On 08 Apr 2015, at 10:14 , CHIRIBOGA Xavier wrote:
> Dear members,
>
>
>
> I am trying a survival analysis , I got:
>
>
>
> Error: could not find function "survreg"
>
>
>
> I already load
>
> the packages: survival
>
> the function: splines
>
Load or install? I suspect you actually
Dear members,
I am trying a survival analysis , I got:
Error: could not find function "survreg"
I already load
the packages: survival
the function: splines
What can I do?
Thanks for you help,
Xavier
__
R-help@r-project.org mailing lis
-- begin included message ---
Hi,
I am fitting a weibull model as follows
my models is
s <- Surv(DFBR$Time,DFBR$Censor)
wei <- survreg(s~Group+UsefulLife,data = DFBR,dist="weibull")
How can i predict the probabilty of failure in next 10 days, for a new data with group =10
and usefuleLife =10
Perhaps:
G10 <- predict(wei, list(Group =10 , UsefulLife =10), type="response")
You should get a single km-type object and will need to extract a time-specific
value.
(You really _should_ post what you have tried.)
--
David
Sent from my iPhone
Sent from my iPhone
> On Mar 14, 2014, at 9:11
Perhaps:
G10 <- predict(wei, list(Group =10 , UsefulLife =10), type="response")
You should get a single km-type object and will need to extract a time-specific
value.
(You really _should_ post what you have tried.)
--
David
Sent from my iPhone
> On Mar 14, 2014, at 7:11 PM, shilpi harpavat
Perhaps:
G10 <- predict(wei, list(Group =10 , UsefulLife =10), type="response")
You should get a single km-type object and will need to extract a time-specific
value.
(You really _should_ post what you have tried.)
--
David
Sent from my iPhone
> On Mar 14, 2014, at 7:11 PM, shilpi harpavat
Hi,
I am fitting a weibull model as follows
my models is
s <- Surv(DFBR$Time,DFBR$Censor)
wei <- survreg(s~Group+UsefulLife,data = DFBR,dist="weibull")
How can i predict the probabilty of failure in next 10 days, for a new data
with group =10 and usefuleLife =100
Thx
shilpi
ents:
> model
> 2.606
>
>
> -Original Message-
> From: Kyle Penner [mailto:kpen...@as.arizona.edu]
> Sent: Wednesday, June 12, 2013 3:49 PM
> To: Terry Therneau
> Cc: r-help@r-project.org
> Subject: Re: [R] survreg with measurement uncertainties
>
> Hi Terry
e standard linear model call:
lm(data ~ model - 1, weights = 1/err^2)
Call:
lm(formula = data ~ model - 1, weights = 1/err^2)
Coefficients:
model
2.606
-Original Message-
From: Kyle Penner [mailto:kpen...@as.arizona.edu]
Sent: Wednesday, June 12, 2013 3:49 PM
To: Terry Therneau
Cc
Hi Terry,
Thanks for your quick reply. I am talking about uncertainty in the
response. I have 2 follow up questions:
1) my understanding from the documentation is that 'id' in cluster(id)
should be the same when the predictors are not independent. Is this
correct? (To be more concrete: my dat
I will assume that you are talking about uncertainty in the response. Then one simple way
to fit the model is to use case weights that are proprional to 1/variance, along with
+cluster(id) in the model statement to get a correct variance for this case. In linear
models this would be called the
survreg allows interval censored data, if that is how you want to represent
measurement uncertainty. See
?Surv
-Original Message-
From: Kyle Penner [mailto:kpen...@as.arizona.edu]
Sent: Tuesday, June 11, 2013 8:02 PM
To: r-help@r-project.org
Subject: [R] survreg with measurement
Hello,
I have some measurements that I am trying to fit a model to. I also
have uncertainties for these measurements. Some of the measurements
are not well detected, so I'd like to use a limit instead of the
actual measurement. (I am always dealing with upper limits, i.e. left
censored data.)
Hello community,
I have a problem with my survreg function in R-Statistics.
My created code is the following:
library(survival)
surv=Surv(IPT_IN_DAYS,Status)
survival.data.weibull=survreg(surv~AGE_DAYS+KM_COUNT+LL_PER_DAY+IPT_HIS+IPT_SDEV,
data=spss, dist="weibull", scale=2.0)
summary(surviva
The short answer is "sort of".
Medium is that survreg implements the model framework found in Kalbfeisch and Prentice,
The statistical analysis of failure time data, chapter, chapter 2.2. The ime variable T has
f(time) ~ X' beta + sigma * W
where W is an error term from some distribution an
On 15/11/12 21:22, David Winsemius wrote:
On Nov 15, 2012, at 5:38 AM, Matthias Ziehm wrote:
Hi all,
Sorry if this has been answered already, but I couldn't find it in the archives
or general internet.
A Markmail/Rhelp search on: gompertz survreg ...brings this link to a reply
by Terry T
On Nov 15, 2012, at 5:38 AM, Matthias Ziehm wrote:
> Hi all,
>
> Sorry if this has been answered already, but I couldn't find it in the
> archives or general internet.
A Markmail/Rhelp search on: gompertz survreg ...brings this link to a reply
by Terry Therneau. Seems to address everything
Hi all,
Sorry if this has been answered already, but I couldn't find it in the
archives or general internet.
Is it possible to implement the gompertz distribution as
survreg.distribution to use with survreg of the survival library?
I haven't found anything and recent attempts from my side wer
Hi,
I need some help to manage frailty in Survreg function; in particular I'm
looking for more information about frailty in survreg function applied to a
loglogistic hazard function.
Actually I need to develope a predictor for frailty random variable realization
(similar to the Proportional Haz
On 04/22/2012 05:00 AM, r-help-requ...@r-project.org wrote:
I am trying to run Weibull PH model in R.
Assume in the data set I have x1 a continuous variable and x2 a
categorical variable with two classes (0= sick and 1= healthy). I fit the
model in the following way.
Test=survreg(Surv(t
Hi all,
I am trying to run Weibull PH model in R.
Assume in the data set I have x1 a continuous variable and x2 a
categorical variable with two classes (0= sick and 1= healthy). I fit the
model in the following way.
Test=survreg(Surv(time,cens)~ x1+x2,dist="weibull")
My questions are
1.
Dear Terry
 here is the survreg line from which I understand that "gender" is significant
survreg(formula = Surv(dias, status) ~ trat * sexo * rep, dist = "weibull")
                  Value    Std. Error     Â
z       p
sexom           -0.2187
--- begin included message ---
Hello R users,
I am analizing survival data (mostly uncensored) and want to extract the
most out of it.
Since I have more than one factor, I?ve read that the survival regression
can help to test the interactions between factors, and then decide how to do
the comparis
Hello R users,
I am analizing survival data (mostly uncensored) and want to extract the
most out of it.
Since I have more than one factor, I´ve read that the survival regression
can help to test the interactions between factors, and then decide how to do
the comparisons using the Log-rank test (sur
On Thu, Dec 8, 2011 at 3:55 AM, ifejobi wrote:
> Hello,
>
> I'm working with some left censored survival data using accelerated failure
> time models. I am interested in fitting different distributions to the data
> but seem to be getting the same results from the model fit using survreg
> regard
Hello,
I'm working with some left censored survival data using accelerated failure
time models. I am interested in fitting different distributions to the data
but seem to be getting the same results from the model fit using survreg
regardless of the assumed distribution.
These two codes seem to
-:begin inclusion ---
There can be some risk in creating your own extraction methods, since
the author of a package may have had reasons for not making it
available, but in this case he did offer a table that includes the
--end ---
You are reading too deeply. When survreg was written 20+ y
There is however the generic vcov(), which has a method for class
"survreg".
One good reason for not having a simple means to extract standard
errors is that considering coefficients in isolation is often unwise
(and has come up on R-help more than once already this month).
The summary() met
On May 14, 2011, at 5:48 AM, andre bedon wrote:
Hi,Just a quick one, does anyone know the command for accessing the
standard errors from a survreg object? I can access the coefficients
by model$coefficients, but I cant seem to find a command to access
the errors. Any help would be greatl
Hi,Just a quick one, does anyone know the command for accessing the standard
errors from a survreg object? I can access the coefficients by
model$coefficients, but I cant seem to find a command to access the errors. Any
help would be greatly appreciated.Regards,Andre
The fire devoured my laptop, but the PC is still working... :)
Sorry for the lazyness, you are totally right.
Here goes a reproducible example, which resembles the main features of our
dataset:
# three experimental factors, 30 subjects
expgrid <- expand.grid(F1=0:1,F2=0:1,F3=0:1,id=1:30)
# set
> I was wondering why survreg (in survival package) can not handle
> three-way interactions. I have an AFT .
You have given us no data to diagnose your problem. What do you mean
by "cannot handle" -- does the package print a message "no 3 way
interactions", gives wrong answers, your laptop
Dear All,
I was wondering why survreg (in survival package) can not handle three-way
interactions. I have an AFT (accelerated failure time) model with a
frailty term. If the model is restricted to two-way interactions, survreg
and gamlss.cens (with random() term) give very similar results, except
Dear Göran
I too have been working on this for the last couple of weeks and this is a
great help thanks. I wonder can you advise on how the covariance matrix of
estimated parameters in phreg relates to those estimated in survreg.
Im hoping you can just put your finger on this. Your eha package
On Fri, Dec 10, 2010 at 5:21 PM, Eleni Rapsomaniki
wrote:
> Dear R-users,
>
> I need to use the aftreg function in package 'eha' to estimate failure times
> for left truncated survival data.
Be careful! This is only possible under strong assumptions about the
(unobserved) covariate vector. This
Dear R-users,
I need to use the aftreg function in package 'eha' to estimate failure times
for left truncated survival data. Apparently, survreg still cannot fit such
models. Both functions should be fitting the accelerated failure time (Weibull)
model. However, as Göran Broström points out in
I a trying to use survreg to fit a Weibull distribution. From the last few
messages I understand how to interpret the parameters. However, how do I get
the covariance of lambda and alpha? Is there a predict command for that?
Thanks
--
View this message in context:
http://n4.nabble.com/Survreg-f
both scale and shape parameters.
Sorry, just forgot to reply to all when replying to your email. Unfortunately,
nobody else has replied yet so I wonder if anybody else could be helpful.
Cheers,
--- On Sat, 11/14/09, David Winsemius wrote:
> From: David Winsemius
> Subject: Re:
On Nov 13, 2009, at 3:17 AM, carol white wrote:
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 t
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,
-
I am trying to estimate by MLE weibull coefficients and deviation. But how
can i estimate deviation? I tried to find at the list and at the internet
this but couldn't make it. Anyone could help me on this?
Thanks.
[[alternative HTML version deleted]]
Dear Terry,
Thanks a lot for your reply. The result of survreg reads like:
Call:
survreg(formula = Surv(end, status) ~ fico_demean, data = raw,
dist = "loglogistic", model = TRUE)
Value Std. Error z p
(Intercept) 2.98365 1.34e-03 2233.2 0.00e+00
fico_demean
---begin included message --
I am trying to use R to do loglogistic hazard estimation. My plan is to
generate a loglogistic hazard sample data and then use survreg to estimate
it. If everything is correct, survreg should return the parameters I have
used to generate the sample data.
I have w
I am trying to use R to do loglogistic hazard estimation. My plan is to
generate a loglogistic hazard sample data and then use survreg to estimate
it. If everything is correct, survreg should return the parameters I have
used to generate the sample data.
I have written the following code to d
Laura Bonnett wrote:
Dear R-listers,
I know that there have been many, many posts on the output from
Survreg. To summarise what I have read, Scale is 1/shape of the
Weibull which is also the standard deviation of the normal
distribution which is also the standard deviation of the log survival
t
Dear R-listers,
I know that there have been many, many posts on the output from
Survreg. To summarise what I have read, Scale is 1/shape of the
Weibull which is also the standard deviation of the normal
distribution which is also the standard deviation of the log survival
time and Intercept is lo
David S. Schwarz wrote:
How does one indicate that a particular survival time is right censored in
the Survreg routine?
Read the documentation? :-)
--
Frank E Harrell Jr Professor and Chair School of Medicine
Department of Biostatistics Vanderbilt Universit
How does one indicate that a particular survival time is right censored in
the Survreg routine?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting gu
> survreg(Surv(times,censor),dist='weibull')
Error in x$terms : $ operator is invalid for atomic vectors
You need a right-hand side for the formula
survreg(Surv(times, censor) ~1, dist='weibull')
In your code the first argument was not recognized as a "formula", and the
program
I think R and survival took a look at what you provided for a formula
to survreg and rejected it because it had no lh-rh separator, "~", and
thus no rh terms. Turns out that it also expects a dataframe even if
both terms for the Surv function are already given and there is
nothing further t
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
Dear all -
Following up my own post, having found a Terry Therneau post about the
value of predict(model,type="quantile"),
http://tolstoy.newcastle.edu.au/R/e4/help/08/03/5335.html
the following code ammendment produces what I was intending to see.
It clear that the 0.5 quantile is the in
Dear all -
I have followed the thread the reply to which was lead by Thomas
Lumley about using pweibull to generate fitted survival curves for
survreg models.
http://tolstoy.newcastle.edu.au/R/help/04/11/7766.html
Using the lung data set,
data(lung)
lung.wbs <- survreg( Surv(time, status)
The combination of survreg + gamma frailty = invalid model, i.e., the example
that you quote.
I did not realize that this had been added to the survreg help file until
very
recently. I will try to fix the oversight. Other, more detailed documentation
states that Gaussian frailty + AIC
On Thu, 17 Apr 2008, Dimitris Rizopoulos wrote:
> Dear R-users,
>
> I have noticed small discrepencies in the reported estimate of the
> variance of the frailty by the print method for survreg() and the
> 'theta' component included in the object fit:
This is the print method for class "survreg.pe
Dear R-users,
I have noticed small discrepencies in the reported estimate of the
variance of the frailty by the print method for survreg() and the
'theta' component included in the object fit:
# Examples in R-2.6.2 for Windows
library(survival) # version 2.34-1 (2008-03-31)
# discrepancy
fit1
endet: Monday, March 10, 2008 8:57 PM
An: r-help@r-project.org
Betreff: [R] survreg asking for parameter "special"
Hello Gurus:
I'm using basic survival analysis, and trying to use survreg, and I provided
the parameter formula= and data=, but I got an error saying:
Error in t
Hello Gurus:
I'm using basic survival analysis, and trying to use survreg, and I provided
the parameter formula= and data=, but I got an error saying:
Error in terms.default(formula, special, data = data) : no terms
component
I'm wondering what it's trying to say. Is it because I did
Dear All - Any help/advice appreciated
OSX 10.5.1, R 2.6.1, survival 2.34
I am trying to fit an interval censored model with cluster or frailty
term (first 20 lines of the data set are below).
---MODEL
m1<-survreg(Surv(start,stop,event,type="interval")~tree+hag+entcir
+depth+
The question (slightly rephrased) was
Why does survreg(Surv(T.1, T.2, type='interval2') ~1, data=x) fail, when x is
T.1 T.2 Status
1 0.000 0.62873036 1
2 0.000 2.07039068 1
3 0.000 1.40297214 1
4 0.000 0.09112871 1
5 0.
0
28 1.6855216 999. 0
29 1.7934033 999. 0
Thanks in advance.
Ehsan
Re: [R] Survreg(), Surv() and interval-censored data
This message: [ Message body ] [ More options ]
Related messages: [ Next message ] [ Previous message ] [ In reply to ] [ Re:
[R] Survreg(),
61 matches
Mail list logo
