python get dictionary key from value is list

Question

I have two dictionaries:

first_dict = {'a': ['1', '2', '3'],
              'b': ['4', '5'],
              'c': ['6'],
              }

second_dict = {'1': 'wqeewe',
               '2': 'efsafa',
               '4': 'fsasaf',
               '6': 'kgoeew',
               '7': 'fkowew'
               }

I want to have a third dict that will contain the key of second_dict and its corresponding value from first_dict's key. This way, I will have :

third_dict = {'1' : 'a',
              '2' : 'a',
              '4' : 'b',
              '6' : 'c',
              '7' : None,

}

here is my way:

def key_return(name):
  for key, value in first_dict.items():
    if name == value:
      return key
    if isinstance(value, list) and name in value:
      return key
  return None

reference: Python return key from value, but its a list in the dictionary

However, I wondering that the another way using dict.get() or something else. Any help would be appreciated. Thanks.

Are all values in the first collection lists? Then your first test is not useful. — Mudassir, Dec 03 '22 at 07:44
As per current use case it is better to `set` instead of `list` because you are using membership test, which is faster on `set` as compared to `list`. — Yaman Jain, Dec 03 '22 at 07:50
@Mudassir I see. It didn't work if there is ['1', '4']. So I need to try another way — matsuri, Dec 03 '22 at 10:25
You can also test `name == value or name in value`. The second test will only happen if the first one is false. — Mudassir, Dec 05 '22 at 08:25

score 1 · Answer 1 · answered Dec 03 '22 at 08:09

you can do it like that:

Code

first_dict = {'a': ['1', '2', '3'],
              'b': ['4', '5'],
              'c': ['6'],
              }

second_dict = {'1': 'wqeewe',
               '2': 'efsafa',
               '4': 'fsasaf',
               '6': 'kgoeew',
               '7': 'fkowew'
               }

third_dict = dict()

for second_key in second_dict.keys():
    found = False
    for first_key, value in first_dict.items():
        if second_key in value:
            third_dict.setdefault(second_key, first_key )
            found = True
    if not found:
        third_dict.setdefault(second_key, None)
                    
print(third_dict)

Output:

{'1': 'a', '2': 'a', '4': 'b', '6': 'c', '7': None}

Hope this helps

score 1 · Accepted Answer · answered Dec 03 '22 at 08:11

version with a_dict.get()

third_dict = {i: {i:k for k,v in first_dict.items() for i in v}.get(i) for i in second_dict.keys()}

this part {i:k for k,v in first_dict.items() for i in v}
creates a dict like {'1': 'a', '2': 'a', '3': 'a', '4': 'b', '5': 'b', '6': 'c'}

score 0 · Answer 3 · answered Dec 03 '22 at 08:17

You can map the values in the first dictionary to their keys with:

values_map = dict([a for k, v in first_dict.items() for a in zip(v, k*len(v))])

then use this map to create the third dictionary:

third_dict = {key: values_map.get(key) for key, value in second_dict.items()}

Since I get that the first_dict may contain single values instead of list you may want first to convert those values to list with:

first_dict = dict(map(lambda x: (x[0], x[1]) if isinstance(x[1], list) else (x[0], [str(x[1])]), first_dict.items()))

score 0 · Answer 4 · answered Dec 03 '22 at 09:13

res = {
    x: k
    for k, xs in first_dict.items()
    for x in xs
    if x in second_dict
}

this creates 1:a , 2:a, 4:b etc. If you also want missing keys like 7:None join it with a dummy dict:

res = {k: None for k in second_dict} | {
    x: k
    for k, xs in first_dict.items()
    for x in xs
    if x in second_dict
}

python get dictionary key from value is list

4 Answers4

Code

Output: