How to select several intervals from a vector?

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Csaba
Csaba 2021 年 11 月 26 日
コメント済み: Csaba 2021 年 11 月 26 日
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.

回答 (1 件)

DGM
DGM 2021 年 11 月 26 日
編集済み: DGM 2021 年 11 月 26 日
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 件のコメント
Csaba
Csaba 2021 年 11 月 26 日
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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