MATLAB Answers

Eike
2

How to multiply a vector with each column of a matrix most efficiently?

Asked by Eike
on 8 May 2011
Latest activity Commented on by Jonathan
on 26 Feb 2019
Hey there,
I have a vector t (nx1) and a matrix A (nxm). I need to multiply t with each column of A element-wise.
Example: t=[ 1; 2 ]; A=[ 1 2; 3 4 ] => Res=[ 1*1 1*2; 2*3 2*4 ]
So far I have tried 4 versions:
% 1. Looping
for j = 1 : m
res( j ) = t .* A( :, j );
end
% 2. Repmat
tmat = repmat( t, 1, m )
res = tmat .* A;
% 3. diag
res = diag( t ) * A;
% 4. Indexing
A = A( : );
t = t( :, ones( 1, m ) );
t = t( : );
res = A .* t;
res = reshape( res, n, m );
I did some testing on these and found #1 to be (by far) the fastest one, followed by #2 and #4 (about 2 to 3 times slower) and then followed by #3 (8 to 9 times slower). This kind of surprised me, since #1 is the only version not taking advantage of matrix operations in some way.
Does anyone know a faster way to do this? Or could anybody give some reasons for the result I found?

  3 Comments

What platform and version of MATLAB are you using?
I'm running MATLAB R2010a on a 64-bit Ubuntu 11.04.
I am sorry, but you haven't defined "m" on the first looping have you ?

Sign in to comment.

6 Answers

Answer by Matt Tearle
on 9 May 2011
 Accepted Answer

The first question has been addressed, so I'll just throw out a couple of points about the second question.
  • The newer the version you're running, the more benefit you'll see from JIT enhancements, so loops aren't as slow and generally evil as they used to be.
  • Methods 2 & 4 both require a lot more memory and related memory operations. Assuming the variables are large, that's a lot of overhead on top of the calculations. (Also remember that MATLAB requires contiguous memory addresses for arrays.)
  • Method 3 not only incurs the same memory-monkeying penalty, but also needs more calculations.

  6 Comments

Loops are fast, but trying this problem out on R2011a, BSXFUN seems to be a better choice, running consistently about 50% faster than a for loop.
For size 2000 matrices,
Loops: 0.0384 sec
BSXFUN: 00251 sec
MATLAB does tend to get more efficient as the releases go on.
Oops. That should be: 0.0251 sec
It should also be mentioned that while using DIAG is very bad, using a sparse diagonal matrix is faster than the loop method, but still a little bit slower than BSXFUN.
sparse(1:m,1:m,t)*A
0.0304 sec

Sign in to comment.


Answer by Matt Fig
on 9 May 2011

I find the looping option fastest on my Win Vista 32 bit. Using r2007b.
>> more_loop_timings
Elapsed time is 0.017479 seconds.
Elapsed time is 0.037864 seconds.
Elapsed time is 0.376887 seconds.
Elapsed time is 0.030641 seconds.
Elapsed time is 0.022277 seconds.
%
%
%
%
function [] = more_loop_timings
% Yet more of these....
t = (1:1000)';
A = rand(length(t),1000); % Large and square
[n,m] = size(A);
%%1. Looping
tic
for jj = m:-1:1
res(:,jj) = t.*A(:,jj); % dynamically pre-allocate
end
toc
%%2. Repmat
tic
res2 = repmat(t,1,m) .* A;
toc
%%3. diag
tic
res3 = diag(t)*A; % Yikes!
toc
% The obvious choice...
tic
res4 = bsxfun(@times,A,t);
toc
%%4. Indexing
tic
t = t(:,ones( 1, m,'single'));
res5 = reshape(A(:).*t(:),n,m);
toc
% isequal(res,res2,res3,res4,res5)

  4 Comments

Show 1 older comment
That pre-allocation is sneaky!
Another fast alternative to the ZEROS function for pre-allocating is my UNINIT function from the FEX:
http://www.mathworks.com/matlabcentral/fileexchange/31362-uninit-create-an-uninitialized-variable-like-zeros-but-faster
I also fancy that prealloc :-P
And thanks for the bsxfun hint, didn't even know that function!
Since I found the timing results to be of pretty high variance, I did some 500 runs (of your code which I copied shamelessly) and normalized the results.
The output is as follows:
Looping - 1.0 (best)
Repmat - 1.9152
diag - 60.6682
bsxfun - 1.4830
indexing - 2.2796
Indeed, looping seems to be the best solution for this problem. Very interesting! :-)

Sign in to comment.


Answer by David
on 21 Jun 2017

I just encountered this problem myself, and found all these answers to be very helpful. However, after poking a little bit more, I discovered that MATLAB 2016b and later allow implicit multiplication to do exactly the same thing as bsxfun, slightly faster.
res = A .* t or res = t .* A are equivalent and solve the problem, as long as t is a vector with the same number of rows OR columns as A.
The other answers were of course correct at the time the question was asked. This page is still being viewed ~1000 times/month, so I wanted to get the current best answer out there.
clearvars;
A = rand(1e4,2e4);
[Nr,Nc] = size(A);
v = rand(Nr,1);
tic;
% answer = zeros(size(A));
for i=Nc:-1:1
answer(:,i) = A(:,i).*v;
end
toc
tic;
answer = A.*(v*ones(1,Nc));
toc
tic;
answer = A.*(repmat(v,1,Nc));
toc
tic;
answer = bsxfun(@times,v,A);
toc
tic;
answer = A.*v;
toc
tic;
answer = v.*A;
toc
Yields the following times
Elapsed time is 0.915789 seconds.
Elapsed time is 0.863713 seconds.
Elapsed time is 0.911458 seconds.
Elapsed time is 0.685757 seconds.
Elapsed time is 0.302117 seconds.
Elapsed time is 0.285025 seconds.
For the last two, I was checking if order matters. It does not; whichever is first is always slightly faster, but moving "answer = v.*A" earlier makes it the slightly slower one.

  2 Comments

I found the same as David on a larger matrix using Ver 2016b.
A=rand(100000,1000);
[Nr,Nc] = size(A);
v = rand(Nr,1);
tic
answer = A.*v(:,ones(1,Nc));
toc
tic
answer = bsxfun(@times,A,v);
toc
tic
answer = v.*A;
toc
Elapsed time is 2.679808 seconds.
Elapsed time is 2.653772 seconds.
Elapsed time is 1.280892 seconds.
"but moving "answer = v.*A" earlier makes it the slightly slower one" - in MATLAB if you run the same computation multiple times it improves the speed on subsequent runs, up to a limit. As v.*A and A.*v are essentially the same then the one run second should be faster.

Sign in to comment.


Answer by Jan
on 9 May 2011

Simplification:
% 4. Indexing
res = t(:, ones(1, m)) .* A;
This is exactly the same effect as performed inside REPMAT. On my Matlab 2009a on a single core BSXFUN is reproducibly the fastest. It is very near to a naively implemented C-Mex version (single threaded, no SSE).
EDITED: Matt's pre-allocation is fast. Matlab 2009a, WinXP 32, 1.5GHz Pentium-M (single core)
t = (1:1000)';
A = rand(1000, 200);
tic; for rep = 1:100; R = func1(t, A); end; toc
function R = func1(t, A) % 0.78 sec
[n, m] = size(A);
R = zeros(n, m);
for i = 1:m
R(:, i) = t .* A(:, i);
end
function R = func2(t, A) % 0.54 sec
[n, m] = size(A);
for i = m:-1:1
R(:, i) = t .* A(:, i);
end
function R = func3(t, A) % 0.54 sec
R = bsxfun(@times, t, A);
function A = func4(t, A) % 0.75 sec
[n, m] = size(A);
for i = 1:m
A(:, i) = t .* A(:, i);
end

  7 Comments

(groan ...) Jan, can you clarify? Did my UNINIT routine fail (which does *not* use mxCreateUninitNumericArray since it is only available from R2008b on), or did your own code fail when you tried to use mxCreateUninitNumericArray? If the former, not sure what the problem is since it worked for me on 32-bit R2006b through R2011a PC WinXP. Unfortunately, I don't have other systems to test with. I haven't tried to figure out how mxFastZeros works yet.
@Jan: The mxFastZeros function signature appears to be this based on some quick testing:
mxArray *mxFastZeros(mxComplexity ComplexFlag, mwSize m, mwSize n);
It returns a double matrix presumably filled with zeros. And it is indeed fast (as fast as my UNINIT function). It does have a drawback in that the returned mxArray is a NORMAL array and not a TEMPORARY array, so you can't return it as-is back to MATLAB without first hacking into it and changing the VariableType field to 4 (TEMPORARY). I got strange results unless I changed the VariableType first. Also, it appears to only return a double matrix. I tried to add a mxClassID argument at the end but it didn't change anything. mxFastZeros appears in all of the MATLAB versions that I have (R2006b through R2011a PC WinXP 32-bit). Don't know about other versions.
@James: Clarification: Your UNINIT function works fine. I played with mxCreateUninitNumericMatrix before, but without success. Most likely I kicked off some bits from the memory by using bad pointers.

Sign in to comment.


Answer by Andrew Newell
on 8 May 2011

I don't find that. If my input is
clear
t=(1:1000)'; A=rand(length(t),200);
[n,m] = size(A);
%%1. Looping
tic
for j = 1 : m
res(:,j) = t .* A( :, j ); % You had an error in this line
end
toc
%%2. Repmat
tic
tmat = repmat( t, 1, m );
res = tmat .* A;
toc
%%3. diag
tic
res = diag( t ) * A;
toc
%%4. Indexing
tic
A = A( : );
t = t( :, ones( 1, m ) );
t = t( : );
res = A .* t;
res = reshape( res, n, m );
toc
The output is
Elapsed time is 0.300359 seconds.
Elapsed time is 0.003952 seconds.
Elapsed time is 0.035276 seconds.
Elapsed time is 0.004409 seconds.

  1 Comment

my research and the result
clear
a = zeros(5,100);
for jj = 1:100
t=rand(1000,1); A=rand(length(t),200);
[n,m] = size(A);
% 1. Looping
tic
for j = 1 : m
res(:,j) = t .* A( :, j );
end
a(1,jj)=toc;
% 2. Repmat
tic
tmat = repmat( t, 1, m );
res = tmat .* A;
a(2,jj)=toc;
% 3. diag
tic
res = diag( t ) * A;
a(3,jj)=toc;
% 4. Indexing
tic
A1 = A( : );
t1 = t( :, ones( 1, m ) );
t1 = t1( : );
res = A1 .* t1;
res = reshape( res, n, m );
a(4,jj)=toc;
% 5. bsxfun
tic
res = bsxfun(@times,t,A );
a(5,jj)=toc;
end
AA = mean(a,2)
AA =
0.0006
0.0023
0.0413
0.0031
0.0015

Sign in to comment.


Answer by Anders Munk-Nielsen on 23 Mar 2012

I'm having similar problems in that I have an NxM matrix, where N is extremely large (5 million) and M is fairly small (below 100) and I have a 1xM vector that I want to multiply onto all N rows. I'm trying to improve on an implementation that uses repmat which is incredibly slow and in my intuition this is due to the huge memory footprint.
My question: Why isn't bsxfun any faster than it is? I would expect it to be waaay faster but it's only somewhat faster. If I were to implement a mex function to do this (a multiplyVectorToMatrix function), what is the crucial feature to get built in? Multithreaded'ness?

  3 Comments

Btw, here is some code to run what I have in mind (By the way, M is K below):
clc; clear all;
N = 300000;
S = 1000;
K = 10;
x = rand(N,K);
B = rand(K,1);
eps = randn(N,1);
Y = x*B + eps;
Bhat = zeros(K,S);
C = randn(N,S);
%% repmat
tic;
px = (x'*x)\x';
Bhat = px*(repmat(Y,1,S)-C);
fprintf('repmat took %1.3g seconds.\n',toc);
%% for loop
tic;
px = (x'*x)\x';
for s=1:S;
%YmC = Y-C(:,s);
Bhat(:,s) = px*(Y-C(:,s));
end;
fprintf('for took %1.3g seconds.\n',toc);
%% parfor loop
tic;
px = (x'*x)\x';
parfor s=1:S;
Bhat(:,s) = px*(Y-C(:,s));
end;
fprintf('Parfor took %1.3g seconds.\n',toc);
%% bsxfun
tic;
px = (x'*x)\x';
Bhat = px*bsxfun(@minus,Y,C);
fprintf('Bsxfun took %1.3g seconds.\n',toc);
bsxfun(@times,yourNxM,yourMx1)
doc bsxfun %for more info.
Your setup for timign this isn't really fair. Look at Matt's framework for a more fair comparison.

Sign in to comment.