Most pythonic way to convert a dict of lists into a list of dicts of all combs of the elements in the lists from the dict?

Question

I have a dictionary of lists. I want to generate all combinations of those lists and return those combinations as a list of dictionaries with the correct key and value pair mappings.

I have achieved this so far by deploying the below code, but I want to be explicit about the key and value pair mappings throughout the process, rather than relying the enumeration approach below. I am concerned given that the dictionary is not ordered that the enumeration approach might be open to not mapping the key value pairs correctly somehow. In any case, I am looking for the most pythonic way to achieve the result.

import itertools

dct_lst = {'Apple':[3,5,7],'Pear':[5,2],'Orange':[10],'Banana':[7,10],'Lemon':[5,7,9,2,1]}

combs = itertools.product(*(dct_lst[key] for key in dct_lst.keys()))

lst_dct = []
for comb in combs:
    dct = {}
    for j,key in enumerate(dct_lst.keys()):
        dct[key] = comb[j]
    lst_dct.append(dct)

lst_dct

[{'Apple': 3, 'Banana': 7, 'Lemon': 5, 'Orange': 10, 'Pear': 5},
 {'Apple': 3, 'Banana': 7, 'Lemon': 7, 'Orange': 10, 'Pear': 5},
 {'Apple': 3, 'Banana': 7, 'Lemon': 9, 'Orange': 10, 'Pear': 5},
 {'Apple': 3, 'Banana': 7, 'Lemon': 2, 'Orange': 10, 'Pear': 5},
 {'Apple': 3, 'Banana': 7, 'Lemon': 1, 'Orange': 10, 'Pear': 5},
 {'Apple': 3, 'Banana': 10, 'Lemon': 5, 'Orange': 10, 'Pear': 5},
 {'Apple': 3, 'Banana': 10, 'Lemon': 7, 'Orange': 10, 'Pear': 5},
 {'Apple': 3, 'Banana': 10, 'Lemon': 9, 'Orange': 10, 'Pear': 5},
 {'Apple': 3, 'Banana': 10, 'Lemon': 2, 'Orange': 10, 'Pear': 5},
 {'Apple': 3, 'Banana': 10, 'Lemon': 1, 'Orange': 10, 'Pear': 5},
 {'Apple': 3, 'Banana': 7, 'Lemon': 5, 'Orange': 10, 'Pear': 2},
 {'Apple': 3, 'Banana': 7, 'Lemon': 7, 'Orange': 10, 'Pear': 2},
 {'Apple': 3, 'Banana': 7, 'Lemon': 9, 'Orange': 10, 'Pear': 2},
 {'Apple': 3, 'Banana': 7, 'Lemon': 2, 'Orange': 10, 'Pear': 2},
 {'Apple': 3, 'Banana': 7, 'Lemon': 1, 'Orange': 10, 'Pear': 2},
 {'Apple': 3, 'Banana': 10, 'Lemon': 5, 'Orange': 10, 'Pear': 2},
 {'Apple': 3, 'Banana': 10, 'Lemon': 7, 'Orange': 10, 'Pear': 2},
 {'Apple': 3, 'Banana': 10, 'Lemon': 9, 'Orange': 10, 'Pear': 2},
 {'Apple': 3, 'Banana': 10, 'Lemon': 2, 'Orange': 10, 'Pear': 2},
 {'Apple': 3, 'Banana': 10, 'Lemon': 1, 'Orange': 10, 'Pear': 2},
 {'Apple': 5, 'Banana': 7, 'Lemon': 5, 'Orange': 10, 'Pear': 5},
 {'Apple': 5, 'Banana': 7, 'Lemon': 7, 'Orange': 10, 'Pear': 5},
 {'Apple': 5, 'Banana': 7, 'Lemon': 9, 'Orange': 10, 'Pear': 5},
 {'Apple': 5, 'Banana': 7, 'Lemon': 2, 'Orange': 10, 'Pear': 5},
 {'Apple': 5, 'Banana': 7, 'Lemon': 1, 'Orange': 10, 'Pear': 5},
 {'Apple': 5, 'Banana': 10, 'Lemon': 5, 'Orange': 10, 'Pear': 5},
 {'Apple': 5, 'Banana': 10, 'Lemon': 7, 'Orange': 10, 'Pear': 5},
 {'Apple': 5, 'Banana': 10, 'Lemon': 9, 'Orange': 10, 'Pear': 5},
 {'Apple': 5, 'Banana': 10, 'Lemon': 2, 'Orange': 10, 'Pear': 5},
 {'Apple': 5, 'Banana': 10, 'Lemon': 1, 'Orange': 10, 'Pear': 5},
 {'Apple': 5, 'Banana': 7, 'Lemon': 5, 'Orange': 10, 'Pear': 2},
 {'Apple': 5, 'Banana': 7, 'Lemon': 7, 'Orange': 10, 'Pear': 2},
 {'Apple': 5, 'Banana': 7, 'Lemon': 9, 'Orange': 10, 'Pear': 2},
 {'Apple': 5, 'Banana': 7, 'Lemon': 2, 'Orange': 10, 'Pear': 2},
 {'Apple': 5, 'Banana': 7, 'Lemon': 1, 'Orange': 10, 'Pear': 2},
 {'Apple': 5, 'Banana': 10, 'Lemon': 5, 'Orange': 10, 'Pear': 2},
 {'Apple': 5, 'Banana': 10, 'Lemon': 7, 'Orange': 10, 'Pear': 2},
 {'Apple': 5, 'Banana': 10, 'Lemon': 9, 'Orange': 10, 'Pear': 2},
 {'Apple': 5, 'Banana': 10, 'Lemon': 2, 'Orange': 10, 'Pear': 2},
 {'Apple': 5, 'Banana': 10, 'Lemon': 1, 'Orange': 10, 'Pear': 2},
 {'Apple': 7, 'Banana': 7, 'Lemon': 5, 'Orange': 10, 'Pear': 5},
 {'Apple': 7, 'Banana': 7, 'Lemon': 7, 'Orange': 10, 'Pear': 5},
 {'Apple': 7, 'Banana': 7, 'Lemon': 9, 'Orange': 10, 'Pear': 5},
 {'Apple': 7, 'Banana': 7, 'Lemon': 2, 'Orange': 10, 'Pear': 5},
 {'Apple': 7, 'Banana': 7, 'Lemon': 1, 'Orange': 10, 'Pear': 5},
 {'Apple': 7, 'Banana': 10, 'Lemon': 5, 'Orange': 10, 'Pear': 5},
 {'Apple': 7, 'Banana': 10, 'Lemon': 7, 'Orange': 10, 'Pear': 5},
 {'Apple': 7, 'Banana': 10, 'Lemon': 9, 'Orange': 10, 'Pear': 5},
 {'Apple': 7, 'Banana': 10, 'Lemon': 2, 'Orange': 10, 'Pear': 5},
 {'Apple': 7, 'Banana': 10, 'Lemon': 1, 'Orange': 10, 'Pear': 5},
 {'Apple': 7, 'Banana': 7, 'Lemon': 5, 'Orange': 10, 'Pear': 2},
 {'Apple': 7, 'Banana': 7, 'Lemon': 7, 'Orange': 10, 'Pear': 2},
 {'Apple': 7, 'Banana': 7, 'Lemon': 9, 'Orange': 10, 'Pear': 2},
 {'Apple': 7, 'Banana': 7, 'Lemon': 2, 'Orange': 10, 'Pear': 2},
 {'Apple': 7, 'Banana': 7, 'Lemon': 1, 'Orange': 10, 'Pear': 2},
 {'Apple': 7, 'Banana': 10, 'Lemon': 5, 'Orange': 10, 'Pear': 2},
 {'Apple': 7, 'Banana': 10, 'Lemon': 7, 'Orange': 10, 'Pear': 2},
 {'Apple': 7, 'Banana': 10, 'Lemon': 9, 'Orange': 10, 'Pear': 2},
 {'Apple': 7, 'Banana': 10, 'Lemon': 2, 'Orange': 10, 'Pear': 2},
 {'Apple': 7, 'Banana': 10, 'Lemon': 1, 'Orange': 10, 'Pear': 2}]

The result is as expected. Though, I am interested in the most robust pythonic way to achieve this result.

Answer 1

You could use itertools.product on just the values of the dict and then dict(zip(...)) the keys and values together to output a list of dicts.

from itertools import product

d = {'Apple':[3,5,7],'Pear':[5,2],'Orange':[10],'Banana':[7,10],'Lemon':[5,7,9,2,1]}

l = [dict(zip(d.keys(), vals)) for vals in product(*d.values())]
print(l)
# [{'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 5}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 7}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 9}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 2}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 1}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 5}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 7}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 9}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 2}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 1}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 5}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 7}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 9}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 2}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 1}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 5}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 7}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 9}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 2}, {'Apple': 3, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 1}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 5}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 7}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 9}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 2}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 1}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 5}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 7}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 9}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 2}, {'Apple': 5, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 1}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 5}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 7}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 9}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 2}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 1}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 5}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 7}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 9}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 2}, {'Apple': 5, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 1}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 5}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 7}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 9}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 2}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 1}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 5}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 7}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 9}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 2}, {'Apple': 7, 'Pear': 5, 'Orange': 10, 'Banana': 10, 'Lemon': 1}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 5}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 7}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 9}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 2}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 7, 'Lemon': 1}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 5}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 7}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 9}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 2}, {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 1}]

Answer 2

You can repeat the key along with the values to build a list of items:

>>> import itertools
>>> dct_lst = {'Apple':[3,5,7],'Pear':[5,2],'Orange':[10],'Banana':[7,10],'Lemon':[5,7,9,2,1]}
>>> L = [[(k, v) for v in vs] for k, vs in dct_lst.items()]
>>> L
[[('Apple', 3), ('Apple', 5), ('Apple', 7)], [('Pear', 5), ('Pear', 2)], [('Orange', 10)], [('Banana', 7), ('Banana', 10)], [('Lemon', 5), ('Lemon', 7), ('Lemon', 9), ('Lemon', 2), ('Lemon', 1)]]

And use itertools.product to create all the combinations:

>>> list (itertools.product(*L))
[(('Apple', 3), ('Pear', 5), ('Orange', 10), ('Banana', 7), ('Lemon', 5)), (('Apple', 3), ('Pear', 5), ('Orange', 10), ('Banana', 7), ('Lemon', 7)), (('Apple', 3), ('Pear', 5), ('Orange', 10), ('Banana', 7), ('Lemon', 9)), ..., (('Apple', 7), ('Pear', 2), ('Orange', 10), ('Banana', 7), ('Lemon', 2)), (('Apple', 7), ('Pear', 2), ('Orange', 10), ('Banana', 7), ('Lemon', 1)), (('Apple', 7), ('Pear', 2), ('Orange', 10), ('Banana', 10), ('Lemon', 5)), (('Apple', 7), ('Pear', 2), ('Orange', 10), ('Banana', 10), ('Lemon', 7)), (('Apple', 7), ('Pear', 2), ('Orange', 10), ('Banana', 10), ('Lemon', 9)), (('Apple', 7), ('Pear', 2), ('Orange', 10), ('Banana', 10), ('Lemon', 2)), (('Apple', 7), ('Pear', 2), ('Orange', 10), ('Banana', 10), ('Lemon', 1))]

The output is made of tuples, but it's easy to create dicts:

>>> list(map(dict, itertools.product(*L)))
[{'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 5}, {'Apple': 3, 'Pear': 5, 'Orange': 10, 'Banana': 7, 'Lemon': 7}, ..., {'Apple': 7, 'Pear': 2, 'Orange': 10, 'Banana': 10, 'Lemon': 1}]

Answer 3

you can use ChianMap to help merge the dicts at the end.

try this:

from itertools import product
from collections import ChainMap

dct_lst = {'Apple':[3,5,7],'Pear':[5,2],'Orange':[10],'Banana':[7,10],'Lemon':[5,7,9,2,1]}

flat_dicts = [[{key: val} for val in lst] for key, lst in dct_lst.items()]
lst_dct = list(map(lambda d: dict(ChainMap(*d)), product(*flat_dicts)))

print(lst_dct)