On 9/3/21 12:59 PM, Bond, Stephen wrote:
>
> I looked at the nocenter and it says (-1,0,1) values but it seems that any
> three-level
> factor is included in that (represented as 1,2,3 in R) .
>
A factor is turned into a set of 0/1 dummy variable, so the nocenter applies.�
I will add
more cla
I looked at the nocenter and it says (-1,0,1) values but it seems that any
three-level factor is included in that (represented as 1,2,3 in R) .
Also, is the baseline curve now showing the reference level and not the
fictional .428 sex? If I predict the risk for a new row, should I multiply the
c
See ?coxph, in particular the new "nocenter" option.
Basically, the "mean" component is used to center later computations.� This can
be
critical for continuous variables, avoiding overflow in the exp function, but
is not
necessary for 0/1 covariates.�� The fact that the default survival curve
Hi,
Please, help me understand what is happening with the means of a Cox model?
I have:
R version 4.0.2 (2020-06-22) -- "Taking Off Again"
Copyright (C) 2020 The R Foundation for Statistical Computing
Platform: x86_64-w64-mingw32/x64 (64-bit)
getOption("contrasts")
unordered ord
Perhaps this package could be considered
https://cran.r-project.org/web/packages/hrIPW/hrIPW.pdf
That packages author also has a 2016 article in Statistics in Medicine on the
properties of estimates from such analyses that might be useful.
—
David Winsemius, MD, MPH
Sent from my iPhone
> On
When running a propensity score weighted analysis using coxph(), are the
weights entered as the log of the weights, or as the weights on the original
scale, i.e. coxph(Surv(time,status)~group,weights=weights ,data=mydata) or
coxph(Surv(time,status)~group,weights=log(weights),data=myda
t)
> # might want to allow different shapes or scales for different events
> for (k in seq(kk)) {
> short[[paste0("time", k)]] <- rweibullPH(nn, shape=2,
> scale=exp(beta[k]*tx)*gamma)
> }
> # might want to allow censoring
>
> long <- reshape(sho
<- coxph(Surv(times) ~ tx, data=long, subset=event==1)
#summary(mod1)
mod2 <- coxph(Surv(times) ~ tx, data=long, subset=event==2)
#summary(mod2)
coef(mod)
coef(mod0)
coef(mod1)
coef(mod2) - coef(mod1)
coef(summary(mod))
coef(summary(mod0))
coef(summary(mod1))
-----Original Message---
On 7/19/19 10:19 AM, Denise b wrote:
Dear R users,
I am interested in estimating the effects of a treatment on two
time-to-event traits (on simulated data), accounting for the dependency
between the two time-to-event outcomes.
I precise that the events are NOT recurrent, NOT competitive, NO
Dear R users,
I am interested in estimating the effects of a treatment on two
time-to-event traits (on simulated data), accounting for the dependency
between the two time-to-event outcomes.
I precise that the events are NOT recurrent, NOT competitive, NOT ordered.
The individuals are NOT related
Dear users,
I'm trying to estimate a conditional logistic model using the
coxph()-function from the survival package. Somehow, the model does not
converge if time is set to the same value for all observations:
library(survival)
set.seed(12345)
n <- 3000
a <- rbinom(n, 1, 0.5
> On Mar 25, 2016, at 8:32 PM, Shelby Leonard
> wrote:
>
>
> I have been working for 12 hours now on trying to fix my cox model. I have a
> large data set (n>300) and In my model I have time in months, status
> (1=dead), and change represents my variable. I am trying to determine if the
> v
Dear Prof Therneau,
I have explored a bit more on this issue - I found that for this specific
problem,
one can get the older convergent behavior with just setting toler.chol to 4
e-12.
This is only a little larger than 2x the current default =
.Machine$double.eps^.75
= 1.8e-12.
Since you have
R throws out this event due to the missing time-dependent covariate at day
6. Is there a way I can keep this event without filling in a covariate
value at day 6?
No.
The Cox model assesses the risk associated with each covariate by comparing, at each event
time, the values of the subject who h
On Jan 13, 2014, at 10:50 AM, Phil Stevenson wrote:
> I am using the coxph() function (from the 'survival' package) for a data
> set that requires one of the covariates to be treated as time-dependent.
> The issue is with this data set is there are some covariate values that are
> missing at the t
I am using the coxph() function (from the 'survival' package) for a data
set that requires one of the covariates to be treated as time-dependent.
The issue is with this data set is there are some covariate values that are
missing at the time point when the event occurs.
For example:
Person A:
Da
I'll re-enter the fray.
The data set is an example where coxph is incorrect; due to round off error it
is treating
a 5 column rank 3 matrix as if it were rank 4. This of course results in 0 digits of
precision.
Immediate fix, for the user, is to add "toler.chol= 1e-10" to their coxph
call.
On Oct 29, 2013, at 4:38 AM, rm wrote:
> Any ideas would be much appreciated; I suspect that this problem of
> constructing the dummies applies not only to function coxph but to other
> regression models in R as well. Effectively, my question is how to better
> control for which dummies and inter
Any ideas would be much appreciated; I suspect that this problem of
constructing the dummies applies not only to function coxph but to other
regression models in R as well. Effectively, my question is how to better
control for which dummies and interactions to include in the model and which
not.
T
The output is as follows. My question is how to include only the interaction
terms where x is equal to "Maintained".
(x has two possible values "Maintained" and "Nonmaintained".)
--
View this message in context:
http://r.789695.n4.nabble.com/coxph-how-to-define-interaction-terms-tp4679162p46
I’m trying to set up Cox Proptional Hazard model with interactions between
time and the covariates (which are categorical). The problem that I face is
that how to define the interactions, i.e. “x+cutStart:x”, properly.
The code below illustrates the problem. R gives the error message ” X matrix
de
Dr. Therneau,
Thank you as always for first writing, and second continuing the Cox model in R
(and earlier I believe in SAS).
While your comments concerning non-proportional hazards is helpful, it does not
fully address the question, "What alternatives do I have if I assume
proportional assump
ards does NOT hold but it may be close enough to be useful.
This is always a problem with goodness-of-fit tests and large datasets.
Chris
-Original Message-
From: Soumitro Dey [mailto:soumitrod...@gmail.com]
Sent: Tuesday, August 13, 2013 10:38 AM
To: Terry Therneau
Cc: r-help@r-pr
Thank you for your response, Terry.
To put the discussion into perspective, my data set is quite large with
over 160,000 samples and 38 variables. The event is true for all samples in
this dataset. The distribution is zero-inflated (i.e. most events occur at
time = 0).
The result of the cox.zph l
That's the primary reason for the plot: so that you can look and think.
The test statistic is based on whether a LS line fit to the plot has zero slope. For
larger data sets you can sometimes have a "significant" p-value but good agreement with
proportional hazards. It's much like an example
Thanks to Bert and Göran for your responses.
To answer Göran's comment, yes I did plot the Schoenfeld residuals using
plot.cox.zph and the lines look horizontal (slope = 0) to me, which makes
me think that it contradicts the results of cox.zph.
What alternatives do I have if I assume proportional
On 08/11/2013 06:14 AM, Soumitro Dey wrote:
Hello all,
This may be a naive question but since I'm new to R/survival models, I
cannot figure it out the problem myself.
I have a coxph model for my data and I am trying to test if the
proportional hazards assumption holds. Using cox.zph on the mo
Although someone on this list may respond, AFAICS this does not seem
to be an R question for R-help.I would suggest that you spend some
time with a local statistician.
A general observation: Statistical model assumptions neither :hold"
nor "don't hold." Quoting George Box, "All models are wrong, b
Hello all,
This may be a naive question but since I'm new to R/survival models, I
cannot figure it out the problem myself.
I have a coxph model for my data and I am trying to test if the
proportional hazards assumption holds. Using cox.zph on the model I get a
global score:
GLOBAL NA 4.20e+02 0
tion but I still need to research that.
Many thanks to Thomas Hielscher who authored the c060 package and was extremely
kind to help me with this.
-Original Message-
From: E Joffe [mailto:ejo...@hotmail.com]
Sent: Sunday, July 07, 2013 10:02 AM
To: 'Marc Schwartz'
Cc: 'r-
rds,
Marc
-Original Message-
From: Marc Schwartz [mailto:marc_schwa...@me.com]
Sent: Saturday, July 06, 2013 8:46 AM
To: E Joffe
Cc: r-help@r-project.org
Subject: Re: [R] coxph won't converge when including categorical (factor)
variables
On Jul 6, 2013, at 7:04 AM, E Joffe wr
I don't know much about your problem, but if you want help you are going to
need to learn to communicate effectively. This may help:
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
---
Je
On Jul 6, 2013, at 7:04 AM, E Joffe wrote:
> Hello,
>
>
>
> [rephrasing and reposting of a previous question (that was not answered)
> with new information]
>
>
>
> I have a dataset of 371 observations.
>
> When I run coxph with numeric variables it works fine.
>
> However, when I try to
Hint: Why do you think no one replied?
1. That it converged with one model/fitting algorithm and not with
another is useless.
2. This cannot possibly be answered without your data.
3. A guess: you are overfitting -- glmnet regularizes and will
therefore fit (highly) correlated regressors coxph w
Hello,
[rephrasing and reposting of a previous question (that was not answered)
with new information]
I have a dataset of 371 observations.
When I run coxph with numeric variables it works fine.
However, when I try to add factor (categorical) variables it returns "Ran
out of iterations a
Hello,
I have a dataset of 371 observations.
When I run coxph with numeric variables it works fine.
However, when I try to add factor variables it returns "Ran out of
iterations and the model did not converge"
There is something very strange with the factors - some of them should
actually be
Dear list,I am quite new to the world of biostatistics and I encounter some
issues in the precise understanding of the coxph function of the survival
package.I have a set of survival data (patient who had (or died from) a
breast cancer) I'd like to see which are the variables that might cause dead
For your first question -- read the manual. ?survfit.coxph will reveal the "censor"
argument, which controls the inclusion of points where the curve does not drop.
For your second, "smooth" is in the eye of the beholder, literally. If the reason for a
smooth curve is to plot it, you need to d
f.int=FALSE)
lines(steps.x, steps.y, col=2)
smoothfun <- approxfun(steps.x, steps.y)
plot(smoothfun, from=0, to=3, add=TRUE, col=3, n=1000, lty=2)
-Original Message-
From: Bond, Stephen [mailto:stephen.b...@cibc.com]
Sent: Thursday, January 17, 2013 9:49 AM
To: r-help@r-project.org
S
I also tried fitting a spline to the resulting survival curve and the result
was horrible.
maybe spline won't work or knots need special handling.
overall, I must have the final point of the smooth survival to be same as the
final point of the raw Cox survival and no flat days, the drops sho
Hello users,
I would like to obtain a survival curve from a Cox model that is smooth and
does not have zero differences due to no events for those particular days.
I have:
> sum((diff(surv))==0)
[1] 18
So you can see 18 days where the survival curve did not drop due to no events.
Is there a wa
Dear Terry,
I agree that this example is highly atypical. Having said that, my experience
with optimization algorithms is that scaling (a.k.a. standardizing) the
continuous covariates is greatly helpful in terms of convergence. Have you
considered automatically standardizing the continuous co
On 09/03/2012 05:00 AM, r-help-requ...@r-project.org wrote:
The coxph function in R is not working for me when I use a continuous predictor
in the model. Specifically, it
> fails to converge, even when bumping up the number of max iterations
or setting reasonable initial values.
> The estimat
The coxph function in R is not working for me when I use a continuous predictor
in the model. Specifically, it fails to converge, even when bumping up the
number of max iterations or setting reasonable initial values. The estimated
Hazard ratio from the model is incorrect (verified by an AFT mod
achieve?
Is it statistically correct to use this option of method=df where I fix the
df to a certain number so that my model converges?
Suggestions (and criticisms :) ) are all welcome. Thank you very much.
Regards,
D
--
View this message in context:
http://r.789695.n4.nabble.com/R-coxph
On Thu, Jul 26, 2012 at 4:19 PM, Thomas Lumley wrote:
> YOu need to update the survival package: from its NEWS file
>
> Changes in version 2.36-14
>A supposedly cosmetic change to coxph in the last release caused
> formulas with a "." on the right hand side to fail. Fix this and add a
> case
YOu need to update the survival package: from its NEWS file
Changes in version 2.36-14
A supposedly cosmetic change to coxph in the last release caused
formulas with a "." on the right hand side to fail. Fix this and add a
case with "." to the test suite.
-thomas
On Thu, Jul 26, 2012 at
Hi all,
I cant' wrap my head around an error from the coxph function (package
survival). Here's an example:
library(survival)
n = 100;
set.seed(1);
time = rexp(n);
event = sample(c(0,1), n, replace = TRUE)
covar = data.frame(z = rnorm(n));
model = coxph(Surv(time, event)~ . , data = covar)
R g
Dear List,
Here is an example of survival data in counting process format
(detailed record of each day)
> data[data$Id == 11,]
# extracted one person's record
Id Event Fup Start Stop sex Drug1
601 11 0 6 01 0 0
602 11 0 6 12 0 0
603 11 0 6 2
As has been answered several times on rhelp ... the baseline hazard is
for a case with the "mean value". It's not a meaningful case with all
factor variables. There can be no cases where fidelity>3 has a
fractional value. You should be using predict() and survfit() to
display estimates for
Hi,
In the following results I interpret exp(coef) as the factor that multiplies
the base hazard rate if the corresponding variable is TRUE. For example,
when the bucket is ks008 and fidelity <= 3, then the rate, compared to the
base rate h_0(t), is h(t) = 0.200 h_0(t). My question is then, to wha
Thanks,
Javier
--
View this message in context:
http://r.789695.n4.nabble.com/COXPH-TIME-DEPENDENT-tp3754837p3756128.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listin
B is the specification for time-varying covariates. Otherwise, your model
will think that each row is one independent observation that either had an
event or was censored at "time" or "total_time."
HTH,
Daniel
javier palacios wrote:
>
> Dear R-community,
>
> which of the following two format
Jacob Brogren brogren.nu> writes:
>
> All,
>
> I rerun once again and managed to reproduce the results from the text book.
> Made no changes to the code. Could
> it be some problem with convergence?
It is possible, but *extremely* unlikely, to get non-deterministic
results from R (i.e. runn
All,
I rerun once again and managed to reproduce the results from the text book.
Made no changes to the code. Could it be some problem with convergence?
Anyhow, now it works!
Cheers
Jacob
ps. I find "The R Book" very useful ds.
28 jun 2011 kl. 15.48 skrev Robert A LaBudde:
> Did you create
Hi,
sorry about that; here is the full output - data set, structure, model and
result.
Cheers
Jacob
> seedlings
cohort death gapsize status
1 September 7 0.5889 1
2 September 3 0.6869 1
3 September12 0.1397 1
4 September 1 0.1921 1
5 September
Did you create the 'status' variable the way indicated on p. 797?
Frequently with Surv() it pays to use syntax such as Surv(death,
status==1) to make a clear logical statement of what is an event
(status==1) vs. censored.
PS. Next time include head(seedlings) and str(seedlings) to make
clear
On Jun 28, 2011, at 6:51 AM, Jacob Brogren wrote:
Hi,
I ran the example on pp. 799-800 from Machael Crawley's "The R Book"
using package survival v. 2.36-5, R 2.13.0 and RStudio 0.94.83. The
model is a Cox's Proportional Hazards model. The result was quite
different compared to the R Boo
Hi,
I ran the example on pp. 799-800 from Machael Crawley's "The R Book" using
package survival v. 2.36-5, R 2.13.0 and RStudio 0.94.83. The model is a Cox's
Proportional Hazards model. The result was quite different compared to the R
Book. I have compared my code to the code in the book but ca
Dear R-users,
I computed a simple coxph model and plotted survival over time with
associated confidence intervals for 2 covariate levels (males and
females).
M1 <- coxph(survobject~sex, data=surv)
M1
survsex <- survfit(survobject~sex,data=surv)
summary(survsex)
plot(survsex, conf.int=T, col=c("b
> Dear users,
>
> In a study with recurrent events:
> My objective is to get estimates of survival (obtained through a Cox
> model) by rank of recurrence and by treatment group.
> With the following code (corresponding to a model with a global
> effect of the treatment=rx), I get no error and m
On May 6, 2011, at 6:22 PM, Eva Bouguen wrote:
Dear users,
In a study with recurrent events:
My objective is to get estimates of survival (obtained through a Cox
model) by rank of recurrence and by treatment group.
With the following code (corresponding to a model with a global
effect of t
Dear users,
In a study with recurrent events:
My objective is to get estimates of survival (obtained through a Cox model) by
rank of recurrence and by treatment group.
With the following code (corresponding to a model with a global effect of the
treatment=rx), I get no error and manage to obtai
Responding to T. Therneau:
> sessionInfo()
R version 2.10.1 (2009-12-14)
i386-pc-mingw32
locale:
[1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252
LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C LC_TIME=English_United States.125
> I have a model with quant vars only and the error message does not
make sense:
Could you tell us what version of S and of the survival package you are
using? You can get this with sessionInfo(), see the posting guide for
details.
This would help me identify the issue. I was planning to pos
plains is missing.
Thanks David.
Stephen
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Thursday, February 03, 2011 3:10 PM
To: Bond, Stephen
Cc: r-help@r-project.org
Subject: Re: [R] coxph fails to survfit
On Feb 3, 2011, at 2:14 PM, Bond, Stephen wrote:
I have
ching time dependent risk to the proper time index of the
survival curve.
Thanks David.
Stephen
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Thursday, February 03, 2011 3:10 PM
To: Bond, Stephen
Cc: r-help@r-project.org
Subject: Re: [R] coxph fails t
On Feb 3, 2011, at 2:14 PM, Bond, Stephen wrote:
I have a model with quant vars only and the error message does not
make sense:
(mod1 <- coxph(Surv(time=strt,time2=stp,event=(resp==1))~ +incpost
+I(amt/1e5)+rate+strata(termfac),
subset=dt<"2010-08-30", data=inc,method="efron"
I have a model with quant vars only and the error message does not make sense:
(mod1 <- coxph(Surv(time=strt,time2=stp,event=(resp==1))~
+incpost+I(amt/1e5)+rate+strata(termfac),
subset=dt<"2010-08-30", data=inc,method="efron"))
Call:
coxph(formula = Surv(time = strt, time2 = stp,
Larry,
You found a data set that kills coxph. I'll have to think about what
to do since on the one hand it's your own fault for trying to fit a very
bad model, and on the other I'd like the routine to give a nice error
message before it dies.
In the data set you sent me the predictor varia
On Nov 25, 2010, at 5:16 PM, Bond, Stephen wrote:
The following fit does not make sense to me, please, correct me if I
have a logical error.
moddowsn
Call:
coxph(formula = Surv(start, stop, resp) ~ sn + matfac2, data = coxsn1,
method = "efron")
coef exp(coef) se(coef)
The following fit does not make sense to me, please, correct me if I have a
logical error.
> moddowsn
Call:
coxph(formula = Surv(start, stop, resp) ~ sn + matfac2, data = coxsn1,
method = "efron")
coef exp(coef) se(coef) z p
sn2 0.0497 1.051 0.02030 2.45
made smart and subsequent calls on the same model will use the
first call to survfit?? It's your call :-)
Kind regards
Stephen B
-Original Message-
From: Terry Therneau [mailto:thern...@mayo.edu]
Sent: Thursday, October 28, 2010 6:39 PM
To: Bond, Stephen; David Winsemius
Cc:
Gentlemen,
I read R-news in batch mode so I'm often a day behind. Let me try to
answer some of the questions.
1. X*beta != linear.predictor.
I'm sorry if the documentation isn't all it could be. Between the book,
tech report, and help I've written about 400 pages, but this particular
topic
comcast.net]
Sent: Wednesday, October 27, 2010 1:15 PM
To: Bond, Stephen
Cc: r-help@r-project.org
Subject: Re: [R] coxph linear.predictors
On Oct 27, 2010, at 12:12 PM, Bond, Stephen wrote:
> I would like to be able to construct hazard rates (or unconditional
> death prob)
Hazards are
sem...@comcast.net]
Sent: Wednesday, October 27, 2010 1:15 PM
To: Bond, Stephen
Cc: r-help@r-project.org
Subject: Re: [R] coxph linear.predictors
On Oct 27, 2010, at 12:12 PM, Bond, Stephen wrote:
> I would like to be able to construct hazard rates (or unconditional
> death prob)
Hazard
riety of formulas, so I guess this is the spirit of SAS
disguising itself as an R package.
Stephen Bond
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Wednesday, October 27, 2010 2:17 PM
To: Bond, Stephen
Cc: r-help@r-project.org
Subject: Re: [R]
lp so
it's quite possible that my errors in this area will get expert review
and correction.
Your comments are appreciated.
Stephen Bond
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Wednesday, October 27, 2010 1:15 PM
To: Bond, Stephen
Cc: r-h
sampled daily this is a day interval.
Your comments are appreciated.
Stephen Bond
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Wednesday, October 27, 2010 1:15 PM
To: Bond, Stephen
Cc: r-help@r-project.org
Subject: Re: [R] coxph linear.predictors
Hi, everybody
>On Wed, Sep 8, 2010 at 5:43 PM, Min-Han Tan > wrote:
David said my R code text attachment got rejected by the mailing list.
Pooh. I don't think that's nice. I don't see anything in the
posting guide about a limit on text attachments.
Well, if you are still trying to understan
On Wed, 8 Sep 2010, Paul Johnson wrote:
run it with factor() instead of ordered(). You don't want the
"orthogonal polynomial" contrasts that result from ordered if you need
to compare against Stata.
If you don't want polynomial contrasts for ordered factors, you can just tell R
not to use th
I look at this question in a different angle. My understanding is:
1. If treat tumor_grade as a numerical variable, you assume the hazard ratio
is invariant between any two adjacent levels of the tumor grade (assuming
invariant covariate patterns of other risks);
2. If you treat the tumor_grade a
run it with factor() instead of ordered(). You don't want the
"orthogonal polynomial" contrasts that result from ordered if you need
to compare against Stata.
I attach an R program that I wrote to explore ordered factors a while
agol I believe this will clear everything up if you study the
exampl
On Sep 8, 2010, at 6:43 PM, Min-Han Tan wrote:
Dear R-help members,
Apologies - I am posting on behalf of a colleague, who is a little
puzzled
as STATA and R seem to be yielding different survival estimates for
the same
dataset when treating a variable as ordinal. Ordered() is used to
r
Dear R-help members,
Apologies - I am posting on behalf of a colleague, who is a little puzzled
as STATA and R seem to be yielding different survival estimates for the same
dataset when treating a variable as ordinal. Ordered() is used to represent
an ordinal variable) I understand that R's coxph
On Aug 9, 2010, at 8:17 PM, Hongying Li wrote:
Hello,
when using coxph function from survival package, what do you do if
you encounter such problem like:
Warning message:
In fitter(X, Y, strats, offset, init, control, weights = weights, :
Loglik converged before variable 3,4 ; beta may
Hello,
when using coxph function from survival package, what do you do if you
encounter such problem like:
Warning message:
In fitter(X, Y, strats, offset, init, control, weights = weights, :
Loglik converged before variable 3,4 ; beta may be infinite.
Thanks!
On Jul 30, 2010, at 11:08 AM, Biau David wrote:
Hello,
I would like to get the likelihood ratio and score tests for
specific variables
in a multivariate coxph model. The default is Wald, so the tests for
each
separate variable is based on Wald's test. I have the other tests
for the full
Hello,
I would like to get the likelihood ratio and score tests for specific variables
in a multivariate coxph model. The default is Wald, so the tests for each
separate variable is based on Wald's test. I have the other tests for the full
model but I don't know how to get them for each variabl
On Jun 15, 2010, at 1:34 PM, Federico Calboli wrote:
Hi everyone,
I'm running a cox ph model on a dataset with a number of variables.
Each variable has a different number of missing data, so that
coxph() drops the individuals who are missing data at one or more
variables. Because of this
On 15 Jun 2010, at 18:34, Federico Calboli wrote:
> I'm running a cox ph model on a dataset with a number of variables. Each
> variable has a different number of missing data, so that coxph() drops the
> individuals who are missing data at one or more variables. Because of this
> dropping (total
Hi everyone,
I'm running a cox ph model on a dataset with a number of variables. Each
variable has a different number of missing data, so that coxph() drops the
individuals who are missing data at one or more variables. Because of this
dropping (totally fine btw) I want to know how many events
I'm doing some coxph() analyses with a large and complex dataset. The
data was collected in different centers, so I am using
strata(centers) to stratify the analysis.
My main issue is, not all centers collected all the variables, so for
a model such as:
coxph(Surv(days, cancer) ~ varA + sex +
Hi everyone,
I'm doing some coxph() analyses with a large and complex dataset. The data was
collected in different centers, so I am using strata(centers) to stratify the
analysis.
My main issue is, not all centers collected all the variables, so for a model
such as:
coxph(Surv(days, cancer)
On Thu, 22 Oct 2009, Mura Tamakou wrote:
Dear All,
I have a question regarding the output of survfit() when I supply a Cox model.
Lets say for example:
we observe that the 95% CIs overlap!! How is this possible since the HR for
spiders is significant.
It's perfectly natural.
To a goo
Dear All,
I have a question regarding the output of survfit() when I supply a Cox model.
Lets say for example:
library(survival)
fit <- coxph(Surv(time, status == 2) ~ factor(spiders), data = pbc)
fit # HR for spiders is significant
newdata <- data.frame(spiders = factor(0:1))
sf <- survfit(f
Dear All,
I am interested to understand better the coxph()
function from the survival R package. There is one
referenced paper there which I am not able to get
hold of and it is
Andersen, P. Gill, R. (1982)
Cox's regression model for counting processes,
a large sample study.
Annals of Statisti
...
> Occasionally, (about 1 in every 100 simulations) I get the following warning:
> Error in coxph(Surv(start, end, censorind) ~ binary + uniform :
> X matrix deemed to be singular; variable 2
It is not uncommon for the X matrix in a Cox model to be close enough to
singular that the p
Hello,
I am currently trying to simulate data and analyze it using the frailty option
in the coxph function. I am working with recurrent event data, using counting
process notation. Occasionally, (about 1 in every 100 simulations) I get the
following warning:
Error in coxph(Surv(start, end,
> It seems that the function anova don't use a robust estimation for the
> analysis of deviance.
You are correct, anova.coxph always uses the partial likelihood
This is a serious oversight, particularly since some of the situations in
which the robust variance is recommended are ones where t
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