提取两个括号之间的字符串，包括python中的嵌套括号

Question

How to extract string between two brackets, including the nested brackets.

There is a string:

""res = sqr(if((a>b)&(a<c),(a+b)*c,(a-b)*c)+if()+if()...)""

How can I extract the all the contents of if() as following:

["if((a>b)&(a<c),(a+b)*c,(a-b)*c)","if()","if()",...]

The format is not fixed, the string may includes multi- if s. So I want to know if there is a pattern that can match the substring. I'll try to give my solution later. thanks.

My solution, if there is any better method, please point out to me:

def extractIfFunc(condStr):

startIndex = [m.start() for m in re.finditer('if\(',condStr)]
parts = []
for index in startIndex:
    current = []
    bracket_level = 0
    for s in condStr[index+3:]:
        if s != '(' and s != ')' and bracket_level >= 0:
            current.append(s)
        elif s == '(':
            current.append(s)
            bracket_level += 1
        elif s == ')':
            bracket_level -= 1 
            if bracket_level < 0:
                current.append(s)
                break
            else:
                current.append(s)     
    parts.append('if('+''.join(current))
return parts  

Answer 1

尝试这个：

st[st.find('(')+1:st.rfind(')')]

Answer 2

>>> import re
>>> s = """res = sqr(if((a>b)&(a<c),(a+b)*c,(a-b)*c)+if()+if()...)"""
>>> re.findall(r'if\((?:[^()]*|\([^()]*\))*\)', s)
['if((a>b)&(a<c),(a+b)*c,(a-b)*c)', 'if()', 'if()']

For such patterns, better to use VERBOSE flag:

>>> lvl2 = re.compile('''
...          if\(            #literal if(
...            (?:           #start of non-capturing group
...             [^()]*       #non-parentheses characters
...             |            #OR
...             \([^()]*\)   #non-nested pair of parentheses
...            )*            #end of non-capturing group, 0 or more times
...          \)              #literal )
...          ''', flags=re.X)
>>> re.findall(lvl2, s)
['if((a>b)&(a<c),(a+b)*c,(a-b)*c)', 'if()', 'if()']

To match any number of nested pairs, you can use regex module, see Recursive Regular Expressions

Answer 3

st = """res = sqr(if((a>b)&(a<c),(a+b)*c,(a-b)*c))"""

print(st[10:][:-1])

OUTPUT:

if((a>b)&(a<c),(a+b)*c,(a-b)*c)

EDIT:

For a generic approach:

import re
st = """res = sqr(if((a>b)&(a<c),(a+b)*c,(a-b)*c))"""
pattern = "\((.*)\)"
print(re.compile(pattern).search(st).group(1))

Where:

\\( matches the character ( literally (case sensitive)

1st Capturing Group (.*)

.* matches any character (except for line terminators)

\\) matches the character ) literally

OUTPUT:

if((a>b)&(a<c),(a+b)*c,(a-b)*c)

regexTester

Answer 4

def extractIfFunc(condStr):
    for i, segment in enumerate(a.split("if")):
        if i == 0:
            continue

        s, n = -1, 0
        for i, c in enumerate(segment):
            if c == '(':
                s = i if s < 0 else s
                n += 1
            elif c == ')':
                n = n - 1 if n > 0 else 0
                if n == 0 and s > -1:
                    yield "if(%s)" % segment[s + 1:i]
                    break


a = """res = sqr(if((a>b)&(a<c),(a+b)*c,(a-b)*c)+if()+if()...)"""

for segment in extractIfFunc(a):
    print(segment)

NOTE:

This is not a real parser. If you want to create a parser that matches LALR(1) grammar , maybe the PLY is what you are looking for. It can help you to build a complete parser.