Hello, I have an array of values in "In", I would like to extract the samples in "In" between the indices specified in "SIdx" and "EIdx". SIdx and EIdx are also arrays.

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Krishna Ghanakota
评论： Voss 2024-3-9
I would like to acheive the functionality without using a for loop. Currently the code is written as
Working Code :
In = 1:100;
SIdx = [1 9 33 76];
EIdx = [5 13 42 83];
Out = {};
for i = 1:length(SIdx)
Out = [Out; {In(SIdx(i):EIdx(i))}];
end
Out =
4×1 cell array
{[ 1 2 3 4 5]}
{[ 9 10 11 12 13]}
{[33 34 35 36 37 38 39 40 41 42]}
{[ 76 77 78 79 80 81 82 83]}
Is there a way to acheive the same functionality without using a for loop.

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采纳的回答

Voss
Voss 2024-3-9
编辑：Voss 2024-3-9
Ran in:
In = 1:100;
SIdx = [1 9 33 76];
EIdx = [5 13 42 83];
siz = zeros(1,2*numel(SIdx)-1);
siz(1:2:end) = EIdx-SIdx+1;
siz(2:2:end) = SIdx(2:end)-EIdx(1:end-1)-1;
Out = mat2cell(In(SIdx(1):EIdx(end)),1,siz).';
Out(2:2:end) = [];
disp(Out)
{[ 1 2 3 4 5]} {[ 9 10 11 12 13]} {[33 34 35 36 37 38 39 40 41 42]} {[ 76 77 78 79 80 81 82 83]}

更多回答（1 个）

Walter Roberson
Walter Roberson 2024-3-9
Ran in:
In = 1:100;
SIdx = [1 9 33 76];
EIdx = [5 13 42 83];
Out = arrayfun(@(S,E) In(S:E), SIdx, EIdx, 'uniform', 0).'
Out = 4×1 cell array
{[ 1 2 3 4 5]} {[ 9 10 11 12 13]} {[33 34 35 36 37 38 39 40 41 42]} {[ 76 77 78 79 80 81 82 83]}
  2 个评论
Krishna Ghanakota
Krishna Ghanakota 2024-3-9
arrayfun(func,A) applies the function func to the elements of A, one element at a time.
The size of "SIdx" can grow very large and I would like to avoid looping.
Is there a way in which "In" can be vector indexed with SIdx and EIdx.
Krishna Ghanakota
Krishna Ghanakota 2024-3-9
The In, SIdx, EIdx values given in the example code are just for illustrating the idea. These are in actual very large arrays and looping takes quite some time.

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