I have a large dataset composed of numerical observations. For this data set, I calculated k-means for which I defined 6 clusters. How can I draw a heatmap, of each cluster? When I try the following, I get an error:
clusters <- kmeans(dataset, 6)
heatmap(clusters$cluster)
So basically, what you can do is to subset further, and use a for loop. The result of clusters is a list, in which centers, or k (clusters), are the sets of observations you specify in the kmeans function.
clusters <- kmeans(dataset, k = 6)
for (i in c(1:k)) {
pheatmap(dataset[names(clusters$cluster[clusters$cluster == i]), ])
}
Here, if k = 6, you get 6 heatmaps, one of each cluster of observations from your dataset.
Related
enter image description hereMy primary objective is to analyze a video by splitting it into frames of RGB images and then converting this RGB image into separate color matrices. I did that in MATLAB and imported it into R for analysis. I have 125 frames each with 3 color matrices (R,G and B). I would like to create a loop to get the summary statistics of each color matrix and also do a K-means cluster analysis on each of the matrices.
I did the summary statistics manually for each matrix, which was fine for this video. However, I have several more so I need to find a way to loop it. I tried to do the cluster analysis on red channel of frame 1 using the package factoextra and cluster.enter image description here
df <- Frame1[["redChannel"]]
fviz_nbclust(df, kmeans, method = "wss")
km <- kmeans(df, centers = 3, nstart = 25)
View(km)
kmRed1 <- kmeans(df, centers = 4, nstart = 25)
View(kmRed1)
View(kmRed1)
kmRed1
K-means clustering with 4 clusters of sizes 4, 886, 8, 14
One of the clusters has 886 data pints and others just have single digits. I don't understand what it means. I would like to loop this as well for all the other matrices and when I try to plot the cluster using this code:
fviz_cluster(km, data = df)
I am getting this error message:
Error in prcomp.default(data, scale = FALSE, center = FALSE) :
cannot rescale a constant/zero column to unit variance
I am new To R and would appreciate any suggestions.
I want to do a Kmeans clustering on a dataset (namely, Sample_Data) with three variables (columns) such as below:
A B C
1 12 10 1
2 8 11 2
3 14 10 1
. . . .
. . . .
. . . .
in a typical way, after scaling the columns, and determining the number of clusters, I will use this function in R:
Sample_Data <- scale(Sample_Data)
output_kmeans <- kmeans(Sample_Data, centers = 5, nstart = 50)
But, what if there is a preference for the variables? I mean that, suppose variable (column) A, is more important than the two other variables?
how can I insert their weights in the model?
Thank you all
You have to use a kmeans weighted clustering, like the one presented in flexclust package:
https://cran.r-project.org/web/packages/flexclust/flexclust.pdf
The function
cclust(x, k, dist = "euclidean", method = "kmeans",
weights=NULL, control=NULL, group=NULL, simple=FALSE,
save.data=FALSE)
Perform k-means clustering, hard competitive learning or neural gas on a data matrix.
weights An optional vector of weights to be used in the fitting process. Works only in combination with hard competitive learning.
A toy example using iris data:
library(flexclust)
data(iris)
cl <- cclust(iris[,-5], k=3, save.data=TRUE,weights =c(1,0.5,1,0.1),method="hardcl")
cl
kcca object of family ‘kmeans’
call:
cclust(x = iris[, -5], k = 3, method = "hardcl", weights = c(1, 0.5, 1, 0.1), save.data = TRUE)
cluster sizes:
1 2 3
50 59 41
As you can see from the output of cclust, also using competitive learning the family is always kmenas.
The difference is related to cluster assignment during training phase:
If method is "kmeans", the classic kmeans algorithm as given by
MacQueen (1967) is used, which works by repeatedly moving all cluster
centers to the mean of their respective Voronoi sets. If "hardcl",
on-line updates are used (AKA hard competitive learning), which work
by randomly drawing an observation from x and moving the closest
center towards that point (e.g., Ripley 1996).
The weights parameter is just a sequence of numbers, in general I use number between 0.01 (minimum weight) and 1 (maximum weight).
I had the same problem and the answer here is not satisfying for me.
What we both wanted was an observation-weighted k-means clustering in R. A good readable example for our question is this link: https://towardsdatascience.com/clustering-the-us-population-observation-weighted-k-means-f4d58b370002
However the solution to use the flexclust package is not satisfying simply b/c the used algorithm is not the "standard" k-means algorithm but the "hard competitive learning" algorithm. The difference are well described above and in the package description.
I looked through many sites and did not find any solution/package in R in order to use to perform a "standard" k-means algorithm with weighted observations. I was also wondering why the flexclust package explicitly do not support weights with the standard k-means algorithm. If anyone has an explanation for this, please feel free to share!
So basically you have two options: First, rewrite the flexclust-algorithm to enable weights within the standard approach. Or second, you can estimate weighted cluster centroids as starting centroids and perform a standard k-means algorithm with only one iteration, then compute new weighted cluster centroids and perform a k-means with one iteration and so on until you reach convergence.
I used the second alternative b/c it was the easier way for me. I used the data.table package, hope you are familiar with it.
rm(list=ls())
library(data.table)
### gen dataset with sample-weights
dataset <- data.table(iris)
dataset[, weights:= rep(c(1, 0.7, 0.3, 4, 5),30)]
dataset[, Species := NULL]
### initial hclust for estimating weighted centroids
clustering <- hclust(dist(dataset[, c(1:4)], method = 'euclidean'),
method = 'ward.D2')
no_of_clusters <- 4
### estimating starting centroids (weighted)
weighted_centroids <- matrix(NA, nrow = no_of_clusters,
ncol = ncol(dataset[, c(1:4)]))
for (i in (1:no_of_clusters))
{
weighted_centroids[i,] <- sapply(dataset[, c(1:4)][cutree(clustering, k =
no_of_clusters) == i,], weighted.mean, w = dataset[cutree(clustering, k = no_of_clusters) == i, weights])
}
### performing weighted k-means as explained in my post
iter <- 0
cluster_i <- 0
cluster_iminus1 <- 1
## while loop: if number of iteration is smaller than 50 and cluster_i (result of
## current iteration) is not identical to cluster_iminus1 (result of former
## iteration) then continue
while(identical(cluster_i, cluster_iminus1) == F && iter < 50){
# update iteration
iter <- iter + 1
# k-means with weighted centroids and one iteration (may generate warning messages
# as no convergence is reached)
cluster_kmeans <- kmeans(x = dataset[, c(1:4)], centers = weighted_centroids, iter = 1)$cluster
# estimating new weighted centroids
weighted_centroids <- matrix(NA, nrow = no_of_clusters,
ncol=ncol(dataset[,c(1:4)]))
for (i in (1:no_of_clusters))
{
weighted_centroids[i,] <- sapply(dataset[, c(1:4)][cutree(clustering, k =
no_of_clusters) == i,], weighted.mean, w = dataset[cutree(clustering, k = no_of_clusters) == i, weights])
}
# update cluster_i and cluster_iminus1
if(iter == 1) {cluster_iminus1 <- 0} else{cluster_iminus1 <- cluster_i}
cluster_i <- cluster_kmeans
}
## merge final clusters to data table
dataset[, cluster := cluster_i]
If you want to increase the weight of a variable (column), just multiply it with a constant c > 1.
It's trivial to show that this increases the weight in the SSQ optimization objective.
I have a dataframe with three features as
library(cluster)
df <- data.frame(f1=rnorm(480,30,1),
f2=rnorm(480,40,0.5),
f3=rnorm(480,50, 2))
Now, I want to do clustering using K-medoids in two steps. In step 1, using some data from df I want to get medoids (cluster centers), and in step 2, I want to use obtained medoids to do clustering on remaining data. Accordingly,
# find medoids using some data
sample_data <- df[1:240,]
sample_data <- scale(sample_data) # scaling features
clus_res1 <- pam(sample_data,k = 4,diss=FALSE)
# Now perform clustering using medoids obtained from above clustering
test_data <- df[241:480,]
test_data <- scale(test_data)
clus_res2 <- pam(test_data,k = 4,diss=FALSE,medoids=clus_res1$medoids)
With this script, I get an error message as
Error in pam(test_data, k = 4, diss = FALSE, medoids = clus_res1$medoids) :
'medoids' must be NULL or vector of 4 distinct indices in {1,2, .., n}, n=240
It is clear that error message is due to the input format of Medoid matrix. How can I convert this matrix to the vector as specified in the error message?
The initial medoids parameter expects index numbers of points in your data set. So 42,17 means to use objects 42 and 17 as initial medoids.
By the definition of medoids, you can only use points of your data set as medoids, not other vectors!
Clustering is unsupervised. No need to split your data in training/test, because there are no labels to overfit to in unsupervised learning.
Notice that in PAM the clustering center is an observation, that is you get 4 observations that each of them is a center of cluster. Demonstration of PAM.
So if you want to try and use the same center, you need to find the observations which are closest to the observations who are the center in your train.
I'm having issue with predicting cluster labeling for a test data, based on a dbscan clustering model on the training data.
I used gower distance matrix when creating the model:
> gowerdist_train <- daisy(analdata_train,
metric = "gower",
stand = FALSE,
type = list(asymm = c(5,6)))
Using this gowerdist matrix, the dbscan clustering model created was:
> sb <- dbscan(gowerdist_train, eps = .23, minPts = 50)
Then I try to use predict to label a test dataset using the above dbscan object:
> predict(sb, newdata = analdata_test, data = analdata_train)
But I receive the following error:
Error in frNN(rbind(data, newdata), eps = object$eps, sort = TRUE,
...) : x has to be a numeric matrix
I can take a guess on where this error might be coming from, which is probably due to the absence of the gower distance matrix that hasn't been created for the test data.
My question is, should I create a gower distance matrix for all data (datanal_train + datanal_test) separately and feed it into predict? how else would the algorithm know what the distance of test data from the train data is, in order to label?
In that case, would the newdata parameter be the new gower distance matrix that contains ALL (train + test) data? and the data parameter in predict would be the training distance matrix, gowerdist_train?
What I am not quite sure about is how would the predict algorithm distinguish between the test and train data set in the newly created gowerdist_all matrix?
The two matrices (new gowerdist for all data and the gowerdist_train) would obviously not have the same dimensions. Also, it doesn't make sense to me to create a gower distance matrix only for test data because distances must be relative to the test data, not the test data itself.
Edit:
I tried using gower distance matrix for all data (train + test) as my new data and received an error when fed to predict:
> gowerdist_all <- daisy(rbind(analdata_train, analdata_test),
metric = "gower",
stand = FALSE,
type = list(asymm = c(5,6)))
> test_sb_label <- predict(sb, newdata = gowerdist_all, data = gowerdist_train)
ERROR: Error in 1:nrow(data) : argument of length 0 In addition:
Warning message: In rbind(data, newdata) : number of columns of
result is not a multiple of vector length (arg 1)
So, my suggested solution doesn't work.
I decided to create a code that would use KNN algorithm in dbscan to predict cluster labeling using gower distance matrix. The code is not very pretty and definitely not programmaticaly efficient but it works. Happy for any suggestions that would improve it.
The pseydocode is:
1) calculate new gower distance matrix for all data, including test and train
2) use the above distance matrix in kNN function (dbscan package) to determine the k nearest neighbours to each test data point.
3) determine the cluster labels for all those nearest points for each test point. Some of them will have no cluster labeling because they are test points themselves
4) create a count matrix to count the frequency of clusters for the k nearest points for each test point
5) use very simple likelihood calculation to choose the cluster for the test point based on its neighbours clusters (the maximum frequency). this part also considers the neighbouring test points. That is, the cluster for the test point is chosen only when the maximum frequency is largest when you add the number of neighbouring test points to the other clusters. Otherwise, it doesn't decide the cluster for that test point and waits for the next iteration when hopefully more of its neighboring test points have had their cluster label decided based on their neighbours.
6) repeat above (steps 2-5) until you've decided all clusters
** Note: this algorithm doesn't converge all the time. (once you do the math, it's obvious why that is) so, in the code i break out of the algorithm when the number of unclustered test points doesn't change after a while. then i repeat 2-6 again with new knn (change the number of nearest neighbours and then run the code again). This will ensure more points are involved in deciding in th enext round. I've tried both larger and smaller knn's and both work. Would be good to know which one is better. I haven't had to run the code more than twice so far to decide the clusters for the test data point.
Here is the code:
#calculate gower distance for all data (test + train)
gowerdist_test <- daisy(all_data[rangeofdataforgowerdist],
metric = "gower",
stand = FALSE,
type = list(asymm = listofasymmvars),
weights = Weights)
summary(gowerdist_test)
Then use the code below to label clusters for test data.
#library(dbscan)
# find the k nearest neibours for each point and order them with distance
iteration_MAX <- 50
iteration_current <- 0
maxUnclusterRepeatNum <- 10
repeatedUnclustNum <- 0
unclusteredNum <- sum(is.na(all_data$Cluster))
previousUnclustereNum <- sum(is.na(all_data$Cluster))
nn_k = 30 #number of neighbourhoods
while (anyNA(all_data$Cluster) & iteration_current < iteration_MAX)
{
if (repeatedUnclustNum >= maxUnclusterRepeatNum) {
print(paste("Max number of repetition (", maxUnclusterRepeatNum ,") for same unclustered data has reached. Clustering terminated unsuccessfully."))
invisible(gc())
break;
}
nn_test <- kNN(gowerdist_test, k = nn_k, sort = TRUE)
# for the TEST points in all data, find the closets TRAIN points and decide statistically which cluster they could belong to, based on the clusters of the nearest TRAIN points
test_matrix <- nn_test$id[1: nrow(analdata_test),] #create matrix of test data knn id's
numClusts <- nlevels(as.factor(sb_train$cluster))
NameClusts <- as.character(levels(as.factor(sb_train$cluster)))
count_clusters <- matrix(0, nrow = nrow(analdata_test), ncol = numClusts + 1) #create a count matrix that would count number of clusters + NA
colnames(count_clusters) <- c("NA", NameClusts) #name each column of the count matrix to cluster numbers
# get the cluster number of each k nearest neibhour of each test point
for (i in 1:nrow(analdata_test))
for (j in 1:nn_k)
{
test_matrix[i,j] <- all_data[nn_test$id[i,j], "Cluster"]
}
# populate the count matrix for the total clusters of the neighbours for each test point
for (i in 1:nrow(analdata_test))
for (j in 1:nn_k)
{
if (!is.na(test_matrix[i,j]))
count_clusters[i, c(as.character(test_matrix[i,j]))] <- count_clusters[i, c(as.character(test_matrix[i,j]))] + 1
else
count_clusters[i, c("NA")] <- count_clusters[i, c("NA")] + 1
}
# add NA's (TEST points) to the other clusters for comparison
count_clusters_withNA <- count_clusters
for (i in 2:ncol(count_clusters))
{
count_clusters_withNA[,i] <- t(rowSums(count_clusters[,c(1,i)]))
}
# This block of code decides the maximum count of cluster for each row considering the number other test points (NA clusters) in the neighbourhood
max_col_countclusters <- apply(count_clusters,1,which.max) #get the column that corresponds to the maximum value of each row
for (i in 1:length(max_col_countclusters)) #insert the maximum value of each row in its associated column in count_clusters_withNA
count_clusters_withNA[i, max_col_countclusters[i]] <- count_clusters[i, max_col_countclusters[i]]
max_col_countclusters_withNA <- apply(count_clusters_withNA,1,which.max) #get the column that corresponds to the maximum value of each row with NA added
compareCountClust <- max_col_countclusters_withNA == max_col_countclusters #compare the two count matrices
all_data$Cluster[1:nrow(analdata_test)] <- ifelse(compareCountClust, NameClusts[max_col_countclusters - 1], all_data$Cluster) #you subtract one because of additional NA column
iteration_current <- iteration_current + 1
unclusteredNum <- sum(is.na(all_data$Cluster))
if (previousUnclustereNum == unclusteredNum)
repeatedUnclustNum <- repeatedUnclustNum + 1
else {
repeatedUnclustNum <- 0
previousUnclustereNum <- unclusteredNum
}
print(paste("Iteration: ", iteration_current, " - Number of remaining unclustered:", sum(is.na(all_data$Cluster))))
if (unclusteredNum == 0)
print("Cluster labeling successfully Completed.")
invisible(gc())
}
I guess you can use this for any other type of clustering algorithm, it doesn't matter how you decided the cluster labels for the train data, as long as they are in your all_data before running the code.
Hope this help.
Not the most efficient or rigorous code. So, happy to see suggestions how to improve it.
*Note: I used t-SNE to compare the clustering of train with the test data and looks impressively clean. so, it seems it is working.
As someone relatively new to R I'm having an issue with creating a for loop.
I have a very large data set with 9000 observations and 25 categorical variables, which I've transformed into binary data and preformed hierarchical clustering. Now I want to try K-Modes clustering to produce an Elbow Plot using the "within-cluster simple-matching distance for each cluster", which is outputted from kmodes$withindiff. I can sum this for each of the k in 1:8 clusters to get the Elbow Plot.
library(klaR)
for(k in 1:8)
{
WCSM[k] <- sum(kmodes(data,k,iter.max=100)$withindiff)
}
plot(1:8,WCSM,type="b", xlab="Number of Clusters",ylab="Within-Cluster
Simple-Matching Distance Summed", main="K-modes Elbow Plot")
My issue is that I want further output from k-modes. For each k in 1:8 I would like to get the vector of integers indicating the cluster to which each object is allocated to given by kmodes$cluster. I need to create a for loop that loops through each k in 1:8 and saves each of the outputs into 8 separate vectors. But I don't know how to do such a for loop. I could just run the 8 lines of code separately but they each take 15mins to run with iter.max=10 so increasing this to iter.max=100 will need to be left running overnight so a loop would be useful.
cl.kmodes2=kmodes(data, 2,iter.max=100)
cl.kmodes3=kmodes(data, 3,iter.max=100)
cl.kmodes4=kmodes(data, 4,iter.max=100)
cl.kmodes5=kmodes(data, 5,iter.max=100)
cl.kmodes6=kmodes(data, 6,iter.max=100)
cl.kmodes7=kmodes(data, 7,iter.max=100)
cl.kmodes8=kmodes(data, 8,iter.max=100)
Ultimately I want to compare the results from the hierarchical binary clustering to the k-modes clustering by getting the Adjusted Rand Index. For example, cutting the tree at k=4 for the hierarchical cluster and comparing this to a 4 cluster solution from k-modes:
dist.binary = dist(data, method="binary")
cl.binary = hclust(dist.binary, method="complete")
hcl.4 = cutree(cl.binary, k = 4)
tab = table(hcl.4, cl.kmodes4$cluster)
library(e1071)
classAgreement(tab)
I agree with Imo, using a list is the best solution.
If you don't want to do that, you could also use assign() to create a new vector in every iteration:
library(klaR)
for(k in 1:8) {
assign(paste("cl.kmodes", k, sep = ""), kmodes(data, k, iter.max = 100))
}
The best method is to put the output from your clusters into a named list:
library(klaR)
myClusterList <- list()
for(k in 1:8) {
myClusterList[[paste0("k.", i)]] <- kmodes(data, i,iter.max=100)
}
You can then pull out the any of the contents easily:
sum(myClusterList[["k.1"]]$withindiff)
or
sum(myClusterList[[1]]$withindiff)
You can also save the list to use in future R sessions, see ?save.