R help - Oct 2008 - svm models in a loop

I want to train svm models on increasingly large training data subsets  
of some zrr as follows:

 > m <- sapply(1:5,function(i)  
svm(person_oid~.,data=zrr[1:100*i,]))    # (*)

However, when I inspect m[1], it literally shows

 > m[1]
[[1]]
svm(formula = person_oid ~ ., data = zrr[1:N, ])

-- as opposed to

 > m1 <- svm(person_oid~.,data=zrr[1:100,])
 > m1
 > m1

Call:
svm(formula = person_oid ~ ., data = zrr[1:100, ])
... -- actual parameters

How do I force actual model evaluation in (*) ?

Cheers,
Alexy

You need to use substitute() on the call.  Something like

sapply(1:5,function(i)
        eval(substitute(svm(person_oid ~ ., data=zrr[1:N,]), list(N=100*i))
        )
On Sun, 12 Oct 2008, Alexy Khrabrov wrote:
> I want to train svm models on increasingly large training data subsets of 
> some zrr as follows:
>
>> m <- sapply(1:5,function(i) svm(person_oid~.,data=zrr[1:100*i,]))   
# (*)
>
> However, when I inspect m[1], it literally shows
>
>> m[1]
> [[1]]
> svm(formula = person_oid ~ ., data = zrr[1:N, ])
I suspect it shows '100*i' not 'N', but in the absence of a
reproducible
example, I cannot check.
> -- as opposed to
>
>> m1 <- svm(person_oid~.,data=zrr[1:100,])
>> m1
>> m1
>
> Call:
> svm(formula = person_oid ~ ., data = zrr[1:100, ])
> ... -- actual parameters
>
> How do I force actual model evaluation in (*) ?
I don't think that is the issue, just the recorded call.
> Cheers,
> Alexy
-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
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