Get row with min values

Question

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?

Asked By: UIC

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Source

Answer 1

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

Answered By: Tyson

Get row with min values

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