Why is this script so slow and how can i make it faster?

Hello,

I am a student at the TU Delft in holland and i have a problem with this matlab script. It takes for ever to run the whole script and the problem is that i have to do it a couple of times. Does someone knows how to make this script faster?

 M=rand(100,110,130);
p=1
for i=1:100;
    for j=1:110;
        for k=1:130;
            if M(i,j,k)<0,5;
                x(p)=i;
                y(p)=j;
                z(p)=k;
                p=p+1
                else
                p=p
            end
        end
    end
end

Thanks for the help

Answer 1

[Jos (10584)]

First, when you use loops, try to pre-allocate the outcome

N = 1000 ;
p = zeros(N,1) ; % pre-allocation
for k=1:N
    p(k) = …
end

Second, matlab can handle matrices at once. You can take advantage of that, rather than going through each element one by one. In your case:

M=rand(5,3,2); % a smaller example
tf = M < 0.5 
idx = find(tf)  % linear index
[x,y,z] = ind2sub(size(M), idx)
% which can be one-lined into [x,y,z] = ind2sub(size(M), find(tf<0.5))

Now these will give the same result:

R1 = M(tf) ;
R2 = M(idx) ;
R3 = zeros(size(x)) ; % see, pre-allocation again
for k=1:numel(x),
  R3(k) = M(x(k),y(k),z(k)) ;
end
disp(R1)
isequal(R1,R2,R3)

[Tobias]

First of all thank you very much, This is indeed much faster. The only problem is that the x,y,z vectors are not lineair. They numbers in this vector have to go from lowest to highest. How can i fix this problem?

[Jos (10584)]

I do not completely get this point, I'm afraid.

Note that x,y, and z are three(!) vectors. Moreover, they are coupled, meaning that an element x(k) belongs to y(k) and z(k). You can sort one of them of course, but then you have to re-arrange the others as well.

Answer 2

[Niklas Nylén]

In addition to Jos' answer, the most time consuming part of your code seems to be printing out p every time it is increased by one. By putting a semicolon after p=p+1 and completely removing the else statement since it does nothing, the execution time is reduced by approximately 95 % (from 20 to 0.5 s on my computer)

Answer 3

[Azzi Abdelmalek]

idx=find(M<0.5);
[ii,jj,kk]=ind2sub(size(M),idx);
[a,idx1]=sortrows([ii jj kk],[1 2 3]);
x=a(:,1)';
y=a(:,2)';
z=a(:,3)';

Answer 4

[Bjorn Gustavsson]

Wouldn't this solve the (homework-) problem:

[x,y,z] = find(M<0.5);

HTH.

[Azzi Abdelmalek]

This doesn't solve his question. You can compare your result and his.

[Bjorn Gustavsson]

True, I didn't even bother reading the help on find, it's in there. But that halfbaked higher-dimension functionality reminds me of 4.2. Now I have to think a bit about shading the builtin find with something along your line.

Answer 5

[Andrei Bobrov]

[i1,i2]=find(M<.5);
c = num2cell(sortrows([i1,rem(i2-1,4)+1,ceil(i2/4)]),1);
[x,y,z] = c{:}

Answer 6

[Azzi Abdelmalek]

Without sortrows. This is much faster

M1=permute(M,[3 2 1]);
idx=find(M1<0.5);
[ii,jj,kk]=ind2sub(size(M1),idx);
x=kk';
y=jj';
z=ii';

Answer 7

[Roger Stafford]

 M=rand(100,110,130);
 [z,y,x] = ndgrid(1:130,1:110,1:100);
 t = M<.5;
 z = z(t);
 y = y(t);
 x = x(t);