Question

Suppose I have a cell

A = {[3,0], [2,1]}

and a cell array

B = {[4,-1],[3,0];      
      [-1,4],[-3,5];
      [3,0],[2,1];
      [2,1],[-1,4]}.

I want to find the indices where both the first or second entry in A shows up in B excluding the row in B where both entries of A show up.

In this example I should get something like [1 4] for the rows in B. I’ve been trying to figure this out using cellfun and cell2mat but keep stumbling.

Answer 1

I would approach this problem by converting my cell arrays to numeric arrays of appropriate dimensions, and then use ismember.

The following example illustrates how this method works on the example cell arrays in the question:

%# Build the example cell arrays
A = {[3,0], [2,1]};
B = {[4,-1],[3,0];      
     [-1,4],[-3,5];
     [3,0],[2,1];
     [3,0],[3,0];
     [2,1],[-1,4]};

%# Get the number of elements in A, and the length of the first element
K = size(A, 2);
J = length(A{1, 1});

%# Convert cell arrays to usefully shaped numerical matrices
ANumVec = cell2mat(A);
ANum = reshape(ANumVec, K, J)';
BNum = cell2mat(B);

%# Find matches of 1*2 vectors in ANum in sets of two columns of BNum 
I1 = ismember(BNum(:, 1:J), ANum, 'rows');
I2 = ismember(BNum(:, J+1:end), ANum, 'rows');
I3 = ismember(BNum, ANumVec, 'rows');

%# Find all indices where there was exactly 1 match (ie omit cases of no matches and cases of 2 matches)
MainIndex = I1 + I2;
MainIndex(I3) = 0;
Soln = find(MainIndex > 0);

Some points:

1) This method finds the indices of all rows in B where an element of A lies in the first or second column of B, excluding the situation where A corresponds exactly to a row of B.

2) This method will fail if there are multiple rows in A. However, it is robust to A being a cell array of size 1*N, where N denotes some arbitrary number of 1*2 numeric vectors. Thus the single row limitation can be circumvented by first reshaping A to a 1*N cell array.

3) Equivalence is tested using the logical operator ==. This can be dangerous with floating point numbers unless you have reason to believe a priori that your inputs will not exhibit any floating point error.

4) I can’t shake the feeling that there is a much more efficient way to solve this problem, but that I’m not seeing it at the moment. 🙂