Comparing two matriced

Hi,

Is there a fast way to compare the differences between two matrices with the same dimension:

example Assume A and B each is a 1000x1000 matrix. Is there a way to find where A and B differ in one step?

Answer 1

[Jan]

And if the matrices are results of floating point computation a certian relative or absolute tolerance might be helpful:

abs_d = (A - B) < Tol
rel_d = ((A - B) ./ min(A, B)) < Tol

And now the same tools as replied by Walter, Sean and Matt can be applied.

Answer 2

[Walter Roberson]

find(A~=B)

[Kiarash Ahi]

good job!

Answer 3

[Sean de Wolski]

The logical matric

 ABne = A~=B

or if you need the indices

 [r c] = find(ABne);

Answer 4

[Matt Fig]

And a visual method:

spy(A~=B)

And a quick method to count the number of locations where A is not equal to B:

nnz(A~=B)

[Sean de Wolski]

Matt, I get NNZ taking significantly longer than sum(:):

A = rand(1000)>.5;

B = rand(1000)>.5;

idx = A~=B;

t1 = 0;

t2 = 0;

for ii = 1:10

tic

S1 = sum(idx(:));

t1 = t1+toc;

tic

S2 = nnz(idx);

t2 = t2+toc;

end

isequal(S1,S2)

t1/t2

ans =

1

ans =

0.13468

[Jan]

@Sean: There have been several discussion in CSSM about SUM versus NNZ. The winner changed with the Matlab versions. To my surprise SUM(SUM(idx)) is 45% faster than SUM(idx(:)) even on my single core processor in Matlab 2009a.

[Sean de Wolski]

Interesting Jan. I have sum(sum()) is about equal on my MAC laptop running 2009b 64bit.

[Matt Fig]

@Sean, yes, NNZ is slower than SUM - I don't recommend using it in a loop.

Who has a guess as to what NNZ is doing with its time under the hood?

[Sean de Wolski]

It was last updated (according to my 2009b) in 2006. Could that have something to do with its sluggishness?

[Matt Fig]

I don't think that is it. I bet TMW knew how to program a simple FOR loop in C back in 2006. I could be wrong of course :).

[Jan]

NNZ is *much* faster in SSE, especially if the data are aligned at 128 bit boundaries.

[Sean de Wolski]

I find it funny how the help for nnz says:

The density of a sparse matrix S is nnz(S)/prod(size(S))"

But M-Lint now says:

numel(S) is faster than prod(size(S))"