search for: varial

Hi everyone, I am trying to write a title for a plot which has a superscript like D^2 and a value of a variable stored in that variable..... i am not sure if i am clear so i will try an example: Suppose i want my title to be like: Family: Gaussian; D^2 = 0.45 where i have the value 0.45 stored in variable d2 (which comes from some previous calculations, and depending which actual variables

...deal but there are two problems. First of all, I have 10000 variables. So the possible network structure is awfully huge, I don't know how long it will take my PC to find the highest-scoring network..........maybe a month? Secondly, I have some prior knowledge that only 500 out of the 10000 variales are possible parents. In another word, only those arrows startting from the 500 variables and pointing to the remaining 99500 variables are allowed in the network. In deal an assignment to "banlist" should help me rule out the impossible arrows. But in my case the number of "im...

...0,1,1,2,4,5,13,16,14), weeks = 1:9,                     treat=c(rep("1mg",3),rep("5mg",3),rep("10mg",3))) and I am using library(splines) to fit glm.m <- glm(count~bs(weeks)+as.factor(treat),family=poisson,data=data1) and I am interested in predicting the count variale for the weeks 10, 11 and 12 with treat 10mg and 15mg. Thanks in advance. Regards, Samuel [[alternative HTML version deleted]]

Am I correct in assuming that the output below essentially translates to "Males have a mean time that is significantly lower than Females"? Is this the correct way to interpret the fact that the coefficient is negative? Assume the variale sex is treated as a factor with Female =0 and Male=1. survmodel<-survreg(survobj~sex,data=data1, dist="weibull") survsum<-summary(survmodel) survsum Value Std. Error z p SexMale -0.47830 0.3745 -1.2770 2.02e-01 -- View th...

...t;-data.frame(fdata) fit<-lm(y~1,data=fdata) f.addterm<-function(x,fit,f.data) { fdata<-f.data fff<-paste('~.',x,sep='+') try(addterm(fit,as.formula(fff))$AIC[2]) } f.addterm(var,fit,fdata) } This function simply add the "x" variale to the linear regression with intercept only using addterm, and return the AIC of the new model. I know I can simply do this with stepAIC of MASS, but for my own situation, I want to try something more complicated. so when I execute the following command, I got an error message: > f.fun(var=&...

Hola: Gracias de nuevo por la ayuda! La solución, como no, funciona. Pero yo quería alguna cosa más flexible y universal que le pudiera pasar como parámetro diferentes opciones de incluir i excluir variables. Si estuviera fuera de la función seria: ===================== DATOS <- data.frame(SE = c("M", "H", "M", "M", "H"),

...'s permission. It seems to fail for me. When I add some key to the agent and logon to remote host, I get this message when ssh tries to authenticate the user: "Agent admitted failure to sign using the key". The key added to agent without the -c switch works, of course. The SSH_ASKPASS variale is set to ssh-askpass, which also works for me. What might be the problem? /S

...000 variables. So the possible network > >> structure is awfully huge, I don't know how long it will > >take my PC to > >> find the highest-scoring network..........maybe a month? Secondly, I > >> have some prior knowledge that only 500 out of the 10000 > >variales are > >> possible parents. In another word, only those arrows startting from > >> the 500 variables and pointing to the remaining 99500 variables are > >> allowed in the network. In deal an assignment to "banlist" should > >> help me rule out the imp...

Hi Timo, when logging out like a001 logout the imap child dies from signal 11. The back trace looks like this: Program received signal SIGSEGV, Segmentation fault. 0xb7ed4991 in strcasecmp () from /lib/tls/i686/cmov/libc.so.6 (gdb) bt #0 0xb7ed4991 in strcasecmp () from /lib/tls/i686/cmov/libc.so.6 #1 0x0806ab6c in command_unregister (name=0x815b9ab "LOGOUT") at commands.c:83 #2

Hello useRs, I have a problem at hand which I'd think is fairly common amongst groups were R is being adopted for Analytics in place of SAS. Users would like to obtain results for logistic regression in R that they have become accustomed to in SAS. Towards this end, I was able to propose the Design package in R which contains many functions to extract the various metrics that SAS reports.

...er.groups = F) > fm1comp = lme(adj.UKV ~ Time + Time.sq, data = comp.adj.UKV, random = list(Patient_no = ~ 1) ) > plot(augPred(fm1comp, level= 1)) Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : variable lengths differ I've checked all the variale lengths, and have also made sure that factors are correctly defined as factors. Is there anything special I need to be doing for augPred to work correctly? I checked the help but didn't find much. cheers, dave David Afshartous, PhD University of Miami School of Business Rm KE-4...

Hi, David. I have recently fixed ferret C sources and successfully compile extension with MSVC.Net The problem was that MS compiler is more stricter that GCC and require that all variables were declared before using. There was ~30 such declaration. I have fixed them all. But I am not sure that it works because tests failed with following error both on clean and patched versions. So seems that it

...(lm(y ~ x+factor(x), data)) The F-value of factor(x) is the lack of fit test, and when lineair, the p value should be <0.05 when using 0.95% significance. Bart Wensui Liu wrote: > > in statistics, is there a measure for curvature? how to detect and > quantify the curvature between 2 variales instead of using > visualization? > thank you so much! > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html...