Sum matrix using recursion in python

Question

I need to create a func that calculate the sum of 2d arr (matrix) for example sum of [[1,1,1],[2,2,2]] would give 9. I tried solving it using two functions - one calls the other but had an error. It seems that when it gets to the second arr and on, it passes an arr in arr like that [[]] so it iterates threw another arr not threw the numbers. I prefer not to use libraries in this case.

This is my code:

def sum_arr(arr):
    s = 0
    if len(arr) == 1:
        s += arr[0]
    else:
        s += (arr[0] + sum_arr(arr[1:]))
    return s

def sum_mat(mtx):
    sm = 0
    if len(mtx) == 1:
        sm += sum_arr(mtx[0])
    else:
        sm += sum_arr(mtx[0]) + sum_arr(mtx[1:])

    return sm

sum_mat([[1, 2, 3],[1,2,4],[7,8,9]])

FWIW, it is probably better to use an empty array as the edge condition and return 0. Currently your sum_arr() function fails on sum_arr([]) instead returning 0, which seems more correct. — Mark
– Mark, Commented Oct 19, 2022 at 20:47
your sum_arr should be return 0 if not arr else arr[0] + sum_arr(arr[1:]) — gog
– gog, Commented Oct 19, 2022 at 21:26

Anthony DiGiovanna · Accepted Answer · 2022-10-19 20:43:37Z

1

I think your problem is that you accidentally are not recursively calling sum_mat. You call sum_arr on arr[0] then call it again on arr[1:]. Try:

def sum_arr(arr):
    s = 0
    if len(arr) == 1:
        s += arr[0]
    else:
        s += (arr[0] + sum_arr(arr[1:]))
    return s

def sum_mat(mtx):
    sm = 0
    if len(mtx) == 1:
        sm += sum_arr(mtx[0])
    else:
        sm += sum_arr(mtx[0]) + sum_mat(mtx[1:]) #changed this line here to fix repeat sum_arr call

    return sm

sum_mat([[1, 2, 3],[1,2,4],[7,8,9]])

answered Oct 19, 2022 at 20:43

Anthony DiGiovanna

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1 Comment

JonathanM Over a year ago

Thank you, you are correct, what a silly mistake I did :)

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