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Build array from descriptive data without a loop

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Gabriel Stanley
Gabriel Stanley el 17 de Feb. de 2023
Comentada: Walter Roberson el 17 de Feb. de 2023
I want to go from
Array1 = [10,3,3;1000,178,4];
to
Array2 = [10;13;16;1000;1178;1356;1534];
without using
Idx2 = 1;
Array2 = zeros(sum(Array1(:,3)),1);
for Idx1 = 1:size(Array1,1)
Array2(Idx2:Idx2+Array1(Idx1,3)-1) = [Array1(Idx1,1),Array1(Idx1,1)+[1:Array1(Idx1,3)-1].*Array1(Idx1,2)];
Idx2 = Idx2+Array1(Idx1,3)-1;
end
Help?

Respuesta aceptada

Walter Roberson
Walter Roberson el 17 de Feb. de 2023
Array1 = [10,3,3;1000,178,4];
Array2 = cell2mat(arrayfun(@(Idx) Array1(Idx,1) + (0:Array1(Idx,3)-1).'*Array1(Idx,2), (1:size(Array1,1)).','uniform', 0))
Array2 = 7×1
10 13 16 1000 1178 1356 1534
  2 comentarios
Gabriel Stanley
Gabriel Stanley el 17 de Feb. de 2023
Kinda disappointed this hasn't ended up being significantly faster in my use-case, but thank you for providing an answer. I never would've thought to use arrayfun (or cellfun & the like) because I've gotten the impression they're generally slower than a loop.
Walter Roberson
Walter Roberson el 17 de Feb. de 2023
You didn't ask for performance, you asked for not using a loop. For most operations (but not all, not if you know the right obscure forms), arrayfun and cellfun are slower than looping.
If you were looking for performance, then your existing code could be tweeked to take advantage of cumsum() instead of calculating the indices each time, and you could use the calculation I used instead of using [original, colon expression] list constructor.

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Más respuestas (1)

Kevin Holly
Kevin Holly el 17 de Feb. de 2023
Array1 = [10,3,3;1000,178,4];
Array2 = cumsum(Array1,2)
Array2 = 2×3
10 13 16 1000 1178 1182
  2 comentarios
Walter Roberson
Walter Roberson el 17 de Feb. de 2023
Tthe third column is the number of elements to generate, with the difference being the second column, and the starting point being the first column.
Kevin Holly
Kevin Holly el 17 de Feb. de 2023
ah, thanks for pointing that out.

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