adding counts of ordered pairs

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Barbara Margolius
Barbara Margolius el 6 de Sept. de 2022
Comentada: Bruno Luong el 6 de Sept. de 2022
I have a sequence of by 3 arrays, say , , , that are generated within a loop. That is, the number of rows of each array varies, but each array has three columns. The first two columns represent ordered pairs. The third column is the count of those ordered pairs. I want to "add" these together so that I get an array that accumulates the counts of these ordered pairs into a new array that is with the same structure. For example, if is
A1=[1 4 3;
3 5 1;
12 4 7;
13 5 2;
14 1 1];
and is
A1=[1 5 1;
13 5 1;
13 7 3;
14 1 5];
then the cumulative matrix should be
A=[1 4 3;
1 5 1;
3 5 1;
12 4 7;
13 5 3;
13 7 3;
14 1 6];
The arrays being "summed" in this way have hundreds of entries and are themselves summaries of arrays with thousands of entries, so efficiency matters.

Respuesta aceptada

Bjorn Gustavsson
Bjorn Gustavsson el 6 de Sept. de 2022
One way to go about this is to use sparse:
A1 = [1 4 3;
3 5 1;
12 4 7;
13 5 2;
14 1 1];
A2 = [1 5 1;
13 5 1;
13 7 3;
14 1 5];
A_all = sparse([A1(:,1);A2(:,1)],[A1(:,2);A2(:,2)],[A1(:,3);A2(:,3)]);
[I1,I2,Val] = find(A_all);
[~,idx1] = sort(I1);
disp([I1(idx1),I2(idx1),Val(idx1)])
HTH
  2 comentarios
Barbara Margolius
Barbara Margolius el 6 de Sept. de 2022
Editada: Barbara Margolius el 6 de Sept. de 2022
I am going to time both your answer and Bruno's to see which works best. I accepted yours because I suspect with my data it will be faster.
Bruno Luong
Bruno Luong el 6 de Sept. de 2022
This will save you a sort
A1 = [1 4 3;
3 5 1;
12 4 7;
13 5 2;
14 1 1];
A2 = [1 5 1;
13 5 1;
13 7 3;
14 1 5];
A_all = sparse([A1(:,2);A2(:,2)],[A1(:,1);A2(:,1)],[A1(:,3);A2(:,3)]);
[I2,I1,Val] = find(A_all);
A = [I1,I2,Val]
A = 7×3
1 4 3 1 5 1 3 5 1 12 4 7 13 5 3 13 7 3 14 1 6

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Más respuestas (1)

Bruno Luong
Bruno Luong el 6 de Sept. de 2022
Editada: Bruno Luong el 6 de Sept. de 2022
A1=[1 4 3;
3 5 1;
12 4 7;
13 5 2;
14 1 1];
A2=[1 5 1;
13 5 1;
13 7 3;
14 1 5];
A12=[A1; A2];
[A12u,~,J]=unique(A12(:,1:2),'rows','stable');
A=[A12u,accumarray(J,A12(:,3))]
A = 7×3
1 4 3 3 5 1 12 4 7 13 5 3 14 1 6 1 5 1 13 7 3
  7 comentarios
Bjorn Gustavsson
Bjorn Gustavsson el 6 de Sept. de 2022
@Bruno Luong: The amount of subconsious/implicit assumptions I make when writing QD-solutions is a bit frightening.
Bruno Luong
Bruno Luong el 6 de Sept. de 2022
That's called "intuition".

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