On Oct 12, 2008, at 2:26 AM, Prof Brian Ripley wrote:
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))
)
Thanks!
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.
Exactly -- I've mixed in an attempt to define N in sapply(...,
function(i) { N <- 100*i, ...}), but now see I need to do it in
substitute().
Cheers,
Alexy
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