On Oct 27, 2010, at 1:45 PM, Bond, Stephen wrote:
coef(fit)*model.matrix(fit)[1,1]
age
11.69021
I don't know what that might be and you are not telling us what you
think it is.
I think this the calculation of the linear predictor, multiplying
(the beta*X)
I expected that coef(fit)*model.matrix(fit)[1,1]= fit
$linear.predictors[1]
And since it does not, I am lost.
Ultimately I need unconditional death probability from day 1 to day
t in the end.
Now it sounds like you do want cap-lambda, the cumulative hazard.
Let's call it Lambda
Since S(t) = exp(-Lambda(t)), you can invert to get Lambda(t) = -
log(S(t))
?survfit
Possibly:
plot(-log( survfit(fit)$surv ))
Survfit takes too much time as it does a lot of things I do not
need. I hope I should be able to get the death prob for every
subject from a single call to survfit and then rescale with
exp(linear.predictor) for each subject.
I have run into slowness with that approach myself. The rms package
provides survest and Predict methods, but if you wanted to stay in
survival, then why not create a survfit()$surv object based on a
vector of means and then calculate an exp(newdata) by which you would
multiply the S(mean|t)? There is also a predict.coxph function with
its type="risk" parameter that you might want to check.
Just a comment: as there is not deltat argument, I think that coxph
assumes deltat=1 so the hazard is the conditional probability over
the next time interval ( t(i-1), t(i) ). With dense data sampled
daily this is a day interval.
That's not the way I understand it. The Lambda() and S() functions
only change at the particular event times on a continuous time so they
are not assumed to have integer values. Pretty sure that Therneau
avoids presenting extraction methods for instantaneous hazards from
Cox models. In his book "Modeling Survival Data" he mostly sticks to
discussing cumulative hazard functions rather than estimating an
interval or instantaneous hazard. And fortunately he visits rhelp so
it's quite possible that my errors in this area will get expert review
and correction.
Your comments are appreciated.
Stephen Bond
-----Original Message-----
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Wednesday, October 27, 2010 1:15 PM
To: Bond, Stephen
Cc: r-help@r-project.org
Subject: Re: [R] coxph linear.predictors
On Oct 27, 2010, at 12:12 PM, Bond, Stephen wrote:
I would like to be able to construct hazard rates (or unconditional
death prob)
Hazards are not probabilities (since probabilities are constrained to
the range [0,1] and hazards are unbounded upward.)
for many subjects from a given survfit.
This will involve adjusting the ( n.event/n.risk)
with (coxph object )$linear.predictors
I must be having another silly day as I cannot reproduce the linear
predictor:
fit <- coxph(Surv(futime, fustat) ~ age, data = ovarian)
fit$linear.predictors[1]
[1] 2.612756
That's the linear predictor (the beta*X) and that particular number
only applies to the first case.
coef(fit)*model.matrix(fit)[1,1]
age
11.69021
I don't know what that might be and you are not telling us what you
think it is.
The above is based on the help listing for coxph.object
coefficients: the coefficients of the linear predictor, which
multiply
the columns of the model matrix. If the model is
over-determined there will be missing values in the vector
corresponding to the redundant columns in the model matrix.
Also, please comment whether n.event/n.risk
The Nelson-Aalen estimator of the cumulative hazard as a function of
intervals prior to t is sum( n-event(t)/ n.risk(t))
gives the baseline hazard exp(alpha) ?
No. The "baseline hazard", as you are calling this, would be an
estimate for persons with all covariates = 0, so in this case is for
women of age=0. (Not a particularly interpretable result in many
situations. The baseline hazard following treated ovarian cancer for
neonates is not medically sensible.)
What is the purpose of this request? Is someone telling you you need
to provide estimates for instantaneous hazards?
--
David Winsemius, MD
West Hartford, CT
David Winsemius, MD
West Hartford, CT
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