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

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Keon Walters on 23 Jun 2022
Commented: Keon Walters on 24 Jun 2022
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.

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
Keon Walters on 24 Jun 2022
Thank you so much Jan! Super helpful i apprieciate the feedback :D

### More Answers (1)

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 CommentsShow NoneHide None
Keon Walters on 23 Jun 2022
The issue is this only compares the first column to every other column instead of every column to every other column. For instance I want to get the difference of the element from column 1 row 1 to the element from column 2 row 1 and column 3 row 1 etc. for all of the columns then compare column 1 row 2 to the elment from column 2 row 2 etc. After all of that I want to compare column 2 row 1 to column 1 row 1 then column 2 row 1 then column 3 row 1 etc.
So in the code I wrote I did:
A.a-A.a(:,i);
Which gets the difference of the first column against every other column and every loop it gets the difference of the next column.
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?

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