adding counts of ordered pairs
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Barbara Margolius
am 6 Sep. 2022
Kommentiert: Bruno Luong
am 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
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1118115/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1118120/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1118125/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1118130/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1118135/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1118140/image.png)
A1=[1 4 3;
3 5 1;
12 4 7;
13 5 2;
14 1 1];
and
is
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1118145/image.png)
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.
0 Kommentare
Akzeptierte Antwort
Bjorn Gustavsson
am 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 Kommentare
Bruno Luong
am 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]
Weitere Antworten (1)
Bruno Luong
am 6 Sep. 2022
Bearbeitet: Bruno Luong
am 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))]
7 Kommentare
Bjorn Gustavsson
am 6 Sep. 2022
@Bruno Luong: The amount of subconsious/implicit assumptions I make when writing QD-solutions is a bit frightening.
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