Extract subvectors from a vector without using loops

Hi, i'm trying to speed up my code and reduce loops in it. My problem is to extract some parts of a vector (original_vector) in new smaller vectors, or in a matrix where every columns is each one of those extracted vectors. So in a variable i have the original vector, and in an other one a "N x 2" matrix where in the first column is stored the start index of the "n-th" subvector to extract, and in the second one the end index of it. I need something faster to substitute the loop:

for k=1:(number_of_extractions) extracted(:,k)=original_vector(index(k,1):index(k,2)); end

P.S. every extracted vector as the same elements

Thanks!

[Walter Roberson]

You refer to original_vector in your description but your code has two layers original(vector(range) ) ?

[Niccoli Francesco]

sorry, wrong digit! The code was this:

for k=1:(number_of_extractions) extracted(:,k)=original_vector(index(k,1):index(k,2)); end

Answer 1

[Walter Roberson]

 d=index(2,1)-index(1,1);
 extracted = original_vector(bsxfun(@plus, (0:d).', index(:, 1).')) ;

[Niccoli Francesco]

Really thanks! I have a problem again, beacause with this kind of formulation it saves me sub_vectors of 1000 elements, but the sub_vectors dimension (definited by index(n,2)-index(n,1)) is different and equal to 90000

Answer 2

[Guillaume]

You can use mat2cell for this but I don't know if it's going to be any faster:

original_vector = 1:100; %demo data
index = [2 15; 30 10; 40 20; 79 5]; %demo data
divisions = [index(:, 1), sum(index, 2)]'; 
divisions = diff([1; divisions(:); numel(original_vector)+1]); %calculate length of each block
subvectors = mat2cell(original_vector, 1, divisions); %this includes spacer blocks (of possible length 0 between each wanted block
subvectors = subvectors(2:2:end);  %get rid of 'spacer blocks'