如何在圆括号之间得到一个表达式



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"

,我想提取包含在最顶层括号中的子字符串。例如,我想获得字符串:"this is (haha) a string(()and it's sneaky)""lorem"

是否有一个很好的python方法来做到这一点?正则表达式显然不能完成这项任务,但也许有一种方法可以让xml解析器完成这项工作?对于我的应用程序,我可以假设括号是格式良好的,即不像(()()。

这是堆栈的标准用例:您按字符读取字符串,每当遇到左括号时,将符号推入堆栈;如果遇到右括号,则将该符号从堆栈中弹出。

因为你只有一种类型的圆括号,你实际上不需要堆栈;相反,只要记住有多少个开括号就足够了。

此外,为了提取文本,我们还记得在第一级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。

可以像这样:

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']

这不是很"python "…但

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

你确定regex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)']

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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