How to generate all combination from values in dict of lists in Python

If you want to keep the key:value in the permutations you can use:

import itertools
keys, values = zip(*my_dict.items())
permutations_dicts = [dict(zip(keys, v)) for v in itertools.product(*values)]

this will provide you a list of dicts with the permutations:

print(permutations_dicts)
[{'A':'D', 'B':'F', 'C':'I'}, 
 {'A':'D', 'B':'F', 'C':'J'},
 ...
 ]

Disclaimer: not exactly what the OP was asking, but google send me here looking for that. If you want to obtain what OP is asking you should just remove the dict part in the list comprehension, i.e.

permutations_dicts = [v for v in itertools.product(*values)]

import itertools as it

my_dict={'A':['D','E'],'B':['F','G','H'],'C':['I','J']}
allNames = sorted(my_dict)
combinations = it.product(*(my_dict[Name] for Name in allNames))
print(list(combinations))

Which prints:

[('D', 'F', 'I'), ('D', 'F', 'J'), ('D', 'G', 'I'), ('D', 'G', 'J'), ('D', 'H', 'I'), ('D', 'H', 'J'), ('E', 'F', 'I'), ('E', 'F', 'J'), ('E', 'G', 'I'), ('E', 'G', 'J'), ('E', 'H', 'I'), ('E', 'H', 'J')]

As a complement, here is code that does it Python so that you get the idea, but itertools is indeed more efficient.

res = [[]]
for _, vals in my_dict.items():
    res = [x+[y] for x in res for y in vals]
print(res)

How about using ParameterGrid from scikit-learn? It creates a generator over which you can iterate in a normal for loop. In each iteration, you will have a dictionary containing the current parameter combination.

from sklearn.model_selection import ParameterGrid

params = {'A':['D','E'],'B':['F','G','H'],'C':['I','J']}
param_grid = ParameterGrid(params)
for dict_ in param_grid:
    # Do something with the current parameter combination in ``dict_``
    print(dict_["A"])
    print(dict_["B"])
    print(dict_["C"])