For循环在解析中返回相同的字符

Question

我正在创建一个程序，它将拆分“单独对”的括号。例如()() 变成|()|()| 但(()) 保持不变，成为|(())|。它不断“获得”相同的字符。我尝试过更改插入位置，例如pos - 1，但仍然无法正常工作。这是我的代码：

def insert(source_str, insert_str, pos):
        return source_str[:pos]+insert_str+source_str[pos:]


x = 0 
rightSideOfEquation = "()bx((x))c(y(c+1)(x)y)"


for pos in range(len(rightSideOfEquation)):
    if x == 0:
        rightSideOfEquation = insert(rightSideOfEquation,'|',pos)
    if rightSideOfEquation[pos] == '(':
        x += 1
    if rightSideOfEquation[pos] == ')':
        x -= 1

print(rightSideOfEquation)

打印|||||||||||||||||||||||()bx((x))c(y(c+1)(x)y)
我希望它打印|()|bx|((x))|c|(y(c+1)(x)y)|

注意：您可以在此处查看： **https://math.stackexchange.com/questions/1682322/recursive-parsing-parenthesis-with-explanation
**我尝试将其更改为pos + 1 和pos -1，但效果不大，除了重复的地方。

Answer 1

在迭代对象时修改它总是会导致灾难。

您需要在迭代旧对象时构建一个新对象：

equation = "a()bx((x))c(y(c+1)(x)y)"
new_equation = []
parens = 0
for ch in equation:
    if ch == '(':
        if parens == 0:
            new_equation.append('|')
        new_equation.append(ch)
        parens += 1
    elif ch == ')':
        new_equation.append(ch)
        parens -= 1
        if parens == 0:
            new_equation.append('|')
    else:
        new_equation.append(ch)
equation = ''.join(new_equation)
print(equation)

这给出了：

a|()|bx|((x))|c|(y(c+1)(x)y)|

Answer 2

你不想改变你正在迭代的东西。我通过创建输入字符串的副本修复了您的代码，它似乎正在工作：

def insert(source_str, insert_str, pos):
    return source_str[:pos]+insert_str+source_str[pos:]


x = 0 
rightSideOfEquation = "a()bx((x))c(y(c+1)(x)y)"
copy = rightSideOfEquation
posincopy = 0

for pos in range(len(rightSideOfEquation)):
    if rightSideOfEquation[pos] == '(':
        x += 1
        if x == 1:
            copy = insert(copy,'|',posincopy)
            posincopy = posincopy + 1
    if rightSideOfEquation[pos] == ')':
        x -= 1
        if x == 0:
            copy = insert(copy,'|',posincopy + 1)
            posincopy = posincopy + 1
    posincopy = posincopy + 1

print(copy)

输出：

a|()|bx|((x))|c|(y(c+1)(x)y)|

Answer 3

在这种情况下，使用“while”语句而不是 for 循环将使您的生活更轻松：

def insert(source_str, insert_str, pos):
    return source_str[:pos]+insert_str+source_str[pos:]

x = 0 
rightSideOfEquation = "a()bx((x))c(y(c+1)(x)y)"
pos = 0

while pos < len(rightSideOfEquation):
  if rightSideOfEquation[pos] == '(':
      if x==0:        
          rightSideOfEquation = insert(rightSideOfEquation,'|',pos)
          pos+=1
      x += 1
  elif rightSideOfEquation[pos] == ')':
      x -= 1
      if x == 0:
          rightSideOfEquation = insert(rightSideOfEquation,'|',pos + 1)
  pos+=1

print(rightSideOfEquation)

这将打印以下内容：

a|()|bx|((x))|c|(y(c+1)(x)y)|

虽然使用递归函数会更简洁、更容易，但我想向您展示如何修复现有代码中的错误，而不是完全改变您的思维过程...