Python summing values in list if it exists in another list

Question

I have a list and a set:

a_list = [[\'1\', 2], [\'2\', 1], [\'1\', 1]]

b_list = {\'1\', \'2\'}

I\'m looking to correspond the items in b_list

Answer 1

You are on the right track! All you have to do is flip the order of your loops. For every value in b_list, you want to sum up all matching values in a_list, so b_list should be the external loop and a_list the internal. Also note your sum variable should be inside the first loop as it is different for every value in b_list.

If you make this change your code works as expected:

a_list = [['1', 2], ['2', 1], ['1', 1]]

b_list = {'1', '2'}

for j in b_list:
    sum = 0
    for i in a_list:
        if i[0] == j:
            sum += i[1]
    print(j, sum)

will give your desired output:

('1', 3)
('2', 1)

---

EDIT: the above solution is a minimal fix to the code posted in the question, however there are more efficient solutions:

Similar to wim's answer, you could use a defaultdictionary, which in this case would be (slightly) more efficient than using the built-in dict class:

from collections import defaultdict
#
a_list = [['1', 2], ['2', 1], ['1', 1]]
b_list = {'1', '2'}

dict = defaultdict(int)

for key, val in a_list:
    if key in b_list:
        dict[key] += val

print([[key, dict[key]] for key in b_list])

** credit to coldspeed for the idea for this second solution.