python - 列表结构比较

Question

任务是编写一个函数same-struct_as，它返回 True 或 False 当需要列表比较嵌套结构时。

例如:

should return True same_structure_as([ 1, 1, 1 ], [ 2, 2, 2 ] ) same_structure_as([ 1, [ 1, 1 ] ], [ 2, [ 2, 2 ] ] )

should return False  same_structure_as([ 1, [ 1, 1 ] ], [ [ 2, 2 ], 2 ] ) same_structure_as([ 1, [ 1, 1 ] ], [ [ 2 ], 2 ] )

我的代码如下所示:

#!/usr/bin/python
# -*- coding: utf-8 -*-


def same_structure_as(original, other):
    count = 0
    if len(original) == len(other):
        for i in range(0, len(original) - 1):
            if isinstance(original[i], int) == isinstance(other[i],
                    int):
                count += 1
            elif len(original[i]) == len(other[i]):
                count += 1
    else:
        return False

    if count == len(original) - 1:
        return True

此代码遍历两个列表的每个元素，并检查它们是否是整数或子列表(具有相同的元素)。当我运行它时，我收到此错误:

elif len(original[i])==len(other[i]):TypeError: object of type 'int' has no len()

Answer 1

在使用 Python 设计程序时，我始终遵循的一条规则是永远不要信任用户。这意味着始终检查您获得的输入的类型，确保您获得的参数具有有效值...

在第一个 if 中，您测试了这两项是否都是 int，但之后您不能假设这两项都不是。如果一个是 int，另一个是列表(这里正是这种情况)怎么办？

您应该添加更多类型检查，如下所示:

def same_structure_as(original, other):
    count = 0
    if len(original) == len(other):
        for i in range(0, len(original) - 1):
            if isinstance(original[i], int) == isinstance(other[i],
                    int):
                count += 1
            elif (isinstance(original[i], list) and
                  isinstance(other[i], list) and
                  len(original[i]) == len(other[i])):
                count += 1
    else:
        return False

    if count == len(original) - 1:
        return True

话虽这么说，有可能(也许更好)避免运行整个循环并在第一次遇到差异时停止:

def same_structure_as(original, other):

    # This avoids to make useless computations when they are not necessary
    if not isinstance(original, list) or not isinstance(other, list)
        # You can even raise a TypeError, to inform the user
        # that they should make sure to provid two lists
        return False

    if len(original) != len(other):
        return False

    for index in range(len(original)):
        if isinstance(original[index], list) and isinstance(other[index], list):
            if type(original[index]) != type(other[index]):
                # This would happen for example with [1, ...] and [[1], ...]
                return False
            if len(original[index]) != len(other[index]):
                # This would happen for example with [[1, 1], ...] and [[1], ...]
                return False

    return True