How to select several intervals from a vector?

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Csaba on 26 Nov 2021
Commented: Csaba on 26 Nov 2021
I have a vector (Y). I want to select a region from this vector. If this is a single region it is easy
What if I have several regions (the number of regions is not fixed)?
So I want to make the vector
[Y(i_from_1:i_to_1) ,Y(i_from_2:i_to_2), ........ ,Y(i_from_n:i_to_n)]
where n is not fixed.
Is there a fast and simple way? i_from and i_to values are in a n*2 matrix.
I can of course do a for cycle, but looking for a simpler method.

Answers (1)

DGM on 26 Nov 2021
Edited: DGM on 26 Nov 2021
Idk. Here's three ways. They all use loops. Is there something more elegant? Prrrrobably. Is it faster? Probably depends. I'm sure there's more to be said about the topic. I'll leave that for others.
A = rand(1,1000);
bex = randi([1 1000],50,2);
timeit(@() loopappending(A,bex))
ans = 1.9935e-04
timeit(@() loopindexing(A,bex))
ans = 1.5171e-04
timeit(@() loopcell(A,bex))
ans = 1.1135e-04
% simply append subvectors
function B = loopappending(A,bex)
B = [];
for b = 1:size(bex,1)
B = [B A(bex(b,1):sign(diff(bex(b,:))):bex(b,2))];
% preallocate and use direct indexing
function B = loopindexing(A,bex)
B = zeros(1,sum(abs(diff(bex,1,2))+1)); % preallocate
endpoints = [0; cumsum(abs(diff(bex,1,2))+1)];
for b = 1:size(bex,1)
B(endpoints(b)+1:endpoints(b+1)) = A(bex(b,1):sign(diff(bex(b,:))):bex(b,2));
% throw output into cell array and then rearrange
function B = loopcell(A,bex)
B = cell(1,size(bex,1));
for b = 1:size(bex,1)
B{b} = A(bex(b,1):sign(diff(bex(b,:))):bex(b,2));
B = horzcat(B{:});
  1 Comment
Csaba on 26 Nov 2021
Yes, cycles are an obvious solution. I am using that recently.
I am looking for a more elegant and faster method.

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