similar to: request for help in R

Hi List! I?m using regsimq() from the ?lmomRFA?-package to calculate error bounds for diverse distributions. For example: regsimq(gumfit$qfunc, nrec = lmom.data$n, f = lcdfgum, boundprob = c(0.025, 0.975)) Several times I got this error massage: Fehler in quantile.default(ou, probs = boundprob, type = 6) : missing values and NaN's not allowed if 'na.rm' is FALSE So my question

Dear List I?m using the ?lmomRFA? package to fit different distributions to my data sample. To calculate the error bounds I used: regsimq(?) and sitequantbounds(?) So my questions are: Are error bounds and confidence intervals the same thing? And: Does regsimq(? boundprob = c(0.05, 0.95)) calculate the 90 or the 95% confidence interval? If error bounds and confidence intervals are

Hello, I am using the lmomco package (lmom.ub and pargev) to compute the GEV parameters (location, scale, and shape), which are used to estimate return values. I was wondering how/if I can calculate upper and lower confidence (CI_u, CI_l) intervals for each return frequency using the GEV parameters to fill-in the table below? Xi (location) = 35.396 Alpha (scale) = 1.726 Kappa (shape) =

Hi Sorry about this 2nd message... I forgot to say, the diff is based on the 2.0.3 tarball from www.networkupstools.org. Jo Turner -xx- -- jT | mail to: hyvan_trant@hotmail.com ** | website: http://www.chiark.greenend.org.uk/~jsturner/

Hello Samba, We tried to use samba-2.2.4-1 with Rational ClearCase and had a problem using WinBind. ClearCase requires the primary group id to be set to the clearcase users group. WinBind sets the primary group to something like DOMAIN_GROUP_RID_USERS. I am including the patch we made to samba-2.2.4 to get things working. I have added two new smb.conf options: winbind primary group //

Andrew, Thank you for your suggestions. The real problem is Rational Technical support. When you pay someone 100k+ they tend to look for any excuse to deem you "unsupported". If you don't have your default group set to the clearcase user group, then technical support will decide that thats where your problem is! I am sure you know how that goes... I have solved problems

Hello: I hope none of you will mind helping a newbie. I'm a student research assistant working with a large data set in which observations are categorized according to two factors. I'm trying to calculate the group mean and variance of a variable (called 'hsgpa' in the example data presented below) to each observation , excluding that observation. For example, if there are

I'm trying to formulate a strategy for our interconnected Asterisk IAX peers to failover to the PSTN in the event of a DDoS. We currently use them like this: DID--->PRI--->Primary Asterisk--->IAX--->On-site Asterisk--->SIP This works fine, and everyone is happy. One of my concerns, though, is if we get DDoS'd - which happens probably once every couple of years. I'd

Dear Wu Gong and Peter Ehlers, thank you very much for your help debugging my script. Now I have a general following up question: Is there a straightforward way to rearrange the following dataset so that all first letters of each column will be combined in one column, all the second letters in a second column, all the third ones in a third column and so on, resulting in 7 columns, i.e. for

Greetings, \Sexpr{} has worked on MikTex with earlier versions of R, I can not seem to get it to work with R-2.4.0. I run Sweave with MikTex using the following statement in Run under Accessories. latex -include-directory="C:\Program Files\R\R-2.4.0\share\texmf" "C:\Documents and Settings\Suzette\Desktop\MyFile\MyProgram.tex" Any advice would be much appreciated, Suzette

I have just started playing with merb. I think this framework has great potential so ATTA BOYS are in order !! Perhaps I missed something in the readme file, but how do I generate a default app in merb. I was looking for something like the cmd below to generate the app hierarchy ?? > merb myapp -Fernand -------------- next part -------------- An HTML attachment was scrubbed... URL:

Dear List, Does somebody know how to get the filename lists in the ziped when "gzfile" or "unz" command is used ? Thanks for your help. Shawn Lee -- The information contained in this communication and any atta...{{dropped}}

Hi, I want to determine the confidence interval on the sum of two sigma's. Is there an easy way to do this in R? I guess I have to use some sort of chisquare convolution algorithm??? Thanx, Roy -- The information contained in this communication and any atta...{{dropped}}

Hi, I am trying to do mixed integer programming using R & Java. My input is a list of flight legs which is proportional to the number of times the loop is executed. The input is fed from java to R. I cannot avoid the for loops as it is part of the logic. I am running into a "cannot allocate a vector of size 5.4mb" error. The system memory available 16Gb and nearly 11Gb of it is

I found it - there has to be a /var/lib/nfs directory with a few subdirectories under it. Once I had created them all, I could run rpc.statd, and that unfroze everything. Thanks to all, esp. Bill for mentioning rpc - that was the clue for me. mhr

Hey, I was wondering if there existed a R function similar to 'fsolve' or 'fzero' Matlab functions? Thanks! Francois Aucoin [[alternative HTML version deleted]]

Hi, Can anyone help me with the following. I have been using R for Monte Carlo simulations and got some results I couldn't explain. Therefor I performed following short test: -------------- mean.sds <- NULL sample.sizes <- 3:30 for(N in sample.sizes){ dum <- NULL for(I in 1:5000){ x <- rnorm(N,0,1) dum <- c(dum,sd(x)) } mean.sds<- c(mean.sds,mean(dum)) }

Dear all, I am trying to create a lattice plot which consists of 1 xyplot and 2 histograms (each for x and y). My first try was like this: x<-rnorm(1000) y<-rnorm(1000) xy <- xyplot(y~x) hist.x <- histogram(x) hist.y <- histogram(y) print(xy, position=c(0, 0.2, 1, 1), more=TRUE) print(hist.x, position=c(0, 0, 1, 0.33),more=T) print(hist.y, position=c(0.8, 0, 1, 1)) Ok, this is

Hi, Can anyone help me with the following (although not directly correlated to R functionality)? I have been looking on the internet but can not find the answer. My question: what is the variation on the mean of a limited distribution (total N points normally distributed), when I have a small sample of that distribution (n < N)? Your help would be very welcome. Thanx, Roy -- The

I am wondering how to estimate the survival curve for a particular case(s) given a coxph model using this example code: #fit a cox proportional hazards model and plot the #predicted survival curve fit <- coxph( Surv(futime,fustat)~resid.ds+strata(rx)+ecog.ps+age,data=ovarian[1:23,]) z <- survfit(fit,newdata=ovarian[24:26,],individual=F) zs <- z$surv zt <-