- 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.
- SQRT is expensive. You want to find the minimum value of an array. Then search the minimum at first and calculate only its SQRT.
- Pre-allocating d is useless and therefore a waste of time: d is not used later, but overwritten.

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

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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.

##### 0 Comments

### Accepted Answer

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

% 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
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))

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