python按“and”和“or”拆分，但不在括号中

Question

我有以下字符串：

(some text) or ((other text) and (some more text)) and (still more text)

我想要一个python正则表达式，将它拆分为

['(some text)', '((other text) and (some more text))', '(still more text)']

我试过了，但没用：

haystack = "(some text) or ((other text) and (some more text)) and (still more text)"
re.split('(or|and)(?![^(]*.\))', haystack) # no worky

感谢您的帮助。你知道吗

Answer 1

此解决方案适用于任意嵌套的括号，正则表达式不能（s是原始字符串）：

from pyparsing import nestedExpr
def lst_to_parens(elt):
    if isinstance(elt,list):
        return '(' + ' '.join(lst_to_parens(e) for e in elt) + ')'
    else:
        return elt

split = nestedExpr('(',')').parseString('(' + s + ')').asList()
split_lists = [elt for elt in split[0] if isinstance(elt,list)]
print ([lst_to_parens(elt) for elt in split_lists])

输出：

['(some text)', '((other text) and (some more text))', '(still more text)']

对于OP的真实测试用例：

s = "(substringof('needle',name)) or ((role eq 'needle') and (substringof('needle',email))) or (job eq 'needle') or (office eq 'needle')"

输出：

["(substringof ('needle' ,name))", "((role eq 'needle') and (substringof ('needle' ,email)))", "(job eq 'needle')", "(office eq 'needle')"]