k-means segmentation for Raman analysis

Good afternoon,

I hope someone can help me with this problem.

I acquired a Raman map XY of a pancreatic tumor cell, so I have a spectrum for each point of this map associate with some coordinates. Then I've realize kmeans clustering with 6 clusters.

Now I have a label (1, 2, 3, 4, 5 or 6) for each coordinates (x and y) and I want to plot them in an image with different colour. Is there a way to do this?

I attach below the coordinates for each group identified by kmeans: the first coloumn is X, while the second is Y.

What I hope to obtain is something like this:

Answer 1

[Mathieu NOE]

hello

I get something quite different from your expectation

first I looked at first group of excel files coordinates_group1.xlsx to coordinates_group6.xlsx

and I got this plot

code

fileDir = pwd; % current directory (or specify which one is the working directory)
S = dir(fullfile(fileDir,'coordinates_group*.xlsx')); % get list of data files in directory
S = natsortfiles(S); % sort file names into natural order , see : 
           %(https://fr.mathworks.com/matlabcentral/fileexchange/47434-natural-order-filename-sort)
figure
hold on

for ck = 1:length(S)
    FileName=S(ck).name;
    fprintf(1,'Now reading %s\n',FileName);
    data = readmatrix( fullfile(fileDir, FileName));
    scatter(data(:,1),data(:,2),'filled')
    FileName = strrep(FileName,'_',' ');
    leg{ck} = FileName;
end

legend(leg);

then I supposed that I was wrong and I had to process the remaining excel file and got somthing quite the same !

what is wrong ??

code

fileDir = pwd;

FileName='coordinates and label.xlsx';
fprintf(1,'Now reading %s\n',FileName);
data = readmatrix( fullfile(fileDir, FileName));
labels = data(:,3);
labels_range = (min(labels):max(labels));

figure
hold on

for ck = 1:numel(labels_range)
    % mask = (labels==ck); % taht works 
    mask = (abs(labels-ck)<1e-6); % this is more robust to round off errors
    leg{ck} = ['label = ' num2str(ck) ];
    scatter(data(mask,1),data(mask,2),'filled')
end
legend(leg);

[Mathieu NOE]

hello again

problem solved ?

[Valeria Iazzetta]

Yes, thank you. It is exactly what I want to obtain when I asked for help, so thanks so much.

Sorry if I didn't answer immediately. To get a result more similar to the image I posted, I implemented a code that associates a color at each pixel of raman map.

[Mathieu NOE]

ok

glad you have a working solution !

Answer 2

[Image Analyst]

I don't know if kmeans is the best, unless you know for a fact that there are definitely exactly 6 classes present in all images. If some images might possibly not have one or more classes in them, then kmeans would be forces to find 6 anyway, and they'd be wrong.

I'd recommend you use the Classification Learner app on the Apps tab of the tool ribbon. You can try out several classification methods and pick the best one. But make sure your training set includes all possible situations including some images with all classes and some with only some of the classes present (if that is a possibility for your data).

I also don't really like kmeans because it's classifications vary from image to image. I think what you'd want is to have one fixed definition of what a class is, and then you can apply it to any image. The classifications won't vary from image to image and it will work regardless of how many classes are in any particular image. For that reason I'd prefer something like Discriminant Analysis, and I'm attaching a demo for that. The demo uses 3 features (predictors) for red, green, and blue, but you could use anything, even a list of 1000 signal values from different wavenumbers.

[Valeria Iazzetta]

Dear Image Analyst, thanks for the answer.

I didn't apply k-means to image, but on Raman spectra. For each spectra of the raman maps that I acquired, I know spatial coordinates which I can associate with the various pixels of the image . Only then do I visualize the spatial distribution of the clusters in the cell.

For this type of analysis I cannot use LDA because I don't have reference spectra for the classes. It is an exploratory analysis.