similar to: axis() default values for "lty", "lwd", and "col"

I am trying to fit a lognormal distribution on interval-censored data. Some of my intervals have a lower bound of zero. Unfortunately, it seems like survreg() cannot deal with lower bounds of zero, despite the fact that plnorm(0)==0 and pnorm(-Inf)==0 are well defined. Below is a short example to reproduce the problem. Does anyone know why survreg() must behave that way? Is there an alternate

R version: 1.7.1 OS: Red Hat Linux 7.2 In this example, I would expect an error for the overly long variable name. This is always reproducable for me. > formula(paste("y~",paste(rep("x",50000),collapse=""))) Segmentation fault Sincerely, Jerome Asselin -- Jerome Asselin (Jérôme), Statistical Analyst British Columbia Centre for Excellence in HIV/AIDS St.

Hi there, I want to justify to right the text of my legend. Consider this short reproducable example. x <- 1:5 y1 <- 1/x y2 <- 2/x plot(rep(x,2),c(y1,y2),type="n",xlab="x",ylab="y") lines(x,y1) lines(x,y2,lty=2) legend(5,2,c("1,000","1,000,000"),lty=1:2,xjust=1,yjust=1)

R version: 1.7.1 OS: Red Hat Linux 7.2 When "covmat" is supplied in princomp(), the output value "center" is all NA's, even though the input matrix was indeed centered. I haven't read anything about this in the help file for princomp(). See code below for an example: pc2$center is all NA's. Jerome Asselin x <- rnorm(6) y <- rnorm(6) X <- cbind(x,y)

I am confused on how the log-likelihood is calculated in a parametric survival problem with frailty. I see a contradiction in the frailty() help file vs. the source code of frailty.gamma(), frailty.gaussian() and frailty.t(). The function frailty.gaussian() appears to calculate the penalty as the negative log-density of independent Gaussian variables, as one would expect: >

I would like to document my findings (with a potential FIX) regarding the issue of calculation of the degrees of freedom with nlme(). The program given at the bottom of this email generates and fit 20 data sets with a mixed-effects LINEAR model, but using the function nlme() instead of lme(). In each case, the correct number of degrees of freedom for the intercept parameter is 12. However, in

The problem occurs when the sample odds ratio is Inf, such as in the following example. Given the fact that both upper bounds of the two 95% confidence intervals are Inf, I would have expected that the two lower bounds be equal, but they aren't. x <- matrix(c(9,4,0,2),2,2) x # [,1] [,2] #[1,] 9 0 #[2,] 4 2 rbind("two.sided.95CI"=fisher.test(x)$conf.int,

Full_Name: Jerome Asselin Version: 1.6.2 OS: RedHat Linux 7.2 Submission from: (NULL) (142.103.173.179) In the following example, the line type lty=2 does not show properly across the entire line. x <- c(seq(0,.5,.001),seq(.6,1,.1)) y <- rep(1,length(x)) plot(x,y,type="s",lty=2) Sincerely, Jerome Asselin

Is there a simpler way then the solution to the one that was posted here? I'm not very proficient with legend, and I don't understand this solution. All I have is two or more lines on one plot that I want to put a legend on and I can't figure out how to do it from the examples. Can you give a very simple example? It does not have to be fancy!! I have never worked with a

R version: 1.7.1 OS: Red Hat Linux 7.2 Hi all, The formula object in model.frame() is not retrieved properly when model.frame() is called from within a function and the "subset" argument is supplied. foo <- function(formula,data,subset=NULL) { cat("\n*****Does formula[-3] == ~y ?**** TRUE *****\n") print(formula[-3] == ~y) cat("\n*****Result of model.frame()

Full_Name: Jerome Asselin Version: 1.7.1 OS: Red Hat Linux 7.2 Submission from: (NULL) (24.77.125.119) Setting the parameter na.action=na.omit should remove incomplete records in princomp. However this does not seem to work as expected. See example below. Sincerely, Jerome Asselin data(USArrests) princomp(USArrests, cor = TRUE) #THIS WORKS USArrests[1,3] <- NA princomp(USArrests, cor =

Full_Name: Jerome Asselin Version: 1.6.2 OS: RedHat Linux 7.2 Submission from: (NULL) (142.103.173.179) Should be an easy fix... Consider the examble below: plot(0,0) legend(0,0,c("Hello!","Hi!"),pch=1:2,lty=1:2,trace=T) It gives the following trace: > plot(0,0) > legend(0,0,c("Hello!","Hi!"),pch=1:2,lty=1:2,trace=T) xchar= 0.05178 ;

Full_Name: Jerome Asselin Version: 1.6.1 OS: RedHat Linux 7.2 Submission from: (NULL) (142.103.173.179) Hello, I get a segmentation fault when I run the following code. I wouldn't expect meaningful results because my response variable contains only missing values. However, I would expect something like a regular error (not a segmentation fault). library(survival) data <-

Full_Name: Jerome Asselin Version: 1.8.0 OS: RedHat Linux 7.2 Submission from: (NULL) (142.103.177.13) I would not expect a segmentation fault; perhaps an error message. > .Call("log") Segmentation fault This is always reproducable for me. Sincerely, Jerome Asselin

I'm rusty, but not *that* rusty here, I hope. If W (=Z*Z' in your case) is singular, it can not have inverse, which by definition also mean that nothing multiply by it will produce the identity matrix (for otherwise it would have an inverse and thus nonsingular). The definition of a generalized inverse is something like: If A is a non-null matrix, and G satisfy AGA = A, then G is called

Hi all, In the (very simple) example below, I have defined a random effect for the residuals in lme(). So the model is equivalent to a FIXED effect model. Could someone explain to me why lme() still gives two standard deviation estimates? I would expect lme() to return either: a) an error or a warning for having an unidentifiable model; b) only one standard deviation estimate. Thank you for your

Hi, >I would like to rank a time-series of data, extract the top ten data items from this series, determine the >corresponding row numbers for each value in the sample, and take a mean of these *row numbers* (not the data). >I would like to do this in R, rather than pre-process the data on the UNIX command line if possible, as I need to >calculate other statistics for the series.

Full_Name: Jerome Asselin Version: 1.6.2 OS: redhat linux 7.2 Submission from: (NULL) (142.103.173.179) Bug or feature? Hard to say... But it sure would be nice if table() would label the frequency of NA's as it does for NaN's. Regards, Jerome > table(c(2,NA,1,1,1),exclude=NULL) 1 2 3 1 1 > table(c(2,NA,1,1,1,NaN),exclude=NULL) 1 2 NaN 3 1 1 1 #For sake of

Full_Name: Jerome Asselin Version: 1.6.2 OS: redhat linux 7.2 Submission from: (NULL) (142.103.173.179) This is certainly not a big problem, but should there really be a warning message when I run this? > x <- c(1,1,2,2,2,2,3,3) > hist(x,sub="Sub Title") Warning messages: 1: parameter "sub" couldn't be set in high-level plot() function 2: parameter

Full_Name: Jerome Asselin Version: 1.3.0 OS: Windows 98 Submission from: (NULL) (24.77.112.193) I have a hard time to magnify the axis annotations in barplot() Here are some examples: #This one has no effect. barplot(1:3,names.arg=1:3,cex.axis=3,xlab="x",ylab="y") #This one magnifies the x and y labels