也许您正在寻找人工分拣(也称为自然排序):
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)
yields
['something1', 'something2', 'something12', 'something17', 'something25', 'something29']
附言。我已经更改了我的答案,以使用 Toothy 的自然排序实现(发布在评论中)here)因为它比我原来的答案要快得多。
如果您希望使用浮点数对文本进行排序,那么您需要将正则表达式从与整数匹配的正则表达式(即(\d+)
) to 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)
yields
['something1', 'something1.0', 'something1.105', 'something1.25', 'something2']