seq(along= surprise
Perhaps this is what was intended?
sims <- list(length=100)
do.call(seq, sims)
seq by itself does not expect a list, but do.call() can create the appropriate call if a list is what you want to pass to the function. Hope this helps, baptiste
On 5 Feb 2009, at 19:46, Uwe Ligges wrote:
Uwe Ligges wrote:
Kjetil Halvorsen wrote:
This surprised me:
reps <- 100 sims <- list(length=reps) sims
$length [1] 100
for(i in seq(along=sims))print(i)
[1] 1 This is R 2.8.1.
What is surprising? sims is now a list that contains 1 element called "length" with a numeric value of 100. Then seq(along=sims) is exactly 1, because sims has length 1. Hence i is printed once (1 iteration of the loop) and is 1 in the first (and only) iteration. Uwe
I should have added that you probably want sims <- vector(mode="list", length=100) Uwe
Kjetil [[alternative HTML version deleted]]
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_____________________________ Baptiste Augui? School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag