Given a 2D list (matrix), how to find the largest element and print its row and column number?

Question

I need to find the row and column number for the largest element of an MxN 2D list. If there is more than on, I want to report the one with the smallest row and column number. If possible, I want to do it without numpy as I am still trying to get the hang of Python by doing these exercises.

Here is an example of what I would like to do:

# m rows and n integers (columns)
m = 3
n = 4
# The 2D-list
arr = [[0, 3, 2, 4],
       [2, 3, 5, 5],
       [5, 1, 2, 3]]

And the output would then need to be:

[1, 2]

As the first 5 is the largest element with the smallest row and column number.

I have gotten as far as finding the largest value per row, but now I just need to compare this per column and the print the indices. This is what I have up to now:

# m rows and n integers (columns)
m = 3
n = 4
MxN = [m, n]
# The 2D-list
arr = [[0, 3, 2, 4],
       [2, 3, 5, 5],
       [5, 1, 2, 3]]

for i in range(MxN[0]):
    row = arr[i]
    maxValue = max(row)
    index = row.index(maxValue)
    print(maxValue, index)

Which gives me the output:

4 3
5 2
5 0

I have no idea how to go on further.

Answer 1

You can just iterate over rows and columns save the index when find a larger number.

# m rows and n integers (columns)
m = 3
n = 4
MxN = [m, n]
# The 2D-list
arr = [[0, 3, 2, 4],
       [2, 3, 5, 5],
       [5, 1, 2, 3]]

max_num = float("-inf")
index = [-1, -1]

for i in range(MxN[0]):
    for j in range(MxN[1]):
        if arr[i][j] > max_num:
            max_num = arr[i][j]
            index = [i, j]

print(index)