Eekhout, I. wrote on 09/19/2011 10:16:17 AM:
>
> Hi all,
>
> I would like to replace the for loop in the code below with a function
> to improve the speed and to make the script more efficient.
> The loop creates a vector of integers (x) with the probability of f for
> each integer.
> The lengt
sorry, there is a superfluous bracket at the end of the line, it should
be read as
sample(1:3,50,prob=c(0.5,0.15,0.35),replace=T)
Am 19.09.2011 19:08, schrieb Eik Vettorazzi:
> Hi Iris,
> maybe I misinterpret this, but I think in the end it all comes down to
> sample(1:3,50,prob=c(0.5,0.15,0.35),r
Hi Iris,
maybe I misinterpret this, but I think in the end it all comes down to
sample(1:3,50,prob=c(0.5,0.15,0.35),replace=T))
cheers
Am 19.09.2011 17:16, schrieb Eekhout, I.:
> Hi all,
>
> I would like to replace the for loop in the code below with a function
> to improve the speed and to mak
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
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
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(
On Wed, Feb 10, 2010 at 9:50 AM, Ivan Calandra
wrote:
> I'm still quite new in R and I don't really understand this whole
> "vectorization" thing.
>
See [1] for an article on vectorisation and loops in R.
Liviu
[1] http://www.r-project.org/doc/Rnews/Rnews_2008-1.pdf
_
Have you considered using a different data structure:
> # change the data structure
> x <- data.frame(
+ type=rep(c('x1', 'x2', 'x3'), each=100),
+ high=c(d[,4], d[,5], d[,6]),
+ value=c(d[,1], d[,2], d[,3]))
> head(x)
type high value
1 x10 0.8936737
2 x10 -1.047298
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