Find row that has a cell containing all ones

Suppose I have a cell array as follows,

B = {[3,2.5,1],[3,-1,3.5];      
     [3,-1,3.5],[3,2.5,2];
     [3,4,-1],[3,3,0.25]};

I find the cells that are the same calling the anonymous function in a for loop. For testing purposes I'm only considering the first two rows of B.

is_same = @(x,y) cellfun(@eq, x, y,'UniformOutput',0);

key = [is_same(B(1, 1), B(2, 1)),...
       is_same(B(1, 1), B(2, 2));...
       is_same(B(1, 2), B(2, 1)),...
       is_same(B(1, 2), B(2, 2))];

This gives me

logic = {[1, 0, 0],[1, 1, 0];...
         [1, 1, 1],[1, 0, 0]};

I now want to find the row in logic, where a cell contains all ones? I want to be able to use this row indice in B to retrieve the row. With this retrieved row, I will apply the same steps above for other rows in B. For now, I just want to get the row indice problem solved.

Answer 1

[Azzi Abdelmalek]

[rows,col]=find(cellfun(@all ,logic))

[Harold]

Nice. I don't know why I didn't think of that. Thanks.

Answer 2

[Jan]

Note: This would be much faster and simpler without anonymous and cellfuns:

key = [isqual(B{1, 1}, B{2, 1}),...
       isqual(B{1, 1}, B{2, 2});...
       isqual(B{1, 2}, B{2, 1}),...
       isqual(B{1, 2}, B{2, 2})];

[Harold]

Thanks Jan, I'll try this also.