similar to: [Fwd: curiousity with hist]

Hello, I was wondering whether a Poisson mixture modeler/cluster analysis package is available for R. I scanned CRAN packages and couldn't find anything but I thought I'd ask. If not could anyone recommend a non-R open source package. I have found 'snob' but this program seems a bit hard to use in an automated, non interactive fashion. regards, Murad -- Murad Nayal M.D. Ph.D.

Hello, I am trying to compare two histograms using barplot. the idea is to plot the histograms as pairs of columns side by side for each x value. I was able to do it using barplot before but I can't remember now for the life of me now how I did it in the past: > d [,1] [,2] -37.5 0.0000000000 2.789396e-05 -32.5 0.0001394700 5.578801e-05 -27.5 0.0019804742

Hello all, This might have been fixed in later versions (I am using R1.7.0), r-help archive contains messages reporting similar problems but no reports of codes fixes. I have encountered a couple of problems using the silhouette function. one occurs when the clustering contains clusters composed of 1 element (Martin Maechler posted code few months ago that fixes a similar problem that occurs

Hello all, since we're on the topic of R-editors. I am using emacs/ess on a unix workstation (to interact with R and have been having a little problem. I usually write the R commands I need to run in a separate buffer then copy and paste them into the *R* buffer for evaluation. The problem is, if any command is spread over multiple lines emacs/R hangs when I paste it in the R buffer for

Hi, I would like to step-through a non-visible function. but apparently I don't know enough about namespaces to get that to work: > methods(predict) ... deleted lines ... [27] predict.rpart* predict.smooth.spline* [31] predict.survreg.penal* Non-visible functions are asterisked > debug(predict.rpart) Error: Object "predict.rpart" not found >

Hi, I am trying to refine models of a continuous response variable and a number of categorical predictor variables. I know of some model refinement tools available in R that help in the selection of model terms like dropterm and addterm from MASS etc. However, I would also like to try to refine the model by 'coalescing' some levels of some of the predictor factors. Is there a standard

When preparing a series of histograms I found that hist was combining the two lowest categories or bins, 1 and 2. Specifying breaks, as illustrated below, resulted in the correct histogram: values <- sample(10,500,replace=TRUE) hist(values) hist(values,breaks = 0:10) Apparently, the number of values strictly less than 1 is shown in the first bin (and since none is less than 1,

Hello, Is there a specific command to clear the graphics window. On occasion I need to construct plots using commands that don't clear the graphics window (like text, lines and points etc.) -only- and hence need to clear the graphics completely before hand. also, is there a way to restore the graphics parameters to default values, say in these cases where you forgot to save the original

Hello all, I am trying to use svm (from the e1071 package) to solve a binary classification problem. The two classes in my particular data set are unequally populated. class 'I' (for important) has about 3000 instances while class "B" (for background) has about 20,000. experimenting with different classifiers I realized that in cases where such an asymmetry exists there is a

> -----Original Message----- > From: Murad Nayal [mailto:mn216 at columbia.edu] > Sent: Tuesday, 24 June 2003 12:02 PM > Cc: r-help at stat.math.ethz.ch > Subject: [R] help on R programming. > > > > > Hello all, > > I am looking for books to help me gain a firmer grasp on the S/R > programming language , programing / data structures etc. it > seems

I have problems getting a histogram with relative frequencies on the y-axis. Here is an example data set: > a <- c(4.626, 4.627, 4.627, 4.628, 4.629, 4.629, 4.630, 4.631, 4.632, 4.632) > d = hist(a,freq=F) > d$density [1] 299.9999 100.0000 200.0000 100.0000 100.0000 200.0000 The obtained densities are given by counts/(total n * bin width), with bin width being

Hello, I have been getting the following intermittent error from kmeans: >str(cavint.p.r) num [1:1967, 1:13] 0.691 0.123 0.388 0.268 0.485 ... - attr(*, "dimnames")=List of 2 ..$ : chr [1:1967] "6" "49" "87" "102" ... ..$ : chr [1:13] "HYD" "NEG" "POS" "OXY" ... > set.seed(34) >

On Sun, 11 Dec 2005 clausen at econ.upenn.edu wrote: > Full_Name: Andrew Clausen > Version: 2.1.0 > OS: Debian GNU/Linux > Submission from: (NULL) (71.242.192.73) > > > Hi, > > When I type > > hist(x, freq=F) > > I get a density function, as I expect. However, if I type > > plot(hist(x, freq=F)) > > then I get the same output as if I had

Hi all I wish to draw an histogram... with dates but the following append, i don't know where is the problem, help(hist.Date) works and i don't see any usefull information on what i'm doing wrong... > hist.Date(dt_cycles) Error: couldn't find function "hist.Date" > hist.date(dt_cycles) Error: couldn't find function "hist.date" > cycles [1] 7 1

Can someone help me with a misunderstanding I'm having with hist? I expected, from the example below, that the number of bins would always be 10 and the length of the counts array the same. According to the help section 'breaks' can be a integer indicating the number of bins. From the example below, the number of bins (length of the counts array) varies. Am I wrong in expecting the

Hi all, I came across some curious behavior today in using hist.Date() and subsequently noted the same behavior in cut.Date(), both of which are using similar code when 'breaks = "months"' or 'breaks = "years"'. I was in the process of creating a histogram of subject enrollment in a clinical trial. The counts needed to be by month, so essentially used:

Full_Name: Alex Deckmyn Version: 2.4.0 OS: linux Submission from: (NULL) (193.190.63.62) specifying the "right" option in hist results in a warning when plot=F. The option is taken into account correctly, but a warning is issued anyway. When plot=T there is no warning. > hist(c(1,1.5),breaks=0:4)$counts [1] 1 1 0 0 > hist(c(1,1.5),breaks=0:4,right=T)$counts [1] 1 1 0 0 >

I discovered a bug in hist(). Try the following: x<-c(-5,-4,-4,-4,-3,-3,-3,-3,-2,-2,0,0,0,0,1,1,1,3,3,5,6) # Note that sum(x)<0: sum(x) # [1] -13 hist(x) # looks ok hist(x,freq=F) # negative bars !! # and finally this gives not 1: sum(hist(x,plot=F)$rel.freqs) # [1] -0.8076923 The reason is, that "sum(x)" is used instead of "length(x)" in the following line near

Hi, This is a very basic question, but I would like to undestand hist(). I thought that the hist( , freq=FALSE) should provide the relative frequencies (probabilities), and so they should sum 1, however: set.seed(2) ah <- hist(rnorm(100), freq=F) sum(ah$intensities) [1] 2 set.seed(2) bh <- hist(rlnorm(100), freq=F) sum(bh$intensities) [1] 0.4999996 I'm getting similar figures with

I just got the same error message with > sessionInfo() R version 3.4.0 (2017-04-21) Platform: x86_64-apple-darwin15.6.0 (64-bit) Running under: macOS Sierra 10.12.4 Matrix products: default BLAS: /System/Library/Frameworks/Accelerate.framework/Versions/A/Frameworks/vecLib.framework/Versions/A/libBLAS.dylib LAPACK: /Library/Frameworks/R.framework/Versions/3.4/Resources/lib/libRlapack.dylib