FINDROW

Find indices of a given row within a matrix when INTERSECT can't help

Description

FINDROW Find indices of a given row within a matrix.

  R = FINDROW(A, B) returns indices of the rows in A that coincide with
  row vector B , otherwise empty.
  [R, I] = FINDROW(A, B) also returns the indices of elements of rows R.

Note that the use of INTERSECT for the same purpose is limited because
the results are sorted and repeated entries are not taken care of.

• MATLAB release: MATLAB 7 (R14)

[Brett Shoelson]

From your description, I would think that for A = magic(4); B = A(3,:); that k=findrow(A,B) would return 3. I'm puzzled by the need for that functionality ([i,k]=intersect(A,B,'rows') returns the same), and MORE puzzled by what it DOES return. Perhaps you can provide an example of its usefulness?

[Mukhtar Ullah]

Thanks for your interest. You are right that my description is not clear, but I have updated it and will apear soon. In fact this function returns k such that A(k)=B (not A(k,:)=B); so in your example it will return k = [3 7 11 15]. The updated function will return two outputs! 3,[3 7 11 15]

[Mukhtar Ullah]

The updated version will no longer require FINDARRAY.

[Brett Shoelson]

So you are jumping through hoops to return j and sub2ind(size(A),A(i)) (Given: [i,j] = intersect(A,B,'rows');)
Why?

[Mukhtar Ullah]

Perhaps you have not read propelly the new description. To get what i want to do: try this
a = pascal(5); b = a(3,:);
[i,j] = intersect(a,b,'rows');
sub2ind(size(a),a(i)) gives [1 1 1 5 15] which is nonsense while
[i,j] = findrow(a,b) gives

i = 3 and j = [3 8 13 18 23]. This j contains indices to 3rd row.

[Mukhtar Ullah]

intersect function has two problems: it sorts the indices which distorts them; and it has no way of dealing with repeated elements.

[Betty]

 

[Jos]

Unfortunately this function errors for non-numeric input.