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
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 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
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
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
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
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
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
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 formula.
cox.xtab_miR=coxph(Surv(time, status) ~ miR + strata(sex,nb
On Feb 23, 2012, at 11:12 AM, R-girl wrote:
Hei,
I have a one simple question which does not seem to be that simple
as I
cannot find any solution/answer:
Is it possible to compare multiple survival curves in R with
survdiff-function when there is interaction term involved in predictor
vari
DEAR R-girl:
On Thu, Feb 23, 2012 at 8:12 AM, R-girl wrote:
> Hei,
>
> I have a one simple question which does not seem to be that simple as I
> cannot find any solution/answer:
>
> Is it possible to compare multiple survival curves in R with
> survdiff-function when there is interaction term inv
Hei,
I have a one simple question which does not seem to be that simple as I
cannot find any solution/answer:
Is it possible to compare multiple survival curves in R with
survdiff-function when there is interaction term involved in predictor
variables (and this interaction is significant)?
Examp
Hello Terri,
Thank you very much. This is the answer I needed. Could you also
tell me how I can calculate 25 and 50% quantiles in R? I can only get median
as far as I know.
Cem
Cem Girit
-Original Message-
From: Terry Therneau [mailto:thern...@mayo.edu]
Sent: Monday, October
On Sun, 2011-10-23 at 12:00 +0200, r-help-requ...@r-project.org wrote:
> The results by the survfit routine do not agree with the
> results of
> these formulae as obtained by SAS.
>
The next question should be "is SAS correct". The answer in this case
is no.
For survival data the
On Oct 20, 2011, at 5:05 PM, Sarah Goslee wrote:
Hi,
Please send your information to the r-help list, not just to me, but
do
note that the list is plain-text only.
But surely all you are looking for is:
dt<-
c
(37,41,40,38,38,37,44,45,48,43,48,46,54,60,32,45,55,62,42,62,62,62,47,42,59,43
1 0.889 0.105 0.706 1.000
>
> 43 8 1 0.778 0.139 0.549 1.000
>
> 47 7 2 0.556 0.166 0.310 0.997
>
>
>
> gtT=Vehicle
>
> time n.risk n.event survival std.err lower
Hi,
On Thu, Oct 20, 2011 at 2:04 PM, Cem Girit wrote:
> Hello,
>
>
>
> I need some results from the survival analysis of my data
> that I do not know whether exist in Survival Package or how to obtain if
> they do:
>
>
>
> 1. The Mean survival time
>
> 2. The standard
Hello,
I need some results from the survival analysis of my data
that I do not know whether exist in Survival Package or how to obtain if
they do:
1. The Mean survival time
2. The standard error of the mean
3. Point and 95% Lower & Upper Confidence Interv
Thanks for taking time to answer this
De: Terry Therneau
Para: David Winsemius
.org>
Enviado: miércoles 28 de septiembre de 2011 23:03
Asunto: Re: [R] survival analysis: interval censored data
David,
Thanks for taking time to answer this.
W
David,
Thanks for taking time to answer this.
When there are interval censored data, computation of the MLE first
finds a minimal number of time points such that every interval with a
death contains one of those points. Those are the points at which the
final curve can have a jump. The alg
On Sep 28, 2011, at 10:56 AM, Ruth Arias wrote:
hallo terry:
I attached araceae data set,
The usual survival analysis via the Kaplan-Meier method only make
estimates at the time of events. When you tabulate your data, you see
that there were no events for the missing (starting) "time"
hallo terry:
I attached araceae data set,
when I use this:
surara<-survfit(Surv(time,time2,event)~categoria)
Call: survfit(formula = Surv(time, time2, event) ~ categoria)
records n.max n.start events median 0.95LCL 0.95UCL
categoria=C 94 63 0 21 NA NA
You have still not given me enough information to reproduce your
problem. "Why doesn't it include all years?" I have no way of knowing,
since we have no data.
--- begin included message --
halo david
when I use type= 'interval'
Call: survfit(formula = Surv(ingreso, fecha, estado, type = "inter
er
Para: r-help@r-project.org
Enviado: martes 27 de septiembre de 2011 7:06
Asunto: Re: [R] survival analysis: interval censored data
Please adhere to the posting guide (i.e., provide a sample of self contained
code and provide the error message). And what does "but it is not working"
me
Please adhere to the posting guide (i.e., provide a sample of self contained
code and provide the error message). And what does "but it is not working"
mean? Is there an error code?
rueu wrote:
>
> hello:
> my data looks like:
> time1 time2 event catagoria
>
> 2004 2006 1
hello:
my data looks like:
time1 time2 event catagoria
2004 2006 1 C
2004 2005 0 C
2005 2010 1 E
2007 2009 1 C
2006 2007 0 E
2008 2010 0 C
2008 2010
6.4.1 Estimation of fixed effects
Heterogeneous team ability is a possible explanation for the result in
Section 6.3. That
result simply indicates that the more goals a team scores, the higher the
probability
that it will score more. However, teams that can score more goals also
indicate teams
with
Hi everybody,
I'm fairly new to survival analysis with R and have some questions how to
apply and interpret the coxph and related functions:
I have time-dependent covariates with several measurements per subject with
constant delta t. The covariates change in each time step.
I fitted the followi
2011/7/21 Abdullah OUESLATI :
>
> [...]
> I do not understand why you expect to get comparable results with SAS
> "discrete" and coxph "exact". They are two different approaches to
> handling ties (as Terry explained; of course, some comparability
> should be expected in "normal" cases).
> [...]
>
[...]
I do not understand why you expect to get comparable results with SAS
"discrete" and coxph "exact". They are two different approaches to
handling ties (as Terry explained; of course, some comparability
should be expected in "normal" cases).
[...]
I didn't know precisely the specifities of
discrete analogue to the continuous time proportional hazards model.
If you want to mimic exactly what SAS (discrete) does in R, try
'toBinary' in eha, and run an ordinary logistic regression on the
result. Should be close to what SAS gives you. One caveat; the result
of 'toBinary&#x
tion power rather than another algorithmic problem ?
--
View this message in context:
http://r.789695.n4.nabble.com/comparing-SAS-and-R-survival-analysis-with-time-dependent-covariates-tp874438p3680340.html
Sent from the R help mailing list archive at Nabble.com.
___
Let me expand a bit on Thomas's answer.
Looking more closely at your data set you have the following:
death time group 0group 1
1.5 0/413/13
3 0/4 5/5
8 4/4 0
At time 1.5 group 1 had 13 deaths out of
On Wed, Jul 20, 2011 at 5:42 AM, AO_Statistics wrote:
>
> Terry Therneau-2 wrote:
>>
>> This query of "why do SAS and S give different answers for Cox models"
>> comes
>> up every so often. The two most common reasons are that
>> a. they are using different options for the ties
>> b.
0 0
run;
proc phreg data=trythis;
model (start, end)*outcome(0)=delir/ ties=discrete;
RUN;
results :
estimate delir : -17.78257
se : 9383
Pr > Khi 2 : 0.9985
convergence status : "Convergence criterion (GCONV=1E-8) satisfied."
So I get 2 different warning messages and wit
Thanks a lot. I used survfit to generate KM plots but I will try to figure
out to find the corresponding returned value for KM estimate. Thank you for
your invaluable help.
On Mon, Jul 18, 2011 at 8:28 PM, David Winsemius wrote:
>
> On Jul 18, 2011, at 10:42 AM, Kishor Tappita wrote:
>
> Ok Davi
Ok David, Thanks for clearing my misconception. Then could you please
suggest me how I could get the KM estimate for a time point. I remember
reading the formula
for calculating the KM. Do, I have to write a method to compute the KM
estimate or will the survival package provide it for me?
Thank y
On Jul 18, 2011, at 10:42 AM, Kishor Tappita wrote:
Ok David, Thanks for clearing my misconception. Then could you
please suggest me how I could get the KM estimate for a time point.
I remember reading the formula
for calculating the KM. Do, I have to write a method to compute the
KM estim
Dear David,
Thank you for your reply. I realized that the summary function can help me
in finding the the order of the curves.
The 0.95UCL (95% upper confidence level) column in particular seems to
follow the same pattern as the curves on the plot.
I am new to survival analysis, so please corre
On Jul 18, 2011, at 10:10 AM, Kishor Tappita wrote:
> Dear David,
>
> Thank you for your reply. I realized that the summary function can
> help me in finding the the order of the curves.
> The 0.95UCL (95% upper confidence level) column in particular seems
> to follow the same pattern as the
On Jul 17, 2011, at 11:27 AM, Kishor Tappita wrote:
Dear List,
I used survival package of R and was able to generate KM plots. Now
I need
help in finding the top curve in a non-visualization way. I mean to
say that
I would like to know the order of curves through any existing
methods r
Dear List,
I used survival package of R and was able to generate KM plots. Now I need
help in finding the top curve in a non-visualization way. I mean to say that
I would like to know the order of curves through any existing methods rather
than looking at the KM plots. I need this information as I
Hello,
I have few questions about recurring events. I would greatly appreciate it
if anyone can assist me.
I have data that consist of approx 1,100 Consecutive patients released from
hospital after first Myocardial infarction (MI). They were followed for 13
years. Recurrent MI and unstable angina
---Begin inclusion --
I need some help to figure out what is the proper model in survival
analysis
for my data.
Subjects were randomized to 3 treatments in trial 1, some of them
experience
the event during the trial;
After period of time those subjects were randomized to 3 treatments
again in
trial
> Date: Fri, 22 Apr 2011 10:00:31 -0400
> From: littleduc...@gmail.com
> To: r-help@r-project.org
> Subject: [R] Survival analysis: same subject with multiple treatments and
> experience multiple events
>
> Hi there,
>
>
Hi there,
I need some help to figure out what is the proper model in survival analysis
for my data.
Subjects were randomized to 3 treatments in trial 1, some of them experience
the event during the trial;
After period of time those subjects were randomized to 3 treatments again in
trial 2, but dif
Hi there
I want to test the expression of a subset of genes for correlation with
patient survival. I found out that the coxph function is appropriate for
doing this since it works with continuous variables such as Affy mRNA
expression values.
I applied the following code:
cp <- coxph(Surv(t.r
Help!
I am unsure if I can analyze data from the following experiment.
Fish were placed in a tank at (t=0)
Measurements of Carbon Dioxide were taken each day for 120 days
(t=0,...120)
A few fish were then randomly pulled out of the tank at different days,
killed and examined for the presence of a
On Sep 16, 2010, at 4:43 PM, David Winsemius wrote:
On Sep 16, 2010, at 12:14 PM, smm7aa wrote:
Help!
I am unsure if I can analyze data from the following experiment.
Fish were placed in a tank at (t=0)
Measurements of Carbon Dioxide were taken each day for 120 days
(t=0,...120)
A few f
On Sep 16, 2010, at 12:14 PM, smm7aa wrote:
Help!
I am unsure if I can analyze data from the following experiment.
Fish were placed in a tank at (t=0)
Measurements of Carbon Dioxide were taken each day for 120 days
(t=0,...120)
A few fish were then randomly pulled out of the tank at differ
Help!
I am unsure if I can analyze data from the following experiment.
Fish were placed in a tank at (t=0)
Measurements of Carbon Dioxide were taken each day for 120 days (t=0,...120)
A few fish were then randomly pulled out of the tank at different days,
killed and examined for the presence of
On Tue, 27 Jul 2010, Christopher David Desjardins wrote:
Hi Charles,
On Fri, 2010-07-23 at 14:40 -0700, Charles C. Berry wrote:
On Fri, 23 Jul 2010, Christopher David Desjardins wrote:
Sorry. I should have included some data. I've attached a subset of my
data (50/192) cases in a Rdata file a
Hi Charles,
On Fri, 2010-07-23 at 14:40 -0700, Charles C. Berry wrote:
> On Fri, 23 Jul 2010, Christopher David Desjardins wrote:
>
> > Sorry. I should have included some data. I've attached a subset of my
> > data (50/192) cases in a Rdata file and have pasted it below.
> >
> > Running anova I g
On Fri, 23 Jul 2010, Christopher David Desjardins wrote:
Sorry. I should have included some data. I've attached a subset of my
data (50/192) cases in a Rdata file and have pasted it below.
Running anova I get the following:
anova(sr.reg.s4.nore)
Df Deviance Resid. Df-2*
Sorry. I should have included some data. I've attached a subset of my
data (50/192) cases in a Rdata file and have pasted it below.
Running anova I get the following:
> anova(sr.reg.s4.nore)
Df Deviance Resid. Df-2*LL P(>|Chi|)
NULL NA NA45 33.89
On Fri, 23 Jul 2010, Christopher David Desjardins wrote:
Hi,
I am trying to fit the following model:
sr.reg.s4.nore <- survreg(Surv(age_sym4,sym4), as.factor(lifedxm),
data=bip.surv)
Next time include a reproducible example. i.e. something we can run.
Now, Google "Hauck Donner Effect" to und
Hi,
I am trying to fit the following model:
sr.reg.s4.nore <- survreg(Surv(age_sym4,sym4), as.factor(lifedxm),
data=bip.surv)
Where age_sym4 is the age that a subject develops clinical thought
problems; sym4 is whether they develop clinical thoughts problems (0 or
1); and lifedxm is mothe
Dear Terry,
Thanks so much for your help; I'm a bit of an R novice at the moment (as you
can probably tell from my failure to use the data argument!) so any help is
most welcome.
I'm hoping to use this model to generate transition probabilities for a
Markov model and, as such, I was wondering if
A few suggestions
1. Use the data argument
wfit <- survreg(Surv(Time, Status) ~ Treatment, data=survdata)
summary(wfit)
2. Use the predict function to look at your survival curve. This
avoids any parameter change mistakes. It also works for any
distribution.
pp <- 0:60/100
wsurv <-
Dear all,
I'm trying to fit a curve to some 1 year failure-time data, so that I can
extrapolate and predict failure rates up to 3 years. The data is in the
general form:
Treatment Time Status
Treatment A28 0
Treat
roject.org] On
Behalf Of FishR
Sent: Wednesday, February 17, 2010 5:33 PM
To: r-help@r-project.org
Subject: Re: [R] Survival analysis
We are looking the extinction of a species of freshwater fish. The logistic
regression was derived by scoring the anecdotal descriptions of the species'
former
On Feb 17, 2010, at 5:32 PM, FishR wrote:
We are looking the extinction of a species of freshwater fish. The
logistic
regression was derived by scoring the anecdotal descriptions of the
species'
former population size (1 for a positive description of the
population e.g.
abundant, and 0 f
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