Build array from descriptive data without a loop
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Gabriel Stanley
il 17 Feb 2023
Commentato: Walter Roberson
il 17 Feb 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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Risposta accettata
Walter Roberson
il 17 Feb 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 Commenti
Walter Roberson
il 17 Feb 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.
Più risposte (1)
Kevin Holly
il 17 Feb 2023
Array1 = [10,3,3;1000,178,4];
Array2 = cumsum(Array1,2)
2 Commenti
Walter Roberson
il 17 Feb 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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