Sum values in a dict of lists

Question

If I have a dictionary such as:

my_dict = {\"A\": [1, 2, 3], \"B\": [9, -4, 2], \"C\": [3, 99, 1]}

How do I create a new dictionary with su

Answer 1

Use sum() function:

my_dict = {"A": [1, 2, 3], "B": [9, -4, 2], "C": [3, 99, 1]}

result = {}
for k, v in my_dict.items():
    result[k] = sum(v)

print(result)

Or just create a dict with a dictionary comprehension:

result = {k: sum(v) for k, v in my_dict.items()}

Output:

{'A': 6, 'B': 7, 'C': 103}

Answer 2

One liners- for the same, just demonstrating some ways to get the expected result,

my_dict = {"A": [1, 2, 3], "B": [9, -4, 2], "C": [3, 99, 1]}

1. Using my_dict.keys():- That returns a list containing the my_dict's keys, with dictionary comprehension.

   res_dict = {key : sum(my_dict[key]) for key in my_dict.keys()}
   

2. Using my_dict.items():- That returns a list containing a tuple for each key value pair of dictionary, by unpacking them gives a best single line solution for this,

   res_dict = {key : sum(val) for key, val in my_dict.items()}
   

3. Using my_dict.values():- That returns a list of all the values in the dictionary(here the list of each lists),

   note:- This one is not intended as a direct solution, just for demonstrating the three python methods used to traverse through a dictionary

   res_dict = dict(zip(my_dict.keys(), [sum(val) for val in my_dict.values()]))
   

The zip function accepts iterators(lists, tuples, strings..), and pairs similar indexed items together and returns a zip object, by using dict it is converted back to a dictionary.

Answer 3

Just a for loop:

new = {}
for key in dict:
    new_d[key]= sum(d[key])

new the dictionary having all the summed values

Answer 4

Try This:

def sumDictionaryValues(d):
    new_d = {}
    for i in d:
        new_d[i]= sum(d[i])
    return new_d