1. The computations "behind the scenes" produce the variance of the cumulative hazard.
This is true for both an ordinary Kaplan-Meier and a Cox model. Transformations to other
scales are done using simple Taylor series.
H = cumulative hazard = log(S); S=survival
var(H) = var(log(S)) = the starting point
S = exp(log(S)), so var(S) is approx [deriv of exp(x)]^2 * var(log(S)) = S^2
var(H)
var(log(log(S)) is approx (1/S^2) var(H)
2. At the time it was written, summary.survfit was used only for printing out the survival
curve at selected times, and the audience for the printout wanted std(S). True, that was
20 years ago, but I don't recall anyone ever asking for summary to do anything else. Your
request is not a bad idea.
Note however that the primary impact of using log(S) or S or log(log(S)) scale is is on
the confidence intervals, and they do appear per request in the summary output.
Terry T.
On 06/28/2014 05:00 AM, r-help-requ...@r-project.org wrote:
Message: 9
Date: Fri, 27 Jun 2014 12:39:29 -0700
From: array chip<arrayprof...@yahoo.com>
To:"r-help@r-project.org" <r-help@r-project.org>
Subject: [R] standard error of survfit.coxph()
Message-ID:
<1403897969.91269.yahoomail...@web122906.mail.ne1.yahoo.com>
Content-Type: text/plain
Hi, can anyone help me to understand the standard errors printed in the output
of survfit.coxph()?
time<-sample(1:15,100,replace=T)
status<-as.numeric(runif(100,0,1)<0.2)
x<-rnorm(100,10,2)
fit<-coxph(Surv(time,status)~x)
??? ### method 1
survfit(fit, newdata=data.frame(time=time,status=status,x=x)[1:5,],
conf.type='log')$std.err
???????????? [,1]??????? [,2]??????? [,3]??????? [,4]?????? [,5]
?[1,] 0.000000000 0.000000000 0.000000000 0.000000000 0.00000000
?[2,] 0.008627644 0.008567253 0.008773699 0.009354788 0.01481819
?[3,] 0.008627644 0.008567253 0.008773699 0.009354788 0.01481819
?[4,] 0.013800603 0.013767977 0.013889971 0.014379928 0.02353371
?[5,] 0.013800603 0.013767977 0.013889971 0.014379928 0.02353371
?[6,] 0.013800603 0.013767977 0.013889971 0.014379928 0.02353371
?[7,] 0.030226811 0.030423883 0.029806263 0.028918817 0.05191161
?[8,] 0.030226811 0.030423883 0.029806263 0.028918817 0.05191161
?[9,] 0.036852571 0.037159980 0.036186931 0.034645002 0.06485394
[10,] 0.044181716 0.044621159 0.043221145 0.040872939 0.07931028
[11,] 0.044181716 0.044621159 0.043221145 0.040872939 0.07931028
[12,] 0.055452631 0.056018832 0.054236881 0.051586391 0.10800413
[13,] 0.070665160 0.071363749 0.069208056 0.066655730 0.14976433
[14,] 0.124140400 0.125564637 0.121281571 0.118002021 0.30971860
[15,] 0.173132357 0.175309455 0.168821266 0.164860523 0.46393111
survfit(fit, newdata=data.frame(time=time,status=status,x=x)[1:5,],
conf.type='log')$time
?[1]? 1? 2? 3? 4? 5? 6? 7? 8? 9 10 11 12 13 14 15
??? ### method 2
summary(survfit(fit, newdata=data.frame(time=time,status=status,x=x)[1:5,],
conf.type='log'),time=10)$std.err
????????????? 1????????? 2????????? 3????????? 4????????? 5
[1,] 0.04061384 0.04106186 0.03963184 0.03715246 0.06867532
By reading the help of ?survfit.object and ?summary.survfit, the standard error provided
in the output of method 1 (survfit()) was for cumulative hazard-log(survival), while the
standard error provided in the output of method 2 (summary.survfit()) was for survival
itself, regardless of how you choose the value for "conf.type" ('log',
'log-log' or 'plain'). This explains why the standard error output is different between
method 1 (10th row) and method 2.
My question is how do I get standard error estimates for log(-log(survival))?
Thanks!
John
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