Conversion of for loop to Vector.

2015 12 16
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更新时间：2015 12 17
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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,,,,

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Guillaume
Guillaume 2015-12-16
An explanation of what the code does (hint: code should have comments) would go a long way towards you getting an answer.
azizullah khan
azizullah khan 2015-12-16
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 2015-12-16
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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Guillaume
Guillaume 2015-12-16

1 个投票

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;

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azizullah khan
azizullah khan 2015-12-16
编辑：azizullah khan 2015-12-16
Sir, i checked the above code and compare the results so didn't get luck. Basically k2 is the image of size 512 x 512 and ll is the number of rows for which pixel value is greater than zero. below is the code
% delete overlays
[a,b]=find(k2>0); % k2 is the image 512 x 512
ll=length(a); % Number of rows in which values are not 0
%c=0;
k3=k2; % image we want in output
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) % ro is standard deviation here it is 4
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,,,
Guillaume
Guillaume 2015-12-16
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 2015-12-17
Thank you, I got luck , my prob is solved!!!

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更多回答（1 个）

Renato Agurto
Renato Agurto 2015-12-16

0 个投票

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;

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azizullah khan
azizullah khan 2015-12-16
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 2015-12-16
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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