How to quickly find the maximum group of same rows in a matrix?

Dear All,

I have a matrix and there exist identical rows in the matrix. I want to find the maximum group of identical rows in this matrix. For example, I have a matrix as follows:

A = [0 1 0 1 0 0 0 0; 1 0 0 0 1 0 0 0; 1 0 0 0 1 0 0 0; 1 1 0 0 0 0 0 0; 0 0 1 0 0 0 0 1; 0 0 0 0 0 1 1 0; 0 1 0 1 0 0 0 0; 0 1 0 1 0 0 0 0].

There are two groups of identical rows: {row 1, row 7 and row 8}, {row 2, row 3}. So the answer is {row 1, row 7, row 8}. But how can I quickly find it?

Best regsards,

Benson

Answer 1

[Star Strider]

Try this:

A = [0     1     0     1     0     0     0     0; 1     0     0     0     1     0     0     0; 1     0     0     0     1     0     0     0; 1     1     0     0     0     0     0     0;  0     0     1     0     0     0     0     1;  0     0     0     0     0     1     1     0;  0     1     0     1     0     0     0     0;  0     1     0     1     0     0     0     0];
[Au,~,idx] = unique(A, 'rows');
tally = accumarray(idx, 1);
Out = A(idx == max(tally),:)                                % The Rows Themselves
RowIdx = find(idx == max(tally))                            % The Row Indices

producing:

RowIdx =
     1
     7
     8

[Benson Gou]

Dear Star,

It works. Thanks a lot.

Benson

[Star Strider]

As always, my pleasure!

[Benson Gou]

Dear Star,

I encountered a weird case which I did not get the correct result.

A = [ 0 1 0 0 1 0 0 0 0;

0 0 0 1 0 1 0 0 0;

1 0 0 0 0 0 1 0 0;

1 0 0 0 0 0 1 0 0;

1 1 0 0 0 0 0 0 0;

1 0 1 0 0 0 0 0 0];

I got

RowIdx = 1.

Could you please help find out the reason? Thanks a lot.

Benson

[Star Strider]

It is strange that my earlier code did not work on the new matrix.

With a slight edit, this version works correctly on both the original ‘A’ and the new ‘A’:

[Au,~,idx] = unique(A, 'rows');
tally = accumarray(idx, 1);
[~,maxidx] = max(tally);
Out = A(idx == maxidx,:)                                % The Rows Themselves
RowIdx = find(idx == maxidx)                            % The Row Indices

producing for the new ‘A’:

Out =
     1     0     0     0     0     0     1     0     0
     1     0     0     0     0     0     1     0     0

RowIdx =
     3
     4

and the correct result (posted previously) for the original ‘A’.

[Benson Gou]

Dear Star,

Thanks a lot for your great help. Now it works very well.

Happy Holidays!

Bension

[Star Strider]

As always, my pleasure!

Happy Holidays to you, too!