在字典中找到一个值，然后获取密钥

Question

I'm using Python 3. I have 3 groups and in these groups I have multiple values.

I have the value that I search and I want to get his group. For example if I have CCC I need to get GROUP 1 and if I have HHH I want the GROUP 3 then do something according to the group.

So I think I gonna create a dict like this (tell me if I'm wrong) :

{
  'group1': {'AAA','BBB','CCC','DDD'}, 
  'group2': {'EEE','FFF','GGG'},
  'group3': {'HHH','JJJ'}
}

So I see that we can revert the dict to get the key from a value so I thought to do this :

dict = {
  'group1': {'AAA','BBB','CCC','DDD'},
  'group2': {'EEE','FFF','GGG'},
  'group3': {'HHH','JJJ'}
}

revdict = dict([(dict[key],key) for key in dict])

group = revdict['CCC']
if group == 'group1':
    # Do something
elif group == 'group2':
    # Do something
elif group == 'group3':
    # Do something

But I don't think it's the good way to do what I want. There is a way to do something like this :

if 'CCC' in dict :
    # Then get the current key. How ?

Or maybe I don't need to create dict but another things ? I open for all your suggestions.

Answer 1

Your reversed dict , commonly called an inverted index, will fail because you are using sets as keys. Sets being mutable, they cannot be hashed to form a dict key. Instead, you probably wanted each element of a group to form a key.

You can rewrite the correct inverted index with a defaultdict .

from collections import defaultdict

groups = {
    'group1': {'AAA', 'BBB', 'CCC', 'DDD'},
    'group2': {'EEE', 'FFF', 'GGG'},
    'group3': {'HHH', 'JJJ'}
}

inverted_index = defaultdict(set)

for name, group in groups.items():
    for element in group:
        inverted_index[element].add(name)

print('group1' in inverted_index['AAA']) # True
print('group1' in inverted_index['EEE']) # False

In an inverted index, an element may have multiple keys pointing to it (even though it is not the case in you data), this is why each value must be a set of keys.

If, as you seem to state it in the comments, your data is assured to only have one-to-one correspondences, then you can simply create a dict .

inverted_index = {element: name for name, group in groups.items() for element in group}
print(inverted_index)

Output

{'AAA': 'group1',
 'BBB': 'group1',
 'CCC': 'group1',
 'DDD': 'group1',
 'EEE': 'group2',
 'FFF': 'group2',
 'GGG': 'group2',
 'HHH': 'group3',
 'JJJ': 'group3'}

Answer 2

You can retrieve the key using list comprehension and only returning the key if the str you are looking for is in its values

d = {
  'group1': {'AAA','BBB','CCC','DDD'},
  'group2': {'EEE','FFF','GGG'},
  'group3': {'HHH','JJJ'}
}

a = [i for i in d if 'CCC' in d[i]]
# ['group1']

Answer 3

I you insist on not using index for the key,

answer = None
for key in d.keys():
   if "CCC" in d[key]:
      answer = key
      break

*edit suggested by @Olivier

Answer 4

for k,v in dict.items():
   if 'CCC' in v:
      print(k)

you can't escape loops or reverse dicts. if you don't have memory constraints use your reverse dict.