"]
x.sort()
x
x.sort(key=str.lower)
x

Note that if your strings are unicode strings, however (like u'some string'), then in Python 2 only (NOT in Python 3 in this case) the above x.sort(key=str.lower) command will fail and output the following error:

TypeError: descriptor 'lower' requires a 'str' object but received a 'unicode'

If you get this error, then either upgrade to Python 3 where they handle unicode sorting, or convert your unicode strings to ASCII strings first, using a list comprehension, like this:

# for Python2, ensure all elements are ASCII (NOT unicode) strings first
x = [str(element) for element in x]  
# for Python2, this sort will only work on ASCII (NOT unicode) strings
x.sort(key=str.lower)

References:

1. https://docs.python.org/3/library/stdtypes.html#list.sort
2. Convert a Unicode string to a string in Python (containing extra symbols)
3. https://www.programiz.com/python-programming/list-comprehension

Answer 8

Try this

def cSort(inlist, minisort=True):
    sortlist = []
    newlist = []
    sortdict = {}
    for entry in inlist:
        try:
            lentry = entry.lower()
        except AttributeError:
            sortlist.append(lentry)
        else:
            try:
                sortdict[lentry].append(entry)
            except KeyError:
                sortdict[lentry] = [entry]
                sortlist.append(lentry)

    sortlist.sort()
    for entry in sortlist:
        try:
            thislist = sortdict[entry]
            if minisort: thislist.sort()
            newlist = newlist + thislist
        except KeyError:
            newlist.append(entry)
    return newlist

---

lst = ['Aden', 'abel']
print cSort(lst)

Output

['abel', 'Aden']