# How vectorize this operation

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Cybernetics on 7 Oct 2019
Commented: Cybernetics on 8 Oct 2019
I habe two vectors and with . I am implemented the following code
z = rand(length(x),1);% Just some fake data to define the size of z. could also be z=zeros(size(x))
for i=m+1:n+1
z(i)=x(i-m:i-1)'*y;
end
Knowing that n get have over a few million elements and m is less than 200, the computations rapidely become slow. How can I vectorize this operation to optimize it?

the cyclist on 7 Oct 2019
Is there a typo in the code you posted? In the first iteration of the loop, i==m+1, therefore
x(i-m:i)'
is
x(1:m+1)'
This vector is one element longer than y, one cannot do the matrix operation inside the for loop.
(Or maybe I messed something up?)
Cybernetics on 7 Oct 2019
Sorry there was a typo in the code. I just changed it
Rik on 7 Oct 2019
This is very close to being a convolution. I can't find out why it isn't, but it is fairly close, as you can see with the example below (the goal would be to make sure the two cols in z_mat are equal).
clearvars,clc
n=20;m=4;
x=(1:n)';y=(1:m)';
z=zeros(n+1,1);
for k=(m+1):(n+1)
z(k)=x((k-m):(k-1))'*y;
end
tmp=conv(x,y,'same');
z_conv=zeros(size(z));
z_conv((m+1):(n+1))=tmp((m-1):(n-1));
z_mat=[z,z_conv];
disp(z_mat)

Bruno Luong on 8 Oct 2019
n=10
m=3
x=rand(n,1)
y=rand(m,1)
z = nan(length(x),1);% Just some fake data to define the size of z. could also be z=zeros(size(x))
for i=m+1:n+1
z(i)=x(i-m:i-1)'*y;
end
z
% My method
z = [nan(m,1); conv(x,flip(y),'valid')]

Rik on 8 Oct 2019
I knew it could be done with a convolution. Can you explain why you need to flip the y to get it to match the loop version?
Bruno Luong on 8 Oct 2019
Well that comes straighforward from the definition of CONV, it performs a slighding sum with one of the array that is straight and another is flipped.
So if one doesn't want to flip dusing the sum, one have to flip the array before calling CONV.
Cybernetics on 8 Oct 2019
Thanks for the the elegant solution.

Daniel M on 7 Oct 2019
Edited: Daniel M on 7 Oct 2019
If you have the signal toolbox, you can use the buffer() command to an array of the x values that you require, then do the matrix multiplication in one shot.
x = 1:50;
m = 6;
L = 15;
n = 48;
z = buffer(x(L+1-m:n),m+1,m,'nodelay'); % size(z) = [7,33]
y = 1:7; % size(y) = [1,7]
vals = y*z; % size = [1,33]
% if no signal toolbox, try
% ind = (L+1:n)-m + (0:m)';
% z = x(ind);
I'm not sure if this will be faster. It could require a lot of memory. But it is vectorized, so test it out and see.

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Cybernetics on 7 Oct 2019
% if no signal toolbox, try
% ind = (L+1:n)-m + (0:m)';
% z = x(ind);
This also induced an error due to the big vector x with 20 million elements
Daniel M on 7 Oct 2019
So do it in chunks. I don't know your computer specifications. It works on my computer fine, and takes 34 seconds to make an array that large. You say the code is slow, but not how slow. You say you want it faster but don't specify what will satisfy your goals.
Are you just impatient? Or is there a strict requirement to compute in a certain time?
I suggest either waiting for the computations to finish, buy a better computer, or reevaluate your code and maybe you don't need to do this calculation in the first place.
Cybernetics on 8 Oct 2019
My computer specifications are: intel core i7 7th generation and 16GB of RAM. The computation in a loop workds fine. I am just wondering, if there a way do avoind the loop, have a more elegent implementation and thus more efficient.