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.

 채택된 답변

Bjorn Gustavsson
Bjorn Gustavsson 2022년 9월 6일

0 개 추천

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

댓글 수: 2

Barbara Margolius
Barbara Margolius 2022년 9월 6일
편집: Barbara Margolius 2022년 9월 6일
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.
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

댓글을 달려면 로그인하십시오.

추가 답변 (1개)

Bruno Luong
Bruno Luong 2022년 9월 6일
편집: Bruno Luong 2022년 9월 6일

1 개 추천

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

댓글 수: 7

Barbara Margolius
Barbara Margolius 2022년 9월 6일
편집: Barbara Margolius 2022년 9월 6일
Thanks! Bjorn's answer also works. I am guessing his will be faster with my data, so I accepted it.
Bruno Luong
Bruno Luong 2022년 9월 6일
편집: Bruno Luong 2022년 9월 6일
I hope you don't have such case where Bjorn's method fails
A1=[1 1 1];
A2=[1 1 -1];
or pairs are not integer >= 1.
Barbara Margolius
Barbara Margolius 2022년 9월 6일
편집: Barbara Margolius 2022년 9월 6일
pairs are all integers (the result of floor function plus 1); third column is countdata, so should be ok.
Bruno Luong
Bruno Luong 2022년 9월 6일
floor can returns 0 no?
Barbara Margolius
Barbara Margolius 2022년 9월 6일
yup, see edit.
Bjorn Gustavsson
Bjorn Gustavsson 2022년 9월 6일
@Bruno Luong: The amount of subconsious/implicit assumptions I make when writing QD-solutions is a bit frightening.
Bruno Luong
Bruno Luong 2022년 9월 6일
That's called "intuition".

댓글을 달려면 로그인하십시오.

카테고리

도움말 센터File Exchange에서 Matrices and Arrays에 대해 자세히 알아보기

제품

릴리스

R2022a

Community Treasure Hunt

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

Start Hunting!

Translated by