# I do have a matrix A(n,n) and a matrix B(m,n) m<n and want to create a third matrix C(n,n) where C(i,j)= k if A(i,j)<=B(k,j) && if A(i,j)>B(k-1,j).

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Youssef on 13 Apr 2017
Commented: Youssef on 13 Apr 2017
The question is to rank the element of each vector (column) in A(n,n) according to the order in B(m,n) and get a new matrix with these ranks. It is a classification of elements by column : example:
A = [0 4 7 ;
1 3 5 ;
3 3 6 ]
B = [0 0 1 ;
3 1 2 ]
A(1,1)<=B(1,1)==>C(1,1)=1<---the index of zero in B;
A(2,1)>B(1,1)=0 && A(2,1)<=B(2,1) ==> C(2,1)=2
A(3,1)==B(2,1)> B(1,1)=0 ==> C(3,1)=2 and go next to the second column, then next one to get
C=[1 2 2, C(:,2),C(:,3)]
Image Analyst on 13 Apr 2017
I don't understand the rule(s). Why does A(1,1) get compared to B(1,1) but the second element of the first column of A get compared to two elements of B, and the third element of the first column of A gets compared to B(2,1) but then B(2,1) gets compared to B(1,1)? Why do you need to do this complicated thing?

Santhana Raj on 13 Apr 2017
You can create three nested loops for i, j and k. and check for your condition:
if ((A(i,j)<=B(k,j) && (A(i,j)>B(k-1,j))
Take care that, when k=1, you cant check both the conditions and only the first conditions is to be applied.
Hope it helps.
Youssef on 13 Apr 2017
Thank you very much. I used a first condition to take care about the first row. I used the same code in 2015b version, and it gave me an error "Subscript indices must either be real positive integers or logicals.". But I run it in an old version (20114b) and it works!!!!