How to manipulate in matrix

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Zhan
Zhan el 22 de Mayo de 2017
Respondida: Andrei Bobrov el 25 de Mayo de 2017
Assume matrix i:
i = [
10010 8
10010 5
10010 2
10065 8
10065 6
10099 4
10099 2
10099 9
10099 4
10099 1
];
The first column is ID. There are three different IDs and repeated for 10 rows. I want to sum the second column corresponding to each ID as follows:
j = [
10010 15
10010 15
10010 15
10065 14
10065 14
10099 20
10099 20
10099 20
10099 20
10099 20
];
  3 comentarios
Zhan
Zhan el 22 de Mayo de 2017
@the cyclist Thank you for your comment. How can I accept an answer?
Guillaume
Guillaume el 22 de Mayo de 2017
How can I accept an answer?
By clicking on the big blue button that says "Accept this answer" above the answer you want to accept.

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Respuestas (3)

Image Analyst
Image Analyst el 22 de Mayo de 2017
Sounds like it might be homework. Here's a start. See if you can finish it.
out = accumarray(i(:, 1), i(:, 2));
out(out==0) = []
Returns [15;14;20]
  1 comentario
Zhan
Zhan el 22 de Mayo de 2017
out = out(i(:,1),:)
% works if IDs are [1,1,1,2,2,3,3,3,3,3]. Any tips for the oroginal ID?

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Akira Agata
Akira Agata el 25 de Mayo de 2017
How about using splitapply function.
A = [
10010 8
10010 5
10010 2
10065 8
10065 6
10099 4
10099 2
10099 9
10099 4
10099 1
];
[G, ID] = findgroups(A(:,1));
B = [ID, splitapply(@sum,A(:,2),G)];
Now, the output matrix B is as follows:
B =
10010 15
10065 14
10099 20

Andrei Bobrov
Andrei Bobrov el 25 de Mayo de 2017
[~,~,c] = unique(ii(:,1));
D = accumarray(c,ii(:,2));
out = [ii(:,1), D(c)];

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