Build array from descriptive data without a loop
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Gabriel Stanley
le 17 Fév 2023
Commenté : Walter Roberson
le 17 Fév 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?
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Walter Roberson
le 17 Fév 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))
2 commentaires
Walter Roberson
le 17 Fév 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.
Plus de réponses (1)
Kevin Holly
le 17 Fév 2023
Array1 = [10,3,3;1000,178,4];
Array2 = cumsum(Array1,2)
2 commentaires
Walter Roberson
le 17 Fév 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.
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