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matlabmathmatrixoctave
necromancer
necromancer

Reputation: 24651

How to express as Matrix operations?

I have a matrix A:

1 3 1
7 5 2
4 3 7
8 2 1
3 9 6
4 5 2

and a matrix B:

2 9 1
4 3 8
9 7 3
4 4 2
6 5 7
2 9 2

I want to compute C:

1*2+3*9+1*1
7*4+5*3+2*8
4*9+3*7+7*3
8*4+2*4+1*2
3*6+9*5+6*7
4*2+5*9+2*2

How can I express this purely using matrix operations? I realize I can do this using . versions but I am interested in pure matrix operations. For example, when I have two vectors x and y, I vastly prefer x'*y over sum(x.*y). Hence I am interested in how to do the above also using matrix operations.

Upvotes: 1

Views: 357

Answers (2)

Phonon
Phonon

Reputation: 12737

You should use vector operations

C = sum( A .* B , 2 );

The .* operator multiplies matrices value-by-value, while sum( <matrix> , 2 ) sums along the rows of a matrix inside the first parameter.

Upvotes: 4

voithos
voithos

Reputation: 70622

If you do not want to use vector operators, you can get the same result by performing a matrix multiplication with the transpose of the second multiplicand (otherwise you'll get a 3x3 result, in this case), and then extracting the diagonal.

Like so: C = diag(A * B')

I'm not quite sure how Octave optimizes this, but it appears to only be slightly slower than the element-wise approach. (at least, for this data set)

function test(func, n, a, b)
    for i = 1:n
        func(a, b);
    endfor
endfunction

octave> tic; test(@(a, b) sum(a.*b, 2), 100000, A, B); toc
Elapsed time is 2.843 seconds.

octave> tic; test(@(a, b) diag(a*b'), 100000, A, B); toc
Elapsed time is 3.2 seconds.

CAUTION: A real life problem shows this to be far slower than the element-wise approach:

octave:100> size(yy)
ans =

   5000     10

octave:101> size(expected)
ans =

   5000     10

octave:102> tic; diag(yy * expected'); toc;
Elapsed time is 0.5447 seconds.
octave:103> tic; sum(yy .* expected, 2); toc;
Elapsed time is 0.0016899 seconds.

Upvotes: 4

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