similar to: Weibull survival regression model with different shape parameters

Hello all, I'm a new user of R and I have used the spdep package to calculate Moran's I estimates. I used the dnearneigh() function and moran.test() function and obtained what I wanted, but now I want to plot my results with the p-values. I regrouped my results for the Moran's I like >m=c(m1$e,m2$e...) #m1 contain the results of the moran.test for lag1

Hello all, I'm a new user on R, and I used the abline function to draw a line on my graph but it doesn't work with xyplot(). With the plot function, abline() is ok but with plot it doesn't make what I want. Maybe it's a little thing to change in plot() but I don't find what! With xyplot I write this : Xyplot(m~ordered(l,levels=1),

Dear R users, I'd like to fit different longevities for males and females (µm vs µf below) with a Weibull interval-censored survival model. Unfortunately more than 50% of the individuals in the experiment died before their sex could be assigned. Hence I was wondering about fitting the proportion of females in the model ( "prop" below) with a likelihood function such as:

Hi all, I am a very recent user of R. Mine is probably a basic question. I would like to know how do we create a certain set of points with geographic coordinates, between two endpoints. Please explain me how to do this using R. Which are the packages to be used for doing this. Thanks in advance, Ravichandra Mondreti. Ravichandra Mondreti PhD student/ Doctorant Spatial

Hello, I'm writing some genetic simulations in R where I would like to place genes along a chromosome proportional to the density of markers in a given region. For example, a chromosome can be presented as a list of marker locations: Chr1<-c(0, 6.5, 17.5, 26.2, 30.5, 36.4, 44.8, 45.7, 47.8, 48.7, 49.2, 50.9, 52.9, 54.5, 56.5, 58.9, 61.2, 64.1) Where the numbers refer to the locations of

Ceci est un message automatique de la passerelle antivirus Guinevere. Un message a ?t? re?u de votre part avec pour objet Re: Thanks! Le message ?tait adress? ? jlrauline@ier.fr. The message probably contains a virus. Il est possible que ce message ait ?t? envoy? par un poste infect? par un virus ayant utilis? votre adresse de messagerie en tant qu'adresse d'exp?dition

Hello I administrate a samba server with two network cards. This server is visible and serve his shares succesfully on two differents subnets. Ports 137,138 & 139 (both tcp & upd) are filtered between those two subnets by cisco switches. The problem is that all my computers on the first subnet are visible from the second subnet in the network neigbourhood. Only their names are

Hi, > In France, the second most important ADSL provider (named "Free") > offers a phone line (which uses VoIP but can only be used as a FXS) > with unlimited free calls to landlines. I was wondering if I would use my Free phone line with Bellster as well, but I am not sure this is authorized by the ISP : http://adsl.free.fr/hd/cgv.html [in French] En particulier,

Hello, I'm trying to migrate openldap users to a samba+ldap solution. I had set up the samba+ldap server succesfully, and restored my accounts via smbldap-populate and an ldif file. But all my LDAP accounts didn't got all the samba field and i'm forced to update it manually via smbldap-usermod -a %user and smbldap -u %user ( and typing a new password ). So my question is : Wich is

Hi, I'm sure this should be simple but I can't figure it out! I want to get the median survival calculated by the survfit function and use the value rather than just be able to print it. Something like this: library(survival) data(lung) lung.byPS = survfit(Surv (time, status) ~ ph.ecog, data=lung) # lung.byPS Call: survfit(formula = Surv(time, status) ~ ph.ecog, data = lung) 1

Hello, is there a R package that provides a log rank trend test for survival data in >=3 treatment groups? Or are there any comparable trend tests for survival data in R? Thanks a lot Markus -- Dipl. Inf. Markus Kreuz Universitaet Leipzig Institut fuer medizinische Informatik, Statistik und Epidemiologie (IMISE) Haertelstr. 16-18 D-04107 Leipzig Tel. +49 341 97 16 276 Fax. +49 341 97 16

Hi, Is it normal to get intercept in the list of covariates in the output of survreg function with standard error, z, p.value etc? Does it mean that intercept was fitted with the covariates? Does Value column represent coefficients or some thing else? Regards, ------------------------------------------------- tmp = survreg(Surv(futime, fustat) ~ ecog.ps + rx, ovarian,

Dear Knowledgeable R Community Members, Please excuse my ignorance, I apologize in advance if this is an easy question, but I am a bit stumped and could use a little guidance. I have a finite mixture modeling problem -- for example, a 2-component Weibull mixture -- where the components have a large overlap, and I am trying to adapt the "mclust" package which concern to normal

Dear all: Could anyone please provide some R codes to plot the below survival data to compare two groups (0 vs 1) after 2 simulations (TRL)? need 95% prediction interval on the plot from these 2 trials. I would like to simulate 1000 trials later. Thanks a lot for your great help and consideration! yan TRL ID ECOG BASE PTR8 GROUP POP ST ind 1 1 1 1 2.2636717 0.255634126 1 1 99.4 F 3 1 2 1

Hi friends, I need to generate a list, that should contain the quantile value of one column in a data frame. while I am compiling this one i am facing one bug,, Really I cannot find out that where the problem has occurred? Could any one help me to come out from this bug?? Here is the code, data <- lapply(comb.data$zFE, function(x) aggregate(x,

I was looking at how the model.frame method for lm works and comparing it to my own for coxph. The big difference is that I try to retain xlevels and predvars information for a new model frame, and lm does not. I use a call to model.frame in predict.coxph, which is why I went that route, but never noted the difference till now (preparing for my course in Nashville). Could someone shed light

Well, I found a solution and posted it on the appdb's page of Red Alert 3. http://appdb.winehq.org/objectManager.php?sClass=version&iId=14371 Here's a copy of the informations I posted on the Red Alert's Appdb webpage : > Thanks to raphael, here's a mini "how to play on LAN to RA3 with Wine + windows users, when the LAN is a VPN" > > My hostname, in

Dear R-users, Using Maximum-likelihood Fitting (fitdistr function) I've got the next error: > fitdistr(datos,"weibull",lower=0) Error in optim(x = c(1.4625e-06, 0.257854, 0.0001217545, 0.11421005, 0.028721576, : L-BFGS-B *needs finite values of 'fn' * where "datos" is a vector of length=1000 between 1.4625e-06 and 0.8867114 I add the lower argument in

Hi, I have a data set which is assumed to follow weibull distr'. How can I find of cdf for this data. For example, for normal data I used (package - lmomco) >cdfnor(15,parnor(lmom.ub(c(df$V1)))) Also, lmomco package does not have functions for finding cdf for some of the distributions like lognormal. Is there any other package, which can handle these distributions?

I need to be able to run a generalized linear model with a log() link and a Weibull family, or something similar to deal with an extreme value distribution. I actually have a large dataset where this is apparently necessary. It has to do with recovery of forensic samples from surfaces, where as much powder as possible is collected. This apparently causes the results to conform to some type