adding counts of ordered pairs

2 Ansichten (letzte 30 Tage)
Barbara Margolius
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
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.

Akzeptierte Antwort

Bjorn Gustavsson
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
Barbara Margolius
Barbara Margolius am 6 Sep. 2022
Bearbeitet: Barbara Margolius am 6 Sep. 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 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]
A = 7×3
1 4 3 1 5 1 3 5 1 12 4 7 13 5 3 13 7 3 14 1 6

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Bruno Luong
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))]
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 Kommentare
Bjorn Gustavsson
Bjorn Gustavsson am 6 Sep. 2022
@Bruno Luong: The amount of subconsious/implicit assumptions I make when writing QD-solutions is a bit frightening.
Bruno Luong
Bruno Luong am 6 Sep. 2022
That's called "intuition".

Melden Sie sich an, um zu kommentieren.

Produkte


Version

R2022a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by