Displaying 20 results from an estimated 93 matches for "hazs".
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2007 Jan 19
1
Error in basehaz function ?
Hello R-users.
I believe that the way basehaz (in the survival package) compute the
baseline hazard function is false.
I come to question this function when it gives me hazard probabilities
greater than 1.
Looking at the code I think I've localised the error :
hazard probability is computed as :
H <- -log(surv)
but it seems to me that hazard probabilities is rather an instantaneous
2008 Sep 26
1
Computing Mean Lifetime from Hazard
Hello,
If all I have access to is an empirically calculated hazard function, is it possible to compute an approximate value for the mean lifetime?
I know that if the hazard function is essentially constant, the mean lifetime is 1/hazard rate. ?But if I'm confident that the empirical hazard function is not constant, I'm not sure how to go about calculating an estimate of mean lifetime.
2009 Jun 20
1
Plotting Cumulative Hazard Functions with Strata
Hello:
So i've fit a hazard function to a set of data using
kmfit<-survfit(Surv(int, event)~factor(cohort))
this factor variable, "cohort" has four levels so naturally the strata
variable has 4 values.
I can use this data to estimate the hazard rate
haz<-n.event/n.risk
and calculate the cumulative hazard function by
H<--log(haz)
Now, I would like to plot this
2013 Jul 16
0
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2010 Mar 16
1
Problems compiling a simple source package on MacOS X
Hello. I wrote a package (that contains C source) that I've been
compiling and running on both Linux and Windows for about a year.
However, that same source package fails to compile on MacOS (10.4.11,
PowerPC G4, Xcode installed, gcc version 4.0.1, make version 3.80, ld
version cctools-590.23.2.obj~17).
There is nothing platform-specific in the code-- just numerical
functions that I
2009 Jun 03
2
Create a time interval from a single time variable
I am trying to set up a data set for a survival analysis with time-varying covariates. The data is already in a long format, but does not have a variable to signify the stopping point for the interval. The variable DaysEnrolled is the variable I would like to use to form this interval. This is what I have now:
ID Age DaysEnrolled HAZ WAZ WHZ Food onARV
2004 May 10
1
Explaining Survival difference between Stata and R
Dear Everybody:
I'm doing my usual "how does that work in R" thing with some Stata
projects. I find a gross gap between the Stata and R in Cox PH models,
and I hope you can give me some pointers about what goes wrong. I'm
getting signals from R/Survival that the model just can't be estimated,
but Stata spits out numbers just fine.
I wonder if I should specify initial
2018 Apr 04
1
parfm unable to fit models when hazard rate is small
Hello, I would like to use the parfm package: https://cran.r-project.org/web/packages/parfm/parfm.pdfhttps://cran.r-project.org/web/packages/parfm/parfm.pdf in my work. This package fits parametric frailty models to survival data. To ensure I was using it properly, I started by running some small simulations to generate some survival data (without any random effects), and analyse the data using
2012 Aug 08
1
basehaz() in package 'Survival' and warnings() with coxph
Hello,
I have a couple of questions with regards to fitting a coxph model to a data
set in R:
I have a very large dataset and wanted to get the baseline hazard using the
basehaz() function in the package : 'survival'.
If I use all the covariates then the output from basehaz(fit), where fit is
a model fit using coxph(), gives 507 unique values for the time and the
corresponding cumulative
2012 Aug 09
1
basehaz() in package survival and warnings with coxph
I've never seen this, and have no idea how to reproduce it.
For resloution you are going to have to give me a working example of the
failure.
Also, per the posting guide, what is your sessionInfo()?
Terry Therneau
On 08/09/2012 04:11 AM, r-help-request at r-project.org wrote:
> I have a couple of questions with regards to fitting a coxph model to a data
> set in R:
>
> I have a
2013 Oct 01
0
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2014 Jan 30
2
objecto ggplot
Daniel,
Disculpa si no entiendo, pero, entonces un objecto ggplot en en realidad
una función
y no un gráfico
El 30 de enero de 2014, 11:09, Marta Garcia <marta000garcia@gmail.com>escribió:
> Sí, devuelve Error: No layers in plot, no le he agregado capas
> por eso el error, la funcion me dice que es un gráfico
>
>
> El 30 de enero de 2014, 10:31, José María Mateos
2006 Jun 12
0
non parametric estimates of the hazard with right censored data
Hi,
I want to plot non parametric estimates of the empirical hazard function for
right censored data. I've tried many functions from different packages
(muhaz, Design, survival, eha, event), but none of them gave me what I
wanted. Am I missing something?
Here's what I want. The data below is the same used by Kiefer (J. Economic
Literature, 1988), which in turns use a subset of the data
2008 Sep 05
1
Confidence Intervals on Hazard Plots
Hello, Is it possible to create confidence intervals for hazard rates? ?I'm creating two muhaz objects:?
haz1 <- muhaz(NumDaysCustomer[cRV=="true"],status[cRV=="true"])
haz2 <- muhaz(NumDaysCustomer[cRV=="false"],status[cRV=="false"])
and plotting them. ?
There are many, many more observations in the cohort cRV=="false" than
2012 Jan 05
1
Fwd: WHO Anthro growth curve macros and R&In-Reply-To=<CAAOCNNZawGtKkWpgFMYADSyxWGTeWEDxqVVHv7=Azo=1G+H9gg@mail.gmail.com>
Dear Gustaf,
I wish you a happy new year!
I am a PhD Staff at LSHTM and I have been trying to use WHO anthro macros in R with the file you posted on internet but when I double check the WHZ, HAZ, WAZ output with output from ENA or STATA, it differs slightly. Do you have any idea why that might be the case?
Many thanks
Severine
2012 Oct 29
0
why isn't integrate function working in a likelihood
Dear R users,
I have been trying to solve for mle's of a function that involves an
integral and I keep getting an error. I created an example to work on first
and even the simple example doesn't give me the mle's. I am getting the
error "Error in integrate(integrand, 0, Inf) : non-finite function value".
I divided my likelihood function into two parts, one part involves
2013 Jul 13
0
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2011 Nov 30
1
Nomogram with stratified cph in rms package, how to get failure probability
Hello,
I am using Dr. Harrell's rms package to make a nomogram. I was able to make
a beautiful one. However, I want to change 5-year survival probability to
5-year failure probability.
I couldn?t get hazard rate from Hazard(f1) because I used cph for the model.
Here is my code:
library(rms)
f1 <- cph(Surv(retime,dfs) ~
age+her2+t_stage+n_stage+er+grade+cytcyt+Cyt_PCDK2 , data=data11,
2009 Nov 30
4
incompatibilidad de tinnR y R
Tengo instalado el R2.9.0 y el TinnR 1.17.2.4 y cuando abro Tinn R no me
detecta q el R este abierto por lo cual no me permite correr ningun
programa, alguien sabe q puede estar pasando?
Saludos
--
José
"... hoy el cambio cualitativo la liberacion, implica, cambios organicos, de
instinto y biologicos, AL MISMO TIEMPO q cambios politicos y sociales" H. M.
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2009 Nov 28
2
R en Windows 7 a 64 bits
Buenas noches para cada uno.
Me alegra saber que superamos los 200 usuarios y tengos expectativas con
respecto a la I Conferencia de R-hispano en Murcia.
No soy usuario de Linux por desconocimiento, algo que lamento. Debo instalar
R en un equipo con Windows 7 a 64 bits. ¿Podrían indicarme por favor cómo
instalarlo? ¿Se hace con el mismo archivo R-2.10.0-win32? ¿Debo cuidar o
cambiar algo antes o