CODEWITHSUNDEEP
pythondictionarycombinations
ibi0tux
ibi0tux

Reputation: 2619

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

I would like to generate all combinations of values which are in lists indexed in a dict:

{'A':['D','E'],'B':['F','G','H'],'C':['I','J']}

Each time, one item of each dict entry would be picked and combined to items from other keys, so I have:

['D','F','I']
['D','F','J']
['D','G','I']
['D','G','J']
['D','H','I']
...
['E','H','J']

I know there is a something to generate combinations of items in list in itertools but I don't think I can use it here since I have different "pools" of values.

Is there any existing solution to do this, or how should I proceed to do it myself, I am quite stuck with this nested structure.

Upvotes: 38

Views: 48734

Answers (5)

Kevin Südmersen
Kevin Südmersen

Reputation: 971

ParameterGrid from scikit-learn creates a generator. Each for loop iteration will give you 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:
    print(dict_)
    

# output
{'A': 'D', 'B': 'F', 'C': 'I'}
...

Upvotes: 15

Max
Max

Reputation: 561

as a complement, here is a 3-liner that does it in pure 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)

Upvotes: 1

Pablo
Pablo

Reputation: 3514

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)]

Upvotes: 63

Maryam
Maryam

Reputation: 476

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')]

Upvotes: 46

Umesh Nepal
Umesh Nepal

Reputation: 35

from itertools import combinations

a=['I1','I2','I3','I4','I5']

list(combinations(a,2))

The output will be:

[('I1', 'I2'),
 ('I1', 'I3'),
 ('I1', 'I4'),
 ('I1', 'I5'),
 ('I2', 'I3'),
 ('I2', 'I4'),
 ('I2', 'I5'),
 ('I3', 'I4'),
 ('I3', 'I5'),
 ('I4', 'I5')]

Upvotes: 1

Related Questions