Question

我有一个字符串，如 '(((a+b)+a)+c)'我想分成两部分，结果将是('((a+b)+a)','c')。

如果我在结果的第一个元素上再次运行它会给我('(a+b)', 'a')

如果我在'(a+b)'再次运行它，它将返回('a', 'b')。

我以为我可以通过正则表达式做到这一点，但我无法弄明白这一点，并且走了许多if语句检查开始和结束括号的路径，但它有点乱“

Answer 1

瞧瞧：

#!/usr/bin/python3.5
def f(s):
    p=s.rsplit('+',1)
    return [p[0][1:],p[1][:-1]]

s='(((a+b)+a)+c)'

for i in range(3):
    k=f(s)
    s=k[0]
    print(k)

输出：

['((a+b)+a)', 'c']
['(a+b)', 'a']
['a', 'b']

Answer 2

以下示例适用于您的示例：

def breakit(s):
    count = 0
    for i, c in enumerate(s):
        if count == 1 and c in '+-':
            return s[1:i].strip(), s[i+1:-1].strip()
        if c == '(': count +=1
        if c == ')': count -= 1
    return s

breakit(s)
>> ('((a+b)+a)', 'c')
breakit(_[0])
('(a+b)', 'a')
breakit(_[0])
('a', 'b')

Answer 3

我以为我也会发布我的答案，不如所选择的解决方案那么优雅，但它有效

def break_into_2(s):

    if len(s) == 1: 
        # limiting case
        return s

    # s[0] can either be a digit or '('
    if s[0].isdigit():
        # digit could be 10,100,1000,...
        idx = 0
        while s[idx].isdigit():
            idx += 1
        a = s[:idx]
        b = s[idx+1:]
        return a, b
    # otherwise, s[0] = '('
    idx = 1
    counter = 1
    # counter tracks opening and closing parenthesis
    # when counter = 0, the left side expression has
    # been found, return the idx at which this happens
    while counter:
        if s[idx] == '(':
            counter+=1
        elif s[idx] == ')':
            counter -=1
        idx +=1
    if s[:idx] == s:
        # this case occurs when brackets enclosing entire expression, s
        # runs the function again with the same expression from idxs 1:-1
        return break_into_2(s[1:-1])
    return s[:idx], s[idx+1:]

在python3中将字符串方程式分为两部分

3 个答案: