Given dictionary with values as list, extract all the possible combinations, both cross keys and with values.
Input : test_dict = {“Gfg” : [4, 5], “is” : [1, 2], “Best” : [9, 4]}
Output : {0: [[‘Gfg’, 4], [‘is’, 1], [‘Best’, 9]], 1: [[‘Gfg’, 4], [‘is’, 1], [‘Best’, 4]], 2: [[‘Gfg’, 4], [‘is’, 2], [‘Best’, 9]], 3: [[‘Gfg’, 4], [‘is’, 2], [‘Best’, 4]], 4: [[‘Gfg’, 5], [‘is’, 1], [‘Best’, 9]], 5: [[‘Gfg’, 5], [‘is’, 1], [‘Best’, 4]], 6: [[‘Gfg’, 5], [‘is’, 2], [‘Best’, 9]], 7: [[‘Gfg’, 5], [‘is’, 2], [‘Best’, 4]]}
Explanation : Prints all possible combination of key with values and cross values as well.Input : test_dict = {“Gfg” : [4], “is” : [1], “Best” : [4]}
Output : {0: [[‘Gfg’, 4], [‘is’, 1], [‘Best’, 4]]}
Explanation : Prints all possible combination of key with values and cross values as well.
Method #1 : Using product() + zip() + loop
The combination of above functions can be used to solve this problem. In this, we perform the first combination of keys with all values using product() and cross key combinations are performed using zip() and loop.
Python3
# Python3 code to demonstrate working of # Dictionary Key Value lists combinations # Using product() + zip() + loop from itertools import product # initializing dictionary test_dict = {"Gfg" : [4, 5, 7], "is" : [1, 2, 9], "Best" : [9, 4, 2]} # printing original dictionary print("The original dictionary is : " + str(test_dict)) temp = list(test_dict.keys()) res = dict() cnt = 0 # making key-value combinations using product for combs in product (*test_dict.values()): # zip used to perform cross keys combinations. res[cnt] = [[ele, cnt] for ele, cnt in zip(test_dict, combs)] cnt += 1 # printing result print("The computed combinations : " + str(res)) |
The original dictionary is : {‘Gfg’: [4, 5, 7], ‘is’: [1, 2, 9], ‘Best’: [9, 4, 2]}
The computed combinations : {0: [[‘Gfg’, 4], [‘is’, 1], [‘Best’, 9]], 1: [[‘Gfg’, 4], [‘is’, 1], [‘Best’, 4]], 2: [[‘Gfg’, 4], [‘is’, 1], [‘Best’, 2]], 3: [[‘Gfg’, 4], [‘is’, 2], [‘Best’, 9]], 4: [[‘Gfg’, 4], [‘is’, 2], [‘Best’, 4]], 5: [[‘Gfg’, 4], [‘is’, 2], [‘Best’, 2]], 6: [[‘Gfg’, 4], [‘is’, 9], [‘Best’, 9]], 7: [[‘Gfg’, 4], [‘is’, 9], [‘Best’, 4]], 8: [[‘Gfg’, 4], [‘is’, 9], [‘Best’, 2]], 9: [[‘Gfg’, 5], [‘is’, 1], [‘Best’, 9]], 10: [[‘Gfg’, 5], [‘is’, 1], [‘Best’, 4]], 11: [[‘Gfg’, 5], [‘is’, 1], [‘Best’, 2]], 12: [[‘Gfg’, 5], [‘is’, 2], [‘Best’, 9]], 13: [[‘Gfg’, 5], [‘is’, 2], [‘Best’, 4]], 14: [[‘Gfg’, 5], [‘is’, 2], [‘Best’, 2]], 15: [[‘Gfg’, 5], [‘is’, 9], [‘Best’, 9]], 16: [[‘Gfg’, 5], [‘is’, 9], [‘Best’, 4]], 17: [[‘Gfg’, 5], [‘is’, 9], [‘Best’, 2]], 18: [[‘Gfg’, 7], [‘is’, 1], [‘Best’, 9]], 19: [[‘Gfg’, 7], [‘is’, 1], [‘Best’, 4]], 20: [[‘Gfg’, 7], [‘is’, 1], [‘Best’, 2]], 21: [[‘Gfg’, 7], [‘is’, 2], [‘Best’, 9]], 22: [[‘Gfg’, 7], [‘is’, 2], [‘Best’, 4]], 23: [[‘Gfg’, 7], [‘is’, 2], [‘Best’, 2]], 24: [[‘Gfg’, 7], [‘is’, 9], [‘Best’, 9]], 25: [[‘Gfg’, 7], [‘is’, 9], [‘Best’, 4]], 26: [[‘Gfg’, 7], [‘is’, 9], [‘Best’, 2]]}
Method #2 : Using product() + loop
The combination of above functions can also be used to solve this problem. In this, we perform the task of performing inner and cross keys combination using product(). Difference is that container of grouping is tuple rather than list.
Python3
# Python3 code to demonstrate working of # Dictionary Key Value lists combinations # Using product() + loop from itertools import product # initializing dictionary test_dict = {"Gfg" : [4, 5, 7], "is" : [1, 2, 9], "Best" : [9, 4, 2]} # printing original dictionary print("The original dictionary is : " + str(test_dict)) res = {} for key, val in test_dict.items(): # for key-value combinations res[key] = product([key], val) # computing cross key combinations res = product(*res.values()) # printing result print("The computed combinations : " + str(list(res))) |
The original dictionary is : {‘Gfg’: [4, 5, 7], ‘is’: [1, 2, 9], ‘Best’: [9, 4, 2]}
The computed combinations : [((‘Gfg’, 4), (‘is’, 1), (‘Best’, 9)), ((‘Gfg’, 4), (‘is’, 1), (‘Best’, 4)), ((‘Gfg’, 4), (‘is’, 1), (‘Best’, 2)), ((‘Gfg’, 4), (‘is’, 2), (‘Best’, 9)), ((‘Gfg’, 4), (‘is’, 2), (‘Best’, 4)), ((‘Gfg’, 4), (‘is’, 2), (‘Best’, 2)), ((‘Gfg’, 4), (‘is’, 9), (‘Best’, 9)), ((‘Gfg’, 4), (‘is’, 9), (‘Best’, 4)), ((‘Gfg’, 4), (‘is’, 9), (‘Best’, 2)), ((‘Gfg’, 5), (‘is’, 1), (‘Best’, 9)), ((‘Gfg’, 5), (‘is’, 1), (‘Best’, 4)), ((‘Gfg’, 5), (‘is’, 1), (‘Best’, 2)), ((‘Gfg’, 5), (‘is’, 2), (‘Best’, 9)), ((‘Gfg’, 5), (‘is’, 2), (‘Best’, 4)), ((‘Gfg’, 5), (‘is’, 2), (‘Best’, 2)), ((‘Gfg’, 5), (‘is’, 9), (‘Best’, 9)), ((‘Gfg’, 5), (‘is’, 9), (‘Best’, 4)), ((‘Gfg’, 5), (‘is’, 9), (‘Best’, 2)), ((‘Gfg’, 7), (‘is’, 1), (‘Best’, 9)), ((‘Gfg’, 7), (‘is’, 1), (‘Best’, 4)), ((‘Gfg’, 7), (‘is’, 1), (‘Best’, 2)), ((‘Gfg’, 7), (‘is’, 2), (‘Best’, 9)), ((‘Gfg’, 7), (‘is’, 2), (‘Best’, 4)), ((‘Gfg’, 7), (‘is’, 2), (‘Best’, 2)), ((‘Gfg’, 7), (‘is’, 9), (‘Best’, 9)), ((‘Gfg’, 7), (‘is’, 9), (‘Best’, 4)), ((‘Gfg’, 7), (‘is’, 9), (‘Best’, 2))]
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