Pixel Extraction From Using Cluster

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rizwan
rizwan 2015 年 3 月 25 日
Hi Experts
Let me explain my problem with details....I have an image from which i need to find the big cluster for this i m using k means and it works fine, now i want to extract the pixels of this big cluster from from the image provided, i m not able to find the pixels of big cluster and then store in a variable ...Please help i have written complete code below with appropriate explanation....
Step 1> I have an image on which i have applied k mean clustering to get clusters, for this what i did is
index_string = int2str(k);
Image_Path = strcat('D:\MS\Research\Classification Model\Research Implementation\EnhancedImage\ROI',index_string,'.jpeg');
I1 = imread(Image_Path);
I=double(I1);
figure
subplot(1,3,1)
imshow(I1)
subplot(1,3,2)
% [idx, C] = kmeans(I(:),4, 'distance','sqEuclidean', 'EmptyAction','singleton', 'replicates',1,'Display','iter');
[idx,C]= kmeans_fast_Color(I,4,'Display', 'iter');
Above code is working fine and doing well
Step two> In this step what i want is to calculate the cluster size and select the big cluster from all four clusters. What i did for this is
a = unique(idx);
out = [a,histc(idx(:),a)];
[a,ix]= max(out(:,2));
It works fine...Up till now i have the max cluster size and its number stored in a and idx respectively....
Step 3> In this step what i want to do is to use the max cluster number and iterate over the idx cluster array to find the locations of the cluster number and store them in a variable. Now i have the locations of the cluster numbers, using these locations i want to extract the pixel numbers from the original Image I and stored in some variable for further use...What i did for this is
[row , col] = find(idx ==ix);
mx_row = max(row);
mx_col = max(col);
for i =1 : max(row)
for j= 1: max(col)
cp = I(i,j);
end
end
cp is the variable in which i want to store the pixels values...but there is a bug which i m not able to find for storing pixel values in cp!!!!!!!!!!
I hope you have now a clear idea of my problem...
Thanks in ADVANCE For assistance!!!
Regards

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