从其他两本字典中创建一本具有相似之处的字典。

Question

我有两本这样的字典：

dict1= {'MO': ['N-2', 'N-8', 'N-7', 'N-6', 'N-9'], 'MO2': ['N0-6'], 'MO3': ['N-2']}
dict2= {'N-2': ['NUMBER1'], 'N0-6': ['NUMBER16'], 'N-9': ['NUMBER33']

我想写一本这样的字典

dict3={'MO'['NUMBER1','NUMBER33'], 'MO2':['NUMBER16']}

因此，我开发了这段代码，但是当我添加这些值时，它就不起作用了。

for ki, vi in dict2.items():
    for key, value in (itertools.chain.from_iterable([itertools.product((k, ), v) for k, v in dict3.items()])):
        if (ki == v):
            print vi

Answer 1

一条单线。:D

{k: [y for x in xs if x in d2 for y in d2[x]] for k, xs in d1.items()}

我们循环遍历d1.

• For中的每个键(k)和值(xs) --在xs中的每个项目x，如果d2包含y ... for y in d2[x].

，则通过d2找到路径，使用y ... for y in d2[x].将结果列表扁平

下面是另一种可能更容易理解的格式化方法：

{
    k: [
        y
        for x in xs if x in d2
        for y in d2[x]
    ]
    for k, xs in d1.items()
}

Answer 2

您只需遍历其中一本字典：

dict1= {'MO': ['N-2', 'N-8', 'N-7', 'N-6', 'N-9'], 'MO2': ['N0-6'], 'MO3': ['N-2']}
dict2= {'N-2': ['NUMBER1'], 'N0-6': ['NUMBER16'], 'N-9': ['NUMBER33']}
dict3 = {}
for key, subkeys in dict1.items():
    for subkey in subkeys:
        dict3.setdefault(key, []).extend(dict2.get(subkey, []))

Answer 3

试一试这个用不合格的

from collections import defaultdict
d =defaultdict(list)
for k in dict1.keys():
    for value in dict1[k]:
        temp = dict2.get(value)
        if temp:
            d[k].append(temp)

输出

defaultdict(list,
            {'MO': [['NUMBER1'], ['NUMBER33']],
             'MO2': [['NUMBER16']],
             'MO3': [['NUMBER1']]})