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

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Tobias
Tobias on 29 Jan 2014
Answered: Roger Stafford on 29 Jan 2014
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

Answers (7)

Jos (10584)
Jos (10584) on 29 Jan 2014
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)
  2 Comments
Jos (10584)
Jos (10584) on 29 Jan 2014
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.

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Niklas Nylén
Niklas Nylén on 29 Jan 2014
Edited: Niklas Nylén on 29 Jan 2014
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)

Azzi Abdelmalek
Azzi Abdelmalek on 29 Jan 2014
Edited: Azzi Abdelmalek on 29 Jan 2014
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)';

Bjorn Gustavsson
Bjorn Gustavsson on 29 Jan 2014
Wouldn't this solve the (homework-) problem:
[x,y,z] = find(M<0.5);
HTH.
  2 Comments
Bjorn Gustavsson
Bjorn Gustavsson on 29 Jan 2014
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.

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Andrei Bobrov
Andrei Bobrov on 29 Jan 2014
[i1,i2]=find(M<.5);
c = num2cell(sortrows([i1,rem(i2-1,4)+1,ceil(i2/4)]),1);
[x,y,z] = c{:}

Azzi Abdelmalek
Azzi Abdelmalek on 29 Jan 2014
Edited: Azzi Abdelmalek on 29 Jan 2014
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';

Roger Stafford
Roger Stafford on 29 Jan 2014
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);

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