正在修复TypeError:类型为“…”的对象没有透镜

Question

这是我的代码。

def areSimilar(item1,item2):
    if(len(item1) == len(item2)):
        for i in range(len(item1)):
            if isinstance(item1[i],list) and isinstance(item2[i],list):
                if areSimilar(item1[i],item2[i])==False:
                    return False

            else:
                if type(item1[i]) != type(item2[i]):
                    return False

    else:
        return False
    return True

以下是想要工作的例子。

>>> print(areSimilar(True, False)) #items are same type and are not lists
True
>>> print(areSimilar(1, 'a') #items are different types
False
>>> print(areSimilar([],[]))                      
True                        
>>> print(areSimilar([],[3])) #list lengths differ
False
>>> print(areSimilar([3.0],[3]))  #lists of same length but index 0 items  are not similar
False
>>> print(areSimilar([5],[3]))  #lists of same length and corresponding lists items are similar
True
>>> print(areSimilar([1,2,['a','b']],[3,4, [1,2,3]])) #items at index 2 are not similar
False
>>> print(areSimilar([1,2,[False, 'b']],[3, 4, [True, 'hello']]))
True
>>> print(areSimilar([[[[],[2],[],['hi', [0]]]]], [[[[],[-2],[],['bye', [1]]]]]))
True
>>> print(areSimilar([[[[],[2],[],['hi', [0]]]]], [[[[],[-2],[],['bye', 0]]]]))
False

除了前两个，所有的都能工作。我到哪里去了

“TypeError:类型为'bool/int'的对象没有len”

我该如何解决这个问题？

Answer 1

只需添加将类型转换为列表的警卫。

def areSimilar(item1,item2):
    if not getattr(item1, '__len__', False):
       item1 = [item1]
    if not getattr(item2, '__len__', False):
       item2 = [item2]
    if(len(item1) == len(item2)):
        for i in range(len(item1)):
            if isinstance(item1[i],list) and isinstance(item2[i],list):
                if areSimilar(item1[i],item2[i])==False:
                    return False

            else:
                if type(item1[i]) != type(item2[i]):
                    return False

    else:
        return False
    return True

然而，解决方案看起来很混乱，整体功能设计需要改变。