Python - Summation Grouping in Dictionary List

Python – Summation Grouping in Dictionary List

Last Updated : 10 May, 2020

Sometimes, while working with Python Dictionaries, we can have a problem in which we need to perform the grouping of dictionaries according to specific key, and perform summation of certain keys while grouping similar key’s value. This is s peculiar problem but can have applications in domains such as web development. Let’s discuss a certain way in which this task can be performed.

Input : test_list = [{‘geeks’: 10, ‘best’: 8, ‘Gfg’: 6}, {‘geeks’: 12, ‘best’: 11, ‘Gfg’: 4}]
Output : {4: {‘geeks’: 12, ‘best’: 11}, 6: {‘geeks’: 10, ‘best’: 8}}
Input : test_list = [{‘CS’: 14, ‘best’: 9, ‘geeks’: 7, ‘Gfg’: 14}]
Output : {14: {‘best’: 9, ‘geeks’: 7}}

Method #1 : Using loop
This is brute force way in which this task can be performed. In this we manually loop through each key from dictionary and perform required grouping according to the key and construct the required grouped and summed dictionary.

# Python3 code to demonstrate working of 
# Summation Grouping in Dictionary List
# Using loop
  
# initializing list
test_list = [{'Gfg' :  1, 'id' : 2, 'best' : 8, 'geeks' : 10}, 
             {'Gfg' :  4, 'id' : 4, 'best':  10, 'geeks' : 12}, 
             {'Gfg' :  4, 'id' : 8, 'best':  11, 'geeks' : 15}]
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing group key 
grp_key = 'Gfg'
  
# initializing sum keys 
sum_keys = ['best', 'geeks']
  
# Summation Grouping in Dictionary List
# Using loop
res = {}
for sub in test_list:
    ele = sub[grp_key]
    if ele not in res:
        res[ele] = {x: 0 for x in sum_keys}
    for y in sum_keys:
        res[ele][y] += int(sub[y])
  
# printing result 
print("The grouped list : " + str(res)) 

Output :

The original list is : [{‘geeks’: 10, ‘id’: 2, ‘best’: 8, ‘Gfg’: 1}, {‘geeks’: 12, ‘id’: 4, ‘best’: 10, ‘Gfg’: 4}, {‘geeks’: 15, ‘id’: 8, ‘best’: 11, ‘Gfg’: 4}]
The grouped list : {1: {‘geeks’: 10, ‘best’: 8}, 4: {‘geeks’: 27, ‘best’: 21}}

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