Python – Sort Dictionary by Values Summation
Give a dictionary with value lists, sort the keys by summation of values in value list.
Input : test_dict = {‘Gfg’ : [6, 7, 4], ‘best’ : [7, 6, 5]}
Output : {‘Gfg’: 17, ‘best’: 18}
Explanation : Sorted by sum, and replaced.Input : test_dict = {‘Gfg’ : [8], ‘best’ : [5]}
Output : {‘best’: 5, ‘Gfg’: 8}
Explanation : Sorted by sum, and replaced.
Method #1 : Using sorted() + dictionary comprehension + sum()
The combination of above functions can be used to solve this problem. In this, we first sum all the list elements using sum() and then next step is to perform sorting of all the keys according to sum extracted using sorted().
Python3
# Python3 code to demonstrate working of # Sort Dictionary by Values Summation# Using dictionary comprehension + sum() + sorted() # initializing dictionarytest_dict = {'Gfg' : [6, 7, 4], 'is' : [4, 3, 2], 'best' : [7, 6, 5]} # printing original dictionaryprint("The original dictionary is : " + str(test_dict)) # summing all the values using sum()temp1 = {val: sum(int(idx) for idx in key) for val, key in test_dict.items()} # using sorted to perform sorting as requiredtemp2 = sorted(temp1.items(), key = lambda ele : temp1[ele[0]]) # rearrange into dictionaryres = {key: val for key, val in temp2} # printing result print("The sorted dictionary : " + str(res)) |
The original dictionary is : {'Gfg': [6, 7, 4], 'is': [4, 3, 2], 'best': [7, 6, 5]}
The sorted dictionary : {'is': 9, 'Gfg': 17, 'best': 18}
Method #2 : Using map() + dictionary comprehension + sorted() + sum()
The combination of above functions can be used to solve this problem. In this, we perform the task of mapping the logic of summation using map().
Python3
# Python3 code to demonstrate working of # Sort Dictionary by Values Summation# Using map() + dictionary comprehension + sorted() + sum() # initializing dictionarytest_dict = {'Gfg' : [6, 7, 4], 'is' : [4, 3, 2], 'best' : [7, 6, 5]} # printing original dictionaryprint("The original dictionary is : " + str(test_dict)) # summing all the values using sum()# map() is used to extend summation to sorted()temp = {key: sum(map(lambda ele: ele, test_dict[key])) for key in test_dict}res = {key: temp[key] for key in sorted(temp, key = temp.get)} # printing result print("The sorted dictionary : " + str(res)) |
The original dictionary is : {'Gfg': [6, 7, 4], 'is': [4, 3, 2], 'best': [7, 6, 5]}
The sorted dictionary : {'is': 9, 'Gfg': 17, 'best': 18}
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