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, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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