similar to: problems with postscript device

I reported this earlier, and have got reports that others have the same promlem on UNIX machines, so it is not only a windows problem S I file a bug report. The function (boot.stat) given at the end produces a postscript file, which cannot be included correctly in LaTeX. Specifically, the image in LaTeX (when translated by dvips to postscript) becomes very small, not using the bounding box, and

I am trying to run the following code in R on a Linux cluster. I would like to use the full processing power (specifying cores/nodes/memory). The code essentially runs predictions based on a GAM regression and saves the results as a CSV file for multiple sets of data (here I only show two). Is it possible to run this code using HPC packages such as Rmpi/snow/doParallel? Thank you!

(earlier I sent it as html by mistake). Hi, How can I get from C an object name used as a function argument? I have sample code in C that gives me access to the name of the function being called: SEXP xname(SEXP x) { const char *fun_name = CHAR(PRINTNAME(CAR(x))); x = CDR(x); const char *arg_name = isNull(TAG(x)) ? "" : CHAR(PRINTNAME(TAG(x)));

Hi all, As I reported earlier (with a typo in the work [BUG]) client certification validation *does not* work even if you do everything exactly according to all documentation and attempts at helpful advice. I have seen this issue with both startssl.com and self-signed certificates, and based on what I've seen from searching the web, this is a problem that has gotten little attention because

All, I'm trying to set up a function which calls the partialPlot function but am getting an error that I can't seem to solve. Here's a simplified version of the function and error... > pplot <- function(rf,pred.var){partialPlot(x=rf,pred.data=acoust,x.var=pred.var)} > > attach(acoust) > acoust.rf <-

I'm having trouble calling randomForest::partialPlot programmatically. It tries to use name of the (R) variable as the data column name. Example: library(randomForest) iris.rf <- randomForest(Species ~ ., data=iris, importance=TRUE, proximity=TRUE) partialPlot(iris.rf, iris, Sepal.Width) # works partialPlot(iris.rf, iris, "Sepal.Width") # works (function(var.name)

Over the years there have been inquiries to these lists as to how to produce plots with the direction(s) of axes reversed. In response to some of these inquiries, I have provided a function call ``rplot'' which automated the procedure for such axis reversal. Recently Herberto Ghezzo of McGill University pointed out to me that there was a bug in my code. I have now fixed this bug (and a

Hi there. I''ve discovered a rather serious security hole in mailx, the good old Berkeley mail program. It''s somehow present at least in the last versions I''ve checked (mailx-8.1.1 in Linux, mailx 5.0 in Solaris). The bug is an exploitable buffer overflow (using the HOME environment variable) that allows any local user to acquire the privileges under which the program

Dear list, can someone explain to me why deparse(substitute(x)) does not seem to work when x is of a user-defined S3 class? In my actual problem, my print method is part of a package, and the method is registered in the NAMESPACE, if that should make a difference. > print.testclass <- function(x,...){ xname <- deparse(substitute(x)) cat("Your object name

Hi all, In the histogram created with the `hist` command, the label of the y axis reads "Frequency ?. Finding that the ylab key is used to change (localize) that word may seem difficult for people who are beginners either in R or in english. I suggest very minor modifications of the file hist.Rd at https://svn.r-project.org/R/trunk/src/library/graphics/man/hist.Rd. 1) make the default value

Hi, I've noticed I get different results fitting a function to some data on my laptop to when I do it on my computer at work. Here's a code snippet of what I do: ##------------------------------------------------------------------ require(circular) ## for Bessel function I.0 ## Data: dd <- c(0.9975948929787, 0.9093316197395, 0.7838819026947, 0.9096108675003, 0.8901804089546,

An HTML attachment was scrubbed... URL: http://lists.digium.com/pipermail/asterisk-users/attachments/20091111/2b828eff/attachment.htm -------------- next part -------------- Hello. I am trying to execute an fax reception script and i am getting the following: [Nov 11 08:40:52] WARNING[12800]: app_system.c:88 system_exec_helper: Unable to execute '/var/lib/asterisk/scripts/mailfax ""

Dear All: I have succeeded in fitting a GAMLSS.dist model to growth data I am working with it. My aim is to create a matrix of predicted percentiles and the corresponding the fitted model's sigma mu nu by agebins. Q: How do it generate these parameters as in L M S per Cole and Green 1992? Here are my working codes. Name of fitted model is gamlssfit > Agebin<-seq(6,36,6)

Hello, I have a problem with the hist() function and showing densities. The densities sum to 50 and not to 1! I use R version 2.9.1 (2009-06-26) and I load the seqinR library. My data is the following vector: [1] 0.1400000 0.2000000 0.2200000 0.2828283 0.1600000 0.1600000 0.3600000 [8] 0.1600000 0.2200000 0.2600000 0.2000000 0.3000000 0.2200000 0.2342342 [15] 0.1800000 0.2200000 0.1600000

Hi there, I hope this question is not as stupid as the one before ... I tried to shorten my histogram (because the distribution is quite skewed and I simply don't want to see the long tail but still use the histogram plot). How can I do something like this? (The example does not work but I don't know why...) data <- rnorm(100) # as example, of course this is not skewed... h <-

Hola, Estoy tratando de crear un método S3 llamado "anthr" dentro del paquete que estoy desarrollando, cuyo argumento principal es "res" que básicamente es una lista con un solo componente. Pero si el segundo argumento llamado "oneSize" es FALSE, "res" es una lista de listas. Lo que he escrito hasta el momento es lo siguiente: anthr <- function(res,

Thanks Bert for all the help. I got the legend figured out Friday but left early becoz of long weekend so didn't get a chance to reply.. I modified the plot margins a little bit and Here's what I finally had... par(mar=c(c(10, 6, 6, 10) + 0.1)); par(xpd=FALSE); with (dataFrame, stripchart(marbles_buried ~ genotype, method="jitter", vertical=TRUE, col = c('blue',

Hola Carlos, Muchas gracias por el enlace, me ha sido de gran ayuda. Ya he entendido cómo funciona el sistema S3. Un saludo, Guillermo > Hola, ¿qué tal? > > Sigue http://www.datanalytics.com/2011/08/04/desarrollo-de-paquetes-con-r-iv-funciones-genericas/ > a rajatabla y lo tendrás. > > Un saludo, > > Carlos J. Gil Bellosta > http://www.datanalytics.com > >

Hello, all; I get an unexpected result when trying to plot a probability histogram with R1.9.1 on windows xp: #with the following code: > x <- runif(100,0,1) > hist(x) > hist(x, freq=F) > h <- hist(x, freq=F) > summary(h) # Length Class Mode #breaks 11 -none- numeric #counts 10 -none- numeric #intensities 10 -none- numeric #density 10

I have two large datasets (156K and 2.06M records). Each row has the hour that an event happened, represented by an integer from 0 to 23. R's histogram is combining some data. Here's the command I ran to get the histogram: > histinfo <- hist(crashes$hour, right=FALSE) Here's histinfo: > histinfo $breaks ?[1] ?0 ?1 ?2 ?3 ?4 ?5 ?6 ?7 ?8 ?9 10 11 12 13 14 15 16 17 18 19 20 21