I'm analyzing a data in R where predictor variables are available but there is no response variable. Using unsupervised learning (k-means) I have identified patterns in the data. But I need to rank the clusters according to their overall performance (example: student's data on exam marks and co-curricular marks). What technique do I use after clustering in R?
The cluster attribute of the kmeans output gives you the index of which cluster each data point is in. Example data taken from kmeans documentation:
nclusters = 5
# a 2-dimensional example
x <- rbind(matrix(rnorm(100, sd = 0.3), ncol = 2),
matrix(rnorm(100, mean = 1, sd = 0.3), ncol = 2))
colnames(x) <- c("x", "y")
cl <- kmeans(x, nclusters, nstart = 25)
Now, your evaluation function (e.g. mean of column values) can be applied to each cluster individually:
for (i in 1:nclusters){
cat(i, apply(x[which(cl$cluster==i),],MARGIN=2,FUN=mean), '\n')
}
Or better still, use some kind of aggregation function, e.g. tapply or aggregate, e.g.:
aggregate(x, by=list(cluster=cl$cluster), FUN=mean)
which gives
cluster x y
1 1 1.2468266 1.1499059
2 2 -0.2787117 0.0958023
3 3 0.5360855 1.0217910
4 4 1.0997776 0.7175210
5 5 0.2472313 -0.1193551
At this point you should be able to rank the values of the aggregation function as needed.
Related
This is an extended question I found from here (Method #1: http://santiago.begueria.es/2010/10/generating-spatially-correlated-random-fields-with-r/) and here (Method #2: https://gist.github.com/brentp/1306786). I know these two sites covered very well (Thanks!) with relatively small size of dimension (e.g., 1000x1). I am trying to generate spatially clustered binary data with large size of dimension like >=100000x1 dimension, for example, c(1,1,1,1,0,1,0,0,0,0, …, 0,0,0,0,0,0,0,0,0,0,0,0) with 1000 times / case study. Here are slightly modified codes from the sites.
# Method #1
dim1 <- 1000
dim2 <- 1
xy <- expand.grid(seq_len(dim1), seq_len(dim2))
colnames(xy) <- c("x", "y")
geo.model <- gstat(formula = z~x+y, locations = ~x+y, dummy = TRUE, beta = 0,
model = vgm(psill = 1,"Exp",
range = dim1), # Range parameter!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
nmax = 30) # Spatial correlation model
sim.mat <- predict(geo.model, newdata = xy, nsim = 1)
sim.mat[,3] <- ifelse(sim.mat[,3] > quantile(sim.mat[,3], .1), 0, 1)
plot(sim.mat[, 3])
# Method #2
# generate autocorrelated data.
nLags = 1000 # number of lags (size of region)
# fake, uncorrelated observations
X = rnorm(nLags)
# fake sigma... correlated decreases distance.
sigma = diag(nLags)
corr = .999
sigma <- corr ^ abs(row(sigma)-col(sigma))
#sigma
# Y is autocorrelated...
Y <- t(X %*% chol(sigma))
y <- ifelse(Y >= quantile(Y, probs=.9), 1, 0)[, 1]
plot(y)
Both methods work very well to generate binary data when dim1 is less than 10000. However, when I tried several hundred thousand (e.g., >= 100,000), it seems to take a long time or memory issue.
For example, when I used “nLags = 50000” in Method #2, I got an error message (“Error: cannot allocate vector of size 9.3 Gb”) after the code “sigma <- corr ^ abs(row(sigma)-col(sigma))”.
I would like to find an efficient (time- and memory-saving) way to generate such a spatially clustered binary data 1000 times (especially, with dim1 >= 100000) per each case study (about 200 cases).
I have thought about applying multiple probabilities in "sample" function or probability distribution. I am not sure how to and beyond my scope.
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'm trying to discretize a numerical variable using Kmeans.
It worked pretty well but I'm wondering how I can find the intervals in my cluster.
I work with FactoMineR to do my kmeans.
I found 3 clusters according to the following graph :
My point now is to identify the intervals of my numerical variable within the clusters.
Is there any option or method in FactoMineR or other package to do it ?
I can do it manually but as I have to do it for a certain amount of variables, I'd like to found an easy way to identify them.
Since you have not provided data I have used the example from the kmeans documentation, which produces two groups for data with two columns x and y. You may split the original data by the cluster each row belongs to and then extract data from each group. I am not sure if my example data resembles your data, but in below code I have simply used the difference between min value of column x and max value of column y as the boundaries of a potential interval (depending on the use case this makes sense or not). Does that help you?
data <- rbind(matrix(rnorm(100, sd = 0.3), ncol = 2),
matrix(rnorm(100, mean = 1, sd = 0.3), ncol = 2))
colnames(data) <- c("x", "y")
cl <- kmeans(data, 2)
data <- as.data.frame(cbind(data, cluster = cl$cluster))
lapply(split(data, data$cluster), function(x) {
min_x <- min(x$x)
max_y <- max(x$y)
diff <- max_y-min_x
c(min_x = min_x , max_y = max_y, diff = diff)
})
# $`1`
# min_x max_y diff
# -0.6906124 0.5123950 1.2030074
#
# $`2`
# min_x max_y diff
# 0.2052112 1.6941800 1.4889688
I want to cluster my data to say 5 clusters, then we need to select 50 individuals with most dissimilar relationship from all the data. That means if cluster one contains 100, two contains 200, three contains 400, four contains 200, and five 100, I have to select 5 from the first cluster + 10 from the second cluster + 20 from the third + 10 from the fourth + 5 from the fifth.
Data example:
mydata<-matrix(nrow=100,ncol=10,rnorm(1000, mean = 0, sd = 1))
What I did till now is clustering the data and rank the individuals within each cluster, then export it to excel and go from there …
That has become became a problem since my data has became really big.
I will appreciate any help or suggestion on how to apply the previous in R
.
I´m not sure if it is exactly what you are searching, but maybe it helps:
mydata<-matrix(nrow=100, ncol=10, rnorm(1000, mean = 0, sd = 1))
rownames(mydata) <- paste0("id", 1:100) # some id for identification
# cluster objects and calculate dissimilarity matrix
cl <- cutree(hclust(
sim <- dist(mydata, diag = TRUE, upper=TRUE)), 5)
# combine results, take sum to aggregate dissimilarity
res <- data.frame(id=rownames(mydata),
cluster=cl, dis_sim=rowSums(as.matrix(sim)))
# order, lowest overall dissimilarity will be first
res <- res[order(res$dis_sim), ]
# split object
reslist <- split(res, f=res$cluster)
## takes first three items with highest overall dissim.
lapply(reslist, tail, n=3)
## returns id´s with highest overall dissimilarity, top 20%
lapply(reslist, function(x, p) tail(x, round(nrow(x)*p)), p=0.2)
regarding you comment, find the code below:
pleas note that the code can be improved in terms of beauty and efficiency.
Further I used a second answer, because otherwise it would be to messy.
# calculation of centroits based on:
# https://stat.ethz.ch/pipermail/r-help/2006-May/105328.html
cl <- hclust(dist(mydata, diag = TRUE, upper=TRUE))
cent <- tapply(mydata,
list(rep(cutree(cl, 5), ncol(mydata)), col(mydata)), mean)
dimnames(cent) <- list(NULL, dimnames(mydata)[[2]])
# add up cluster number and data and split by cluster
newdf <- data.frame(data=mydata, cluster=cutree(cl, k=5))
newdfl <- split(newdf, f=newdf$cluster)
# add centroids and drop cluster info
totaldf <- lapply(1:5,
function(i, li, cen) rbind(cen[i, ], li[[i]][ , -11]),
li=newdfl, cen=cent)
# calculate new distance to centroits and sort them
dist_to_cent <- lapply(totaldf, function(x)
sort(as.matrix(dist(x, diag=TRUE, upper=TRUE))[1, ]))
dist_to_cent
for calculation of centroids out of hclust see R-Mailinglist
I want to cluster my data to say 5 clusters, then we need to select 50 individuals with most dissimilar relationship from all the data. That means if cluster one contains 100, two contains 200, three contains 400, four contains 200, and five 100, I have to select 5 from the first cluster + 10 from the second cluster + 20 from the third + 10 from the fourth + 5 from the fifth.
Data example:
mydata<-matrix(nrow=100,ncol=10,rnorm(1000, mean = 0, sd = 1))
What I did till now is clustering the data and rank the individuals within each cluster, then export it to excel and go from there …
That has become became a problem since my data has became really big.
I will appreciate any help or suggestion on how to apply the previous in R
.
I´m not sure if it is exactly what you are searching, but maybe it helps:
mydata<-matrix(nrow=100, ncol=10, rnorm(1000, mean = 0, sd = 1))
rownames(mydata) <- paste0("id", 1:100) # some id for identification
# cluster objects and calculate dissimilarity matrix
cl <- cutree(hclust(
sim <- dist(mydata, diag = TRUE, upper=TRUE)), 5)
# combine results, take sum to aggregate dissimilarity
res <- data.frame(id=rownames(mydata),
cluster=cl, dis_sim=rowSums(as.matrix(sim)))
# order, lowest overall dissimilarity will be first
res <- res[order(res$dis_sim), ]
# split object
reslist <- split(res, f=res$cluster)
## takes first three items with highest overall dissim.
lapply(reslist, tail, n=3)
## returns id´s with highest overall dissimilarity, top 20%
lapply(reslist, function(x, p) tail(x, round(nrow(x)*p)), p=0.2)
regarding you comment, find the code below:
pleas note that the code can be improved in terms of beauty and efficiency.
Further I used a second answer, because otherwise it would be to messy.
# calculation of centroits based on:
# https://stat.ethz.ch/pipermail/r-help/2006-May/105328.html
cl <- hclust(dist(mydata, diag = TRUE, upper=TRUE))
cent <- tapply(mydata,
list(rep(cutree(cl, 5), ncol(mydata)), col(mydata)), mean)
dimnames(cent) <- list(NULL, dimnames(mydata)[[2]])
# add up cluster number and data and split by cluster
newdf <- data.frame(data=mydata, cluster=cutree(cl, k=5))
newdfl <- split(newdf, f=newdf$cluster)
# add centroids and drop cluster info
totaldf <- lapply(1:5,
function(i, li, cen) rbind(cen[i, ], li[[i]][ , -11]),
li=newdfl, cen=cent)
# calculate new distance to centroits and sort them
dist_to_cent <- lapply(totaldf, function(x)
sort(as.matrix(dist(x, diag=TRUE, upper=TRUE))[1, ]))
dist_to_cent
for calculation of centroids out of hclust see R-Mailinglist