How to select several intervals from a vector?

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Csaba
Csaba 2021-11-26
评论： Csaba 2024-4-26
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.

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采纳的回答

Kristoffer
Kristoffer 2023-10-10
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))
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显示 1更早的评论
Tony
Tony 2024-4-26
Ran in:
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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更多回答（1 个）

DGM
DGM 2021-11-26
编辑：DGM 2021-11-26
Ran in:
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
  1 个评论
Csaba
Csaba 2021-11-26
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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