从以列表为值的字典中获取键

Question

Consider the following dictionary.

dict = {
    'key_1': ['name1', 'name2', 'name3'],
    'key_2': ['name4', 'name5', 'name6']
}

Given a string called "name6", how to easy figure that this is belongs to "key_2" key with lesser looping. I have the following code, how can I optimize for time constraint of this. Above code is just an example, there are several such keys present in the dictionary.

dict = {
    'key_1': ['name1', 'name2', 'name3'],
    'key_2': ['name4', 'name5', 'name6']
}

output_key = None

for key in dict:
    if 'name6' in dict[key]:
        output_key = key
        break

Answer 1

In the current form of this dict, you cannot find the key that stores "name6" without iterating over the keys.

You need a different data structure, you can create another dict where the keys are each unique string from the original dict and values are list of keys that hold that string in the original dict, that way you can check for that string in the new dict in O1, and get a list of keys that all contain that string in the old dict.

The creation of that new dict is O(n) (where n is the number of strings the original dict holds), but since its also the cost of the search you performed, that is still a win, if you need to lookup m values, you has to do O(m*n) before, but after its O(max(m,n))

dictionary = {
'key_1': ['name1', 'name2', 'name3'],
'key_2': ['name4', 'name5', 'name6']
}

new_dict = dict()
for key in dictionary:
    for value in dictionary[key]:
        if value in new_dict:
            new_dict[value].append(key)
        else:
            new_dict[value] = [key]

If you know from the start that each value can only be in one key (in the original dict) there is no need for the lists, but it's nice to have just in case;)

Answer 2

This removes the loop and according to other answers seems to be faster, specially for small dictionaries, but ( disclaimer ) I did not benchmark or perform any performance testing. Also if the key is not in the list you'll get an index out of bonds error that you need to handle (or make sure the list returned is not empty, ie there is a result, before accessing element [0] ):

mydict = {
    'key_1': ['name1', 'name2', 'name3'],
    'key_2': ['name4', 'name5', 'name6']
}

output_key = next((mykey for mykey,myval in mydict.items() if 'name6' in myval), None)

print(output_key) # key_2

Edit
Replaced with next to get the first value that matches, or default value ( None ).