How to subtract a vector from each row of a matrix? [duplicate]

This question already has answers here: Closed 11 years ago.

Possible Duplicate:

How can I divide each row of a matrix by a fixed row?

I'm looking for an elegant way to subtract the same vector from each row of a matrix. Here is a non elegant way of doing it.

a = [1 2 3];
b = rand(7,3);
c(:,1) = b(:,1) - a(1);
c(:,2) = b(:,2) - a(2);
c(:,3) = b(:,3) - a(3);

Also, the elegant way can't be slower than this method.

I've tried

c = b-repmat(a,size(b,1),1); 

and it seems slower.

EDIT: The winner is this method.

c(:,1) = b(:,1) - a(1);
c(:,2) = b(:,2) - a(2);
c(:,3) = b(:,3) - a(3);

EDIT: More methods, and tic toc results:

n = 1e6;
m = 3;
iter = 100;
a = rand(1,m);
b = rand(n,m);

tic
c = zeros(size(b));
for i = 1:iter
    c(:,1) = b(:,1) - a(1);
    c(:,2) = b(:,2) - a(2);
    c(:,3) = b(:,3) - a(3);
end
toc

tic
c = zeros(size(b));
for i = 1:iter
    c(:,1) = b(:,1) - a(1);
    c(:,2) = b(:,2) - a(2);
    c(:,3) = b(:,3) - a(3);
end
toc

tic
c = zeros(size(b));
for i = 1:iter
    for j = 1:3
        c(:,j) = b(:,j) - a(j);
    end
end
toc

tic
for i = 1:iter
    c = b-repmat(a,size(b,1),1);
end
toc

tic
for i = 1:iter
    c = bsxfun(@minus,b,a);
end
toc

tic
c = zeros(size(b));
for i = 1:iter
    for j = 1:size(b,1)
        c(j,:) = b(j,:) - a;
    end
end
toc

results

Elapsed time is 0.622730 seconds.
Elapsed time is 0.627321 s开发者_JAVA技巧econds.
Elapsed time is 0.713384 seconds.
Elapsed time is 2.621642 seconds.
Elapsed time is 1.323490 seconds.
Elapsed time is 17.269901 seconds.

---

Here is my contribution:

c = b - ones(size(b))*diag(a)

Now speed testing it:

tic
for i = 1:10000
    c = zeros(size(b));
    b = rand(7,3);
    c = b - ones(size(b))*diag(a);
end
toc

The result:

Elapsed time is 0.099979 seconds.

Not quite as fast, but it is clean.

---

There are only three obvious answers, and you gave two of them in your question.

The third is by row,

c(1,:) = b(1,:) - a; %...

but I'd expect that to be slower than your by-column processing for large matrixes since it accesses elements out of memory order.

If you turn your by-column processing into a for loop in a *.m file or subfunction, is it still faster than the repmat version?

One other thing you might test for speed: Try preallocating c.

c = zeros(size(b));
c(:,1) = b(:,1) - a(1); %...