image help

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What is the meaning of the following code and why it is running in infinite loop??

image=imread('pp.jpg'); [rows,cols]=size(image); pixels=[];
for i=1:rows
  for j=1:cols
    pixels=[pixels;image(i,j)];
  end
end

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Answer 1

Image Analyst 2012-3-20

在 MATLAB Online 中打开

1 个投票

Just use

pixels = reshape(yourImageArray, [1 numel(yourImageArray)]);

or

pixels = yourImageArray(:);

to get a 1D list of all the pixels. I think your code takes nearly forever because it needs to reallocate a new array every time you do pixels=[pixels;image(i,j)]; and that array gets bigger and bigger each time. And like the others said, don't use "image" for the name of any variable because you're blowing away the built-in function of that name if you do.

Tip: In the future, make sure there is a blank line before your code, then highlight all your code and click the "*{}Code*" icon above your edit box to format your text like code so that we can read it better.

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pdp rpa 2012-3-22

Thank you so much sir for your kind response. Sory again i m seekig for your kind help. i have the following code of dbscan. but the problem is that i want to use this code for image segmentation. so plz help me to do so.

function [class,type]=dbscan(X,K,d)

% [class,type]=dbscan(X,K,d)

% X: matrix

% K: minimal points

% d: maximal distance

% class: assignments

% type: core: 1, border: 0, outlier: -1

if nargin<3

d=epsilon(X,K);

end

m=size(X, 1);

class=zeros(m,1);

type=zeros(m,1);

touched=zeros(m,1);

c=1;

for i=1:m

if touched(i), continue; end

indices=find(distances(X(i,:),X)<=d);

k=length(indices);

if k==1

type(i)=-1;

class(i)=-1;

touched(i)=1;

elseif k<=K

type(i)=0;

class(i)=0;

else

type(i)=1;

class(indices)=c;

while ~isempty(indices)

index=indices(1);

touched(index)=1;

I=find(distances(X(index,:),X)<=d);

if length(I)>1

class(I)=c;

if length(I)<=K;

type(index)=0;

else

type(index)=1;

end

for i=1:length(I)

if touched(I(i)), continue; end

touched(I(i))=1;

indices(end+1)=I(i);

class(I(i))=c;

end

indices=indices(2:end);

end

c=c+1;

end

I=find(class==0);

class(I)=-1;

type(I)=-1;

function d=epsilon(x,k)

[m,n]=size(x);

d=((prod(max(x)-min(x))*k*gamma(.5*n+1))/(m*sqrt(pi.^n))).^(1/n);

function D=distances(x,X)

D=sqrt(sum((ones(size(X, 1),1)*x-X).^2, 2));

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Answer 2

Jakob Sørensen 2012-3-20

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It shouldn't be infinite, but very possibly it will take a long time to run it. It seems to plot the image in a very weird and time consuming way, over and over again. I can't think of any reason to use that kind of code, but maybe someone else can. What is it you want to do?

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Jakob Sørensen 2012-3-20

Also just noted another thing, which explain why it seemed so confusing. "image" is the name of a function Matlab used to plot images (see "help image"), so using it as a variable name, why not disastrous, isn't very smart.

pdp rpa 2012-3-22

thank you sir...actually i have found this piece of code for image segmentation using dbscan. plz can you suggest some good code of dbscab for image segmentation. I will be very thankful to you.

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