Python – All combination Dictionary List

Last Updated : 08 Mar, 2023

Given 2 lists, form all possible combination of dictionaries that can be formed by taking keys from list1 and values from list2.

Input : test_list1 = [“Gfg”, “is”, “Best”], test_list2 = [4]
Output : [{‘Gfg’: 4, ‘is’: 4, ‘Best’: 4}]
Explanation : All combinations dictionary list extracted.

Input : test_list1 = [“Gfg”, “is”, “Best”], test_list2 = [5, 6]
Output : [{‘Gfg’: 5, ‘is’: 5, ‘Best’: 5}, {‘Gfg’: 5, ‘is’: 5, ‘Best’: 6}, {‘Gfg’: 5, ‘is’: 6, ‘Best’: 5}, {‘Gfg’: 5, ‘is’: 6, ‘Best’: 6}, {‘Gfg’: 6, ‘is’: 5, ‘Best’: 5}, {‘Gfg’: 6, ‘is’: 5, ‘Best’: 6}, {‘Gfg’: 6, ‘is’: 6, ‘Best’: 5}, {‘Gfg’: 6, ‘is’: 6, ‘Best’: 6}]
Explanation : All combinations dictionary list extracted.

Method #1 : Using product() + dictionary comprehension + list comprehension

In this, we get all the combination possible using product() and dictionary comprehension generates each dictionary, list comprehension is used to iterate combinations generated.

Python3

# Python3 code to demonstrate working of 
# All combination Dictionary List
# Using product() + list comprehension + dictionary comprehension
from itertools import product
 
# initializing lists
test_list1 = ["Gfg", "is", "Best"]
test_list2 = [4, 5]
 
# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
 
# generating combinations
temp = product(test_list2, repeat = len(test_list1))
 
# constructing dicts using combinations
res = [{key : val for (key , val) in zip(test_list1, ele)} for ele in temp]
 
# printing result 
print("The combinations dictionary : " + str(res))

Output:

The original list 1 is : [‘Gfg’, ‘is’, ‘Best’] The original list 2 is : [4, 5] The combinations dictionary : [{‘Gfg’: 4, ‘is’: 4, ‘Best’: 4}, {‘Gfg’: 4, ‘is’: 4, ‘Best’: 5}, {‘Gfg’: 4, ‘is’: 5, ‘Best’: 4}, {‘Gfg’: 4, ‘is’: 5, ‘Best’: 5}, {‘Gfg’: 5, ‘is’: 4, ‘Best’: 4}, {‘Gfg’: 5, ‘is’: 4, ‘Best’: 5}, {‘Gfg’: 5, ‘is’: 5, ‘Best’: 4}, {‘Gfg’: 5, ‘is’: 5, ‘Best’: 5}]

Time Complexity: O(n*n)
Auxiliary Space: O(n)

Method #2 : Using permutations() + dictionary comprehension + list comprehension [ For unique elements ]

If we require values to be all unique, permutations() can be used inplace of product() to achieve this task.

Python3

# Python3 code to demonstrate working of
# All combination Dictionary List
# Using product() + list comprehension + dictionary comprehension
from itertools import permutations
 
# initializing lists
test_list1 = ["Gfg", "is", "Best"]
test_list2 = [4, 5, 6]
 
# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
 
# generating combinations
temp = list(permutations(test_list2, len(test_list1)))
 
# constructing dicts using combinations
res = [{key: val for (key, val) in zip(test_list1, ele)} for ele in temp]
 
# printing result
print("The combinations dictionary : " + str(res))

Output:

The original list 1 is : [‘Gfg’, ‘is’, ‘Best’] The original list 2 is : [4, 5, 6] The combinations dictionary : [{‘Gfg’: 4, ‘is’: 5, ‘Best’: 6}, {‘Gfg’: 4, ‘is’: 6, ‘Best’: 5}, {‘Gfg’: 5, ‘is’: 4, ‘Best’: 6}, {‘Gfg’: 5, ‘is’: 6, ‘Best’: 4}, {‘Gfg’: 6, ‘is’: 4, ‘Best’: 5}, {‘Gfg’: 6, ‘is’: 5, ‘Best’: 4}]

Time Complexity: O(n!) where n is the number of elements in the list “test_list”. permutations() + dictionary comprehension + list comprehension performs n! number of operations.
Auxiliary Space: O(n), extra space of size n is required

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