On Aug 21, 2011, at 7:03 AM, Salvo Mac wrote:

 I've   attached a sample of the data sets, this is my full code.

#R 2.13
#library(survival)
#library(Hmisc)
#library(splines)
library(rms)
train<-as.data.frame( train<-read.csv("G:\\train.txt", header=T, sep="\t")) test<-as.data.frame( test<-read.table("G:\\test.txt", header=T, sep="\t"))
f.1<-cph(Surv(time,event)~age, x=T, y=T,surv=T, data=train)
val.surv(f.1, newdata=test, u=10)


#plot(calibrate(f.1, u=30, B=20))

Salvo;

I'm not sure where the error is coming from (although I will share my speculation below). I modified your code somewhat. I was under the impression that read.csv had sep="," hardcoded and the read.csv( ... seo="\t") would throw and error. The as.data.fame around either of the read.* statements is completely unnecessary:

Input code:
train<-read.table("~/train.txt", header=T, sep="\t")
 test<-read.table("~/test.txt", header=T, sep="\t")

I thought the newdata=test argument should succeed, but it does throw the error that you report:

 f.1<-cph(Surv(time,event)~age, x=T, y=T,surv=T, data=train)
 val.surv(f.1, newdata=test, u=10)
Error in val.surv(f.1, newdata = test, u = 10) :
  dims [product 210] do not match the length of object [314]
In addition: Warning message:
In est.surv + S[, 1] :
  longer object length is not a multiple of shorter object length

I tried sampling from test with replacement to generate a dataframe of equal extent and with that object I do not get the same errors:

> extend <- test[sample(1:nrow(test), 314, replace=TRUE), ]
>  f.1<-cph(Surv(time,event)~age, x=T, y=T,surv=T, data=train)
> val.surv(f.1, newdata=extend, u=10)

Validation of Predicted Survival at Time= 10    n= 314 , events= 64

hare fit:

dim A/D   loglik       AIC        penalty
                                min    max
  1 Add   -485.24    976.22  271.33     Inf
  2 Del   -349.57    710.64    5.67  271.33
  3 Del   -346.74    710.72    0.96    5.67
  4 Del   -346.26    715.51    0.01    0.96
  5 Add   -346.25    721.25    0.00    0.01

the present optimal number of dimensions is 2.
penalty(AIC) was 5.75, the default (BIC), would have been 5.75.

  dim1           dim2           beta        SE         Wald
Constant                             -10       0.55  -18.14
Time        43                      0.15      0.015   10.30

Function used to transform predictions:
function (p)  log(-log(p))

Mean absolute error in predicted probabilities: 0.0331
0.9 Quantile of absolute errors               : 0.0613

I have also tried looking at the code and adding options(error=utils::recover) to see if I can identify the point where the length mismatch is being generated, (but I am NOT an ace debugger). I can see that est.surv is created. I can also get the predict(fit, newdata, type = "lp") call to run with train and give sensible numbers. You did not create (or at least say so) a datadist. I tried that when I saw that "lim" was dependent on datadist limits.

ddT <- datadist(train); options(datadist="ddT")

Seeing the usehare was set to TRUE; I submitted the code section that was intended for that situation to a browser session and the the first line throws the error that I see at the console:

Enter a frame number, or 0 to exit

1: val.surv(f.1, newdata = test, u = 10)

Selection: 1
Called from: top level
Browse[1]> i <- !is.na(est.surv + S[, 1] + S[, 2])
Error during wrapup: dims [product 210] do not match the length of object [314]

As I read this that line is supposed to create an index of valid rows in newdata and the fitted values but is failing in the situation where the nrows of the newdata does not match that of the fit.

At this point one option might be trying to generate a structure that matches the val.surv output by going through the code and building it up bit by bit:

 w <- structure(list(harefit = f, p = est.surv, actual = actual,
            pseq = pseq, actualseq = actualseq, u = u, fun = fun,
n = nrow(S), d = sum(S[, 2]), units = units), class = "val.survh")
        return(w)

But a much better option would be to report the error to Frank Harrell. I'm copying him since I think your .txt files probably reached the list as well as my mailbox.

--
David.






________________________________
From: David Winsemius <dwinsem...@comcast.net>
To: Salvo Mac <salvo_...@yahoo.com>
Cc: "r-help@R-project.org" <r-help@r-project.org>
Sent: Sunday, August 21, 2011 5:29 AM
Subject: Re: [R] val.surv


On Aug 20, 2011, at 10:25 PM, Salvo Mac wrote:

The test and train are like split data sets, contain similar variables but from different countries so the two sets are somehow independent. And yes it is a data frame.

What is a data.frame? test and train may be dataframes, but test[, "age"] is not a dataframe.

So I extracted age, time and event.

Code? The code you offered before would have created a newdata object (yes, a data.frame) with a single column bearing the same name as the vector argument ,,,,, "test1". Not named "age". Try it. do str on such an object:

str(dataframe(test1))


So test is data frame,(age, time, event). does that suffice?

It certainly does not allow me to reproduce the error you got (which I still think is probably related to the structure of your argument to newdata.)

That's all I can say without data and code.

--David.





________________________________
From: David Winsemius <dwinsem...@comcast.net>

Cc: "r-help@R-project.org" <r-help@r-project.org>
Sent: Sunday, August 21, 2011 3:19 AM
Subject: Re: [R] val.surv


On Aug 20, 2011, at 8:08 PM, Salvo Mac wrote:

Thanks David

However, I tried your trick on val.surv with newdata=test['age'] but still didn't work.
Still gives the same error message:

Error in val.surv(f.1, newdata = test1["age"], u = 10) :
    dims [product 1797] do not match the length of object [2496]
In addition: Warning message:
In est.surv + S[, 1] :
    longer object length is not a multiple of shorter object length


As I said (and you did
not act  upon):

The fundamental thing you are doing wrong for q1 is failing to unambiguously describe the test object.

I said it was a guess. Now stop wasting our time and offer what is needed.

--david.

Salvo
________________________________
From: David Winsemius <dwinsem...@comcast.net>

Cc: "r-help@R-project.org" <r-help@r-project.org>
Sent: Sunday, August 21, 2011 12:55 AM
Subject: Re: [R] val.surv


On Aug 20, 2011, at 3:32 PM, Salvo Mac wrote:

    Dear R-users,

I  have two questions regarding
validation and calibration of Survival regression models.

1. I am trying to calibrate and validate a cox model using val.surv.
here is my code:
    f.1<-cph(Surv(time,event)~age, x=T, y=T, data=train)
    test1<-test[,"age"]
    val.surv(f.1, newdata=data.frame(test1), u=10)

    but I get an error message:

    Error in val.surv(f.1, newdata = data.frame(testi), u = 10) :
     dims [product 1797] do not match the length of object [2496]
    In addition: Warning message:
In est.surv + S[, 1] :
longer object length is not a multiple of shorter object length

I ran the example in the r-documentation but couldn't extract dxy from result.

    What am I doing wrong?

The fundamental thing you are doing wrong for q1 is failing to unambiguously describe the test object. I would think that if test were a dataframe then wrapping data.frame around a vector might not get it named correctly as 'age'. You might try newdata= test['age']. Just a guess.


2. In validate and calibrate cph functions. If it is frailty fit, does the the bootstrap resample clusters or just individuals

The code above appears to be dependent on the rms package. The frailty function is part of the underlying survival package and I do not see it mentioned in the index for Harrell's RMS text. You will probably need to wait until Frank comes across this. He is generally very good about correction my errors and knowledge gaps.




David Winsemius, MD
West Hartford, CT<train.txt><test.txt>______________________________________________
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David Winsemius, MD
West Hartford, CT

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