如何正确排序含数字的字符串?
我有一个包含数字的字符串列表,但我找不到一个好的方法来对它们进行排序。
比如我得到的内容是这样的:
something1
something12
something17
something2
something25
something29
使用 sort()
方法。
我知道我可能需要以某种方式提取出这些数字,然后再对列表进行排序,但我不知道最简单的方法该怎么做。
1 个回答
357
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)
结果是
['something1', 'something2', 'something12', 'something17', 'something25', 'something29']
附注:我已经把我的回答改成使用Toothy的自然排序实现(在评论中发布的 这里),因为它比我原来的答案快很多。
如果你想对包含小数的文本进行排序,那么你需要把正则表达式从匹配整数的(也就是 (\d+)
)改成 匹配小数的正则表达式:
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)
结果是
['something1', 'something1.0', 'something1.105', 'something1.25', 'something2']