Question

我正在编写一个函数，该函数验证作为参数传递的两个列表，如果两个列表的结构相同，则返回True，否则返回False。

尝试在代码的多个位置使用打印语句，但没有发现问题。对于不同的结构列表，其他语句按预期方式输出“ False”，但是奇怪的是，该函数返回True，尽管它应该返回False。

def same_structure_as(original,other):
    if isinstance(original, list) and isinstance(other, list):
        if len(original) == len(other):
            for el in original:
                if isinstance(el, list):
                    orig_new = original[original.index(el)]
                    other_new = other[original.index(el)]
                    same_structure_as(orig_new,other_new)
            return True
        else:
            return False
    else:
        print("False")
        return False

same_structure_as([1,[1,1]],[[2,2],2])

由于两个输入列表的结构不同，因此代码应返回False。 print语句正确打印“ False”，但即使我给出“ return False”也返回“ True”

Answer 1

如果内部列表不匹配，则不会返回False：

def same_structure_as(original,other):
    if isinstance(original, list) and isinstance(other, list):
        if len(original) == len(other):
            for el in original:
                if isinstance(el, list):
                    orig_new = original[original.index(el)]
                    other_new = other[original.index(el)]
                    same_structure_as(orig_new,other_new) # here - the result is ignored
            return True
        else:
            return False
    else:
        print("False")
        return False

您需要

def same_structure_as(original,other):
    if isinstance(original, list) and isinstance(other, list):
        if len(original) == len(other):
            for el in original:
                if isinstance(el, list):
                    orig_new = original[original.index(el)]
                    other_new = other[original.index(el)]
                    if not same_structure_as(orig_new,other_new): # early exit if not same 
                        return False
                     # else continue testing (no else needed - simply not return anything)
            return True
        else:
            return False
    else:
        print("False")
        return False

否则，您会“检测/打印”错误，但永远不会采取行动。

Answer 2

最简单的解决方法是简单地return递归函数（我确定这是您的意图）：

def same_structure_as(original, other):
    if isinstance(original, list) and isinstance(other, list):
        if len(original) == len(other):
            for el in original:
                if isinstance(el, list):
                    orig_new = original[original.index(el)]
                    other_new = other[original.index(el)]
                    return same_structure_as(orig_new, other_new) # just add a return here
            return True
        else:
            return False
    else:
        print("False")
        return False

print(same_structure_as([1,[1,1]],[[2,2],2]))

正确无误：

错误

错误

我drew a diagram在上一篇文章中解释了这种情况

Python：else语句返回True，尽管其中提到返回False

2 个答案: