Filtering the common rows between two matrices
7 views (last 30 days)
Show older comments
I have two matrices A and B. Matrix A consists of 3 columns and matrix B consists of 4 columns.
A = [1 2 3; 4 5 6; 7 8 9];
B = [1 2 3 90; 3 1 2 88; 4 5 6 17; 6 5 4 19; 7 8 9 12; 15 18 22 20];
And I want to filter the common rows between matrix A and B according to each row of A and store the common rows of matrix B in a cell. This cell consists of multiple arrays equal to the number of the rows of matrix A.
The expected result should be something like that:
matching_cell{1} = [1 2 3 90; 3 1 2 88];
matching_cell{2} = [4 5 6 17; 6 5 4 19];
matching_cell{3} = [7 8 9 12];
Note: I do not want to use neasted for loop with ismember as the matrices dimensions are large and it will take a lot of time to run.
Thanks.
3 Comments
Stephen23
on 1 Feb 2024
Moved: Dyuman Joshi
on 25 Feb 2024
Note that SPLITAPPLY will fail if any row of A does not exist in B (or produce a shorter output array):
A = [1 2 3; 0 0 0; 4 5 6];
B = [1 2 3 90; 3 1 2 88; 4 5 6 17; 6 5 4 19; 7 8 9 12; 15 18 22 20];
C = sort(B,2);
[idx1,idx2]=ismember(C(:,1:size(A,2)), A, 'rows');
out = splitapply(@(x) {x}, B(idx1,:), idx2(idx1))
Dyuman Joshi
on 25 Feb 2024
Moved: Dyuman Joshi
on 25 Feb 2024
Yes, Thank you for pointing it out, I am aware of the requirement of spliapply() as stated in the error message.
As of now, I can't seem think of a solution that would work for all cases, so I am moving my answer to a comment, and accepting your answer as it provides a robust solution.
Accepted Answer
Stephen23
on 1 Feb 2024
Edited: Stephen23
on 1 Feb 2024
A = [1,2,3; 4,5,6; 7,8,9]
B = [1,2,3,90; 3,1,2,88; 4,5,6,17; 6,5,4,19; 7,8,9,12; 15,18,22,20]
[X,Y] = ismember(sort(B(:,1:3),2),A,'rows');
F = @(n)B(n==Y,:);
C = arrayfun(F,1:size(A,1),'uni',0)
C{:}
This will produce a consistent 1*size(A,1) output cell array, even if there are rows that do not match.
0 Comments
More Answers (0)
See Also
Categories
Find more on Logical in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!