Python - Nested dictionary Combinations

Sometimes, while working with Python dictionaries, we can have a problem in which we need to construct all the combination of dictionary keys with different values. This problem can have application in domains such as gaming and day-day programming. Lets discuss certain way in which we can perform this task.

Input : test_dict = {'gfg': {'is' : [6], 'for' : [10], 'best': [4]}} Output : {'gfg0': {'for': 10, 'best': 4, 'is': 6}} Input : test_dict = {'gfg': {'best' : [10]}} Output : {'gfg0': {'best': 10}}

Method : Using product() + dictionary comprehension + zip() The combination of above functions can be used to solve this problem. In this, we extract all possible combinations using product, zip() performs the task of pairing keys to values filtered and dictionary comprehension is used to store all the constructed dictionaries. 

# Python3 code to demonstrate working of 
# Nested dictionary Combinations
# Using product() + dictionary comprehension + zip()
from itertools import product

# initializing dictionary
test_dict = {'gfg': {'is' : [6, 7, 8], 'best': [1, 9, 4]}}

# printing original dictionary
print("The original dictionary : " + str(test_dict))

# Nested dictionary Combinations
# Using product() + dictionary comprehension + zip()
res = { key + str(j) : dict(zip(val.keys(), k))
        for key, val in test_dict.items()
        for j, k in enumerate(product(*val.values()))}

# printing result 
print("The possible combinations : " + str(res)) 

Using nested for loops to iterate over keys and values:

Approach:

Initialize an empty dictionary res to store the flattened dictionary.
Loop through each key-value pair in the input dictionary d using the items() method.
For each key-value pair, loop through each key-value pair in the value using the items() method and enumerate them.
Concatenate the parent key with the index and child key to create a new key for the flattened dictionary.
Assign the corresponding child value to the new key.
Add the new key-value pair to the res dictionary.
Return the flattened dictionary res.

def flatten_dict1(d):
    res = {}
    for k, v in d.items():
        for i, (k2, v2) in enumerate(v.items()):
            res[f"{k}{i}"] = {k2: v2[0]}
    return res
test_dict = {'gfg': {'is': [6], 'for': [10], 'best': [4]}}
print(flatten_dict1(test_dict))

{'gfg0': {'is': 6}, 'gfg1': {'for': 10}, 'gfg2': {'best': 4}}

Time Complexity: O(n^2) where n is the number of keys in the input dictionary
Auxiliary Space: O(n)