Well actually, in that case, I don't think it is really necessary to
replace the loop because it is quite fast with the small size I have.
But I thought it would help me to understand how would vectorization
work and how to use the *apply functions.
Thanks anyway for your piece of advice!
Ivan
Le 2/10/2010 17:03, Petr PIKAL a écrit :
Hi
how many objects do you have in a list? Loop is ineffective if you use
several nested loops and/or there is some unnecessary mimic of simple
function.
e.g. you can use this
set.seed(666)
x<- runif(10)
vysled<- 0
for(i in 1:length(x)) {
if(vysled> x[i]) vysled<- vysled else vysled<- x[i]
}
vysled
but I would prefer max(x)
If you do not perceive performance issues there is usually no need to
elaborate *apply.
Regards
Petr
r-help-boun...@r-project.org napsal dne 10.02.2010 11:59:21:
After reading the R news, I've tried this code and it works:
> rapply(list(names(test),test), write.csv, file="filename.csv",
append=T, row.names=F)
However, the output is structured like this:
names(test)[[1]]
names(test)[[2]]
etc...
test[[1]]
test[[2]]
etc...
I would like to alternate names(test) and test in the output. The
desired output would be structured like this:
names(test)[[1]]
test[[1]]
names(test)[[2]]
test[[2]]
etc...
Can someone guide me for that step? Better solutions for the whole thing
are of course welcomed!
Thanks a lot
Ivan
Le 2/10/2010 10:50, Ivan Calandra a écrit :
Hi everybody!
I'm still quite new in R and I don't really understand this whole
"vectorization" thing.
For now, I use a for loop (and it works fine), but I think it would be
useful to replace it.
I want to export the result of a test statistic, which is stored as a
list, into a csv file. I therefore have to export each element of the
list separately.
Here is the code:
----
str(test)
List of 3
$ output : num [1:15, 1:6] 1 2 3 4 5 6 7 8 9 10 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : NULL
.. ..$ : chr [1:6] "con.num" "psihat" "p.value" "p.crit" ...
$ con : num [1:6, 1:15] 1 -1 0 0 0 0 1 0 -1 0 ...
$ num.sig: int 0
for (i in 1:3){
write.csv(test[[i]], file="filename.csv", append=T, quote=F,
row.names=T)
}
----
As I said, I don't completely understand, but I think one of these
"apply" function might do what I need.
Thanks in advance
Ivan
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