Filtering the common rows between two matrices

9 Ansichten (letzte 30 Tage)
Bassem Mokhtar
Bassem Mokhtar am 1 Feb. 2024
Verschoben: Dyuman Joshi am 25 Feb. 2024
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 Kommentare
Stephen23
Stephen23 am 1 Feb. 2024
Verschoben: Dyuman Joshi am 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))
Error using splitapply
For N groups, every integer between 1 and N must occur at least once in the vector of group numbers.
Dyuman Joshi
Dyuman Joshi am 25 Feb. 2024
Verschoben: Dyuman Joshi am 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.

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Stephen23
Stephen23 am 1 Feb. 2024
Bearbeitet: Stephen23 am 1 Feb. 2024
A = [1,2,3; 4,5,6; 7,8,9]
A = 3×3
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]
B = 6×4
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 = 1×3 cell array
{2×4 double} {2×4 double} {[7 8 9 12]}
C{:}
ans = 2×4
1 2 3 90 3 1 2 88
ans = 2×4
4 5 6 17 6 5 4 19
ans = 1×4
7 8 9 12
This will produce a consistent 1*size(A,1) output cell array, even if there are rows that do not match.

Weitere Antworten (0)

Kategorien

Mehr zu Resizing and Reshaping Matrices finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by