如何获得平衡括号之间的表达式

Question

假设我得到以下类型的字符串：

"(this is (haha) a string(()and it's sneaky)) ipsom (lorem) bla"

我想提取最顶层括号中包含的子字符串。 IE。我想获取字符串："this is (haha) a string(()and it's sneaky)" 和 "lorem".

是否有一个很好的 pythonic 方法来做到这一点？正则表达式显然不能胜任这项任务，但也许有办法让 xml 解析器来完成这项工作？对于我的应用程序，我可以假设括号格式正确，即不是类似 (()(().

Answer 1

您确定正则表达式不够好吗？

>>> x=re.compile(r'\((?:(?:\(.*?\))|(?:[^\(\)]*?))\)')
>>> x.findall("(this is (haha) a string(()and it's sneaky)) ipsom (lorem) bla")
["(this is (haha) a string(()and it's sneaky)", '(lorem)']
>>> x.findall("((((this is (haha) a string((a(s)d)and ((it's sneaky))))))) ipsom (lorem) bla")
["((((this is (haha) a string((a(s)d)and ((it's sneaky))", '(lorem)']

Answer 2

这不是很 "pythonic"...但是

def find_strings_inside(what_open,what_close,s):
    stack = []
    msg = []
    for c in s:
        s1=""
        if c == what_open:
           stack.append(c)
           if len(stack) == 1:
               continue
        elif c == what_close and stack:
           stack.pop()
           if not stack:
              yield "".join(msg)
              msg[:] = []
        if stack:
            msg.append(c)

x= list(find_strings_inside("(",")","(this is (haha) a string(()and it's sneaky)) ipsom (lorem) bla"))

print x

Answer 3

这是堆栈的标准用例：按字符读取字符串，每当遇到左括号时，就将符号压入堆栈；如果遇到右括号，则从堆栈中弹出符号。

因为你只有一种括号，所以你实际上不需要堆栈；相反，只要记住有多少个左括号就足够了。

此外，为了提取文本，我们还会记住第一层括号打开时部分开始的位置，并在遇到匹配的右括号时收集结果字符串。

这可能是这样的：

string = "(this is (haha) a string(()and it's sneaky)) ipsom (lorem) bla"

stack = 0
startIndex = None
results = []

for i, c in enumerate(string):
    if c == '(':
        if stack == 0:
            startIndex = i + 1 # string to extract starts one index later

        # push to stack
        stack += 1
    elif c == ')':
        # pop stack
        stack -= 1

        if stack == 0:
            results.append(string[startIndex:i])

print(results)
# ["this is (haha) a string(()and it's sneaky)", 'lorem']

Answer 4

这或多或少重复了已经说过的内容，但可能更容易阅读：

def extract(string):
    flag = 0
    result, accum = [], []
    for c in string:
        if c == ')':
            flag -= 1
        if flag:
            accum.append(c)
        if c == '(':
            flag += 1
        if not flag and accum:
            result.append(''.join(accum))
            accum = []
    return result

>> print extract(test)
["this is (haha) a string(()and it's sneaky)", 'lorem']

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