Convert List To Dictionary With Duplicate Keys Using Dict Comprehension

Good day all, I am trying to convert a list of length-2 items to a dictionary using the below: my_list = ['b4', 'c3', 'c5'] my_dict = {key: value for (key, value) in my_list} The

Solution 1:

For one key in a dictionary you can only store one value.

You can chose to have the value as a list.

{'b': ['4'], 'c': ['3', '5']}

following code will do that for you :

new_dict = {}
for (key, value) in my_list:
    if key in new_dict:
        new_dict[key].append(value)
    else:
        new_dict[key] = [value]
print(new_dict)
# output: {'b': ['4'], 'c': ['3', '5']}

Same thing can be done with setdefault. Thanks @Aadit M Shah for pointing it out

new_dict = {}
for (key, value) in my_list:
    new_dict.setdefault(key, []).append(value)
print(new_dict)
# output: {'b': ['4'], 'c': ['3', '5']}

Same thing can be done with defaultdict. Thanks @MMF for pointing it out.

from collections import defaultdict
new_dict = defaultdict(list)
for (key, value) in my_list:
    new_dict[key].append(value)
print(new_dict)
# output: defaultdict(<class 'list'>, {'b': ['4'], 'c': ['3', '5']})

you can also chose to store the value as a list of dictionaries:

[{'b': '4'}, {'c': '3'}, {'c': '5'}]

following code will do that for you

new_list = [{key: value} for (key, value) in my_list]

Solution 2:

If you don't care about the O(n^2) asymptotic behaviour you can use a dict comprehension including a list comprehension:

>>> {key: [i[1] for i in my_list if i[0] == key] for (key, value) in my_list}
{'b': ['4'], 'c': ['3', '5']}

or the iteration_utilities.groupedby function (which might be even faster than using collections.defaultdict):

>>> from iteration_utilities import groupedby
>>> from operator import itemgetter
>>> groupedby(my_list, key=itemgetter(0), keep=itemgetter(1))
{'b': ['4'], 'c': ['3', '5']}

Solution 3:

You can use defaultdict to avoid checking if a key is in the dictionnary or not :

from collections import defaultdict

my_dict = defaultdict(list)
for k, v in my_list:
    my_dict[k].append(v)

Output :

defaultdict(list, {'b': ['4'], 'c': ['3', '5']})