image help

1 次查看（过去 30 天）
pdp rpa
pdp rpa 2012-3-20
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

请先登录，再回答此问题。

采纳的回答

Image Analyst
Image Analyst 2012-3-20
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.
  1 个评论
pdp rpa
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
end
indices=indices(2:end);
end
c=c+1;
end
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));

请先登录，再进行评论。

更多回答（1 个）

Jakob Sørensen
Jakob Sørensen 2012-3-20
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?
  3 个评论
显示 1更早的评论
Jakob Sørensen
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
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.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Image Arithmetic 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by