提取和连接嵌套圆括号中的文本

如果您的目标只是删除所有括号，那么您可以使用replace函数

string = '(Order Filled (today))'
string.replace('(','').replace(')','')

更新：只删除那些有结尾的括号

我知道你要求most efficient way，但当你使用replace函数时，它也会删除non nested parentheses，所以我有一个只删除嵌套括号的函数：

def markup(message):
msg = message
result = ""
markups = "()"
num = 0
# Check if the parentheses
def has_ending(_str, i, msg):
if _str == "(":
return ")" in msg[i+1:]
elif _str == ")":
return False
else:
return False

for i in range(len(msg)) :
t = msg[i]
s = t
# if the char is '(' or ')'
if t in markups:
if has_ending(t, i, msg) and num%2 == 0 :
s = ""
num += 1 
elif t == ")" and num%2 == 1 :
s = ""
num += 1 
# else
else :
s = t
result += s
return result
print(markup('Hello )(O) (O)( (O)'))
# prints 'Hello )O O (O'