[R] val.surv
David Winsemius
dwinsemius at comcast.net
Sun Aug 21 18:35:45 CEST 2011
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 <dwinsemius at comcast.net>
> To: Salvo Mac <salvo_mac at yahoo.com>
> Cc: "r-help at R-project.org" <r-help at 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 <dwinsemius at comcast.net>
>>
>> Cc: "r-help at R-project.org" <r-help at 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 <dwinsemius at comcast.net>
>>>
>>> Cc: "r-help at R-project.org" <r-help at 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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