speed up for long list matching

Is there a better way to do this as the length of idtsall is more than 30k? Thanks in advance.

                for n = 1:length(idtsall)

                                ind = find(ismember(idts, idtsall{n}));
                                if(ind > 0)
                                        ranktsall(n) = rankts(ind);
                                else
                                        ranktsall(n) = 0;
                                end
                end

Answer 1

[Walter Roberson]

The two-output ismember will return the index in the second output, allowing you to skip the find() step. The first output of insmember() will indicate whether it was found or not.

[foundit, ind] = ismember(idts, idtsall{n});
if foundit
  ranksall(n) = rankts(ind);
else
  ranksall(n) = 0;
end

Now, to check: idtsall is a cell array of vectors? All the same size or different sizes?

[Eric]

Thanks, Walter. The idtsall is a vector of cell array. The string is of the same size.

[Jan]

Which string is of the same size as what other string?

[Eric]

The strings in the cell array of both idtsall and idts are of the same size.

Answer 2

[Jan]

What type and dimension is idts?

ranktsall = zeros(1, length(idtsall);   % Pre-allocate!
for n = 1:length(idtsall)
  ind = find();
  if all(ismember(idts, idtsall{n}))   % Explicite ALL
     ranktsall(n) = rankts(ind);
  end
end

With a pre-allocation the processing is much faster and the else branch is not required anymore.

There can be much faster methods depending of the size and dimensions of idts and the contents of idtsall.

[Eric]

idts is a subset of idtsall.

Answer 3

[Eric]

I'm thinking.. could we have something like this..

ranktsall = zeros(length(idtsall), 1);
[idx, val] = matching(idts, idtsall);        % I have no idea about this part
ranktsall(idx) = val;

This would be perfect! idts is a subset of idtsall.

CAN SOMEONE HELP???