On 30.05.2010 19:23, Alan Lue wrote:
Is there a performance advantage to doing this, as opposed to growing
the vector within the loop? I suppose R could have to dynamically
reallocate memory at some point?
Right, but that takes time since memory management is always expensive
(and this way you need memory management in each iteration rather than
once) and your objects may be copied around all the time.
Uwe Ligges
Alan
2010/5/30 Uwe Ligges<lig...@statistik.tu-dortmund.de>:
On 26.05.2010 08:52, Alan Lue wrote:
Come to think of it, we can't save the output of each invocation and
concatenate it later, since we need the output as input for the next
iteration.
Yes, but you can do it a bit cleverer than before by initializing to the
fill length as in:
r.seq<- numeric(nrow(d))
r.seq[1]<- 2 * (1 / d$Dt[1] - 1)
for (i in 2:nrow(d)) {
r.seq[i]<- uniroot(bdt.deviation, interval = c(0, 1),
D.T = d$Dt[i], r.prior = r.seq[i-1])$root
}
Uwe Ligges
Alan
On Tue, May 25, 2010 at 11:43 PM, Alan Lue<alan....@gmail.com> wrote:
Since `for' loops are slow in R, and since `apply' functions are
faster, I was wondering whether there were a way to use an apply
function—or to otherwise avoid using a loop—when iterating over a
statement that updates its input.
For example, here's some such code:
r.seq<- 2 * (1 / d$Dt[1] - 1)
for (i in 2:nrow(d)) {
rf<- uniroot(bdt.deviation, interval=c(0, 1), D.T=d$Dt[i],
r.prior=r.seq)
r.seq<- append(r.seq, rf$root)
}
The call to `uniroot()' both updates `r.seq' and reads it as input.
We could save the output of each invocation of `uniroot()' and
concatenate it later, but is there a better way to write this (i.e.,
to execute more quickly) while updating `r.seq' in each iteration?
Alan
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