Vectorize comparing a column vector in a matrix to all other vectors in the matrix.

6 views (last 30 days)
Hello,
I am trying to take the difference between a column in a matrix to the rest of the matrix but I am trying to do it without using a loop is this possible? Additionally, would doing it without the loop be more efficient, or would you have to spend a bunch of time in overhead to set it up so that you can do it vectorized?
Example Code:
% Variable Definitions
A.a = rand(10,11);
A.b = rand(10,11);
A.c = rand(10,11);
% Memory Pre-allocation
d = zeros(10,11);
e = zeros(1,11);
% Loop
for i = 1:size(A.a,2)
a = A.a - A.a(:,i);
b = A.b - A.b(:,i);
c = A.c - A.c(:,i);
d = sqrt(a.^2 + b.^2 + c.^2);
e(i) = max(max(d));
end
% The line below shows that knowing the column
% of that the maximum values are in is important.
[~, index] = min(e);
%% Alternatively it could look something like the below
for i = 1:size(A.a,2)
a(:,:,i) = A.a - A.a(:,i);
b(:,:,i) = A.b - A.b(:,i);
c(:,:,i) = A.c - A.c(:,i);
end
d = sqrt(a.^2 + b.^2 +c.^2);
e = max(max(d));
% But I want to do it without having to do the loop? Is this possible.

Accepted Answer

Jan
Jan on 23 Jun 2022
Why do you want to vectorize this code? I assume, this will cause a slow down, because this creates large temporary array, which are not needed.
If you want to speed it up, other methods are more useful
  1. Accessing a field of a struct costs time. Create a temporary variable instead. This does not duplicate the data, but creates a shared data copy.
  2. SQRT is expensive. You want to find the minimum value of an array. Then search the minimum at first and calculate only its SQRT.
  3. Pre-allocating d is useless and therefore a waste of time: d is not used later, but overwritten.
% Memory Pre-allocation
e = zeros(1,11);
Aa = A.a; % Abbreviations
Ab = A.b;
Ac = A.c;
% Loop
for i = 1:size(Aa,2)
a = Aa - Aa(:,i);
b = Ab - Ab(:,i);
c = Ac - Ac(:,i);
d = (a.^2 + b.^2 + c.^2);
e(i) = sqrt(max(max(d)));
end

More Answers (1)

Michael
Michael on 23 Jun 2022
You can use repmat.m to make a matrix of the same size.
a = A.a - repmat(A.a(:,1),1,size(A.a,2))
b = A.b - repmat(A.b(:,1),1,size(A.b,2))
c = A.c - repmat(A.c(:,1),1,size(A.c,2))
  2 Comments
Keon Walters
Keon Walters on 23 Jun 2022
I edit the question to give an alternative example but it still has the loop which I think is less efficent but am not sure?

Sign in to comment.

Categories

Find more on General Applications in Help Center and File Exchange

Products


Release

R2019b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by