For a matrix, I want to get the row (if it exists) which has the lesser values for each column - strictly dominated by all other rows. For example,

A = [[7,5,8,2]
    [10,3,7,8]
    [6,2,6,1]]

B = [[7,5,8,2]
    [10,3,7,8]
    [9,5,6,7]
    [6,2,6,1]]

For matrix A, row [6,2,6,1] has the minimum values. A[2][i] < A[1][i] and A[2][i] < A[0][i] for all values of i. So, its index should be returned.

For matrix B, no row with the lowest values exists (6 in column 2 is not less than other values in that column).

Is there a cleaner way to do that, other than to loop each column individually?

CodePudding user response：

Here you go. See the comments for what each line does.

import numpy as np

def get_index_of_minimum_row(arr):
    """ returns the index of the minimum row or None if it does not exist """

    #Get the minimum value of each column.
    column_minimums = arr.min(axis=0)

    #Create a mask of the array for the minimum values (True where it is the minimum, False where it is not).
    column_minimums_mask = arr <= column_minimums

    #Ensure that the minimum value is the only minimum value in its column.
    has_only_one_minimum_per_column = column_minimums_mask.sum() == arr.shape[1]
    if not has_only_one_minimum_per_column:
        return None

    #Ensure that all minimums are in the same row.
    all_minimums_are_in_the_same_row = column_minimums_mask.all(axis=1)
    if not all_minimums_are_in_the_same_row.any():
        return None

    #Get the index of the minimum row and return it
    return np.squeeze(np.argwhere(all_minimums_are_in_the_same_row))

A = np.array([
    [7,5,8,2],
    [10,3,7,8],
    [6,2,6,1],
])

B = np.array([
    [7,5,8,2],
    [10,3,7,8],
    [9,5,6,7],
    [6,2,6,1],
])

print(get_index_of_minimum_row(A)) #returns 2
print(get_index_of_minimum_row(B)) #returns None