Conversion of for loop to Vector.

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azizullah khan
azizullah khan el 16 de Dic. de 2015
Comentada: azizullah khan el 17 de Dic. de 2015
I have two for loops and it's processing is very slow.I need to speed up the task. Any help would be highly appreciated.
for i=1:ll
for j=1:ll
X=[a(i),b(i);a(j),b(j)];
d = pdist(X,'euclidean');
if d>0 && d<(4.1*ro)
if k2(a(i),b(i))>k2(a(j),b(j))
k3(a(j),b(j))=0;
end;
if k2(a(i),b(i))<k2(a(j),b(j))
k3(a(i),b(i))=0;
end;
end;
end;
end;
Thank you,,,,
  3 comentarios
azizullah khan
azizullah khan el 16 de Dic. de 2015
Basically the above code is for feature selection on the basis of euclidean distance. K2, K3 are matrixes of images, and we are making the elements of K3 to zero on the basis of euclidean distance. thx
Guillaume
Guillaume el 16 de Dic. de 2015
What I meant is that you should have a description of the purpose of each line of code. Ideally, this should be comments in the code itself.
I can guarantee you won't have a clue how the code above does its job if you come back to it in a year.

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Respuesta aceptada

Guillaume
Guillaume el 16 de Dic. de 2015
Here is something that I believe does the same as your code. Added bonus, it's got comments:
ll = 20; %size of matrices, maximum value of a and b vectors
%generate some demo data, since you didn't provide any:
a = randperm(ll); %a is row coordinates of matrices corresponding to ... something
b = randperm(ll); %b is column coordinates of matrices corresponding to ... something
k2 = rand(ll); %k2 is a matrix corresponding to ...?
k3 = ones(ll); %don't know what that is.
ro = 3; %threshold scale below which something (?) happen
%compute euclidean distance between all the points made by corresponding (a,b) pairs:
d = hypot(bsxfun(@minus, b, b'), bsxfun(@minus, a, a')); %equivalent to your pdist
%find which distances are within the threshold
[row, col] = find(d>0 & d<4.1*ro); %correspond to your if d ...
%find which of the two pixels of k2 defined by [a(row), b(row)] and [a(col), b(col)] is the smallest:
s = sign(k2(sub2ind(size(k2), a(row), b(row))) - k2(sub2ind(size(k2), a(col), b(col)))); %equivalent to your two if k2...
%s is 1 when it's the column pixel that is smaller, -1 when it's the row pixel, and 0 when both equal
%set corresponding pixel to 0 in k3:
k3(sub2ind(size(k3), a(col(s == 1)), b(col(s == 1)))) = 0;
k3(sub2ind(size(k3), a(row(s == -1)), b(row(s == -1)))) = 0;
  3 comentarios
Guillaume
Guillaume el 16 de Dic. de 2015
I've just tested my code on a 256x256 image. It produces the exact same result as yours but significantly faster (less than a second vs over 20 minutes), so I'm unclear what the problem is.
Attached the code and image I used for testing.
Output:
Elapsed time is 1218.195877 seconds.
Elapsed time is 0.800296 seconds.
Both results are the same
1218 seconds is 20.3 minutes!
azizullah khan
azizullah khan el 17 de Dic. de 2015
Thank you, I got luck , my prob is solved!!!

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Más respuestas (1)

Renato Agurto
Renato Agurto el 16 de Dic. de 2015
Hello,
I would try this under the assumjption that:
pdist(X1,'euclidean') == pdist(X2,'euclidean')
if:
X1 = [a(i),b(i);a(j),b(j)];
X2 = [a(j),b(j);a(i),b(i)];
Code:
for i=1:ll
for j=i+1:ll
X=[a(i),b(i);a(j),b(j)];
d = pdist(X,'euclidean');
if d>0 && d<(4.1*ro)
if k2(a(i),b(i))>k2(a(j),b(j))
k3(a(j),b(j))=0;
elseif k2(a(i),b(i))<k2(a(j),b(j))
k3(a(i),b(i))=0;
end;
end;
end;
end;
  2 comentarios
azizullah khan
azizullah khan el 16 de Dic. de 2015
Basically i want to remove the for loops, coz everytimg it's calculating the euclidean distance and calculating euclidean distance is very time consuming. So, i would really appreciate that if sort out with like removing the for loop will be better, thank you,
Guillaume
Guillaume el 16 de Dic. de 2015
As I've shown in my answer you can calculate the euclidean distance between all points in one go with just one line:
d = hypot(bsxfun(@minus, b, b'), bsxfun(@minus, a, a'))
The loops, the pdist, the if, all of this is unnecessary.

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