How can I speedup this for loop and ismember?

Due to the large size of the arrays, the following suffers from extremly slow execution. Is it possible to speedup this code. Vout need to be modified by Vin according to specific locations pointed to by indx1 and indx2. In order to retrieve back the values of Vin from Vout, modification need to be down sequentialy starting from 1 to L Where some of these indices are distributed randomly between indx1 and indx2.

e=1; a=1; b=1; N=4;
L=262144;  K=786432;
Vin=rand(1,L);
Vout=rand(1,K); 
Q=randperm(K);
for r=Q,
  if ismember(e,indx1)
    Vout(r)=Vin(indx1(a));
    a=a+1;
    e=e+1;
  elseif ismember(e,indx2) 
    Vout(r)=Vin(indx2(b));
    b=b+1;
    e=e+N;
  else
    Vout(r)=1;
    e=e+N;
  end
  if e>L
    break;
  end
end

[Paul Hoffrichter]

What are sizes of indx1 and indx2 and how defined?

If this script were a function, what outputs do you require?

[Omar Ali Muhammed]

Dear, size(indx1)+size(indx2) Less than K. They contain indices of certain location inside the vector Vin. indx1 and indx2 are derived based on certain conditions on the vector elements Vin. Some values of Vout is replaced by Vin. Vout is the output of this function.

[Jan]

Please post some meaningful input data. I cannot guess, what indx1 and indx2 are.

[Omar Ali Muhammed]

Dear First of all thank you very much. indx1 and indx2 contains integer values in the range 1 to possibly 1000000. It depends upon the size of Vin.

Regards

[Jan]

Are indx1 and indx2 pairwise distinct or sorted? Are they unique?

Without meaningful input data, an optimization of the code is not possible. Perhaps a suggestion works, when the guesses are correct, but this is no trustworthy way of programming.

[Omar Ali Muhammed]

Yes dear indx1 and indx2 are unique distinct and sorted vectors of indices.

[Paul Hoffrichter]

L=262144; K=786432; % Notice that K > L

>> size(indx1)+size(indx2) Less than K

If length( indx1 ) = length( indx1 ) = K/2 - 1, then your restriction on size is satisfied, But since K > L, then the indx1 and indx2 would have duplicate values. But you wrote:

"indx1 and indx2 are unique distinct and sorted vectors of indices."

So, maybe my understanding of the problem is wrong. Please clarify.

[Omar Ali Muhammed]

Dear elements of indx1 and indx2 are in the range of 1 to L.

Regards

[Paul Hoffrichter]

But in your example, K/L = 3. As I already mentioned, if, for example, length( indx1 ) = length( indx1 ) = K/2 - 1, this satisifies your requirement that size(indx1)+size(indx2) Less than K. In this case, the length of indx1 is larger than L. If all the values in indx1 is in the range of 1 to L, then there must be duplicates. But you said that indx1 has unique distinct and sorted vectors of indices.

The model you are presenting to us does not appear to be consistent for these reasons.

[Omar Ali Muhammed]

Dear I clearly stated that elements of indx1 and indx2 are in the range of 1 to L. So all indices of indx1 and indx2 not duplicated sorted and subset of K. e.g let us say K=100, then Vout has 100 elements, L=20, Vin has ten elements so indx1 may take values like 4, 7, 55, 89 and indx2 may take values like 1, 34, 67, 90, 97. Regards

Answer 1

[Jan]

Without having working inputs, we cannot run the function. Then it is not possible to improve the code reliably. Except for one detail: "e" is a scalar. Then ismember is an overkill. Faster:

if any(e == indx1)    

If the elements of indx1 and indx2 are "small" integers, a lookup table can be faster:

LUT = zeros(1, L, 'uint8');
LUT(indx1) = 1;
LUT(indx2) = 2;   % do indx1 and indx2 share elements?! 

for r = Q
  switch LUT(e)
      case 0
          Vout(r) = 1;
          e       = e + N;
      case 1
          Vout(r) = Vin(indx1(a));
          a = a + 1;
          e = e + 1;
      case 2
          Vout(r) = Vin(indx2(b));
          b = b + 1;
          e = e + N;
  end
end

As said already, I cannot check if this is faster. Are indx1 and indx2 sorted? Are they unique integers?

I assume there is a vectorized approach also.