Python将2个字典连接到第3个2D字典中，其中字典2的键是字典1中的列表值

Question

I couldn't find this particular python dictionary question anywhere.

I have two dictionaries:

dict1 = {'key1':['val1','val2','val3']}
dict2 = {'val1':['a','b','c']}

I want a 3rd, 2D dictionary with:

dict3 = {'key1': {'val1':['a','b','c']} }

So, joining 2 dictionaries where the key of the second dictionary is a list value of the first dictionary.

I was trying some nested looping along the lines of:

for key1, val1 in dict1.items():
    for key2, in val2 in dict2.items():
        # do something here

I am not sure if that is the best way to do this.

Answer 1

You can use a dictionary comprehension and then check if the final result contains only one dictionary. If the latter is true, then a dictionary of dictionaries will be the final result; else, a listing of dictionaries will be stored for the key:

dict1 = {'key1':['val1','val2','val3']}
dict2 = {'val1':['a','b','c']}
new_dict = {a:[{i:dict2[i]} for i in b if i in dict2] for a, b in dict1.items()}
last_result = {a:b if len(b) > 1 else b[0] for a, b in new_dict.items()}

Output:

{'key1': {'val1': ['a', 'b', 'c']}}

Answer 2

This is best done by iterating over dict1 and looking for matching values in dict2 :

result = {}
for key, value_list in dict1.items():
    result[key] = subdict = {}

    for value in value_list:
        try:
            subdict[value] = dict2[value]
        except KeyError:
            pass

Result:

{'key1': {'val1': ['a', 'b', 'c']}}

Answer 3

dict1 = {
    'key1':['val1','val2','val3']
}
dict2 = {
    'val1':['a','b','c']
}
dict3 = {
    key : { val_key : dict2[val_key] 
        for val_key in val_list if (val_key in dict2.keys())
    } for key, val_list in dict1.items() 
}

Answer 4

You can try this solution .

dict1 = {'key1':['val1','val2','val3']}
dict2 = {'val1':['a','b','c']}

join_dict={}

for i,j in dict1.items():
    for sub_l,sub_value in dict2.items():
        if sub_l in j:
            join_dict[i]={sub_l:sub_value}

print(join_dict)

output:

{'key1': {'val1': ['a', 'b', 'c']}}