问题:如何对字符串列表进行排序?

在Python中创建按字母顺序排序的列表的最佳方法是什么?

What is the best way of creating an alphabetically sorted list in Python?


回答 0

基本答案:

mylist = ["b", "C", "A"]
mylist.sort()

这会修改您的原始列表(即就地排序)。要获得列表的排序副本,而无需更改原始副本,请使用以下sorted()函数:

for x in sorted(mylist):
    print x

但是,上面的示例有些天真,因为它们没有考虑区域设置,而是执行区分大小写的排序。您可以利用可选参数key指定自定义排序顺序(使用的替代方法cmp是不推荐使用的解决方案,因为它必须多次评估- key每个元素仅计算一次)。

因此,要根据当前语言环境进行排序,并考虑到特定于语言的规则(这cmp_to_key是functools的帮助函数):

sorted(mylist, key=cmp_to_key(locale.strcoll))

最后,如果需要,您可以指定自定义语言环境进行排序:

import locale
locale.setlocale(locale.LC_ALL, 'en_US.UTF-8') # vary depending on your lang/locale
assert sorted((u'Ab', u'ad', u'aa'),
  key=cmp_to_key(locale.strcoll)) == [u'aa', u'Ab', u'ad']

最后要注意的是:您将看到使用该lower()方法的不区分大小写的排序示例-这些是不正确的,因为它们仅适用于ASCII字符子集。对于任何非英语数据,这两个错误:

# this is incorrect!
mylist.sort(key=lambda x: x.lower())
# alternative notation, a bit faster, but still wrong
mylist.sort(key=str.lower)

Basic answer:

mylist = ["b", "C", "A"]
mylist.sort()

This modifies your original list (i.e. sorts in-place). To get a sorted copy of the list, without changing the original, use the sorted() function:

for x in sorted(mylist):
    print x

However, the examples above are a bit naive, because they don’t take locale into account, and perform a case-sensitive sorting. You can take advantage of the optional parameter key to specify custom sorting order (the alternative, using cmp, is a deprecated solution, as it has to be evaluated multiple times – key is only computed once per element).

So, to sort according to the current locale, taking language-specific rules into account (cmp_to_key is a helper function from functools):

sorted(mylist, key=cmp_to_key(locale.strcoll))

And finally, if you need, you can specify a custom locale for sorting:

import locale
locale.setlocale(locale.LC_ALL, 'en_US.UTF-8') # vary depending on your lang/locale
assert sorted((u'Ab', u'ad', u'aa'),
  key=cmp_to_key(locale.strcoll)) == [u'aa', u'Ab', u'ad']

Last note: you will see examples of case-insensitive sorting which use the lower() method – those are incorrect, because they work only for the ASCII subset of characters. Those two are wrong for any non-English data:

# this is incorrect!
mylist.sort(key=lambda x: x.lower())
# alternative notation, a bit faster, but still wrong
mylist.sort(key=str.lower)

回答 1

还值得注意的sorted()功能:

for x in sorted(list):
    print x

这将返回列表的新排序版本,而不更改原始列表。

It is also worth noting the sorted() function:

for x in sorted(list):
    print x

This returns a new, sorted version of a list without changing the original list.


回答 2

list.sort()

真的就是这么简单:)

list.sort()

It really is that simple :)


回答 3

字符串排序的正确方法是:

import locale
locale.setlocale(locale.LC_ALL, 'en_US.UTF-8') # vary depending on your lang/locale
assert sorted((u'Ab', u'ad', u'aa'), cmp=locale.strcoll) == [u'aa', u'Ab', u'ad']

# Without using locale.strcoll you get:
assert sorted((u'Ab', u'ad', u'aa')) == [u'Ab', u'aa', u'ad']

前面的示例mylist.sort(key=lambda x: x.lower())对于仅ASCII上下文适用。

The proper way to sort strings is:

import locale
locale.setlocale(locale.LC_ALL, 'en_US.UTF-8') # vary depending on your lang/locale
assert sorted((u'Ab', u'ad', u'aa'), cmp=locale.strcoll) == [u'aa', u'Ab', u'ad']

# Without using locale.strcoll you get:
assert sorted((u'Ab', u'ad', u'aa')) == [u'Ab', u'aa', u'ad']

The previous example of mylist.sort(key=lambda x: x.lower()) will work fine for ASCII-only contexts.


回答 4

请在Python3中使用sorted()函数

items = ["love", "like", "play", "cool", "my"]
sorted(items2)

Please use sorted() function in Python3

items = ["love", "like", "play", "cool", "my"]
sorted(items2)

回答 5

但是,这如何处理特定于语言的排序规则?是否考虑到语言环境?

不,list.sort()是通用排序功能。如果要根据Unicode规则进行排序,则必须定义一个自定义的排序键函数。您可以尝试使用pyuca模块,但我不知道它的完整性。

But how does this handle language specific sorting rules? Does it take locale into account?

No, list.sort() is a generic sorting function. If you want to sort according to the Unicode rules, you’ll have to define a custom sort key function. You can try using the pyuca module, but I don’t know how complete it is.


回答 6

这是一个老问题,但是如果您想在不进行设置的情况下进行 locale.LC_ALL感知区域设置的排序,则可以按照此答案的建议使用PyICU库

import icu # PyICU

def sorted_strings(strings, locale=None):
    if locale is None:
       return sorted(strings)
    collator = icu.Collator.createInstance(icu.Locale(locale))
    return sorted(strings, key=collator.getSortKey)

然后用例如:

new_list = sorted_strings(list_of_strings, "de_DE.utf8")

这对我有用,而无需安装任何语言环境或更改其他系统设置。

(这已经在上面的评论中建议,但是我想让它更加突出,因为我一开始就很想念它。)

Old question, but if you want to do locale-aware sorting without setting locale.LC_ALL you can do so by using the PyICU library as suggested by this answer:

import icu # PyICU

def sorted_strings(strings, locale=None):
    if locale is None:
       return sorted(strings)
    collator = icu.Collator.createInstance(icu.Locale(locale))
    return sorted(strings, key=collator.getSortKey)

Then call with e.g.:

new_list = sorted_strings(list_of_strings, "de_DE.utf8")

This worked for me without installing any locales or changing other system settings.

(This was already suggested in a comment above, but I wanted to give it more prominence, because I missed it myself at first.)


回答 7

假设 s = "ZWzaAd"

要在字符串上方排序,简单的解决方案将是在字符串下方。

print ''.join(sorted(s))

Suppose s = "ZWzaAd"

To sort above string the simple solution will be below one.

print ''.join(sorted(s))

回答 8

或许:

names = ['Jasmine', 'Alberto', 'Ross', 'dig-dog']
print ("The solution for this is about this names being sorted:",sorted(names, key=lambda name:name.lower()))

Or maybe:

names = ['Jasmine', 'Alberto', 'Ross', 'dig-dog']
print ("The solution for this is about this names being sorted:",sorted(names, key=lambda name:name.lower()))

回答 9

l =['abc' , 'cd' , 'xy' , 'ba' , 'dc']
l.sort()
print(l1)

结果

[‘abc’,’ba’,’cd’,’dc’,’xy’]

l =['abc' , 'cd' , 'xy' , 'ba' , 'dc']
l.sort()
print(l1)

Result

[‘abc’, ‘ba’, ‘cd’, ‘dc’, ‘xy’]


回答 10

很简单:https : //trinket.io/library/trinkets/5db81676e4

scores = '54 - Alice,35 - Bob,27 - Carol,27 - Chuck,05 - Craig,30 - Dan,27 - Erin,77 - Eve,14 - Fay,20 - Frank,48 - Grace,61 - Heidi,03 - Judy,28 - Mallory,05 - Olivia,44 - Oscar,34 - Peggy,30 - Sybil,82 - Trent,75 - Trudy,92 - Victor,37 - Walter'

得分= scores.split(’,’)for x in sorted(scores):print(x)

It is simple: https://trinket.io/library/trinkets/5db81676e4

scores = '54 - Alice,35 - Bob,27 - Carol,27 - Chuck,05 - Craig,30 - Dan,27 - Erin,77 - Eve,14 - Fay,20 - Frank,48 - Grace,61 - Heidi,03 - Judy,28 - Mallory,05 - Olivia,44 - Oscar,34 - Peggy,30 - Sybil,82 - Trent,75 - Trudy,92 - Victor,37 - Walter'

scores = scores.split(‘,’) for x in sorted(scores): print(x)


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