也许您正在寻找人工排序（也称为自然排序）：
import redef atoi(text): return int(text) if text.isdigit() else textdef 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']
PS。我已经更改了答案，以使用Toothy的自然排序实现（在此处发表评论），因为它比我的原始答案快得多。
如果您希望使用浮点数对文本进行排序，则需要将正则表达式从与整数（即
(d+)）匹配的正则表达式更改为与浮点数匹配的正则表达式：
import redef atof(text): try: retval = float(text) except ValueError: retval = text return retvaldef 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']
