比如有一个字符串 s:
s = "((abc)((123))())blabla"
我们知道s的开头是"(",我们想要找到它的对立面,在"blabla"之前的")",在Python中应该如何做?
有没有可能以一种简单直观的方式做到这一点,而不使用状态机?或者有没有任何库可以实现这个功能?
>>> s = "((abc)((123))())blabla"
>>> count = 0
>>> for i,e in enumerate(s):
if e == '(':
count += 1
elif e == ')':
count -= 1
if not count:
break
>>> s[:i + 1]
'((abc)((123))())'
>>>
通过代码,您可以实现以下操作:
from collections import defaultdict
opens = defaultdict(int)
open_close_pair = []
s = '((abc)((123))())blabla'
openc, closec = '(', ')'
for c in range(0, len(s)):
if s[c] == openc:
# +1 in every entry
for key, val in opens.items():
opens[key] += 1
opens[c] += 1
elif s[c] == closec:
# -1 in every entery
for key, val in opens.items():
opens[key] -= 1
else:
pass
for key, val in opens.items():
if val == 0:
# open the entry to the open close pairs
open_close_pair.append( (key, c))
# the bracket is close so can be removed from the counter
del opens[key]
for x in open_close_pair:
print " %s %s " % (s[x[0]], s[x[1]])
print open_close_pair
print opens
输出结果为:
( )
( )
( )
( )
( )
[(1, 5), (7, 11), (6, 12), (13, 14), (0, 15)]
defaultdict(<type 'int'>, {})
算法如下:
pyparsing
做类似这样的事情。 - mgilson