Octave - find the min (or max) value of a Matrix, and the associated row and column

octave_tip.md

Whilst working through the many (Octave) coding assignment from Andrew Ng's Stanford Machine Learning course, a common problem that I have to solve revolves around this:

Given a Matrix A with m rows, and n columns find the mininum (or maximum) value and the associated row and column number

This article summarises my solution to this problem (which, hopefully this will also come in hadny to you!). Note that Octave index start from 1 (instead of 0).

Sample Matrix

Say we have a Matrix A that look like this:

octave:69> A = rand(3,4);
octave:70> A
A =

   0.128245   0.453621   0.205679   0.139355
   0.151691   0.431844   0.822562   0.044340
   0.736356   0.056389   0.893922   0.347978

Get Min Value

The minimum value may be found easily by doing this:

octave:71> min(min(A))
ans =  0.044340

(Note: the assiciated location is row 2, column 4 - if you scan through the matrix manually).

To find the associated row and column programmatically, just simply do this...

Find associated row

The associated row number is 2, as per followings:

octave:76> [minval, row] = min(min(A,[],2));
octave:77> row
row =  2

Find associated column

The associated column number is 4, as per followings:

octave:72> [minval, col] = min(min(A,[],1));
octave:75> col
col =  4

Do the same for max value

Simple, do the same but with max() instead of min().

Good luck!

Another useful way to slice a matrix is to return, from a specific column, a value that corresponds to the row of the minimal value in another column.

For example, to obtain the best lambda for the minimal validation cost:

c = 2;
A = [lambda_vec, error_val];                                           % simplified output: usually contains error_train as 2nd col then error_val
[best_lambda, least_cost] = A( A(:,c)==min(A(:,c)), : )     % w/o above simplification, set c=3

If a matrix has not only one but two or more equal minimum (or maximum) values , then the first method may fail, as 'min(min(A,[],2));' and 'min(min(A,[],1))' may return a row and a column indeces for two different minimum (or maximum) locations. So 'A(row,col)' will not be any of the minimums (or maximums), but an other value.

The '[i, j] = find(X == minVal);' will return the vectors of row/col indeces of the similar minimum values.

To find the row and column corresponding to a minimum element of a matrix is way simpler than the above
[m,ind]=min(A(:));
[r,c]=ind2sub(size(A),ind); %r and c are the numbers you are looking for.

Typical way is to:
first, get min, or max from matrix, using min(matrix, [],1) for column min, min(matrix, [ ] , 2) for row min. the same to max.
and then use [i,j] = find(matrix == min or max) to find the indices
simple and effective.

just be careful when you have multiple min or max values, then you need to subset i, or j, as i, j returned will be vectors.