similar to: Strange error message

Displaying 20 results from an estimated 2000 matches similar to: "Strange error message"

2008 Apr 07
2
basehaz and newdata
I am unable to get the basehaz function to apply a proportional hazards model to a new data frame. I replicated my specific situation with the example for coxph in the help, where I changed the x value of the first record from 0 to 1. Is there something incorrect in the syntax that I am using? Thanks in advance! test1 <- list(time= c(4, 3,1,1,2,2,3), status=c(1,NA,1,0,1,1,0),
2008 Jun 07
1
expected risk from coxph (survival)
Hello, When I try to to obtain the expected risk for a new dataset using coxph in the survival package I get an error. Using the example from ?coxph: > test1 <- list(time= c(4, 3,1,1,2,2,3),+ status=c(1,NA,1,0,1,1,0),+ x= c(0, 2,1,1,1,0,0),+ sex= c(0, 0,0,0,1,1,1))> cox<-coxph( Surv(time, status) ~ x + strata(sex), test1)
2011 Mar 09
2
Anomaly with unique and match
I stumbled onto this working on an update to coxph. The last 6 lines below are the question, the rest create a test data set. tmt585% R R version 2.12.2 (2011-02-25) Copyright (C) 2011 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: x86_64-unknown-linux-gnu (64-bit) # Lines of code from survival/tests/singtest.R > library(survival) Loading required package: splines
2007 Dec 09
2
Getting estimates from survfit.coxph
Dear all, I'm having difficulty getting access to data generated by survfit and print.survfit when they are using with a Cox model (survfit.coxph). I would like to programmatically access the median survival time for each strata together with the 95% confidence interval. I can get it on screen, but can't get to it algorithmically. I found myself examining the source of print.survfit to
2007 Feb 05
2
Two ways to deal with age in Cox model
I hope one and all will allow a stats question: When running a cox proportional hazards model ,there are two ways to deal with age, including age as a covariate, or to include age as part of the follow-up time, viz, Age as a covariate: tetest1 <- list(time= c(4, 3,1,1,2,2,3), status=c(1,NA,1,0,1,1,0), age= c(0, 2,1,1,1,0,0),
2002 Jan 09
1
bug in read.table?
Hello, in the new Relase (1.4) i get a different (worser) result for read.table with as.is=T: it crash! Input file (t.txt, with a name, 5 character and a numeric column) Name short kind logable use save lag m "mo" "x" "n" "1" "n" 0 Ptp "PT" "l" "y" "m" "n" 0 R-Code
2010 Feb 08
2
Error on start R in server
Hello all, (Thank for your reply) I have a web-application in Apache Tomcat, when i start R in this application, I used packe RSJava Code ROmegahatInterpreter interp; String [] rargs = {"--no-save"}; REvaluator e; interp = new ROmegahatInterpreter ( ROmegahatInterpreter.fixArgs(rargs),false ); e =new REvaluator (); Errors ================ R version 2.10.1 (2009-12-14)
2010 Jan 21
2
"stack imbalance in ..." when loading a workspace
Hi all, I just failed in loading a saved wordspace (13MB of size), and received these errors: Warning: stack imbalance in 'missing', 52 then 51 Warning: stack imbalance in 'if', 50 then 53 Warning: stack imbalance in 'as.environment', 57 then 59 Warning: stack imbalance in 'ls', 54 then 53 Warning: stack imbalance in '.Internal', 54 then 53 Warning: stack
2009 May 08
1
anyDuplicated(incomp=NA) fails
With today's R 2.10.0(devel) I get: > anyDuplicated(c(1,NA,3,NA,5), incomp=NA) # expect 0 Warning: stack imbalance in 'anyDuplicated', 20 then 21 Warning: stack imbalance in '.Internal', 19 then 20 Warning: stack imbalance in '{', 17 then 18 [1] 0 > anyDuplicated(c(1,NA,3,NA,3), incomp=NA) # expect 5 Warning: stack imbalance in 'anyDuplicated', 20 then 21
2007 Feb 23
1
help with RMySQL
Hi R users, I am using RMySQL to connect to a database in MySQL. I have 3 questions. 1)When I give the following command dbListTables(con) I get the output stack imbalance in .Call, 142 then 143 stack imbalance in <-, 140 then 141 stack imbalance in {, 138 then 139 stack imbalance in standardGeneric, 126 then 127 stack imbalance in class, 121 then 122 stack imbalance in
2001 Dec 25
1
read.table (PR#1227)
Full_Name: Antonio Possolo Version: 1.4.0 OS: Linux Submission from: (NULL) (24.25.141.205) ## inputFile has the following three lines (each without the initial ``## '') ## a,b ## m,1 ## n,3 ## and resides in $HOME/R fileHOME <- paste(Sys.getenv("HOME"), "/R/", sep="") fileNAME <- "inputFile" ## Given the command, z <-
2002 Jan 15
1
Error message in R: stack imbalance
Hello... I just installed R version 1.4.0 on my laptop running Redhat Linux 7.1. I downloaded the RPM from CRAN and when I try to use a function I have used on other Redhat systems, I get the following error message. -------- > library(qtl) > gastritis <- read.cross(format="csv",dir="/home/sen/qtl/data/gastritis", +
2012 Dec 11
1
Debian packaging and openblas related crash when profiling in R
Hello R-sig-debian and (hopefully) Dirk: On Debian wheezy, I have the R packaging that CRAN (you) provide. I run into a little trouble while trying to fiddle with alternative BLAS. I know you and I went around on this last year and I think perhaps I've found something wrong in the framework, or I've just done something wrong. I installed the packages openblas-base and openblas-dev, and
2005 May 03
2
comparing lm(), survreg( ... , dist="gaussian") and survreg( ... , dist="lognormal")
Dear R-Helpers: I have tried everything I can think of and hope not to appear too foolish when my error is pointed out to me. I have some real data (18 points) that look linear on a log-log plot so I used them for a comparison of lm() and survreg. There are no suspensions. survreg.df <- data.frame(Cycles=c(2009000, 577000, 145000, 376000, 37000, 979000, 17420000, 71065000, 46397000,
2002 Jan 03
1
type.convert() (PR#1236)
Full_Name: Don Sun Version: 1.4.0 OS: Red Hat Linux release 6.2 Submission from: (NULL) (204.178.20.14) > type.convert(letters[1:2],as.is=T) stack imbalance in internal type.convert, 7 then 6stack imbalance in .Internal, 6 then 5 [1] "a" "b" Error: unprotect(): stack imbalance > version _ platform i686-pc-linux-gnu arch i686 os
2006 Oct 01
1
stack imbalance in contour
I'm not sure if this has much significance or not -- but it sounds rather ominous. It doesn't appear to be new as it happens with 2.0.0 in Linux (but the formatting of the warning messages has improved). > contour(array(3e10, c(10,10), list(1:10, 11:20))) Warning: stack imbalance in 'contour', 20 then 24 Warning: stack imbalance in '.Internal', 19 then 23 Warning:
2008 Apr 25
3
Use of survreg.distributions
Dear R-user: I am using survreg(Surv()) for fitting a Tobit model of left-censored longitudinal data. For logarithmic transformation of y data, I am trying use survreg.distributions in the following way: tfit=survreg(Surv(y, y>=-5, type="left")~x + cluster(id), dist="gaussian", data=y.data, scale=0, weights=w) my.gaussian<-survreg.distributions$gaussian
2010 Nov 15
1
interpretation of coefficients in survreg AND obtaining the hazard function
1. The weibull is the only distribution that can be written in both a proportional hazazrds for and an accelerated failure time form. Survreg uses the latter. In an ACF model, we model the time to failure. Positive coefficients are good (longer time to death). In a PH model, we model the death rate. Positive coefficients are bad (higher death rate). You are not the first to be confused
2012 Jan 26
1
3-parametric Weibull regression
Hello, I'm quite new to R and want to make a Weibull-regression with the survival package. I know how to build my "Surv"-object and how to make a standard-weibull regression with "survreg". However, I want to fit a translated or 3-parametric weibull dist to account for a failure-free time. I think I would need a new object in survreg.distributions, but I don't know how
2009 Mar 08
2
survreg help in R
Hey all, I am trying to use the survreg function in R to estimate the mean and standard deviation to come up with the MLE of alpha and lambda for the weibull distribution. I am doing the following: times<-c(10,13,18,19,23,30,36,38,54,56,59,75,93,97,104,107,107,107) censor<-c(1,0,0,1,0,1,1,0,0,0,1,1,1,1,0,1,0,0) survreg(Surv(times,censor),dist='weibull') and I get the following