What is the best way to define vectors based on a condition?

Hello,

So I do have a function, which has the (same sized) vectors x,y,z as input parameters and a same sized k as an output parameter.

Now I want to calculate k with a simple for loop. As I come from other languages I could get it to run. However it is pretty slow and I know MATLAB has a lot of tricks for these kind of stuff.

So the minimal running example code is:

A=2.*p+1;
Bm=p+1;
Bs=p+2;
x_i=(x+y)*0.01;
x_s=x-x_i;
[r,c] = size(T);
k = zeros(r,c);
for i=1:r
    if x(i)>=x_i(i)
        k(i)=A(i)+Bm(i).*x_s(i);
    elseif x(i)<x_i(i)
        k(i) = A(i)+Bs(i).*x_s(i);
    end
    if k(i)==0
        k(i) = 0.001;
    end
end

Does MATLAB have a more efficient way to implement this?

Thanks in advance

Answer 1

[KSSV]

A=2.*p+1;
Bm=p+1;
Bs=p+2;
x_i=(x+y)*0.01;
x_s=x-x_i;
[r,c] = size(T);
k = 0.001*ones(r,1);
% condition 1
idx1 = x>=x_i ; 
k(idx1)=A(idx1)+Bm(idx1).*x_s(idx1);
% condition 2 
idx2 = x<x_i ;
k(idx2) = A(idx2)+Bs(idx2).*x_s(idx2);

[Christian Berwanger]

Thanks. Just one question:

In the conditions:

idx1 = x(i)>=x_i(i) ; 

Do I need the (i) as an index? How does matlab know what the i should be?

[KSSV]

Ohhh...no...index i is not required....I copied the lines and forgot to remove those. Edited the answer.

[Christian Berwanger]

So. Do I understand this correctly?

It is possible that k still will be zero (As a result of adding 0 during one consition)

Fixing this would be

k(k==0)=0.001;

Is that correct?

[KSSV]

k is defined as

k = 0.001*ones(r,1);

So, already where elements are not changed, it remains 0.001. So this condition is already taken care.

[Christian Berwanger]

Yes. I know. But if, for some reason (which rarely might occur (As A,Bm and Bs are defined in a different way. However it would too complex for just a minimal working example), within the condition, k will be 0, I want to catch it. I just want to be sure as I later divide by k.

[KSSV]

Okay, then you may go ahead with that.