similar to: degrees of freedom in nlme() (PR#2384)

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.

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 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)

Hi, I would like to recommend a minor modification in axis() which I believe can simplify the making of plots for publications. I am trying to define default values for par() in order to make labels bigger and lines thicker, so that the resulting plots look good when resized for publication purposes. I ran into the following problem... axis() does not use par() values as default for

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: >

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)

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,

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()

Dear list members, I have 982 quotations of a given stock index and I want to run a Ljung-Box test on these data to test for autocorrelation. Later on I will estimate 8 coefficients. I do not know how many degrees of freedom should I assume in the formula for Ljung-Box test. Could anyone tell me please? Below the formula: Box.test(x, lag = ????, type = c("Ljung-Box"), fitdf = 0)

Hello everyone, I am trying to regress applicants' performance in an assessment center (AC) on their gender (individual level) and the size of the AC (group level) with a multi-level model: model.0 <- lme(performance ~ ACsize + gender, random = ~1 | ACNumber, method = "ML", control = list(opt = "optim")) I have 1047 applicants in 118 ACs: >

Hello, I want to compute a correlation test but I do not want to use the degrees of freedom that are calculated by default but I want to set a particular number of degrees of freedom. I looked in the manual, different other functions but I did not found how to do it Thanks in advance for your answers Yours Florence Dufour PhD Student AZTI Tecnalia - Spain

Dear all, How can I extract the total and residual d.f. from a gnls object? I have tried str(summary(gnls.model)) and str(gnls.model) as well as gnls(), but couldn?t find the entry in the resulting lists. Many thanks! Best wishes Christoph -- Dr. rer.nat. Christoph Scherber University of Goettingen DNPW, Agroecology Waldweg 26 D-37073 Goettingen Germany phone +49 (0)551 39 8807 fax +49

Hi there, some statistical programs (e.g. SPSS) calculate a correction of the degrees of freedom in a repeated measure analysis of variance (see Greenhouse-Geisser (1958) or Huynh-Feld (1976)) by a factor epsilon. This factor is used to correct the deg. of freedom to get a corrected f-test. Is this also possible with R? Thanks, Sven P.S.: I read in the lm help page: singular.ok logical,

Hi all, Could someone please help me to calculate the P-value by using Chi-square test with a certain number of degrees of freedom? I have a data set to be calculated here: observed: 224, 64, 6 expected: 222.9, 66.2, 4.9 degrees of freedom: 1 I have been reading the documentations for three days, and can't find the answers. Please help.Thanks in advance. Regards, Frank

Hi, I have trouble to figure out how the df is derived in LME. Here is my model, lme(y~x+log(den)+sex+dep,data=lwd,random= list(group=~x)) Number of total samples (N) is 3237 number of groups (J) is 26 number of level-1 variables (Q1) is 3, i.e., x, log(den) and sex number of level-2 variables (Q2) is 1, i.e., dep x and den are continuous variable sex is associated with individual samples

I'm having hard time understanding the computation of degrees of freedom when runing nlme () on the following model: > formula(my data.gd) dLt ~ Lt | ID TasavB<- function(Lt, Linf, K) (K*(Linf-Lt)) my model.nlme <- nlme (dLt ~ TasavB(Lt, Linf, K), data = my data.gd, fixed = list(Linf ~ 1, K ~ 1), start = list(fixed = c(70, 0.4)), na.action= na.include,

Hello all, I'm trying to do a nested linear model with a dataset that incorporates an observation for each of several classes within each of several plots. I have 219 plots, and 17 classes within each plot. data.frame has columns "plot","class","age","dep.var" With lm(dep.var~class*age), The summary(lm) function returns t-test and F-test values

After an RSiteSeach("Box.test") I found some discussion regarding the degrees of freedom in the computation of the Ljung-Box test using Box.test(), but did not find any posting about the proper degrees of freedom. Box.test() uses "lag=number" as the degrees of freedom. However, I believe the correct degrees of freedom should be "number-p-q" where p and q are

I wonder whether any of you know of an efficient way to calculate the approximate degrees of freedom of a gamm() fit. Calculating the smoother/projection matrix S: y -> \hat y and then its trace by sum(eigen(S))$values is what I've been doing so far- but I was hoping there might be a more efficient way than doing the spectral decomposition of an NxN-matrix. The degrees of freedom

You could calculate the confidence interval of the correlation at your desired df: http://davidmlane.com/hyperstat/B8544.html The below code takes as arguments the observed correlation, N, and alpha, calculates the confidence interval and checks whether this includes 0. cor.test2=function(r,n,a=.05){ phi=function(x){ log((1+x)/(1-x))/2 } inv.phi=function(x){