Filtering the common rows between two matrices

Question

Bassem Mokhtar 2024-2-1

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2077086-filtering-the-common-rows-between-two-matrices

移动：Dyuman Joshi 2024-2-25

采纳的回答： Stephen23

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.

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Stephen23 2024-2-1

移动：Dyuman Joshi 2024-2-25

在 MATLAB Online 中打开

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 2024-2-25

移动：Dyuman Joshi 2024-2-25

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.

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Answer 1

Stephen23 2024-2-1

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2077086-filtering-the-common-rows-between-two-matrices#answer_1401076

编辑：Stephen23 2024-2-1

在 MATLAB Online 中打开

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