减去两个词典的值

Question

我想减去dict1 - dict2.如果结果是狗的负数并不重要,我只是有兴趣学习这个概念:)

owner = ['Bob', 'Sarah', 'Ann']

dict1 = {'Bob': {'shepherd': [4,6,3],'collie': [23, 3, 45], 'poodle':[2,0,6]}, {'Sarah': {'shepherd': [1,2,3],'collie': [3, 31, 4], 'poodle': [21,5,6]},{'Ann': {'shepherd': [4,6,3],'collie': [23, 3, 45], 'poodle': [2,10,8]}}


 dict2 = {'Bob': {'shepherd': 10,'collie': 15,'poodle': 34},'Sarah': {'shepherd': 3,'collie': 25,'poodle': 4},'Ann': {'shepherd': 2,'collie': 1,'poodle': 0}}

我想要:

wanted =   dict1 = {'Bob': {'shepherd': [-6,-4,-7],'collie': [8, -12, 30], 'poodle':[-32,-34,-28]}, {'Sarah': {'shepherd': [-2,-1,0],'collie': [-22, 6, -21], 'poodle': [17,1,2]},{'Ann': {'shepherd': [2,4,1],'collie': [22, 2, 44], 'poodle': [2,10,8]}}

我仍然是字典的新手,所以这里是我一直在尝试但得到NoneType错误.我假设它是因为dict1有比dict2更多的值,但我找不到执行上述计算的方法.

wanted = {}

for i in owner:
    wanted.update({i: dict1.get(i) - dict2.get(i)})

Answer 1

您可以使用嵌套字典理解.我稍微调整了你的字典,因为它的语法不正确(有一些额外的花括号).不需要所有者数组,因为我们可以只使用字典中的键.

dict1 = {
    'Bob': {
        'shepherd': [4, 6, 3],
        'collie': [23, 3, 45],
        'poodle': [2, 0, 6],
    },
    'Sarah': {
        'shepherd': [1, 2, 3],
        'collie': [3, 31, 4],
        'poodle': [21, 5, 6],
    },
    'Ann': {
        'shepherd': [4, 6, 3],
        'collie': [23, 3, 45],
        'poodle': [2, 10, 8],
    }
}
dict2 = {
    'Bob': {'shepherd': 10, 'collie': 15, 'poodle': 34},
    'Sarah': {'shepherd': 3, 'collie': 25, 'poodle': 4},
    'Ann': {'shepherd': 2, 'collie': 1, 'poodle': 0},
}
wanted = {
    owner: {
        dog: [num - dict2[owner][dog] for num in num_list]
        for dog, num_list in dog_dict.iteritems()
    } for owner, dog_dict in dict1.iteritems()
}
print wanted

输出:

{
    'Sarah': {
        'shepherd': [-2, -1, 0],
        'collie': [-22, 6, -21],
        'poodle': [17, 1, 2]
    },
    'Bob': {
        'shepherd': [-6, -4, -7],
        'collie': [8, -12, 30],
        'poodle': [-32, -34, -28]
    },
    'Ann': {
        'shepherd': [2, 4, 1],
        'collie': [22, 2, 44],
        'poodle': [2, 10, 8]
    }
}

这假设dict2具有与dict1相同的所有键.如果不是这种情况,您需要设置一些默认值以避免a KeyError.

编辑:这里是一个简短的解释,说明如何解释那些不熟悉字典理解的字典理解.如果你对列表推导更熟悉,字典理解非常相似,只有你创建一个字典而不是列表.所以,{i: str(i) for i in xrange(10)}会给你一本字典{0: '0', 1: '1', ..., 9: '9'}.

现在我们可以将其应用于解决方案.这段代码:

wanted = {
    owner: {
        dog: [num - dict2[owner][dog] for num in num_list]
        for dog, num_list in dog_dict.iteritems()
    } for owner, dog_dict in dict1.iteritems()
}

相当于:

wanted = {}
for owner, dog_dict in dict1.iteritems():
    wanted[owner] = {}
    for dog, num_list in dog_dict.iteritems():
        wanted[owner][dog] = [num - dict2[owner][dog] for num in num_list]