How to apply Silhouette Score for optimal K in MATLAB

Hello,I Hope you are doing well. I have the following dataset i want to apply Silhouette Score from scrath.

I have the Python Script for that. Can anybody help me in implmenting it into MATLAB

The following code is in Python

costs = []
for p in range(10):
    kmeans = K_Means(k=p,data = data[0],centeriod_init='random')
    centroids, cluster_assignments, iters, orig_centroids = kmeans.fit(data[0])
    X = data[0]
    dist_ji = 0
    a = 0
    s=0
    for i in range(len(data[0])):
        for j in range(p):
            dist_ji += euclidean_dist(centroids[j,:],X[i,:])
            #print(dist_ji)
    dist_ji -= sum(cluster_assignments[:,1])/len(data[0])
    a = sum(cluster_assignments[:,1])/(len(data[0])-1)
    s = (dist_ji - a)/max(dist_ji,a)
    s = np.array(s)
    s =  s.item()
    costs.append(s)
x = np.arange(10)
plt.plot(x,costs)
plt.title("Silhoutte Score")
plt.xlabel("K -->")
plt.ylabel("Dispersion")

[Med Future]

@Walter Roberson @Image Analyst @DGM

Answer 1

[Walter Roberson]

I do not know enough python to know how to convert this code.

I suspect: call kmeans() with p as the number of centroids, getting back indices and centroid locations. Then take

nearest_center = CentroidLocations(CentroidIdx,:);
a = mean((data - nearest_center).^2,2);

or something like that.

[Med Future]

@Walter Roberson

I have write the following code in MATLAB but some command i am unable to understand how they used in MATLAB

like

dist_ji += euclidean_dist(centroids[j,:],X[i,:])

#print(dist_ji)

dist_ji -= sum(cluster_assignments[:,1])/len(data[0])

a = sum(cluster_assignments[:,1])/(len(data[0])-1)

s = (dist_ji - a)/max(dist_ji,a)

costs=[]
for p=1:10
    [idx,C,sumd]=kmeans(dataset1,p);
    X=dataset1;
    dist_ji=0;
    a=0
    s=0
    for i=1:length(dataset1)
        for j=1:10
        
        end
    end
end

[Walter Roberson]

euclidean_dist(centroids[j,:],X[i,:])

in MATLAB could be written as

pdist2(centroids(j,:), X(i,:))

but that requires the statistics toolbox, and is more overhead than writing

sum((centrodis(j,:) - X(i,:)).^2)

If you make a small modification of the code I posted earlier

nearest_center = CentroidLocations(CentroidIdx,:);
distance = sum((data - nearest_center).^2,2);

then distance would be a column vector of respective euclidean distance between each point and its associated centroid .The python code then goes on to calculate the sum of those and divide the total by the number of entries, which in MATLAB could written as mean(distance). What I posted earlier is probably slightly wrong for the situation; I probably should have posted

a = mean(sum((data - nearest_center).^2,2));

and that line together with the assignment to nearest_center would replace the block of python code

X = data[0]

dist_ji = 0

a = 0

s=0

for i in range(len(data[0])):

for j in range(p):

dist_ji += euclidean_dist(centroids[j,:],X[i,:])

#print(dist_ji)

dist_ji -= sum(cluster_assignments[:,1])/len(data[0])

a = sum(cluster_assignments[:,1])/(len(data[0])-1)

[Med Future]

@Walter Roberson CentroidLocations is not any command in MATLAB

[Walter Roberson]

N = 10;
s = zeros(size(data,1), N);

for p = 1 : N
    [CentroidIdx, CentroidLocations] = kmeans(data, p);  %random initialization is default
    nearest_center = CentroidLocations(CentroidIdx,:);
    dist_ji = sum((data - nearest_center).^2,2);
    a = mean(dist_ji);
    s(:,p) = (dist_ji - a)./max(dist_ji,a);
end

plot(s)