Python analyse entre crochets blocs

Pseudocode:

For each string in the array:
    Find the first '{'. If there is none, leave that string alone.
    Init a counter to 0. 
    For each character in the string:  
        If you see a '{', increment the counter.
        If you see a '}', decrement the counter.
        If the counter reaches 0, break.
    Here, if your counter is not 0, you have invalid input (unbalanced brackets)
    If it is, then take the string from the first '{' up to the '}' that put the
     counter at 0, and that is a new element in your array.

Je suis une sorte de Python, alors allez facile sur moi, mais ici, c'est une application qui fonctionne:

def balanced_braces(args):
    parts = []
    for arg in args:
        if '{' not in arg:
            continue
        chars = []
        n = 0
        for c in arg:
            if c == '{':
                if n > 0:
                    chars.append(c)
                n += 1
            elif c == '}':
                n -= 1
                if n > 0:
                    chars.append(c)
                elif n == 0:
                    parts.append(''.join(chars).lstrip().rstrip())
                    chars = []
            elif n > 0:
                chars.append(c)
    return parts

t1 = balanced_braces(["{{ a } { b } { { { c } } } }"]);
print t1
t2 = balanced_braces(t1)
print t2
t3 = balanced_braces(t2)
print t3
t4 = balanced_braces(t3)
print t4

De sortie:

['{ a } { b } { { { c } } }']
['a', 'b', '{ { c } }']
['{ c }']
['c']

Vous pouvez également analyser tout à la fois, bien que je trouve le {a} à dire "a" plutôt que ["a"] un peu bizarre. Si j'ai bien compris le format correctement:

import re
import sys
_mbrack_rb = re.compile("([^{}]*)}") # re.match doesn't have a pos parameter
def mbrack(s):
"""Parse matching brackets.
>>> mbrack("{a}")
'a'
>>> mbrack("{{a}{b}}")
['a', 'b']
>>> mbrack("{{a}{b}{{{c}}}}")
['a', 'b', [['c']]]
>>> mbrack("a")
Traceback (most recent call last):
ValueError: expected left bracket
>>> mbrack("{a}{b}")
Traceback (most recent call last):
ValueError: more than one root
>>> mbrack("{a")
Traceback (most recent call last):
ValueError: expected value then right bracket
>>> mbrack("{a{}}")
Traceback (most recent call last):
ValueError: expected value then right bracket
>>> mbrack("{a}}")
Traceback (most recent call last):
ValueError: unbalanced brackets (found right bracket)
>>> mbrack("{{a}")
Traceback (most recent call last):
ValueError: unbalanced brackets (not enough right brackets)
"""
stack = [[]]
i, end = 0, len(s)
while i < end:
if s[i] != "{":
raise ValueError("expected left bracket")
elif i != 0 and len(stack) == 1:
raise ValueError("more than one root")
while i < end and s[i] == "{":
L = []
stack[-1].append(L)
stack.append(L)
i += 1
stack.pop()
stack[-1].pop()
m = _mbrack_rb.match(s, i)
if m is None:
raise ValueError("expected value then right bracket")
stack[-1].append(m.group(1))
i = m.end(0)
while i < end and s[i] == "}":
if len(stack) == 1:
raise ValueError("unbalanced brackets (found right bracket)")
stack.pop()
i += 1
if len(stack) != 1:
raise ValueError("unbalanced brackets (not enough right brackets)")
return stack[0][0]
def main(args):
if args:
print >>sys.stderr, "unexpected arguments: %r" % args
import doctest
r = doctest.testmod()
print r
return r[0]
if __name__ == "__main__":
sys.exit(main(sys.argv[1:]))

Solution de nettoyage. Cela permettra de trouver le retour de la chaîne entre la limite extérieure de support. Si Aucun n'est retourné, il n'y a pas de match.

def findBrackets( aString ):
if '{' in aString:
match = aString.split('{',1)[1]
open = 1
for index in xrange(len(match)):
if match[index] in '{}':
open = (open + 1) if match[index] == '{' else (open - 1)
if not open:
return match[:index]

Voici une solution je suis venu avec un semblable cas d'utilisation. Cela a été vaguement basé sur le acceptées pseudo code de réponse. Je ne voulais pas ajouter toutes les dépendances de bibliothèques externes:

def parse_segments(source, recurse=False):
"""
extract any substring enclosed in parenthesis
source should be a string
"""
unmatched_count = 0
start_pos = 0
opened = False
open_pos = 0
cur_pos = 0
finished = []
segments = []
for character in source:
#scan for mismatched parenthesis:
if character == '(':
unmatched_count += 1
if not opened:
open_pos = cur_pos
opened = True
if character == ')':
unmatched_count -= 1
if opened and unmatched_count == 0:
segment = source[open_pos:cur_pos+1]
segments.append(segment)
clean = source[start_pos:open_pos]
if clean:
finished.append(clean)
opened = False
start_pos = cur_pos+1
cur_pos += 1
assert unmatched_count == 0
if start_pos != cur_pos:
#get anything that was left over here
finished.append(source[start_pos:cur_pos])
#now check on recursion:
for item in segments:
#get rid of bounding parentheses:
pruned = item[1:-1]
if recurse:
results = parse_tags(pruned, recurse)
finished.expand(results)
else:
finished.append(pruned)
return finished