Cycle of operations on vector sections

I'm trying to make a cycle that would process a section of the vector and return this section to me changed

x=rand(1,2302)
y=rand(1,2302)
step=100;
for i=1:size(x,2)
    i1=y(i:i+step);
    i2=x(i:i+step);
    a(i)=myfun(i1,i2);
end
%% for example my function
function [m,s] = myfun(x,y)
    n = length(x);
    m = x-0.26;
    s= y-0.77
end

thank you in advance

[Davide Masiello]

Please share the code for myfun.

[Lev Mihailov]

@Davide Masiello I added an example function

Answer 1

[Davide Masiello]

The problem is myfun spits out two vectors of size 1 x step+1.

You could assign this vectors to the rows of a new matrix (see a and b in the example below) .

x = rand(1,2302);
y = rand(1,2302);
step = 100;

for i = 1:size(x,2)-step
    i1 = y(i:i+step);
    i2 = x(i:i+step);
    [a(i,:),b(i,:)] = myfun(i1,i2);
end


function [m,s] = myfun(x,y)
n = length(x); % <--- not needed
m = x-0.26;
s= y-0.77;
end

[Lev Mihailov]

@Davide Masiello Thank you, but I have a question, could you help me, if in my function the values ​​are reversed (1:size(x) to 1:size(x),1) how can I get back a vector of length 1:size(x)

%%% somewhere in my function
s=s';

[Davide Masiello]

In that case you can replace

[a(i,:),b(i,:)]

with

[a(:,i),b(:,i)]

so each output of myfun is stored as a column in the matrices a and b.

[Lev Mihailov]

@Davide Masiello just now I noticed this, and my function returns me the same vector (but changed), but why is it a matrix and not a vector?