Is it possible to multiply a 3D matrix with a coumn vector?

I have a 3D matrix with three rows and three columns. I want to multiply this matrix with a column vector of 3 rows. How can it be done?

Answer 1

[Jan]

With Matlab >= 2016b:

A = rand(3, 3, 1000);
b = rand(3, 1);
C = squeeze(sum(A .* reshape(b, 1, 3), 2));

With older versions:

C = squeeze(sum(bsxfun(@times, A, reshape(b, 1, 3)), 2))

[Andrei Bobrov]

+1

[Jan]

The three versions have advantages and disadvantages:

• KSSV: This is clean and simple. During debugging the intention is directly clear. The loop will need some time.
• Andrei: This is efficient because it uses the fast built-in matrix multiplication. For large inputs permute needs time.
• Mine: This does need a copy of the input data, but a temporary array also before creating the sum.

Timings:

A = rand(3, 3, 1e6);
B = rand(3, 1);
tic, C = zeros(3, size(A, 3));
for i = 1:size(A, 3)
  C(:,i) = A(:,:,i)*B ;
end, toc
tic; C = reshape(reshape(permute(A,[2,1,3]),3,[]).'*B,3,[]); toc
tic; C = squeeze(sum(bsxfun(@times, A, reshape(B, 1, 3)), 2)); toc

 Elapsed time is 1.605281 seconds.  % Loop
 Elapsed time is 0.069983 seconds.  % permute
 Elapsed time is 0.098582 seconds.  % sum(times())

(Matlab 2009a! Test this on a modern version also)

I would insert KSSV's loop as a comment and Andrei's method for computations.

[Richard]

A = rand(3, 3, 1e6);
B = rand(3, 1);
tic, C = zeros(3, size(A, 3));
for i = 1:size(A, 3)
  C(:,i) = A(:,:,i)*B ;
end, toc
tic; C = reshape(reshape(permute(A,[2,1,3]),3,[]).'*B,3,[]); toc
tic; C = squeeze(sum(bsxfun(@times, A, reshape(B, 1, 3)), 2)); toc

Elapsed time is 0.891301 seconds. % Loop

Elapsed time is 0.082350 seconds. % permute

Elapsed time is 0.088938 seconds. % sum(times())

Matlab 2019b Update 3

Answer 2

[Andrei Bobrov]

C = reshape(reshape(permute(A,[2,1,3]),3,[]).'*B,3,[]);

[Jan]

+1: This is the fastest solution.

Answer 3

[KSSV]

A = rand(3,3,3) ;
B = rand(3,1) ;

C = zeros(3,3) ;
for i = 1:3
    C(:,i) = A(:,:,i)*B ;
end

[Jan]

+1: Clean and clear.

Answer 4

[Tyler R]

I got confused because some of the dimensions are size 3, but there is also a 3rd dimension, so for generalization's sake:

N = 150;
K = 20;
T = 30;
A = rand(N,K,T);
B = rand(K,1);
C = zeros(N,T);

Andrei's method:

C = reshape(reshape(permute(A,[2 1 3]),K,[]).'*B,N,[]);

Jan's method:

C = squeeze(sum(A .* reshape(B, 1, K), 2));

KSSV's method:

for t = 1:T
    C(:,t) = A(:,:,t)*B ;
end

[Roman Gorlov]

When B is rand(K, P), with P > 1, then both proposed methods don't work. I've tried different permuations, but no luck replicating the for loop result.