Python – Summation of Custom nested keys in Dictionary
Given dictionary with keys as nested dictionaries, find the sum of values of certain custom keys inside nested dictionary.
Input : test_dict = {‘Gfg’ : {1 : 6, 5: 9, 9: 12}, ‘is’ : {1 : 9, 5: 7, 9: 2}, ‘best’ : {1 : 3, 5: 4, 9: 14}}, sum_key = [1]
Output : 18
Explanation : 6 + 9 + 3 = 18, only values with key 1 are summed.Input : test_dict = {‘Gfg’ : {1 : 6, 5: 9, 9: 12}, ‘is’ : {1 : 9, 5: 7, 9: 2}, ‘best’ : {1 : 3, 5: 4, 9: 14}}, sum_key = [5, 9]
Output : 48
Explanation : Keys 5, 9 are summed.
Method #1 : loop
This is brute way in which this problem can be solved. In this, we employ a loop for all the list elements and keep updating sum value from all the nested dictionaries.
Python3
# Python3 code to demonstrate working of # Summation of Custom nested keys in Dictionary # Using loop # initializing dictionary test_dict = { 'Gfg' : { 1 : 6 , 5 : 9 , 9 : 12 }, 'is' : { 1 : 9 , 5 : 7 , 9 : 2 }, 'best' : { 1 : 3 , 5 : 4 , 9 : 14 }} # printing original dictionary print ( "The original dictionary is : " + str (test_dict)) # initializing sum keys sum_key = [ 1 , 9 ] sum = 0 for ele in sum_key: for key, val in test_dict.items(): # extracting summation of required values sum = sum + val[ele] # printing result print ( "The required summation : " + str ( sum )) |
The original dictionary is : {‘Gfg’: {1: 6, 5: 9, 9: 12}, ‘is’: {1: 9, 5: 7, 9: 2}, ‘best’: {1: 3, 5: 4, 9: 14}}
The required summation : 46
Method #2 : Using list comprehension + sum()
The combination of above functions can also be employed to solve this problem. In this, we perform the task of summation using sum() and rest logic is also encapsulated as one-liner using list comprehension.
Python3
# Python3 code to demonstrate working of # Summation of Custom nested keys in Dictionary # Using list comprehension + sum() # initializing dictionary test_dict = { 'Gfg' : { 1 : 6 , 5 : 9 , 9 : 12 }, 'is' : { 1 : 9 , 5 : 7 , 9 : 2 }, 'best' : { 1 : 3 , 5 : 4 , 9 : 14 }} # printing original dictionary print ( "The original dictionary is : " + str (test_dict)) # initializing sum keys sum_key = [ 1 , 9 ] # sum() used to get cumulative summation res = sum ([val[ele] for ele in sum_key for key, val in test_dict.items()]) # printing result print ( "The required summation : " + str (res)) |
The original dictionary is : {‘Gfg’: {1: 6, 5: 9, 9: 12}, ‘is’: {1: 9, 5: 7, 9: 2}, ‘best’: {1: 3, 5: 4, 9: 14}}
The required summation : 46
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