# adding counts of ordered pairs

2 views (last 30 days)
Barbara Margolius on 6 Sep 2022
Commented: Bruno Luong on 6 Sep 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.

Bjorn Gustavsson on 6 Sep 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 CommentsShowHide 1 older comment
Bruno Luong on 6 Sep 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

### More Answers (1)

Bruno Luong on 6 Sep 2022
Edited: Bruno Luong on 6 Sep 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
Bruno Luong on 6 Sep 2022
That's called "intuition".

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