similar to: Attributes or list programming "efficiency"

Dear R Fellow-Travellers: What is your recommended way of dealing with a left-censored response (non-detects) in (linear Gaussian) mixed effects models? Specifics: Response is a numeric positive measurement (of volume, actually); but when it falls below some unknown and slightly random value (depending on how the sample is prepared and measured), it cannot be measured and is recorded as 0.

What would be nice is to have something like a "robust" task view... Andy From: Berton Gunter > > Many thanks for this Martin. There now are several packages > with what appear to be overlapping functions (or at least > algorithms). Besides those you mentioned, "robust" and > "roblm" are at least two others. Any recommendations about > how or

>From R-help, subject "Variable number of loops" I've opened a new thread, moving from R-help to R-devel .. >>>>> Jim Lemon <drjimlemon at gmail.com> >>>>> on Sun, 17 May 2015 09:19:06 +1000 writes: > Hi all, Given the number of help requests that involve > permutations/combinations, and the less than obvious > naming

I tried command as.numeric(Qty) #converting column"Qty" in object "demand" to numeric. However it remains a non-numeric value. What is the best way to convert a column of data to a numeric value in R? What is the best way to convert dates into a numeric format (or some format able to be used in an x,y scatter) while still maintaining its inherent "date" property?

I am looking for a package that will allow me to choose R (a real number) that minimizes the correlation of y and x^R, i.e. find R such that corr(y,x^R) is minimized. Any suggestions for packages I might look at would be helpful. Thanks, John John David Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology Baltimore VA

Folks: In the past 2 days I have seen a large increase of spam getting into R-help. Are others experiencing this problem? If so, has there been some change to the spam filters on the R-servers? If not, is the problem on my end? Feel free to reply privately. Thanks. Bert Gunter Genentech Nonclinical Statistics South San Francisco, CA 94404 650-467-7374

Dear All, would you please provide your thoughts on the following: let us say I have: a <-c(1,5,8,15,32,69) b <-c(8.5,33) and I would like to extract from "a" the two values that are closest to the values in "b", where the length of this vectors may change but b will allways be shorter than "a". So at the end based on this example I should have the result

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Folks: Problem: [<-.data,frame() is losing attributes under certain curcumstances shown below. I think this is a bug, at least in documentation, as I was unable to find explicit documentation of the behavior. Indeed, the only documentation I found told me attributes are preserved. Here is a detailed description: > df <- data.frame(a=1:3, b=letters[1:3]) > attr(df,"foo")

suppose I have two sets of vectors: x1,x2,...,xN and y1,y2,...,yN. I want N correlations: cor(x1,y1), cor(x2,y2), ..., cor(xN,yN). my sets of vectors are arranged as data frames x & y (vector=column): x <- data.frame(a=rnorm(10),b=rnorm(10),c=rnorm(10)) y <- data.frame(d=rnorm(10),e=rnorm(10),f=rnorm(10)) cor(x,y) returns a _matrix_ of all pairwise correlations: cor(x,y)

Peter You question is difficult to answer without more information about the distribution of your residuals. Different residual patterns call for different transformations to stabilize the variance. One very common form of heterocedasticity is increasing variance with increasing values of an independent predictor, i.e. the variance of the residuals of y=x increase as x increases. In this case a

Hi Folks: **Off Topic** Those interested in clinical trials may find the following of interest: http://www.nytimes.com/2010/09/19/health/research/19trial.html It concerns the ethicality of randomizing those with life-threatening disease to relatively ineffective SOC when new "biologically targeted" therapies "appear" to be more effective. While the context may be new, the

Hi folks, I am working on my student internship to USA, california and would like to attend a R Advanced Programming course while I am in USA. Where can I find such a course. I am new to this list. Trevor O'brien >From Ireland. [[alternative HTML version deleted]]

Newbies (and others!) may find useful the R Reference Card made available by Tom Short and Rpad at http://www.rpad.org/Rpad/Rpad-refcard.pdf or through the "Contributed" link on CRAN (where some other reference cards are also linked). It categorizes and organizes a bunch of R's basic, most used functions so that they can be easily found. For example, paste() is under the

Folks: I would like to write two lines of text in two different font sizes (or faces or ...) as the title ("main") of a trellis plot. The following code does it, but not well: xyplot((0:1)~(0:1), main = textGrob(lab=c("Some Text","Some More Text"),y=c(.95,.8), gp=gpar(cex=c(1.2,1.0))) ) There is too much space between the title text and the plot. I assume

Hi, You could try: A<- matrix(unlist(read.table(text=" 1 2 3 4 5 6 7 8 9 9 8 7 6 5 4 3 2 1 ",sep="",header=FALSE)),ncol=3,byrow=FALSE,dimnames=NULL) library(matrixStats) ?res1<-t(sapply(split(as.data.frame(A),as.numeric(gl(nrow(A),2,6))),colProds)) ?res1 #? [,1] [,2] [,3] #1??? 4?? 10?? 18 #2?? 63?? 64?? 63 #3?? 18?? 10??? 4

Hi all, I am looking for a command for doing 2D filtering (rectangular or Gaussian) in R... I have looked at ksmooth, filter and convolve but they seem to be 1D... Any thoughts? Thanks a lot! [[alternative HTML version deleted]]

Hello, I am brand new in this site as well as in R. Please accept my early apology if it's not the right place to ask such a question. I have the following equation: b=[2(1-a)(1+2a)^0.5]/(1+3a) How can I solve the above equation for 'a' using R? Thanks in advance. KD [[alternative HTML version deleted]]

Dear all, I noticed that a data.frame has four attributes: - names - row.names - class - variable.labels While one can use the first three (i.e. names(foo) or class(foo)), the fourth one can only be used via: attributes(foo)$variable.labels (which is kind of a tedious thing to type) Is it or would be possible to simply use: variable.labels(foo) like the first three attributes? I tried: varlab

Folks: delta <- 1:5 I would like to put 5 separate lines of text of the form "10 %+-% delta[i]" into a lattice key legend, where ""%+-%" is the plotmath plus/minus symbol and delta[i] is the ith value of delta. The construct: lapply(delta,function(d)bquote(10%+-%.(d))) appears to produce a list of expressions of the correct form, and, indeed, if I assign the above