Calculate the average of list of lists based on two elements in the list?

I have the following list:

  mylist: [[(5, 1, 11), (5, 2, 13), (5, 3, 26),
           (3, 1, 60), (3, 2, 40), (3, 3, 70), 
           (6, 1, 30), (6, 2, 80), (2, 3, 80)],
           [(5, 1, 7), (5, 2, 8), (5, 3, 6),
           (3, 1, 50), (3, 2, 44), (3, 3, 44), 
           (6, 1, 20), (6, 2, 40), (2, 3, 50)],
           [(5, 1, 22), (5, 2, 18), (5, 3, 60),
           (3, 1, 10), (3, 2, 20), (3, 3, 30), 
           (6, 1, 60), (6, 2, 20), (2, 3, 30)]]

I want to calculate the average of the items which have the same “first and the second elements”. E.g., from the below example, I want to take the average of the elements which have ‘5’ and ‘1’ in the first two elements of the list. So, my desired output should be like this:

 output: [(5, 1, 13.3), (5, 2, 25.6), (5, 3, 30.6),
           (3, 1, 40), (3, 2, 34.6), (3, 3, 48), 
           (6, 1, 36.6), (6, 2, 46.6), (6, 3, 56.6)]
        

If I have only two items in the lists like:

   mylist: [[(1, 11), ( 2, 13), ( 3, 26),
           [( 1, 60), ( 2, 40), ( 3, 70)],...]

I could easily calculate the average by the below code:

  np.mean(mylist, axis=0)

Answer

see below

from collections import defaultdict

lst = [[(5, 1, 11), (5, 2, 13), (5, 3, 26),
        (3, 1, 60), (3, 2, 40), (3, 3, 70),
        (6, 1, 30), (6, 2, 80), (2, 3, 80)],
       [(5, 1, 7), (5, 2, 8), (5, 3, 6),
        (3, 1, 50), (3, 2, 44), (3, 3, 44),
        (6, 1, 20), (6, 2, 40), (2, 3, 50)],
       [(5, 1, 22), (5, 2, 18), (5, 3, 60),
        (3, 1, 10), (3, 2, 20), (3, 3, 30),
        (6, 1, 60), (6, 2, 20), (2, 3, 30)]]

data = defaultdict(list)
for ex_entry in lst:
    for in_entry in ex_entry:
        data[(in_entry[0], in_entry[1])].append(in_entry[2])
for key, value in data.items():
    print(f'{key} -> {sum(value) / len(value)}')

output

(5, 1) -> 13.333333333333334
(5, 2) -> 13.0
(5, 3) -> 30.666666666666668
(3, 1) -> 40.0
(3, 2) -> 34.666666666666664
(3, 3) -> 48.0
(6, 1) -> 36.666666666666664
(6, 2) -> 46.666666666666664
(2, 3) -> 53.333333333333336