How can I make iteration of 2 for loops of 2 variables of same length, so that output will be the same length of input variables, not their multiplication.

Hello here I am attaching the code I am running. If I iterate only one variable say LL keeping another variable mm constant value, then the outpot ,that is S_s runs well with the same length of input vector LL. But if I iterate both input vector LL and mm then the output gets LL*mm dimension. actually LL and mm are related, that is for every LL value there is a mm value. I just want when the code is running for one particular LL value, then just pick that corresponding mm value. Then go for another LL and mm values and so on, and store the output S_s for every set of LL mm pair. Thanks.

interploation_Points = 100;

a= [2, 4, 10, 133, 211, 314, 1000,1.3*10^3];

L= linspace(min(a),max(a),interploation_Points);

M= [-0.617030303030302 + 0.396013468013468i,-75.9774643403734 + 9.26989211985172i,-151.337898377717 + 18.1437707716900i];

m = linspace(min(M),max(M),interploation_Points); 

i=1;
% j=1;

% for La = min(L):interploation_Points: max(L)
for LL = 1:length(L)

%      mm = -0.617030303030302 + 0.396013468013468i;
  for mm= 1:length(m)
k = (2* pi ./LL);

x = (5 .* k);

z =10.* pi.*(mm ./LL);

   n_max = round(x + 4.05 .*(x) .^(1/3) + 2);
   n = 1:n_max; 
  s_x =  besselj(n+0.5,x);
  s_z =  besselj(n+0.5,z);
  a = (s_z - mm .* s_x )./(s_x - mm .* s_z);
  S_s(i) = 2*pi./ k .^2 .* (sum ((2*n+1) .* (abs(a) .* abs(a))));

LL = LL + 1;
mm=mm+1;
i = i+1;
% j = j+1;
 end
 end
plot(LL, S_s);

[az]

I am attaching the mat file as the code seems not looking good in the question. Thanks

Answer 1

[Walter Roberson]

for K = 1 : length(LL)
  this_LL = LL(K);
  this_mm = mm(K);
  ....
  output(K) = result
end

[az]

Thank you Mr. Roberson for quick response. Is this for K = 1 : length(LL) will be length of L instead of LL, ie K = 1 : length(L)

[az]

As it is saying LL is undefined function, I am little confused as should I put only for loop ( omitting my for loops) this way that you suggested, or insert your command inside my for loop?

[Walter Roberson]

"If I iterate only one variable say LL"

LL is the variable you are iterating over the content of.

In terms of your code,

for K = 1 : length(L)
  this_L = L(K);
  this_m = m(K);
  ....
  output(K) = result
end

[az]

It works. So there is no need to use K= 1 before starting the for loop right? And outpot(K) with only one suffix 'K' will store the iterate L m pair values right? Thanks a lot.

[Walter Roberson]

There is no need to use K=1 before the for loop. Both variables will be stepped at the same time.

[az]

Thank you Mr. Roberson.