You saved me. . ! Thank you so much. . . There is one small error in the formula you have shown, it should be (Y - B)^2 and not (X -B)^2. . . you solved it correctly but the formula isnt correct. . . Thank you so much, once again. . . Distance = sqrt [(x2−x1)^2+(y2−y1)^2]
The video explanation is incorrect. In Average linkage clustering, the distance between two clusters is defined as the average of distances between all pairs of objects, where each pair is made up of one object from each group. In the average linkage method, D(r,s) is computed as D(r,s) = Trs / ( Nr * Ns) Where Trs is the sum of all pairwise distances between cluster r and cluster s. Nrand Ns are the sizes of the clusters r and s, respectively.
I find this video explanation incorrect. The average between a cluster with 2 points and the second cluster with 3 points the average of all five points. But in your video, you are teaching as take the average of the first cluster with 2 points and then take the average of the second cluster with 3 points and get the answer. Now average both the averages from above result to get the answer. This will not give the correct answer. It can only satisfy when we have the same points in both the comparing clusters. For example avg(1,2) = 1.5 avg(3,4,5) = 4 avg(1,2,3,4,5) = 3 avg(avg(1,2), avg(3,4,5)) = 2.75 which is not equal to 3. Correct me if I am wrong.
Yes you are right. In Average linkage clustering, the distance between two clusters is defined as the average of distances between all pairs of objects, where each pair is made up of one object from each group. In the average linkage method, D(r,s) is computed as D(r,s) = Trs / ( Nr * Ns) Where Trs is the sum of all pairwise distances between cluster r and cluster s. Nr and Ns are the sizes of the clusters r and s, respectively.
Thanks for video, i have one question, in the first distance matrix : if we would have (P3-P6) = 0.11 and (P3-P4) = 0.11 , in this case what we would do? then in the next step would we take avarage of for example (P1-P3) + (P1-P4) + (P1-P6) / 3 ? or we should just choose one of the same distance to make first cluster then keep continue ?
In euclidean distance , u must add other co-ordinates variables in similar manner of x and y A=sqrt[(x2-x1)^2+(y2-y1)^2+(z2-z1)^2+ ...] remaining procedure will be the same.
Thanks for the explanation. I was replicating the exercise in Stata and the result of the last cluster gives 0.279 instead of 0.30. Do you know why this difference between the manual method and Stata can be presented? Thanks for your attention
![K-Mean Clustering [Single Dataset]](http://i.ytimg.com/vi/rESDfP2ik3k/mqdefault.jpg)
![K-Mean Clustering [Single Dataset]](/img/tr.png)
![Hierarchical Agglomerative Clustering [HAC - Single Link]](http://i.ytimg.com/vi/RdT7bhm1M3E/mqdefault.jpg)
Thank you! I just can say that, you literally save my work!
Thanks
Have an exam on multivariate methods soon and your clustering videos have helped me so much! Thank you, I really appreciate your work.
You saved me. . ! Thank you so much. . . There is one small error in the formula you have shown, it should be (Y - B)^2 and not (X -B)^2. . . you solved it correctly but the formula isnt correct. . . Thank you so much, once again. . .
Distance = sqrt [(x2−x1)^2+(y2−y1)^2]
Thank you so much for these amazing videos! Helped me a lot during exams. Keep making more such amazing videos!
Thank you! You've helped me in my study to Data Mining exam
It's literally the best👍
Great Job mam , Thankyou
at 11:19 it should be dist( (P3,P6,P4,P2,P5),P1) = 1/2 x [dist((P3,P6,P4),P1) + dist((P2,P5),P1)] , please check it , in LHS you missed P1
Great lecture.Preparation for combiniing the points and then update the real distance between objects and clusters.
The video explanation is incorrect.
In Average linkage clustering, the distance between two clusters is defined as the average of distances between all pairs of objects, where each pair is made up of one object from each group.
In the average linkage method, D(r,s) is computed as
D(r,s) = Trs / ( Nr * Ns)
Where Trs is the sum of all pairwise distances between cluster r and cluster s. Nrand Ns are the sizes of the clusters r and s, respectively.
The way my lecturer gave me this exact example as an assignment xD
I find this video explanation incorrect.
The average between a cluster with 2 points and the second cluster with 3 points the average of all five points. But in your video, you are teaching as take the average of the first cluster with 2 points and then take the average of the second cluster with 3 points and get the answer. Now average both the averages from above result to get the answer. This will not give the correct answer. It can only satisfy when we have the same points in both the comparing clusters.
For example
avg(1,2) = 1.5
avg(3,4,5) = 4
avg(1,2,3,4,5) = 3
avg(avg(1,2), avg(3,4,5)) = 2.75 which is not equal to 3.
Correct me if I am wrong.
Yes you are right.
In Average linkage clustering, the distance between two clusters is defined as the average of distances between all pairs of objects, where each pair is made up of one object from each group.
In the average linkage method, D(r,s) is computed as
D(r,s) = Trs / ( Nr * Ns)
Where Trs is the sum of all pairwise distances between cluster r and cluster s. Nr and Ns are the sizes of the clusters r and s, respectively.
Thank you soooo much for this video! Life saving indeed!
Good explanation ! Easily understood
Fantastic, thank you
mam, the explanation was crystal clear. tq.
Simple but awesome explanation
are you taking rounding, for example (0.3+0.29)/2 = 0.59 , so you take rounding to 0.6?
Thanks
Thanks for video, i have one question, in the first distance matrix : if we would have (P3-P6) = 0.11 and (P3-P4) = 0.11 , in this case what we would do? then in the next step would we take avarage of for example (P1-P3) + (P1-P4) + (P1-P6) / 3 ? or we should just choose one of the same distance to make first cluster then keep continue ?
thankyou mam, it was very nice and clear explaination.
Thank you Ma'am! Well explained…
Where do we use Single Linkage, Complete linkage and Average Linkage? I mean which approach to use where?
thank u so much mam...i am very thankful to you..you saved me.....once again thanks a lot
Mam, after building model how this predict works like this predict([50,69]) ......plz reply mam
It was very helpful, Thank you so much
But I have one doubt; how to tackle the problem if there are more co-ordinates apart from x and y ?
In euclidean distance , u must add other co-ordinates variables in similar manner of x and y
A=sqrt[(x2-x1)^2+(y2-y1)^2+(z2-z1)^2+ ...] remaining procedure will be the same.
Thank u for saving us 🙏
Very easy and helpful
Thank u very much.
Thanks for the explanation. I was replicating the exercise in Stata and the result of the last cluster gives 0.279 instead of 0.30. Do you know why this difference between the manual method and Stata can be presented?
Thanks for your attention
thank you so much! thats really helpful!!
Very Useful :) Thank you
Very clear, thanks a lot.
what is the logic behind single and complete about their naming convention?
Would average link clustering and centroid clustering in this case be the same?
Where is the HAC- Single Link? vdo
thank you so much!
Thank u so much maam..
very helpfull. thank you.
Kindly you explain the distance between centroid technique in HAC
Thanks!
thank you sooooooooooooooo much
can you pls share the ppt used ?
Awesome
thank you so much mam, this video really helpful.
can you please make a video on ward's method also?
thanks
Saied Akbar Ahmady For sure
@@AnuradhaBhatia Ma'am is there any video for Ward's Method we can refer?
This explanation is incorrect