如何正确排序字符串与一个数字内？

也许你正在寻找 人类分类(也称为 自然分类) :

import re


def atoi(text):
return int(text) if text.isdigit() else text


def natural_keys(text):
'''
alist.sort(key=natural_keys) sorts in human order
http://nedbatchelder.com/blog/200712/human_sorting.html
(See Toothy's implementation in the comments)
'''
return [ atoi(c) for c in re.split(r'(\d+)', text) ]


alist=[
"something1",
"something12",
"something17",
"something2",
"something25",
"something29"]


alist.sort(key=natural_keys)
print(alist)

产量

['something1', 'something2', 'something12', 'something17', 'something25', 'something29']

附言。我已经更改了我的答案，使用 Toothy 的自然排序实现(发布在评论 给你中) ，因为它比我原来的答案快得多。

如果您希望使用 float 对文本进行排序，那么您需要将匹配 int (即 (\d+))的 regex 更改为 匹配浮点数的正则表达式:

import re


def atof(text):
try:
retval = float(text)
except ValueError:
retval = text
return retval


def natural_keys(text):
'''
alist.sort(key=natural_keys) sorts in human order
http://nedbatchelder.com/blog/200712/human_sorting.html
(See Toothy's implementation in the comments)
float regex comes from https://stackoverflow.com/a/12643073/190597
'''
return [ atof(c) for c in re.split(r'[+-]?([0-9]+(?:[.][0-9]*)?|[.][0-9]+)', text) ]


alist=[
"something1",
"something2",
"something1.0",
"something1.25",
"something1.105"]


alist.sort(key=natural_keys)
print(alist)

产量

['something1', 'something1.0', 'something1.105', 'something1.25', 'something2']