similar to: Help needed to clarify hclust and cutree algorithms

Hi, I have the distance matrix computed and I feed it to hclust function. The plot function produces a dense dendrogram as well. But, the cutree function applied does not produce the desired list. Here is the code x=data.frame(similarity_matrix) colnames(x) = c(source_tags_vec) rownames(x) = c(source_tags_vec) clust_tree=hclust(as.dist(x),method="complete") plot(clust_tree)

Hi! I just discovered that cutree() and cut.dendrogram() do not assign the same cluster numberings when called on the same tree. More specifically, cutree() assigns cluster numbers by order of appearance in the data, while cut.dendrogram() sorts clusters by height (see example below). I guess this is for historical reasons? I'm hit by this difference when I want to get a vector of cluster

I am making heatmaps for a dataset (~ 300*600 matrix) with the following R script (I am not familiar with R and this is the first time I am using it). library("gplots") library("Cairo") mydata <- read.csv(file="data.csv", header=TRUE, sep=",") rownames(mydata)=mydata$Name mydata <- mydata[,2:297] mydatamatrix <- data.matrix(mydata) mydatascale

Full_Name: Anja von Heydebreck Version: 1.3.0 OS: Alpha Unix Submission from: (NULL) (141.14.19.61) Hi, I repeatedly obtained meaningless results from the function 'cutree' in the 'mva' package, when the argument 'h' was greater or equal to the maximum height occuring: > library('mva') > y [,1] [,2] [,3] [,4] [1,] 0 1 -1 1 [2,] 0 -1

I have been attempting to do some work using hclust, and have run into a (possibly subtle) problem. The background is that I constructed a dissimilarity matrix ``d1'' (it involved something called the ``Jaccard similarity coefficient''; I won't go into the details unless requested). I then did d2 <- as.dist(d1) try <- hclust(d2,method=ward)

I've found that while cutree() and rect.hclust() make the same classes for a given height in the dendrogram, the actual labeling of the classes is different. For example, both produce the same 4 classes but class 1 according to cutree() is class 4 according to rect.hclust(). Would it be possible that future versions provide the same labeling? rect.hclust() is useful to display the classes

Hello, I would like to cut a hclust tree into several groups at a specific similarity. I assume this can be achieved by specifying the "h" argument with the specified similarity, e.g.: clust<-hclust(dist,"average") cut<-cutree(clust,h=0.65) Now, I would like to draw rectangles around the branches of the dendrogram highlighting the corresponding clusters, as is done by

Hi Everyone, I am doing hierarchical clustering analysis and have a question regarding "cutree". I am doing things like this: hc <- hclust(dist(X)) a <- cutree(hc, k=2) Basically "a" is a vector containing the assignments of 1 or 2 for each sample. May I know how "cutree" decides to assign 1 and 2's to each sample (in other words, how clusters 1 and 2

Hi! In the example: hc <- hclust(dist(USArrests), "ave") dend1 <- as.dendrogram(hc) dend2 <- cut(dend1, h=70) Do the branches "Branch 1", "Branch 2", "Branch 2"...in dend2$upper str(dend2$upper) --[dendrogram w/ 2 branches and 4 members at h = 152] |--[dendrogram w/ 2 branches and 2 members at h = 77.6] | |--leaf "Branch 1" (h=

Good afternoon, After cuting a hierarchical tree using cutree(), how to check correspondances between classes and branches? This is what we do: srndpchc <- hclust(dist(srndpc$x[1:1000,1:3]),method="ward") #creation of hierarchical tree plclust(srndpchc,hmin=20000) #visualisation srndpchc20000 = cutree(srndpchc,h=20000) #returns 4 classes table(srndpchc20000 ) srndclass20000 =

I had a look at the online documentation, and didn't see from that what is my problem. If I should have, pardon me. Here is my session. As I understand the documentation, this should work with only an hclust object. I get a similar error when in include a FUN argument. I am using V2.4.0. > hc Call: hclust(d = dist(mtx2, method = "manh"), method =

Hi, I have a gene expression experiment with 20 samples and 25000 genes each. I'd like to perform clustering on these. It turned out to become much faster when I transform the underlying matrix with t(matrix). Unfortunately then I'm not anymore able to use cutree to access individual clusters. In general I do something like this: hc <- hclust(dist(USArrests), "ave")

I have the following code that perform hiearchical clustering and plot them in heatmap. __ library(gplots) set.seed(538) # generate data y <- matrix(rnorm(50), 10, 5, dimnames=list(paste("g", 1:10, sep=""), paste("t", 1:5, sep=""))) # the actual data is much larger that the above # perform hiearchical clustering and plot heatmap test <- heatmap.2(y)

Hello, I'm having trouble figuring out how to see resulting groups (clusters) from my hclust() output. I have a very large matrix of 4371 plots and 29 species, so simply looking at the graph is impossible. There must be a way to 'print' the results to a table that shows which plots were in what group, correct? I've attached the matrix I'm working with (the whole thing

Hi, this is rather a (presumed) bug report than a question because I can solve my personal statistical problem by working with hclust instead of agnes. I have done a complete linkage clustering on a dist object dm with 30 objects with agnes (R 1.8.0 on RedHat) and I want to obtain the partition that results from a cut at height=0.4. I run > cl1a <- agnes(dm, method="complete")

Hi, this is rather a (presumed) bug report than a question because I can solve my personal statistical problem by working with hclust instead of agnes. I have done a complete linkage clustering on a dist object dm with 30 objects with agnes (R 1.8.0 on RedHat) and I want to obtain the partition that results from a cut at height=0.4. I run > cl1a <- agnes(dm, method="complete")

Hi. I did write those functions, and sent them (I thought) to one of the R maintainers to see whether they would be appropriate for inclusion (because I'd seen some requests on the mailing lists). However, I'm happy to post them -- I should have thought of it before. WARNING: I've tested these functions on some data arising in my work and also on the USArrests data that comes with

Hello all, A question/suggestion: I was wondering if there is a chance of changing stats::cutree to be S3 and use cutree.hclust? For example: cutree <- function(tree, k = NULL, h = NULL,...) { UseMethod("cutree") } cutree.hclust <- stats::cutree # This will obviously need the actual content of stats::cutree This would be nicer for people like me to add new methods to

howdy R friends, I've searched CRAN but to no avail... I'm trying to use mva's hclust and print out for say 10 clusters in batch. How do I do this? It's unclear if I can use cutree. thanks, John Strumila -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send

I generated a heatmap in R using the following commands: > mydata <- read.csv(file="Data.csv", header=TRUE, sep=",") > mydata <- mydata[rowSums(mydata[,-1]^2) >0, ] > rownames(mydata)=mydata$Name > mydata <- mydata[,2:253] > mydatamatrix <- data.matrix(mydata) > mydatascale <- t(scale(t(mydatamatrix))) > hr <-