Effacer les filtres
Effacer les filtres

Marix manipulation

2 vues (au cours des 30 derniers jours)
Charles
Charles le 5 Août 2011
How do I perform this operation? I have a data set with recurring values at a given co-ordinate. I need to sum values of the repeating coordinates. I initially used a for loop to solve the problem but it is too slow. Here is a simple example of the problem:
val = rand(1);
a = [1 1 1 val;
1 2 1 val;
1 1 1 val];
Since the 1st and 3rd rows have the same co-ordinates, I need to add their values, delete the entries and replace with the new summed value. I know it should be possible using sub2ind and a cumsum operation, but I haven't figure exactly how to combine them yet. If I haven't written an answer at the time you are reading this, it means I still haven't got the combination right. Thanks.
  1 commentaire
Pierre
Pierre le 5 Août 2011
>> _Since the 1st and 3rd rows have the same co-ordinates, I need to add their values, delete the entries and replace with the new summed value._
I wonder, if anybody can understand what you're trying to do there... I can't. :S
Please post your actual solution (with the for loop), this would allow us to see what the output's supposed to look like, would you?

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Daniel Shub
Daniel Shub le 5 Août 2011
My solution has a for loop.
[b, i, j] = unique(a(:, 1:(end-1)), 'rows');
c = b;
for ii = 1:length(i)
c(ii, 4) = sum(a(j==ii, 4));
end
My guess is that it can be eliminated (and potentially made faster) with a bsxfun or arrayfun. In order to optimize it, we would need to know the sizes and how often things repeat.
  3 commentaires
Charles
Charles le 5 Août 2011
ACCUMARRAY might be the better option as it doesn't involve a loop.
Daniel Shub
Daniel Shub le 5 Août 2011
While I am happy to have my answer accepted, my guess is that if you un-accept my answer (not even sure you can do that), once the people in the US wake up, you will get a bsxfun/arrayfun/accumarray solution.

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