why setdiff have time longer than for ?

*hi,

Itried to make my code be faster , so i used setdiff instead of a part in my code

I used:*

                  v=(1:18);
                   buf=setdiff(v,unique_x);
                  relevant_item(ttt(i),1:length(buf))=buf;

*instead of *

                  for v=1:18
                    b2=find(unique_x==v);
                   if isempty(b2)
                       relevant_item(ttt(i),h)=v;
                       h=h+1;
                   end
               end

*why the running time of this part is less than of setdiff?

thanks*

Answer 1

[Walter Roberson]

setdiff() essentially has to sort each time -- or at least to check if the set is ordered.

Are the unique_x restricted to being in the allowed range for v, 1:18 ? If so, then how about

buf = 1:10;
buf(unique_x) = [];
relevant_item(ttt(i),1:length(buf))=buf;

[huda nawaf]

thanks ,yes it is best in terms of time

[huda nawaf]

Walter, if buf =[ 3 49 10 4 7 20];

I mean not 1:18, can find way to run the above code neither use setdiff , nor for?

thanks

[Jan]

@huda: Look into the code of setdiff. You find the relevant part in the fast MEX function ismembc2, which expects pre-sorted data. So sort your data explicitly, call this function and use the replied indices. Note that ismembc2 is not documented, but you find enough information about it in the net.