Implementation of kmeans clustering using R
$begingroup$
I have implemented kmeans clustering on iris dataset (inbuilt dataset) in R. The code is given below:
X=as.matrix(iris[-5]);
K=3;
prevCentroids=matrix(0,K,dim(X)[2]);
centroids=X[sample(1:dim(X)[1],K),];
dot=numeric(3);
C=numeric(150);
while(!isTRUE(all.equal(centroids,prevCentroids)))
{
for(i in 1:dim(X)[1])
{
for(j in 1:dim(centroids)[1])
{
dot[j]=(X[i,]-centroids[j,])%*%(X[i,]-centroids[j,]);
}
C[i]=which.min(dot);
}
prevCentroids=centroids;
for(k in 1:K)
{
centroids[k,]=colMeans(X[which(C==k),]);
}
}
print(cbind(iris,C));
Sometimes, with this code, I get 85% clustering correct. But sometimes, it is just 37% as correct, if I compare it with the already clustered inbuilt iris dataset.
Could anyone please tell me where I am going wrong?
k-means
edited Jan 11 '18 at 7:25
Toros91
1,9762628
asked Jan 10 '18 at 16:54
user44436user44436
61
$endgroup$
add a comment |
$begingroup$
I have implemented kmeans clustering on iris dataset (inbuilt dataset) in R. The code is given below:
X=as.matrix(iris[-5]);
K=3;
prevCentroids=matrix(0,K,dim(X)[2]);
centroids=X[sample(1:dim(X)[1],K),];
dot=numeric(3);
C=numeric(150);
while(!isTRUE(all.equal(centroids,prevCentroids)))
{
for(i in 1:dim(X)[1])
{
for(j in 1:dim(centroids)[1])
{
dot[j]=(X[i,]-centroids[j,])%*%(X[i,]-centroids[j,]);
}
C[i]=which.min(dot);
}
prevCentroids=centroids;
for(k in 1:K)
{
centroids[k,]=colMeans(X[which(C==k),]);
}
}
print(cbind(iris,C));
Sometimes, with this code, I get 85% clustering correct. But sometimes, it is just 37% as correct, if I compare it with the already clustered inbuilt iris dataset.
Could anyone please tell me where I am going wrong?
k-means
edited Jan 11 '18 at 7:25
Toros91
1,9762628
asked Jan 10 '18 at 16:54
user44436user44436
61
$endgroup$
$begingroup$
why didn't you use set.seed(), so that the same sample is extracted every-time you run the code. since you haven't enabled it, your outcome changes every-time . Why are you trying to implement everything by yourself rather than using KMeans directly in R?
$endgroup$
– Toros91
Jan 11 '18 at 7:47
$begingroup$
@toros91 probably because implementing an algorithm is the best way to understand how it works. You've never implemented kmeans?
$endgroup$
– David Marx
Jan 11 '18 at 8:17
$begingroup$
@DavidMarx: Probably what you said is true, I did but long time ago.
$endgroup$
– Toros91
Jan 11 '18 at 8:25
add a comment |
$begingroup$
I have implemented kmeans clustering on iris dataset (inbuilt dataset) in R. The code is given below:
X=as.matrix(iris[-5]);
K=3;
prevCentroids=matrix(0,K,dim(X)[2]);
centroids=X[sample(1:dim(X)[1],K),];
dot=numeric(3);
C=numeric(150);
while(!isTRUE(all.equal(centroids,prevCentroids)))
{
for(i in 1:dim(X)[1])
{
for(j in 1:dim(centroids)[1])
{
dot[j]=(X[i,]-centroids[j,])%*%(X[i,]-centroids[j,]);
}
C[i]=which.min(dot);
}
prevCentroids=centroids;
for(k in 1:K)
{
centroids[k,]=colMeans(X[which(C==k),]);
}
}
print(cbind(iris,C));
Sometimes, with this code, I get 85% clustering correct. But sometimes, it is just 37% as correct, if I compare it with the already clustered inbuilt iris dataset.
Could anyone please tell me where I am going wrong?
k-means
edited Jan 11 '18 at 7:25
Toros91
1,9762628
asked Jan 10 '18 at 16:54
user44436user44436
61
$endgroup$
I have implemented kmeans clustering on iris dataset (inbuilt dataset) in R. The code is given below:
X=as.matrix(iris[-5]);
K=3;
prevCentroids=matrix(0,K,dim(X)[2]);
centroids=X[sample(1:dim(X)[1],K),];
dot=numeric(3);
C=numeric(150);
while(!isTRUE(all.equal(centroids,prevCentroids)))
{
for(i in 1:dim(X)[1])
{
for(j in 1:dim(centroids)[1])
{
dot[j]=(X[i,]-centroids[j,])%*%(X[i,]-centroids[j,]);
}
C[i]=which.min(dot);
}
prevCentroids=centroids;
for(k in 1:K)
{
centroids[k,]=colMeans(X[which(C==k),]);
}
}
print(cbind(iris,C));
Sometimes, with this code, I get 85% clustering correct. But sometimes, it is just 37% as correct, if I compare it with the already clustered inbuilt iris dataset.
Could anyone please tell me where I am going wrong?
k-means
k-means
edited Jan 11 '18 at 7:25
Toros91
1,9762628
asked Jan 10 '18 at 16:54
user44436user44436
61
edited Jan 11 '18 at 7:25
Toros91
1,9762628
asked Jan 10 '18 at 16:54
user44436user44436
61
edited Jan 11 '18 at 7:25
Toros91
1,9762628
edited Jan 11 '18 at 7:25
Toros91
1,9762628
edited Jan 11 '18 at 7:25
Toros91
1,9762628
1,9762628
asked Jan 10 '18 at 16:54
user44436user44436
61
asked Jan 10 '18 at 16:54
user44436user44436
61
asked Jan 10 '18 at 16:54
user44436user44436
61
61
$begingroup$
why didn't you use set.seed(), so that the same sample is extracted every-time you run the code. since you haven't enabled it, your outcome changes every-time . Why are you trying to implement everything by yourself rather than using KMeans directly in R?
$endgroup$
– Toros91
Jan 11 '18 at 7:47
$begingroup$
@toros91 probably because implementing an algorithm is the best way to understand how it works. You've never implemented kmeans?
$endgroup$
– David Marx
Jan 11 '18 at 8:17
$begingroup$
@DavidMarx: Probably what you said is true, I did but long time ago.
$endgroup$
– Toros91
Jan 11 '18 at 8:25
add a comment |
$begingroup$
why didn't you use set.seed(), so that the same sample is extracted every-time you run the code. since you haven't enabled it, your outcome changes every-time . Why are you trying to implement everything by yourself rather than using KMeans directly in R?
$endgroup$
– Toros91
Jan 11 '18 at 7:47
$begingroup$
@toros91 probably because implementing an algorithm is the best way to understand how it works. You've never implemented kmeans?
$endgroup$
– David Marx
Jan 11 '18 at 8:17
$begingroup$
@DavidMarx: Probably what you said is true, I did but long time ago.
$endgroup$
– Toros91
Jan 11 '18 at 8:25
$begingroup$
why didn't you use set.seed(), so that the same sample is extracted every-time you run the code. since you haven't enabled it, your outcome changes every-time . Why are you trying to implement everything by yourself rather than using KMeans directly in R?
$endgroup$
– Toros91
Jan 11 '18 at 7:47
$begingroup$
why didn't you use set.seed(), so that the same sample is extracted every-time you run the code. since you haven't enabled it, your outcome changes every-time . Why are you trying to implement everything by yourself rather than using KMeans directly in R?
$endgroup$
– Toros91
Jan 11 '18 at 7:47
$begingroup$
@toros91 probably because implementing an algorithm is the best way to understand how it works. You've never implemented kmeans?
$endgroup$
– David Marx
Jan 11 '18 at 8:17
$begingroup$
@toros91 probably because implementing an algorithm is the best way to understand how it works. You've never implemented kmeans?
$endgroup$
– David Marx
Jan 11 '18 at 8:17
$begingroup$
@DavidMarx: Probably what you said is true, I did but long time ago.
$endgroup$
– Toros91
Jan 11 '18 at 8:25
$begingroup$
@DavidMarx: Probably what you said is true, I did but long time ago.
$endgroup$
– Toros91
Jan 11 '18 at 8:25
add a comment |
1 Answer
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$begingroup$
Actually, this is the expected behavior for running a k-means algorithm on the iris dataset. The iris dataset contains only two distinct clusters. So if you randomly set up three centroids, the third centroid will either end up on the right or on the wrong cluster, causing the algorithm to split that cluster into two (see the left picture).
Your outcome with 37% accuracy will most probably look somewhat similar to the left picture where the left cluster has been split into two clusters. The 85% accuracy will most probably come from those times in which the right cluster has been split into two clusters (you cannot be sure though without printing the results).
![Chire [Public domain], from Wikimedia Commons](https://i.stack.imgur.com/KygZy.png)
For testing purposes, you can simply use set.seed(123) before running the algorithm, so that you always get the same results. However, this doesn't solve the problem that the iris dataset isn't ideal for clustering purposes. I would suggest you check out other datasets or you give the R-package cluster.datasets a try.
answered 11 hours ago
georg_ungeorg_un
1
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georg_un is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$begingroup$
Actually, this is the expected behavior for running a k-means algorithm on the iris dataset. The iris dataset contains only two distinct clusters. So if you randomly set up three centroids, the third centroid will either end up on the right or on the wrong cluster, causing the algorithm to split that cluster into two (see the left picture).
Your outcome with 37% accuracy will most probably look somewhat similar to the left picture where the left cluster has been split into two clusters. The 85% accuracy will most probably come from those times in which the right cluster has been split into two clusters (you cannot be sure though without printing the results).
For testing purposes, you can simply use set.seed(123) before running the algorithm, so that you always get the same results. However, this doesn't solve the problem that the iris dataset isn't ideal for clustering purposes. I would suggest you check out other datasets or you give the R-package cluster.datasets a try.
answered 11 hours ago
georg_ungeorg_un
1
New contributor
georg_un is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Actually, this is the expected behavior for running a k-means algorithm on the iris dataset. The iris dataset contains only two distinct clusters. So if you randomly set up three centroids, the third centroid will either end up on the right or on the wrong cluster, causing the algorithm to split that cluster into two (see the left picture).
Your outcome with 37% accuracy will most probably look somewhat similar to the left picture where the left cluster has been split into two clusters. The 85% accuracy will most probably come from those times in which the right cluster has been split into two clusters (you cannot be sure though without printing the results).
For testing purposes, you can simply use set.seed(123) before running the algorithm, so that you always get the same results. However, this doesn't solve the problem that the iris dataset isn't ideal for clustering purposes. I would suggest you check out other datasets or you give the R-package cluster.datasets a try.
answered 11 hours ago
georg_ungeorg_un
1
New contributor
georg_un is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Actually, this is the expected behavior for running a k-means algorithm on the iris dataset. The iris dataset contains only two distinct clusters. So if you randomly set up three centroids, the third centroid will either end up on the right or on the wrong cluster, causing the algorithm to split that cluster into two (see the left picture).
Your outcome with 37% accuracy will most probably look somewhat similar to the left picture where the left cluster has been split into two clusters. The 85% accuracy will most probably come from those times in which the right cluster has been split into two clusters (you cannot be sure though without printing the results).
For testing purposes, you can simply use set.seed(123) before running the algorithm, so that you always get the same results. However, this doesn't solve the problem that the iris dataset isn't ideal for clustering purposes. I would suggest you check out other datasets or you give the R-package cluster.datasets a try.
answered 11 hours ago
georg_ungeorg_un
1
New contributor
georg_un is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Actually, this is the expected behavior for running a k-means algorithm on the iris dataset. The iris dataset contains only two distinct clusters. So if you randomly set up three centroids, the third centroid will either end up on the right or on the wrong cluster, causing the algorithm to split that cluster into two (see the left picture).
Your outcome with 37% accuracy will most probably look somewhat similar to the left picture where the left cluster has been split into two clusters. The 85% accuracy will most probably come from those times in which the right cluster has been split into two clusters (you cannot be sure though without printing the results).
For testing purposes, you can simply use set.seed(123) before running the algorithm, so that you always get the same results. However, this doesn't solve the problem that the iris dataset isn't ideal for clustering purposes. I would suggest you check out other datasets or you give the R-package cluster.datasets a try.
answered 11 hours ago
georg_ungeorg_un
1
New contributor
georg_un is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 11 hours ago
georg_ungeorg_un
1
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answered 11 hours ago
georg_ungeorg_un
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answered 11 hours ago
georg_ungeorg_un
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$begingroup$
why didn't you use set.seed(), so that the same sample is extracted every-time you run the code. since you haven't enabled it, your outcome changes every-time . Why are you trying to implement everything by yourself rather than using KMeans directly in R?
$endgroup$
– Toros91
Jan 11 '18 at 7:47
$begingroup$
@toros91 probably because implementing an algorithm is the best way to understand how it works. You've never implemented kmeans?
$endgroup$
– David Marx
Jan 11 '18 at 8:17
$begingroup$
@DavidMarx: Probably what you said is true, I did but long time ago.
$endgroup$
– Toros91
Jan 11 '18 at 8:25