How to select several intervals from a vector?

I have a vector (Y). I want to select a region from this vector. If this is a single region it is easy

X=Y(i_from:i_to);

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.

Answer 1

[Kristoffer]

You can make a vector containing the values specified by the intervals in Y using:

cell2mat(arrayfun(@(A,B) A:B, Y(:,1)', Y(:,2)', 'uniform', 0))

[Csaba]

This is not what I want.

Let say I have

Y=1:2:30;
intervals=[1,3;...
           11,15];

Then the desired output would be:

[1,3,5,21,23,25,27,29]

[Tony]

If you change Y to intervals in @Kristoffer's expression, we get your desired output

Y=1:2:30;
intervals=[1,3;...
           11,15];
Y(cell2mat(arrayfun(@(A,B) A:B, intervals(:,1)', intervals(:,2)', 'uniform', 0)))

ans = 1x8

1 3 5 21 23 25 27 29

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[Csaba]

OK, this works. Thank you!

Answer 2

[DGM]

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))];
    end
end

% 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));
    end
end

% 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));
    end
    B = horzcat(B{:});
end

[Csaba]

Yes, cycles are an obvious solution. I am using that recently.

I am looking for a more elegant and faster method.