Hi - I am using the survfit() function to produce Kaplan-Meier
survival curves for several different groups.
survfit (Surv() ~ cohort, data=d)
Everything works fine, but I'd like to do something different.
I have a 12 month survival curve (base) and I also have two survival
curves based on some
www.9folders.com/>
From: Marc Schwartz
Sent: Friday, 4 September 2020 18:32
To: POLWART, Calum (SOUTH TEES HOSPITALS NHS FOUNDATION TRUST)
Cc: R-help; Terry Therneau
Subject: Re: [R] Survival Object - is 12month survival = 365days
On Sep 4, 2020, at 11:45 AM, POLWART,
On Sep 4, 2020, at 11:45 AM, POLWART, Calum (SOUTH TEES HOSPITALS NHS
FOUNDATION TRUST) via R-help wrote:
>
> Using survfit I can get the '1 year' Survival from this dataset which holds
> survival in days:
>
> require (survival)
> survfit( Surv(time, status) ~sex, data=colon)
> summary (fit, 3
Using survfit I can get the '1 year' Survival from this dataset which holds
survival in days:
require (survival)
survfit( Surv(time, status) ~sex, data=colon)
summary (fit, 365)
My current real world data I'm calculating time using lubridate to calculate
time and since it made the axis easy I j
;
> From: Ergin Artun
> Sent: Thursday, May 21, 2020 3:08 PM
> To: PIKAL Petr
> Subject: Re: [R] survival anaylsis with tabulated data
>
> Dear Pical,
> I'm also not an expert and had a typo error while translating colum names
> at code.
>
> At my data set
t knowing details I wonder if anybody is able to help you.
Cheers
Petr
From: Ergin Artun
Sent: Thursday, May 21, 2020 3:08 PM
To: PIKAL Petr
Subject: Re: [R] survival anaylsis with tabulated data
Dear Pical,
I'm also not an expert and had a typo error while translating colum names at
cod
or not properly structured data.
This is plain text mail list, please do not use HTML formating.
Cheers
Petr
> -Original Message-
> From: R-help On Behalf Of Ergin Artun
> Sent: Wednesday, May 20, 2020 2:01 PM
> To: r-help@r-project.org
> Subject: [R] survival anaylsis
Dear R Friends,
I'm a medical doctor with some knowledge about statistic and programming.
I want to analysis and compare data of different countries with survival or
popEpi tools in R.
I couldn't able to make tables with one row for every people of countries
without illness. I can made a data ta
cpolwart wrote:
> the event can only happen once
>
1. Yes, I'm not interested in repeat events.
2. I just had an interest in learning the code for survival analysis
in case there are: event, left censored (not left truncating) and
right censored, nothing more! But I no longer want to burden anyone
On 2020-04-26 10:48, Medic wrote:
Very grateful for the all comments!
My data contains:
• left censored
• right censored
• events
(interval censored does not contain!)
(P.S. I understood, that the code with "type = 'left'" is not
suitable, because is ONLY for left-censored.)
I wanted to get th
Very grateful for the all comments!
My data contains:
• left censored
• right censored
• events
(interval censored does not contain!)
(P.S. I understood, that the code with "type = 'left'" is not
suitable, because is ONLY for left-censored.)
I wanted to get the appropriate code for my so mix dat
Dear dr Medic,
Den 2020-04-17 kl. 23:03, skrev Medic:
On 2020-04-17 20:06, Medic wrote:
I can't understand how to do a survival analysis (?Surv ()) when some
event occurred before the start of observation (left censored). If I
understand correctly, there are two methods. I chose a method with:
On 2020-04-17 20:06, Medic wrote:
> I can't understand how to do a survival analysis (?Surv ()) when some
> event occurred before the start of observation (left censored). If I
> understand correctly, there are two methods. I chose a method with: 1)
> time from the start of treatment to the event a
On 2020-04-17 20:06, Medic wrote:
I can't understand how to do a survival analysis (?Surv ()) when some
event occurred before the start of observation (left censored). If I
understand correctly, there are two methods. I chose a method with: 1)
time from the start of treatment to the event and 2)
I can't understand how to do a survival analysis (?Surv ()) when some
event occurred before the start of observation (left censored). If I
understand correctly, there are two methods. I chose a method with: 1)
time from the start of treatment to the event and 2) the indicator of
the event. I did (i
Dear list users,
I need some guidelines to run a conditional logistic regression with
fixed effects.
Let me give you some background.
I have a survey where each respondent is asked the same question once
a year for three consecutive years. There is no apriori on the extent
to which choice at time
Dear list,
I'm trying to understand what kind of uncertainty intervals are
plotted for survival::survfit.coxph when 'conf.int=TRUE' is called,
e.g.:
f <- coxph(Surv(futime, fustat) ~ age, data = ovarian)
s <- survfit(f, newdata=data.frame(age=c(30, 60)))
plot(s, conf.int=TRUE)
Are these bands th
Dear list,
I'm trying to understand what kind of uncertainty intervals are
plotted for survival::survfit.coxph when 'conf.int=TRUE' is called,
e.g.:
f <- coxph(Surv(futime, fustat) ~ age, data = ovarian)
s <- survfit(f, newdata=data.frame(age=c(30, 60)))
plot(s, conf.int=TRUE)
Are these bands th
Thank You,Eric. I run this code without id="id" and it worked!
Eric Berger :
Medic via R - help < r - help @ r -project.org >
wrote:
addicts.cp=survSplit(addicts, cut=addicts$survt[addicts$status==1],
> end="survt", event="status",start="start", id="id")
>ERROR in survSplit(addicts, cut
You need to read the Help page for the function survSplit
?survSplit
states that the id argument is a
"character string with the name of new id variable to create (optional).
This can be useful if the data set does not already contain an identifier"
The addicts data.frame has as its first column
When I run the code from the book Kleinbaum "Survival analysis" (7. Running an
extended cox model. Page 646) I got an ERROR. What do you think is the reason?
I attach a file with few lines of dataset "addicts-dput.r".
> library(cmprsk)
> addicts = read.csv ("addicts.csv")
id clinic status su
> On Nov 6, 2017, at 5:45 AM, Meghna Govil wrote:
>
> Thanks David. Could you show me how to do that in my example ?
Possibly:
kmsurvival_np <- survfit(Surv( tte, censored) ~ . , data=train1_na)
I say "possibly" because I don't know whether all the columns of `train1_na`
were included in th
Try putting your data.frame (train_na) as argument to the data=
parameter and then in the formula put the actual variable names but
without the train_na$ prefix.
On 06/11/2017 13:45, Meghna Govil via R-help wrote:
Thanks David. Could you show me how to do that in my example ?
Thanks,
Meghna
Thanks David. Could you show me how to do that in my example ?
Thanks,
Meghna
> On Nov 6, 2017, at 12:58 AM, David Winsemius wrote:
>
> You should stop trying to use matrices on the RHS and using separate vectors
> to Surv. Instead use a data argument and have the names in your formula refer
You should stop trying to use matrices on the RHS and using separate vectors to
Surv. Instead use a data argument and have the names in your formula refer to
column names.
—
David
Sent from my iPhone
> On Nov 5, 2017, at 7:21 PM, Meghna Govil via R-help
> wrote:
>
>
> Hi - Below is my co
Hi - Below is my code and then the error when I run the last line.
time_np <- train1_na$tte
event_np <- train1_na$censored
X_np <- cbind(
train1_na$AMT,
train1_na$DISCOUNT_AMT,
train1_na$high_price_pcnt,
train1_na$EM_RECEIVED,
train1_na$DM_RECEIVED,
train1_na$TXN_WITH_RINGCODE,
> On Oct 16, 2017, at 8:28 AM, Meghna Govil wrote:
>
> Hi
>
> I'm trying to predict the values for a survreg object called
> loglogistic_na. Here is the definition of loglogistic_na and following that
> the syntax used for the predict function. But upon execution I don't get
> any output. Not s
> Hi
>
> I'm trying to predict the values for a survreg object called loglogistic_na.
> Here is the definition of loglogistic_na and following that the syntax used
> for the predict function. But upon execution I don't get any output. Not sure
> what I'm doing wrong:
>
> loglogistic_na <- sur
Hi
I'm trying to predict the values for a survreg object called
loglogistic_na. Here is the definition of loglogistic_na and following that
the syntax used for the predict function. But upon execution I don't get
any output. Not sure what I'm doing wrong:
loglogistic_na <- survreg(Surv(time_na,ev
,
Venkat
-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
Sent: Friday, April 28, 2017 6:46 PM
To: Henric Winell; Therneau, Terry M., Ph.D.; R-help
Cc: Seshan, Venkatraman E./Epidemiology-Biostatistics
Subject: Re: [R] survival package can't find Ccoxfit6
On
On 28/04/2017 5:37 PM, Henric Winell wrote:
On 2017-04-26 22:17, Duncan Murdoch wrote:
On 26/04/2017 2:51 PM, Therneau, Terry M., Ph.D. wrote:
A user contacted me directly about this, I answered with my best
understanding of the
recent R-help discussion of the issue, and their response to my
r
On 2017-04-26 22:17, Duncan Murdoch wrote:
On 26/04/2017 2:51 PM, Therneau, Terry M., Ph.D. wrote:
A user contacted me directly about this, I answered with my best
understanding of the
recent R-help discussion of the issue, and their response to my
response shows that I'm
not quite right.
I
Since my coding philosophy is "why compute something that is not needed" I
don't have timing data comparing coxph.fit to the stripped down version. I will
try to come up with a test suite.
I do work under Linux (the initial Windows output was because I had both 3.3.1
and 3.4.0 on that machine).
Thank you Drs. Therneau and Murdoch.
"Why not use coxph.fit?" -- My use case scenario is that I needed the Cox model
coefficients for resampled data. I was trying to reduce the computational
overhead of coxph.fit (since it will repeated a large number of times) by
stripping all the parts that I
On 27/04/2017 10:53 AM, sesh...@mskcc.org wrote:
Thank you Drs. Therneau and Murdoch.
"Why not use coxph.fit?" -- My use case scenario is that I needed the Cox model
coefficients for resampled data. I was trying to reduce the computational overhead of
coxph.fit (since it will repeated a large
On 04/27/2017 09:53 AM, sesh...@mskcc.org wrote:
Thank you Drs. Therneau and Murdoch.
"Why not use coxph.fit?" -- My use case scenario is that I needed the Cox model
coefficients for resampled data. I was trying to reduce the computational overhead of
coxph.fit (since it will repeated a larg
Let me summarize rather than repeat the entire thread:
An error report from a user (seshan) stumped me, and I asked for help here.
Duncan Murdoch picked up on fine details of the error message, i.e., that the error did
NOT come from within the survival package. That changes the whole tenor of
om]
Sent: Wednesday, April 26, 2017 4:18 PM
To: Therneau, Terry M., Ph.D.; R-help
Cc: Seshan, Venkatraman E./Epidemiology-Biostatistics
Subject: Re: [R] survival package can't find Ccoxfit6
On 26/04/2017 2:51 PM, Therneau, Terry M., Ph.D. wrote:
> A user contacted me directly about this,
])), oo$coefficients,
> control)
[1] -0.4838181
Venkat
From: Duncan Murdoch [murdoch.dun...@gmail.com]
Sent: Wednesday, April 26, 2017 4:17 PM
To: Therneau, Terry M., Ph.D.; R-help
Cc: Seshan, Venkatraman E./Epidemiology-Biostatistics
Subject: Re: [R] su
From: Duncan Murdoch [murdoch.dun...@gmail.com]
Sent: Wednesday, April 26, 2017 4:17 PM
To: Therneau, Terry M., Ph.D.; R-help
Cc: Seshan, Venkatraman E./Epidemiology-Biostatistics
Subject: Re: [R] survival package can't find Ccoxfit6
On 26/04/2017 2:51 PM, Therneau, Terry M., Ph.D.
Thanks,
Venkat
-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
Sent: Wednesday, April 26, 2017 4:18 PM
To: Therneau, Terry M., Ph.D.; R-help
Cc: Seshan, Venkatraman E./Epidemiology-Biostatistics
Subject: Re: [R] survival package can't find Ccoxfit6
On 26/04/
On 26/04/2017 2:51 PM, Therneau, Terry M., Ph.D. wrote:
A user contacted me directly about this, I answered with my best understanding
of the
recent R-help discussion of the issue, and their response to my response shows
that I'm
not quite right.
I am emphatically not an MS Windows user so am
A user contacted me directly about this, I answered with my best understanding of the
recent R-help discussion of the issue, and their response to my response shows that I'm
not quite right.
I am emphatically not an MS Windows user so am asking for help -- which I will cut/paste
to this user a
Hi Folks,
Hopefully this question has enough R and not too much stats to be
appropriate for this list. Based on,* Hosmer et al. 2013. Logistic
regression for matched case-control studies. Applied Logistic
Regression *(eqtn.
7.8)*, *I am assessing GOF of conditional (or matched) logistic regression
Hi!
Does anyone know about survival analysis using interval censored data and
the intcox package?
I fed different ant colonies with different food (called treatment) and I
measured every 2 days the death of ants. I would like to see if the
treatment can explain a different death of individuals.
S
Thanks Terry!
I thought that since I was providing survConcordance with the model object
that the same formula would be applied. But I was obviously wrong. I just
ran survConcordance with the addition of the strata argument, as you
suggested, and got the same answer as summary(fit)with the sam
I read the digest form which puts me behind, plus the last 2 days have been solid meetings
with an external advisory group so I missed the initial query. Three responses.
1. The clogit routine sets the data up properly and then calls a stratified Cox model. If
you want the survConcordance ro
r(site), method ="efron",
> data = dat) # warning
> fit <- clogit(resp ~ x1 + x2 + strata(ID) + cluster(site), method
> ="efron", data = dat) # no warning
> summary(fit)
>
> Chris
>
> -Original Message-
> From: Joe Ceradini [mailto:j
er(site), method ="efron",
data = dat) # no warning
summary(fit)
Chris
-Original Message-
From: Joe Ceradini [mailto:joecerad...@gmail.com]
Sent: Tuesday, January 19, 2016 12:48 PM
To: r-help@r-project.org
Subject: [R] Survival::coxph (clogit), survConcordance vs. summary(fit)
Hi,
I'm running conditional logistic regression with survival::clogit. I have
"1-1 case-control" data, i.e., there is 1 case and 1 control in each strata.
Model:
fit <- clogit(resp ~ x1 + x2, strata(ID), cluster(site), method ="efron",
data = dat)
Where resp is 1's and 0's, and x1 and x2 are both
Hi Terry,
I already suspected it had something to do with missing values in one of
the covariates. I couldn't get the na.action="na.exclude" to work (perhaps
this is because I am relatively unexperienced with R), but I managed to
solve the problem by using the following command with "complete.case
I expect that reading the result of print(fit.weib) will answer your question. If there
were any missing values in the data set, then the fit.weib$linear.predictors will be
shorter than the original data set,
and the printout will have a note about "...deleted due to missing".
The simplest sol
Hi,
I am fitting an AFT model assuming a Weibull distribution and I would like
to check the residuals compared to the Kaplan Meier residuals, but when I
try to create the Kaplan Meier residuals I get an error: Time and status
are different lengths.
I am using the following script:
# Fitting the
It depends on several factors. You need answers to all these questions: How
many events occurred, ... and was the period of observation long enough to
cover a significant fraction of the life expectancy, …. and is there external
evidence or theory that will help establish that this process shou
On 2015-08-18 01:44, David Winsemius wrote:
On Aug 17, 2015, at 1:51 PM, David Winsemius wrote:
On Aug 17, 2015, at 12:10 PM, survivalUser wrote:
Dear All,
I would like to build a model, based on survival analysis on some
data, that is able to predict the /*expected time until death*/
fo
I read this list a day late as a digest so my answers are rarely the first. (Which is
nice as David W answers most of the survival questions for me!)
What you are asking is reasonable, and in fact is common practice in the realm of
industrial reliability, e.g., Meeker and Escobar, Statistical
Thank you David for your answer.
Some follow-up questions:
- So, do you think that try to estimate the life expectancy would be risky
and probably not justifiable? Is there some sort of 'confidence' that the
model could give me for a prediction?
- type=response - I found it here:
https://stat.e
Ooops. I meant to drop that other message but hit the send icon instead.
On Aug 17, 2015, at 3:39 PM, Bert Gunter wrote:
> David:
>
> I may have misunderstood you here, specifically:
>
> "As such I would ask if you really wanted to use a parametric survival
> model in the first place? "
>
>
On Aug 17, 2015, at 1:51 PM, David Winsemius wrote:
>
> On Aug 17, 2015, at 12:10 PM, survivalUser wrote:
>
>> Dear All,
>>
>> I would like to build a model, based on survival analysis on some data, that
>> is able to predict the /*expected time until death*/ for a new data
>> instance.
>
> A
On Aug 17, 2015, at 3:39 PM, Bert Gunter wrote:
> David:
>
> I may have misunderstood you here, specifically:
>
> "As such I would ask if you really wanted to use a parametric survival
> model in the first place? "
>
> The K-M curve is , of course, a **non-parametric** fit, and that is
> why t
David:
I may have misunderstood you here, specifically:
"As such I would ask if you really wanted to use a parametric survival
model in the first place? "
The K-M curve is , of course, a **non-parametric** fit, and that is
why there can be no mean survival time unless the last point is a
death.
On Aug 17, 2015, at 12:10 PM, survivalUser wrote:
> Dear All,
>
> I would like to build a model, based on survival analysis on some data, that
> is able to predict the /*expected time until death*/ for a new data
> instance.
Are you sure you want to use life expectancy as the outcome? In order
Dear All,
I would like to build a model, based on survival analysis on some data, that
is able to predict the /*expected time until death*/ for a new data
instance.
Data
For each individual in the population I have the, for each unit of time, the
status information and several continuous covariat
better.
Your code is very interesting. Especially the parametric part. Nice to get an
actual p-value for the test.
Thanks,
Paul
On Mon, 7/21/14, Andrews, Chris wrote:
Subject: RE: [R] Survival Analysis with an Historical Control
Cc: "
Chris
-Original Message-
From: Paul Miller [mailto:pjmiller...@yahoo.com]
Sent: Thursday, July 10, 2014 8:59 AM
To: Andrews, Chris
Cc: r-help@r-project.org
Subject: RE: [R] Survival Analysis with an Historical Control
Hi Chris,
Thanks for pointing out the use of "View page source".
Thanks,
Paul
On Wed, 7/9/14, Andrews, Chris wrote:
Subject: RE: [R] Survival Analysis with an Historical Control
r-project.org>
Received: Wednesday, July 9, 2014, 11:26 AM
The code is actually
available at the websites you provide. Try &quo
Hi Dr. Therneau,
Thanks for your response. This is very helpful.
My historical control value is 16 weeks. I've been having some trouble though
determining how this value was obtained. Are you able to indicate how people
normally go about determining a value for the historical control? Or do you
You are asking for a one sample test. Using your own data:
connection <- textConnection("
GD2 1 8 12 GD2 3 -12 10 GD2 6 -52 7
GD2 7 28 10 GD2 8 44 6 GD2 10 14 8
GD2 12 3 8 GD2 14 -52 9 GD2 15 35 11
GD2 18 6 13 GD2 20 12 7 GD2 23 -7 13
GD2 24 -52 9 GD2 26 -52 12
er...@yahoo.com]
Sent: Tuesday, July 08, 2014 12:00 PM
To: r-help@r-project.org
Subject: [R] Survival Analysis with an Historical Control
Hello All,
I'm trying to figure out how to perform a survival analysis with an historical
control. I've spent some time looking online and in my boook
Hello All,
I'm trying to figure out how to perform a survival analysis with an historical
control. I've spent some time looking online and in my boooks but haven't found
much showing how to do this. Was wondering if there is a R package that can do
it, or if there are resources somewhere that s
Hi all, let's say we can fit a Cox model with a numeric variable "x" as the
independent variable. The we can calculate, say 10-year survival, for any given
value of "x" (0 to 10 in increment of 0.1 in the example below):
> fit <- coxph(Surv(time, event)~x,dat)
> surv10yr<-
summary(survfit(fit,n
Hi all, let's say we can fit a Cox model with a numeric variable "x" as the
independent variable. The we can calculate, say 10-year survival, for any given
value of "x" (0 to 10 in increment of 0.1 in the example below):
> fit <- coxph(Surv(time, event)~x,dat)
> surv10yr<-
summary(survfit(fit,ne
Hi,
I am trying to predict the time to certain type of failure given the
following data on Certain Factory Equipments. The data I have are
readings collected every day for sensor installed on those equipments
.On Same day, an equipment can have different Repaires performed,and on
some days no
On Jan 28, 2014, at 4:19 PM, Dennis Murphy wrote:
> Hi David:
>
>>> Do you meant [sic] the (cumulative) hazard function that is 1 minus the
>>> survival function?
>
> Um, 1 - S(t) = F(t), the cumulative *distribution function* of the
> random variable T. The cumulative hazard function is -ln S
On Jan 28, 2014, at 7:58 AM, Endy BlackEndy wrote:
> Dear R users, how I can implement Breslow and Tarone-Ware tests for
> survival analysis?
Tests ... of what?
Perhaps you should learn to search:
install.packages('sos')
library(sos)
findFn("Tarone-Ware")
The 'comp' function in {survMisc} pac
Dear R users, how I can implement Breslow and Tarone-Ware tests for
survival analysis?
Also, is there any way I can estimate and plot the hazard function,along
the lines of the survival function in Survival Package? (The package muhaz
seems that does not do what it promises, except if I could not u
Hi,
Thanks for your response.
On Thu, Nov 14, 2013 at 4:44 PM, Terry Therneau wrote:
> I think that your data is censored, not truncated.
> For a fault introduced 1/2005 and erased 2/2006, duration = 13 months
> For a fault introduced 4/2010 and still in existence at the last
> observation 1
I think that your data is censored, not truncated.
For a fault introduced 1/2005 and erased 2/2006, duration = 13 months
For a fault introduced 4/2010 and still in existence at the last observation 12/2010,
duration> 8 months.
For a fault introduced before 2004, erased 3/2005, in a machine
Hi,
I would like to know how to handle truncated data.
My intend is to have the survival curve of a software fault in order
to have some information
about fault lifespan.
I have some observations of a software system between 2004 and 2010.
The system was first released in 1994.
The event consider
On 06/05/2013 12:33 AM, r-help-requ...@r-project.org wrote:
Dear friends - I'm on windows 7, R 2.15.2
when I run the example for aareg in survival package I see this:
plot(lfit[4], ylim=c(-4,4))
error in xy.coords(x, y, xlabel, ylabel, log) :
'x' is a list, but does not have components
On Jun 4, 2013, at 1:36 PM, Troels Ring wrote:
Dear friends - I'm on windows 7, R 2.15.2
when I run the example for aareg in survival package I see this:
plot(lfit[4], ylim=c(-4,4))
error in xy.coords(x, y, xlabel, ylabel, log) :
'x' is a list, but does not have components 'x' and 'y'
Is th
Dear friends - I'm on windows 7, R 2.15.2
when I run the example for aareg in survival package I see this:
plot(lfit[4], ylim=c(-4,4))
error in xy.coords(x, y, xlabel, ylabel, log) :
'x' is a list, but does not have components 'x' and 'y'
Is that a matter of an old R?
Best wishes
Troels Ri
It is not OK.
Post to a statistical list, like stats.stackexchange.com , instead.
-- Bert
On Fri, May 24, 2013 at 1:55 PM, Troels Ring wrote:
> Dear friends - I hope it is OK to ask a question on the principles of
> survival analysis. I'm a medical person in this project, but wonders about
> t
Dear friends - I hope it is OK to ask a question on the principles of
survival analysis. I'm a medical person in this project, but wonders
about the statistical methods proposed. We have a huge dataset and
assumption of proportional hazard is said to be fulfilled if survival
time is fractionate
Thanks for your thoughts.
1. The function is 'somewhat' linear. Small curvature but beginning and end are
similar. Off course I could fit more complex functions of time.
2. Do I really have to build (start, stop) datasets as I have time-varying
covariate effect but not time-varying covariate?
Dear All,
I have built a survival cox-model, which includes a covariate * time
interaction. (non-proportionality detected)
I am now wondering how could I most easily get survival predictions from my
model.
My model was specified:
coxph(formula = Surv(event_time_mod, event_indicator_mod) ~ Se
Dear All,
I have built a survival cox-model, which includes a covariate * time
interaction. (non-proportionality detected)
I am now wondering how could I most easily get survival predictions from my
model.
My model was specified:
coxph(formula = Surv(event_time_mod, event_indicator_mod) ~ Sex +
Dear all
I have a fairly funky problem that I think demands some sort of survival
analysis. There are two Red List assessments for mammals: 1986 and 2008.
Some mammals changed their Red List status between those dates. Those
changes can be regarded as "events" and are "interval censored" in the sen
e to get the survival
curves for testX? Thanks in advance. DK
CC: r-help@r-project.org
From: dwinsem...@comcast.net
To: dkrsta...@hotmail.com
Subject: Re: [R] survival survfit with newdata
Date: Thu, 17 May 2012 00:52:55 -0400
On May 16, 2012, at 5:08 PM, Damjan Krstajic wrote:
Dea
oject.org
> From: dwinsem...@comcast.net
> To: dkrsta...@hotmail.com
> Subject: Re: [R] survival survfit with newdata
> Date: Thu, 17 May 2012 00:52:55 -0400
>
>
> On May 16, 2012, at 5:08 PM, Damjan Krstajic wrote:
>
> >
> > Dear all,
> >
> > I am confused wit
On May 16, 2012, at 5:08 PM, Damjan Krstajic wrote:
Dear all,
I am confused with the behaviour of survfit with newdata option.
Yes. It has the same behavior as any other newdata/predict from
regression. You need to supply a dataframe with the same names as in
the original formula. Doesn
Dear all,
I am confused with the behaviour of survfit with newdata option.
I am using the latest version R-2-15-0. In the simple example below I am
building a coxph model on 90 patients and trying to predict 10 patients.
Unfortunately the survival curve at the end is for 90 patients. Could som
I forgot to mention, the Design package is deprecated (which suggest
your version of R is rather dated, current is 2.15). The rms package
replaces Design.
On Thu, May 10, 2012 at 10:23 PM, Joshua Wiley wrote:
> Hi Grace,
>
> I seem to have sent an empty draft before. Anyway, something like
> th
Hi Grace,
I seem to have sent an empty draft before. Anyway, something like
this might be an approach (untested):
require(rms)
set.seed(10)
dat <- data.frame(
age = rnorm(500, 40, 10),
race = factor(sample.int(2,500,TRUE), labels = c("a", "b")))
X <- model.matrix(~ age + race, data = dat)
b
On Thu, May 10, 2012 at 7:41 PM, Grace Ma wrote:
> 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 usin
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
http://r.789695.n4.nabble.com/file/n4619765/survival_file.png
Hi everyone,
This is what my data looks like, I haven't included the covariates, the
example would get too large.
uid id date feverstart
dumfever
130 75 3464652011-04-11
Dear R users,
I fear this is terribly trivial but I'm struggling to get my head around it.
First of all, I'm using the "survival" package in R 2.12.2 on Windows Vista
with the RExcel plugin. You probably only need to know that I'm using
"survival" for this.
I have data collected from 180 or so
Dear All
Does any one know if the following survival analysis exists, and if so, is it
available in R?
I am trying to parametrically model survival where some events happen at time =
0. I am particularly interested in generating the 95% confidence interval
around the survival function over its
On 04/18/2012 05:00 AM, r-help-requ...@r-project.org wrote:
Hi,
I'm trying to use the R Survival analysis on a windows 7 system.
The input data format is described at the end of this mail.
1/ I tried to perform a survival analysis including stratified variables
using the following fo
1 - 100 of 272 matches
Mail list logo
