similar to: Problem with locfit( ... , family="hazard")

Displaying 20 results from an estimated 1000 matches similar to: "Problem with locfit( ... , family="hazard")"

2005 Oct 05
1
how do I write Rd file for this?
Dear R-devel, I'm working on Prof. Loader's new version of locfit to try to get it pass R CMD check. I'm almost there, but I have a problem with some Rd files that I hope some one can help me resolve. Here's an example: In the package there's a function called locfit.censor(). This function can be used in a few different ways: locfit.censor(x, y, cens, ...)
1998 Mar 26
1
R-beta: problem with locfit
I installed the locfit package under Linux (gcc 2.7.2). Installation was ok but > x <- runif(200) > y.compl <- 10*x*x*rgamma(200,3) > med.y <- median(y.compl) > cens <- ifelse(y.compl<=med.y,1,0) > y <- cens * y.compl + (1-cens)*med.y > library(locfit) > m <- locfit(y~x,cens=cens,family="gamma") /usr/local/src/R-0.61.1/bin/R.binary: can't
2004 Jun 20
1
problem locfit
I have a problem with the use of locfit with censured data, when I carry out locfit by: fitbmt<-locfit(~recur,data=BMTAGE11,cens=df.status,family="hazard",alpha=0.5) it does not give me any message, but if I want to obtain the graph or even if I ask for (fitbmt) made it gives me the following message: > fitbmt31 Problem: Object "fitbmt31" not found, while calling
2007 Aug 28
5
ERROR: While executing gem ... (Gem::Installer::ExtensionBu
Hi all, Sorry if this is answered somewhere -- I am new to ruby and to linux, and can''t figure it out: When I try to install ferret (see below), I get ERROR: While executing gem ... (Gem::Installer::ExtensionBuildError). Same thing happens for any version I pick from th list. I am using: gcc (GCC) 4.1.2 (Ubuntu 4.1.2-0ubuntu4) ruby 1.8.5 (2006-08-25) [i486-linux] gem 0.9.4 No idea
2010 Jan 04
1
no "rcorrp.cens" in hmisc package
Dear, I wanna to compare AUC generated by two distribution models using the same sample. I tried improveProb function's example code below. set.seed(1) library(survival) x1 <- rnorm(400) x2 <- x1 + rnorm(400) d.time <- rexp(400) + (x1 - min(x1)) cens <- runif(400,.5,2) death <- d.time <= cens d.time <- pmin(d.time, cens) rcorrp.cens(x1, x2, Surv(d.time, death))
2009 Sep 08
1
rcorrp.cens and U statistics
I have two alternative Cox models with C-statistics 0.72 and 0.78. My question is if 0.78 is significantly greater than 0.72. I'm using rcorrp.cens. I cannot find the U statistics in the output of the function. This is the output of the help example: > x1 <- rnorm(400) > x2 <- x1 + rnorm(400) > d.time <- rexp(400) + (x1 - min(x1)) > cens <- runif(400,.5,2) > death
2011 May 22
1
How to calculate confidence interval of C statistic by rcorr.cens
Hi, I'm trying to calculate 95% confidence interval of C statistic of logistic regression model using rcorr.cens in rms package. I wrote a brief function for this purpose as the followings; CstatisticCI <- function(x) # x is object of rcorr.cens. { se <- x["S.D."]/sqrt(x["n"]) Low95 <- x["C Index"] - 1.96*se Upper95 <- x["C
2005 Jul 11
1
validation, calibration and Design
Hi R experts, I am trying to do a prognostic model validation study, using cancer survival data. There are 2 data sets - 1500 cases used to develop a nomogram, and another of 800 cases used as an independent validation cohort. I have validated the nomogram in the original data (easy with the Design tools), and then want to show that it also has good results with the independent data using 60
2006 Apr 21
1
rcorrp.cens
Hi R-users, I'm having some problems in using the Hmisc package. I'm estimating a cox ph model and want to test whether the drop in concordance index due to omitting one covariate is significant. I think (but I'm not sure) here are two ways to do that: 1) predict two cox model (the full model and model without the covariate of interest) and estimate the concordance index (i.e. area
2006 Jul 25
1
HELP with NLME
Hi, I was very much hoping someone could help me with the following. I am trying to convert some SAS NLMIXED code to NLME in R (v.2.1), but I get an error message. Does anyone have any suggestions? I think my error is with the random effect "u" which seems to be parametrized differently in the SAS code. In case it's helpful, what I am essentially trying to do is estimate parameters
2011 Mar 01
1
which does the "S.D." returned by {Hmisc} rcorr.cens measure?
Dear R-help, This is an example in the {Hmisc} manual under rcorr.cens function: > set.seed(1) > x <- round(rnorm(200)) > y <- rnorm(200) > round(rcorr.cens(x, y, outx=F),4) C Index Dxy S.D. n missing uncensored Relevant Pairs Concordant Uncertain 0.4831 -0.0338 0.0462 200.0000
2009 Mar 09
1
rcorr.cens Goodman-Kruskal gamma
Dear r-helpers! I want to classify my vegetation data with hierachical cluster analysis. My Dataset consist of Abundance-Values (Braun-Blanquet ordinal scale; ranked) for each plant species and relev?. I found a lot of r-packages dealing with cluster analysis, but none of them is able to calculate a distance measure for ranked data. Podani recommends the use of Goodman and Kruskals' Gamma for
2012 Aug 31
3
fitting lognormal censored data
Hi , I am trying to get some estimator based on lognormal distribution when we have left,interval, and right censored data. Since, there is now avalible pakage in R can help me in this, I had to write my own code using Newton Raphson method which requires first and second derivative of log likelihood but my problem after runing the code is the estimators were too high. with this email ,I provide
2004 Apr 21
1
Boot package
Dear mailing list, I tried to run the example for the conditional bootstap written in the help file of censboot. I got the following result: STRATIFIED CONDITIONAL BOOTSTRAP FOR CENSORED DATA Call: censboot(data = aml, statistic = aml.fun, R = 499, F.surv = aml.s1, G.surv = aml.s2, strata = aml$group, sim = "cond") Bootstrap Statistics : original bias std. error t1*
2006 May 30
1
position of number at risk in survplot() graphs
Dear R-help How can one get survplot() to place the number at risk just below the survival curve as opposed to the default which is just above the x-axis? I tried the code bellow but the result is not satisfactory as some numbers are repeated several times at different y coordinates and the position of the n.risk numbers corresponds to the x-axis tick marks not the survival curve time of
2012 Apr 11
1
R-help; generating censored data
Hello, ?can i implement this as 10% censored data where t gives me failure and x censored. Thank you p=2;b=120 n=50 set.seed(132); r<-sample(1:50,45) t<-rweibull(r,shape=p,scale=b) t set.seed(123);? cens <- sample(1:50, 5)? x<-runif(cens,shape=p,scale=b)? x Chris Guure Researcher, Institute for Mathematical Research UPM
2012 May 11
2
survival analysis simulation question
Hi, I am trying to simulate a regression on survival data under a few conditions: 1. Under different error distributions 2. Have the error term be dependent on the covariates But I'm not sure how to specify either conditions. I am using the Design package to perform the survival analysis using the survreg, bj, coxph functions. Any help is greatly appreciated. This is what I have so far:
2011 Aug 25
1
survplot() for cph(): Design vs rms
Hi, in Design package, a plot of survival probability vs. a covariate can be generated by survplot() on a cph object using the folliowing code: n <- 1000 set.seed(731) age <- 50 + 12*rnorm(n) label(age) <- "Age" sex <- factor(sample(c('male','female'), n, TRUE)) cens <- 15*runif(n) h <- .02*exp(.04*(age-50)+.8*(sex=='Female')) dt <-
2012 Dec 03
1
Confidence bands with function survplot
Dear all, I am trying to plot KM curves with confidence bands with function survplot under package rms. However, the following codes do not seem to work. The KM curves are produced, but the confidence bands are not there. Any insights? Thanks in advance. library(rms) ########data generation############ n <- 1000 set.seed(731) age <- 50 + 12*rnorm(n) label(age) <- "Age"
2011 May 08
1
question about val.surv in R
Dear R users: I tried to use val.surv to give an internal validation of survival prediction model. I used the sample sources. # Generate failure times from an exponential distribution set.seed(123) # so can reproduce results n <- 1000 age <- 50 + 12*rnorm(n) sex <- factor(sample(c('Male','Female'), n, rep=TRUE, prob=c(.6, .4))) cens <- 15*runif(n) h