在python中返回1而不是true

Question

我试图在python中返回一个而不是true.

我正在处理的代码是:

delimiters = ( '()', '[]', '{}', "''", '""' )
esc = '\\'

def is_balanced(s, delimiters=delimiters, esc=esc):
    stack = []
    opening = tuple(str[0] for str in delimiters)
    closing = tuple(str[1] for str in delimiters)
    for i, c in enumerate(s):
        if len(stack) and stack[-1] == -1:
            stack.pop()
        elif c in esc:
            stack.append(-1)
        elif c in opening and (not len(stack) or opening[stack[-1]] != closing[stack[-1]]):
            stack.append(opening.index(c))
        elif c in closing:
            if len(stack) == 0 or closing.index(c) != stack[-1]:
                return False
            stack.pop()

    return len(stack) == 0

num_cases = raw_input()
num_cases = int(num_cases)
for num in range(num_cases):
    s = raw_input()
    print is_balanced(s)

它基本上检查键入的字符串是否平衡.如果平衡,应返回1,如果不是0.

我试过这个:

1
Test string
True

它返回true.我希望它返回1.我该怎么办？

Answer 1

或者,您可以将布尔值转换为int:

>>>myBoolean = True
>>>int(myBoolean)
1
>>>myBoolean = False
>>>int(myBoolean)
0

Answer 2

咦？你改变了代码:

代替

return False

写

return 0

而不是

return len(stack) == 0

写

if len(stack) == 0:
  return 1
return 0

后一个3-liner可以在一行上重写,但为了清楚起见,我选择了上面的内容.