Filtering the common rows between two matrices

11 ビュー (過去 30 日間)
Bassem Mokhtar
Bassem Mokhtar 2024 年 2 月 1 日
移動済み: Dyuman Joshi 2024 年 2 月 25 日
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 件のコメント
Stephen23
Stephen23 2024 年 2 月 1 日
移動済み: Dyuman Joshi 2024 年 2 月 25 日
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 2024 年 2 月 25 日
移動済み: Dyuman Joshi 2024 年 2 月 25 日
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.

サインインしてコメントする。

採用された回答

Stephen23
Stephen23 2024 年 2 月 1 日
編集済み: Stephen23 2024 年 2 月 1 日
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.

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeResizing and Reshaping Matrices についてさらに検索

タグ

Community Treasure Hunt

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

Start Hunting!

Translated by