finding similar rows in matrices

5 visualizzazioni (ultimi 30 giorni)
Dany
Dany il 11 Feb 2014
Hi, i have several matrices that has 3 columns each, but the number of rows is not equal. im trying to find a fast way of comparing the first two columns in all 3 matrices and to find the rows (column 1 and 2) that are identical in all 3 of them. i've tried "ismember", it works ok but when dealing with lots of matrices its too slow.
can anyone recomend a different function? thank you for your help.
  3 Commenti
Dany
Dany il 11 Feb 2014
Modificato: the cyclist il 11 Feb 2014
a=[1,2,3;
1,3,5;
2,5,7];
b=[1,2,6;
1,4,8;
2,5,8];
c=[1,4,7;
1,5,3;
2,5,9];
those three matrices have only one row (column 1 and 2) in common. the third one ...... thats basicaly what i want, the index of the common rows in all matrices.
SIVAKUMAR KARURNKARAN
SIVAKUMAR KARURNKARAN il 24 Apr 2015
thankyou sir i have one doubt in my coding.how to remove a duplicate matrices in the set of all matrices.

Accedi per commentare.

Risposta accettata

the cyclist
the cyclist il 11 Feb 2014
Modificato: the cyclist il 11 Feb 2014
I think your best bet for something fast is the intersect() command.
doc intersect
for details. Be sure to use the "rows" argument.
Here is some code that I believe will do what you want, but I have to admit I did not test much:
a=[1,2,3;
1,3,5;
2,5,7];
b=[1,2,6;
1,4,8;
2,5,8];
c=[1,4,7;
1,5,3;
2,5,9];
[ab,ia,ib] = intersect(a(:,1:2),b(:,1:2),'rows')
[abc,iab,ic] = intersect(ab(:,1:2),c(:,1:2),'rows')
a_row_idx = ia(iab)
b_row_idx = ib(iab)
c_row_idx = ic

Più risposte (2)

Jos (10584)
Jos (10584) il 11 Feb 2014
INTERSECT or ISMEMBER come into mind:
tf = ismember(C, A(ismember(A,B,'rows')), 'rows')
result = C(tf,:)

Dany
Dany il 11 Feb 2014
thank you. i will try the "intersect" command.
is it possible to put several commads of "ismember" one into another? can those functions work on several matrices at once? more then two?

Prodotti

Community Treasure Hunt

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

Start Hunting!

Translated by