从Python格式解析

Question

Python中是否有任何方法可以反转通过"%"运算符完成的格式化操作？

formated = "%d ooo%s" % (12, "ps")
#formated is now '12 ooops'
(arg1, arg2) = theFunctionImSeeking("12 ooops", "%d ooo%s")
#arg1 is 12 and arg2 is "ps"

编辑 Regexp可以解决这个问题,但是它们更难写,我怀疑它们更慢,因为它们可以处理更复杂的结构.我真的很喜欢sscanf.

Answer 1

使用正则表达式(re模块):

>>> import re
>>> match = re.search('(\d+) ooo(\w+)', '12 ooops')
>>> match.group(1), match.group(2)
('12', 'ps')

正则表达式就像你可以做你想做的那样近.没有办法使用相同的格式字符串('%d ooo%s').

编辑:正如@Daenyth建议的那样,你可以用这种行为实现自己的函数:

import re

def python_scanf(my_str, pattern):
    D = ('%d',      '(\d+?)')
    F = ('%f', '(\d+\.\d+?)')
    S = ('%s',       '(.+?)')
    re_pattern = pattern.replace(*D).replace(*F).replace(*S)
    match = re.match(re_pattern, my_str)
    if match:
        return match.groups()
    raise ValueError("String doesn't match pattern")

用法:

>>> python_scanf("12 ooops", "%d ooo%s")
('12', 'p')
>>> python_scanf("12 ooops", "%d uuu%s")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 10, in python_scanf
ValueError: String doesn't match pattern

当然,python_scanf不会使用像%.4f或更复杂的模式%r.