问题:不区分大小写的列表排序,而不降低结果大小?

我有一个这样的字符串列表:

['Aden', 'abel']

我要对项目排序,不区分大小写。所以我想得到:

['abel', 'Aden']

但与sorted()或相反list.sort(),因为大写字母先于小写字母。

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。

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.


回答 0

在Python 3.3+中,有str.casefold一种专为无条件匹配而设计的方法:

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lower方法。

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并不是这样工作的:您无法比较字节字符串和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。

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

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']

回答 1

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

在Python 3中str是unicode,但在Python 2中,您可以使用这种更通用的方法,该方法对str和都适用unicode

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

In Python 3 str is unicode but in Python 2 you can use this more general approach which works for both str and unicode:

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

回答 2

您也可以尝试使用此方法对列表进行就地排序:

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

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

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

回答 3

这在Python 3中有效,并且不涉及小写结果(!)。

values.sort(key=str.lower)

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

values.sort(key=str.lower)

回答 4

在python3中,您可以使用

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

In python3 you can use

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

回答 5

我是通过Python 3.3做到的:

 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]

然后,您可以调用此函数:

sortCaseIns(yourListToSort)

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)

回答 6

不区分大小写的排序,在Python 2 OR 3中对字符串进行排序(在Python 2.7.17和Python 3.6.9中测试):

>>> 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']

关键是key=str.lower。这些命令只是这些命令的外观,以便于复制粘贴,因此您可以对其进行测试:

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

请注意,但是,如果您的字符串是unicode字符串(如u'some string'),则仅在Python 2中(在这种情况下,在Python 3中不是),上述x.sort(key=str.lower)命令将失败并输出以下错误:

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

如果出现此错误,请升级到Python 3来处理unicode排序,或者先使用列表推导将unicode字符串转换为ASCII字符串,如下所示:

# 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)

参考文献:

  1. https://docs.python.org/3/library/stdtypes.html#list.sort
  2. 将Unicode字符串转换为Python中的字符串(包含多余的符号)
  3. https://www.programiz.com/python-programming/list-comprehension

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

回答 7

试试这个

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)

输出量

['abel', 'Aden']

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']


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