similar to: AW: boxplot a list of objects

Hi, Just something I don't understand: data <- data.frame(V1=c(1:12),F1=c(rep("a",4),rep("b",4),rep("c",4))) data_ac <- data[which(data$F1 !="b"), ] levels(data_ac$F1) Why the level "b" is always present ? thanks Tristan, R 2.0.1 for Linux Fedora 3 -- ------------------------------------------------------------ Tristan

Hi I use RSvgDevice to output plot, and modify them using svg editor (inkscape or sodipodi on Linux). Some month ago, results were perfect. I did exactly the same analysis today on the same data, and unfortunatly the results are different. While looking to the svg file, it seems that all information concerning "fill" and "stroke" of objects are lost. The consequence is that

Hi list, #Imagine we have vectors of different length (in practice 100 vectors): a<-c(1:10) b<-c(1:20) c<-c(1:30) #then we got a list of the names of those objects: list<-c("a","b","c") #I don't find how to boxplot them using a less stupid way than : boxplot(get(list[1]),get(list[2]),get(list[3])) Thanks for any advice ! --

Hello all, I am planning to redeploy my workstation under KUBUNTU. Does any body has any r experience installing/using r on this platform? Best regards. -- Constant Depi??reux Managing Director Applied QUality Technologies Europe sprl Rue des D??port??s 123, B-4800 Verviers (Tel) +32 87 292175 - (Fax) +32 87 292171 - (Mobile) +32 475 555 818 (Web) http://www.aqte.be - (Courriel)

Hello all, I am planning to redeploy my workstation under KUBUNTU. Does any body has any r experience installing/using r on this platform? Best regards. -- Constant Depi??reux Managing Director Applied QUality Technologies Europe sprl Rue des D??port??s 123, B-4800 Verviers (Tel) +32 87 292175 - (Fax) +32 87 292171 - (Mobile) +32 475 555 818 (Web) http://www.aqte.be - (Courriel)

Dear all, Thanks to those 3 people who sent me answers to my question. Got the problem solved. Great! Now, another question of mine is: I would like to run an R script from the Linux prompt. Is there any way possible to do this? The reason is, the calculation that I'm doing takes a few hours, and I would like to automatize it. Or does it mean that I have to run source within the R

Hi, I would like to open several tables with a loop, using something like : ----------------- $ ls 1.txt 2.txt 3.txt 4.txt $ R > for (i in 1:4) tabi<-read.table("i.txt") Error in file(file, "r") : unable to open connection In addition: Warning message: cannot open file `i.txt' ------------------ thanks for any help Tristan Lefebure

Dear R users, I face to a nested pattern and despite the numerous examples in the help I am still confused. I sampled bugs in different habitats within sites which were within rivers themselves within different regions. The habitat correspond to different substrata (not systematically present in all sites). For rivers and sites, I have environemental variables (e.g. altitude and slope of

Dear all, I think I'm nearly there in writing R code which will read in files with two variable parts to the file name and then assigning these file names to objects, which also have two variable parts. I have got the code running without encountering errors, however, I receive 50+ of the same warnings: 1: In assign(paste("Fekete_", index$year, index$month, sep = ""),

Dear all, as you already know, it is possible to send short winpopup messages to Win boxes using the command smbclient -M host. If a Win user wants to reply, it should be possible to use the Windows' WinPopUp GUI to reply, but the message gets lost somewhere. For example: smbclient -M winbox Added interface ip=192.1.1.125 bcast=192.1.1.255 nmask=255.255.255.0 Connected. Type your message,

Howdee, My question appears at #6 below: 1. I want to model the growth of each of a large number of individuals using a 4-parameter logistic growth curve. 2. nlme does not converge with the random structure that I want to use. 3. nlsList does not converge for some individuals. 4. I decided to go around nlsList using: t(sapply(split(data, list(data$id)), function(subd){coef(nls(mass ~

Dear all, curve(x^3*(1-x)^7, from = 0, to = 1) works as expected but, omitting the "xlim" or the "to" and "from" arguments and calling "curve" more than once: par(mfrow = c(2,2)) for (i in 1:4) curve(x^3*(1-x)^7) gives an expected (al least to me) result. Note also that a "pu" object is returned by curve > pu [1] -0.1802445 1.1802445

I have received the below notice about the rsync list. There is a worm among us. ------- Forwarded message follows ------- Return-Path: <postmaster@innevi.com> Received: from bleep.innevi.com ([64.30.26.9]) by mail.dubois-king.COM (8.11.6/8.11.6) with ESMTP id i1K7n3p14977 for <ppalumbo@dubois-king.com>; Fri, 20 Feb 2004 02:49:03 -0500 Content-Type: text/plain;

sorry for my previus (WRONG) answer, as someone already pointed out a solution could be subset(data, PUNTAR==c("IX49","IX48")) > -----Messaggio originale----- > Da: Guazzetti Stefano > Inviato: venerd?? 23 luglio 2004 15.55 > A: 'Luis Rideau Cruz'; r-help at stat.math.ethz.ch > Oggetto: R: [R] retrieve rows from frame assuming criterion > >

Howdee, *** I know that the lmList() function exists, yet I don't want to use it. *** Would anyone be kind enough to tell how I can apply the function lm() to each level of a given factor so to obtain the intercept and slope for each factor level within a matrix? For instance, suppose a dataframe containing 3 variables: id, x and y. I want to compute the function lm() for each level

Dear expeRts, I would like to use a oneway_test (from package coin) to test whether two groups differ on various variables. The variables are encoded within a data frame. Unfortunately, I obtained an error, that I don't understand. Could you please help me ? Example: library(coin) y <- as.data.frame(matrix(rnorm(200), ncol=2)) group <- as.factor(unif(100)) lapply(y,

Dear useRs, I have written a function that implements a Bayesian method to compare a patient's score on two tasks with that of a small control group, as described in Crawford, J. and Garthwaite, P. (2007). Comparison of a single case to a control or normative sample in neuropsychology: Development of a bayesian approach. Cognitive Neuropsychology, 24(4):343?372. The function (see

Dear useRs, I have written a very simple function to compute some probabilities on words (function is below). The function includes a which() statement applied to a vector of characters (word.split): sapply (word.split, function(x) which(letters==x)). This statement worked as expected when used outside the global function : > word <- "hello" > (word.split <-

Hi, I am running cwrsync. 2004/05/11 [93] rsync error: some files could not be transferred (code 23) at /home/lapo/packaging/tmp/rsync-2.5.7/main.c(383) I am uncertain as to what this path statement is about in the above error message. /home/lapo/packaging/tmp/rsync-2.5.7/main.c(383) Thanks, Pat -- Patricia Palumbo DuBois & King, Inc. ppalumbo@dubois-king.com 802-728-4113 | ext 322

Dear R-users, I am wondering whether R implements a function returning the non central parameter of a t distribution (equivalent of the TnoncT function from SAS), given /x/ a value from a t distribution, /df /the degrees of freedom and /p/ the probability of x under this distribution. Thanks a lot, Matthieu -- Matthieu Dubois, /Ph.D/. /Student/ Cognitive Neuroscience Unit, UCL,