Help on code optimization

I have two matrices A (519840x5) and B (319966x5). Columns 2 to 4 for both contain "day of the year (format: 1-366) ", "hour (format:0-23) " and "minute (format: 0-59)". I would like for each day, hour and minute of matrix B to get a value for the same day, hour and minute, from matrix A and insert it in A, in case they don't have a common day-hour-minute then this value to be -999. I have done using the following code, but since it is too slow i would appreciate any help for speed optimization

for i=1:size(B,1)
   id=find(A(:,2)==B(i,2)&A(:,3)==B(i,3)&A(:,4)==B(i,4));
   if size(id,1)==1 %check if condition exists
       tot=A(id,5); % if yes then get this value
       else
       tot=-999; %otherwise tot=-999
   end
B(i,6)=tot; %insert to last comlum the tot value
end

Thank you

Answer 1

[Walter Roberson]

Unfortunately I do not have access at the moment to test this:

[uAkey, mA, nA] = unique(A(:,2)*10000 + A(:,3)*100 + A(:,4));
[uBkey, mB, nB] = unique(B(:,2)*10000 + B(:,3)*100 + B(:,4));
[tf, idx] = ismember(uvB, uvA);
B(~tf,6) = -999;  %the ones not found
B(tf,6) = A(mA(idx(tf)),5);

Yes, this is a bit tricky in the name of efficiency.

A less tricky and less efficient version would be:

Akey = A(:,2)*10000 + A(:,3)*100 + A(:,4);
Bkey = B(:,2)*10000 + B(:,3)*100 + B(:,4);
[tf, idx] = ismember(Bkey, Akey);
B(~tf,6) = -999;
B(tf,6) = A(idx,5);

In either case, if there are duplicate B then they will all be assigned the same value. If there are duplicate A keys, both versions of the code will use the id from the last entry in A with that key.

[Kostas]

To be total honest i am still trying to understand how your code works, but it 's really great, it did what i wanted in just few seconds while in my way it took almost 20 minutes. Thanks a lot

Answer 2

[bym]

you might start by pre-allocating the new size of B before the for loop:

B = [B,zeros(319955,1)];
for i = 1:length(B)
id=find(A(:,2)==B(i,2)&A(:,3)==B(i,3)&A(:,4)==B(i,4));
 if size(id,1)==1 %check if condition exists
     tot=A(id,5); % if yes then get this value
     else
     tot=-999; %otherwise tot=-999
 end
B(i,6)=tot; %insert to last comlum the tot value
end

[Walter Roberson]

pre-allocating is not really relevant here. If the 6th column does not exist then the first assignment to that column will create the entire column (all rows); in very new versions it might even be able to the extension in-place.