R help - Oct 2007 - some question about partial prediction in survival

Hi there:
i got a problem to get the prediction from a model recently.  for 
example if i use a survival analysis to predict the risk. i use the code 
like below: i found the the prediction is not equal to (coef * x + coef 
* sex) , could someone help me with why this happened? and  can someone 
explain to me how this command "predict(f, type="terms")"
works? is
every partial prediction equal to coef*x=predict(f,type="terms")[,1]
and
coef*sex=predict(f,type="terms")[,2]?  it looks like they did not
match.
however 
"predict(f)=predict(f,type="terms")[,1]+predict(f,type="terms")[,2]"
how should i understand this. thanks so much for your help

test1 <- list(time=  c(4, 1,1,2,2,3),
                status=c(1,1,0,1,1,0),
                x=     c(0,1,1,1,0,0),
                sex=   c(0,0,0,1,1,1))
f<-coxph( Surv(time, status) ~ x + sex, test1)

Call:
coxph(formula = Surv(time, status) ~ x + sex, data = test1)


      coef exp(coef) se(coef)      z    p
x    1.713     5.546     1.34  1.282 0.20
sex -0.154     0.857     1.45 -0.106 0.92

Likelihood ratio test=1.85  on 2 df, p=0.397  n= 6

bests;
T.D

On Sun, 14 Oct 2007, coldeyes.Rhelp wrote:
> Hi there:
> i got a problem to get the prediction from a model recently.  for
> example if i use a survival analysis to predict the risk. i use the code
> like below: i found the the prediction is not equal to (coef * x + coef
> * sex) , could someone help me with why this happened?
The intercept is not identifiable in a Cox model, and the code takes 
advantage of this to center the variables.  The predicted values are the 
values you expected, minus the mean.

For the example in ?predict.coxph
R> fit<-coxph(Surv(time,status)~x,data=aml)
R> predict(fit)
          1          2          3          4          5          6          7
-0.4776692 -0.4776692 -0.4776692 -0.4776692 -0.4776692 -0.4776692 -0.4776692
          8          9         10         11         12         13         14
-0.4776692 -0.4776692 -0.4776692 -0.4776692  0.4378634  0.4378634  0.4378634
         15         16         17         18         19         20         21
  0.4378634  0.4378634  0.4378634  0.4378634  0.4378634  0.4378634  0.4378634
         22         23
  0.4378634  0.4378634
R> sum(predict(fit))
[1] 7.21645e-16
R> a<-coef(fit)*(aml$x=="Nonmaintained")
R> a-mean(a)
  [1] -0.4776692 -0.4776692 -0.4776692 -0.4776692 -0.4776692 -0.4776692
  [7] -0.4776692 -0.4776692 -0.4776692 -0.4776692 -0.4776692  0.4378634
[13]  0.4378634  0.4378634  0.4378634  0.4378634  0.4378634  0.4378634
[19]  0.4378634  0.4378634  0.4378634  0.4378634  0.4378634

>							 and  can someone
> explain to me how this command "predict(f,
type="terms")" works? is
> every partial prediction equal to
coef*x=predict(f,type="terms")[,1] and
> coef*sex=predict(f,type="terms")[,2]?  it looks like they did not
match.
> however
>
"predict(f)=predict(f,type="terms")[,1]+predict(f,type="terms")[,2]"
The same thing happens here. Each column of predict(,type="terms")
sums to
zero.  In this case it is the same behaviour as  lm() and glm().

 	-thomas

For numerical accuracy, the coxph routine centers each covariate before doing 
the computation.  All of the downstream results (predict, survfit, etc) use this
centered data.  
 	Terry Therneau