MATLAB: How do i replace this code to fast up

rbm

Variable poshidstates have 4 possible values(0,1,2,3). Everytime, poshidstates has to select 1 value out of this four.In this below code, i=100,j=500. Because of below mentioned code part , my program run more than 8 hour instead of just 1 hour(I have to call this code for 600 batches and each batch for 50 times). how can i replace this?

   val_vect=[0 1 2 3];
  for i=1:numcases;
      for j=1:numhid;
          prob=[poshidprobs0(i,j),poshidprobs1(i,j),poshidprobs2(i,j),poshidprobs3(i,j)];
          K=find(mnrnd(1,prob)==1);
          poshidstate(i,j)=val_vect(K);
      end
  end

Best Answer

Ok, now I see why it takes so long. You're repeating that calculation 30,000 times.

But seeing what you're doing, this might be faster:

 temp = rand(100, 500);
 poshidstate = (temp > poshidprobs0) + (temp > (poshidprobs0 + poshidprobs1)) + (temp > (poshidprobs0 + poshidprobs1 + poshidprobs2));

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MATLAB: Avarage running time problem

Here are three ways to use tic/toc to measure the timing of a loop. Each method has pros and cons.

Method 1: Least overhead & resolution

The tic() & toc() functions require processing time, albeit a very small amount. Storing the results of toc() also requires processing time. This method contains the least overhead but you don't get the single-iteration times.

start = tic(); 
for i=1:100
    1+1;
    2+2;
end
k= toc(start)/100;

Method 2: Single iterations times

This method gives you the single-iteration times but the processing time is added by tic/toc and storing the toc output.

t = nan(1, 100); 
for i=1:100
    start = tic(); 
    1+1;
    2+2;
    t(i)=toc(start);
end
k=mean(t)

Method 3: Single iterations times, less overhead

This method removes tic() from the loop which doesn't reduce the overhead by much but it's worth sharing. The single-iteration times are computed outside of the loop.

t = nan(1, 101); 
start = tic(); 
t(1) = toc(start); 
for i=1:100
    1+1;
    2+2;
    t(i+1)=toc(start);
end
k=mean(diff(t))
% Compare that to (t(end)-t(1))/100

Reasons your timing was off

tic() returns an integer (uint64) that has meaning only for the toc function. Example: t = tic(); t = 1943576784277
toc() returns the elapsed time since the previous tic() which is very quick in your loop so you're subracting a tiny amount from a large, meaningless integer.
The 1+1 and 2+2 lines display the result of each iteration in the command window since those lines are not supressed with a semicolon (;). That takes a lot of time.
The t variable was not pre-allocated so the length of t grows with each iteration. That also takes a lot of time.

Since the computations are so simple, you'll find a wide range of mean(t) if you run that script many times. Consider using the timeit function.

MATLAB: What does this section of code do

It looks like:

i & j are generated as a random point on the board.

dx and dy are generated as a random direction (they are either -1, 0, or 1)

dx and dy are then checked to see that, when 4 is added to i and j in the dx and dy directions, the result is still on the board. If everything is still on the board, then the while loop exits. Otherwise, new random dx and dy directions are generated for testing. Also, at least one of dx or dy must be non-zero for the while loop to exit.

Then the board i and j position as well as three additional positions in the direction of dx and dy are assigned a value of 1.

Best Answer

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MATLAB: Avarage running time problem

MATLAB: What does this section of code do

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