Multiply each element of a vector with a matrix

I want to multiply each element of a vector with a matrix such that I end up with a 3D matrix (or higher dimentions).

For example if A is a vector and B is a matrix I would write:

for indx=1:length(A)
    result(:,:,indx)=A(indx).*B
end

Is for-loops the way to go here or are there any better solutions?

Answer 1

[Sean de Wolski]

BSXFUN!!!

B = [1 2; 3 4];
A = 1:5;

D = bsxfun(@times,B,reshape(A,1,1,numel(A)));

kron is notorious (at least in older versions) for being very slow.

And obligatory timings:

B = repmat([1 2; 3 4],10,10);
A = 1:50;

t1 = 0;
t2 = 0;
t3 = 0;

for ii = 1:100

      tic
      C = reshape(kron(A,B),[size(B),numel(A)]);
      t1=t1+toc;
      tic
      D = bsxfun(@times,B,reshape(A,1,1,numel(A)));
      t2=t2+toc;
      tic
      for indx = length(A):-1:1  % Backwards for pre-allocation
          result(:, :, indx) = A(indx) .* B;
      end
      t3=t3+toc;
  end
  isequal(C,D,result)
  [t1 t2 t3]

%{
ans =
     1
ans =
    0.0128    0.0069    0.0098
%}

[Jan]

+1. BSXFUN. Even the three exclamation marks are no exaggeration.

[Friedrich]

+1 Nice one and its a way faster than kron and reshape.

[Lars Kluken]

Thank you! For my code I found that this was about 10 times faster than the for-loop.

[Lars Kluken]

I was also wondering if there is something similar for multiplicating a 2d matrix with slices 3d matrix?

[Sean de Wolski]

for matrix multiplication, use a for-loop or James Tursa's mtimex on the FEX.

for element by element:

bsxfun(@times,magic(3),rand(3,3,10));

Answer 2

[James Tursa]

Yet another method for completeness using an outer product formulation. May not be any faster than bsxfun et al:

result = reshape(B(:)*A(:).',[size(B) numel(A)]);

(This tip actually comes from Bruno via another thread)

[Sean de Wolski]

Nice

[Lars Kluken]

This is also a good solution. Sometimes a little faster than the bsxfun, and sometimes a little slower. It seems to depend on the sizes of the matrix and the vector.

Answer 3

[Friedrich]

Hi,

try kron and reshape:

B = [1 2; 3 4]
A = 1:5

reshape(kron(A,B),[size(B),numel(A)])

[Lars Kluken]

Thanks for a quick answer.

I tried this, but it was actually slower than the for loop.

[Friedrich]

Every other way I can think of is more complex and not very fast. So its seems that a for loop is the best way here.

[Jan]

You can look into the code of KRON. As far as I remember, this is a FOR loop also.

[Friedrich]

No there isnt. Its meshrgrid and smart indexing

[Jan]

I feel so blind without a Matlab installation. Is MESHGRID free of FOR loops now? I've read all the basic function in Matlab 4, 5.3, 6.5, 2008b, 2009a. Now I'm starting to get confused when I read them in 2011b again. It would be so nice to have a list of changes for each function...

[Sean de Wolski]

Looking in meshgrid in 2012a it is just a series of reshapes and ones() used for indexing.

Answer 4

[Jan]

Did you pre-allocate the output?

result = zeros([size(B), length(A)]);
for indx = 1:length(A)
  result(:, :, indx) = A(indx) .* B;
end

Or:

for indx = length(A):-1:1  % Backwards for pre-allocation
  result(:, :, indx) = A(indx) .* B;
end

[Lars Kluken]

Yes I did. In my case the for-loop is fast enough. I was just hoping for a neater solution. I never thought of backwards for-loops for pre-allocation.