yes. you can use the elbow and silhouette methods for any partitioning algorithm that requires you to provide the value of k first eg in k means, k-medoids clustering etc. So assuming you have data, you can run it with let's say k = 10 or 20 first, applying the methods to choose the optimal number for k. With this optimal k value, say k = 3, you would then redo the clustering to get your final clustered data (without applying the methods).
I'm a bit confused.To apply the k-means algorithm, the number of clusters k must be determined. This can be accomplished through the elbow method or the silhouette method. However, each of these methods involves enumerating the values of k = 1, 2, 3,..... (
yes. you can use the elbow and silhouette methods for any partitioning algorithm that requires you to provide the value of k first eg in k means, k-medoids clustering etc. So assuming you have data, you can run it with let's say k = 10 or 20 first, applying the methods to choose the optimal number for k. With this optimal k value, say k = 3, you would then redo the clustering to get your final clustered data (without applying the methods). A computer would normally do all k's at once, you don't need to enumerate each k.
Why did you chose the point X=5 (4:30)? Why not 3 or 7? Besides, as far as I understand you made a decision visually, but what if we calculate it on PC (with no visualisation)?
Why does the elbow method tell us that 5 is the optimum number of clusters, while the silhouette says 3? Why not 3 for the inertia as well, if there is also a bend at point 3?
watch out: the b(i) coefficient presents a misleading definition. It is the average distance between the point i and the points in the NEAREST cluster.
K should be 5 not 3.
That was a typo
K should be 5 for Elbow method
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@@51_sajalgupta84 bro why you need nudes???
@@51_sajalgupta84 have you never been taught manner by your parents?
this is the best video i've seen on elbow and silhouette method !! Thank you so much !!
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thank you sir this helped me a lot!
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Both the methods needs to form clusters right? So does we have to use k Means Algorithm to form clusters for both the Elbow and Shillote methods?
yes. you can use the elbow and silhouette methods for any partitioning algorithm that requires you to provide the value of k first eg in k means, k-medoids clustering etc. So assuming you have data, you can run it with let's say k = 10 or 20 first, applying the methods to choose the optimal number for k. With this optimal k value, say k = 3, you would then redo the clustering to get your final clustered data (without applying the methods).
The legend pic in black chasma..
Thanks
very helpful
I'm a bit confused.To apply the k-means algorithm, the number of clusters k must be determined. This can be accomplished through the elbow method or the silhouette method. However, each of these methods involves enumerating the values of k = 1, 2, 3,..... (
yes. you can use the elbow and silhouette methods for any partitioning algorithm that requires you to provide the value of k first eg in k means, k-medoids clustering etc. So assuming you have data, you can run it with let's say k = 10 or 20 first, applying the methods to choose the optimal number for k. With this optimal k value, say k = 3, you would then redo the clustering to get your final clustered data (without applying the methods). A computer would normally do all k's at once, you don't need to enumerate each k.
Why did you chose the point X=5 (4:30)? Why not 3 or 7? Besides, as far as I understand you made a decision visually, but what if we calculate it on PC (with no visualisation)?
there is process in ml to visualize it
Why does the elbow method tell us that 5 is the optimum number of clusters, while the silhouette says 3? Why not 3 for the inertia as well, if there is also a bend at point 3?
same question
How many centroids should we choose for each value of k
1 centroid per cluster
Fo example: if you want 5 clusters hgen you need to select 5 centroids
watch out: the b(i) coefficient presents a misleading definition. It is the average distance between the point i and the points in the NEAREST cluster.
thanks
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