Effacer les filtres
Effacer les filtres

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.

1 vue (au cours des 30 derniers jours)
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.

Réponse acceptée

Voss
Voss le 9 Mar 2024
Modifié(e) : Voss le 9 Mar 2024
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]}

Plus de réponses (1)

Walter Roberson
Walter Roberson le 9 Mar 2024
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 commentaires
Krishna Ghanakota
Krishna Ghanakota le 9 Mar 2024
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 le 9 Mar 2024
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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