从“if”语句而不是“else”语句返回结果的递归函数

Question

我是 Python 的新手，我正在尝试编写一个简单的递归函数：

def bugged_recursion(inp_value,list_index=0):
    '''Define a recursive function that tag lists according to one parameter.
    '''
    #check if a criterion is true at position 0 in the list
    if list_index == 0:
        if inp_value[list_index] == 'valid':
            status = 'valid inp_value'
        #if the criterion is false call the function at the next index
        else:
            status = 'invalid inp'
            list_index +=1
            bugged_recursion(inp_value,list_index=list_index)
    #check if a criterion is true at position 1 in the list
    else:
        if inp_value[list_index] == 'valid':
            status = 'index {} is a valid inp_value'.format(list_index)
        else:
            status = 'index is never a valid inp_value'
    print(status)
    #return the input and its status
    return (inp_value,status)

if __name__ == '__main__':
    inp_value = ['invalid','invalid']
    bugged_recursion(inp_value)

我不明白为什么这个函数会返回 if 语句中的状态，而不是返回最后一个 else 语句中包含的状态。 对我来说，最奇怪的是它会在某些时候打印正确的状态，但不会返回。

我无法理解为什么...我真的很好奇如何使用递归函数来执行此任务。

Answer 1

哇哇，这是多么的折磨。

def bugged_recursion(inp_value, list_index=0):
    # i don't get why you compare list_index here
    if list_index == 0:
        # you'll get an IndexError if list_index > len(inp_value)
        if inp_value[list_index] == 'valid':
            status = 'valid inp_value'
        else:
            status = 'invalid inp'
            # there is only one place where you increment list_index
            # i suppose there is something wrong here
            list_index +=1
            # you forgot to return here
            return bugged_recursion(inp_value, list_index=list_index)
    else:
        if inp_value[list_index] == 'valid':
            status = 'index {} is a valid inp_value'.format(list_index)
        else:
            status = 'index is never a valid inp_value'
    return (inp_value,status)

除此之外，人们通常倾向于尽可能避免recursion（例如Dive into Python）。

这是否满足您的需求？

def no_recursion(inp_value):
    for i, val in enumerate(inp_value):
        # you probably have a good reason to test the index
        if i == 0:
            if val == 'valid':
                yield 'valid inp_value: %s' % val
            else:
                yield 'invalid inp: %s' % val
        else:
            yield 'index %s is %s valid inp_value' % (
                i,
                'a' if val == 'valid' else 'never'
            )

print tuple(no_recursion(inp_value))

给：('invalid inp: invalid', 'index 1 is never valid inp_value')