The S3 class model isn't really a class model. It's more a way of
overloading functions. So it's rather simple, and there's less there
than you might be expecting. I always thought the "Object Oriented
Programming" chapter of the R language definition manual that ships
with R was a very good treatm
Hi Caitlin and Ben,
Thanks for your responses! My issue is that I'd like to create one
continuous line, rather than 3 lines overlayed.
The code I've attached works for a population of 400 and samples 100 times.
I'd like to extend this to 300 samples and 3 populations. So, the x-axis
would range f
Hi R Users,
I am relatively new to programming in R … so I apologise if my questions appear
‘dumb’.
I am using a package that defines a number of S3 classes. I want to create an
S3 child class of one of these classes. The parent class has a contractor with
many arguments. I have been having di
Thanks for your answer.
However, my variable is simulated from the cumulative distribution function of
the Poisson distribution. So, the pattern obtained from the function "qpois" is
not the same as the observed pattern (i.e., obtained from the function "ppois")
set.seed(5)
mortality_probabili
If you run it under the profiler in RStudio, you will see that the 'lm'
call is taking about 2 seconds longer in the function which might have to
do with resolving the reference. So it is probably the function call in
'lapply' vs. the in-line statement in the 'for' loop that account for the
differ
I haven't tried your code, but I've seen far too many attempts like this where
there is no simple call to the function with starting parameters to see if a
sensible answer is returned. If you get NaN or Inf or ... that isn't sensible,
then you know it is not a good idea to try further.
Beyond thi
This is an inappropriate question for this list. See the posting guide
below and list archives for what is appropriate.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom Coun
Dear Jesus,
The difference is marginal when each code chunk does the same things. Your
for loop does not yields the same output as the lapply. Here is the cleaned
version of your code.
n<-1
set.seed(123)
x<-rnorm(n)
y<-x+rnorm(n)
rand.data<-data.frame(x,y)
k<-100
samples <- split(sample(n), r
A Google search on "lapply vs for r" or "lapply vs loop r" might have saved you
some trouble. Many people have debunked this myth. Strangely they all start out
with "everyone knows" or "it is commonly said that." I'm sure someone must have
said it, but no one seems to be able to provide an autho
Hmmm.
If I understand you correctly, your question has to do with adding lines to
your graph? If so, my ggplot2 skills are sort of floppy, but you could append
your sampling results to your data frame (one for each sample set) and then
simply add layers. Sort of like this.
N <- 10
x <- 1:N
d
The lapply loop and the for loop have very similar speed characteristics.
Differences seen are almost always due to how you use memory in the body of the
loop. This fact is not new. You may be under the incorrect assumption that
using lapply is somehow equivalent to "vectorization", which it is
Hi!
I am doing a lapply and for comparaison and I get that for is faster than
lapply.
What I have done:
n<-10
set.seed(123)
x<-rnorm(n)
y<-x+rnorm(n)
rand.data<-data.frame(x,y)
k<-100
samples<-split(sample(1:n),rep(1:k,length=n))
res<-list()
t<-Sys.time()
for(i in 1:100){
modelo<-lm(y
> How can I draw a Hypercube sample for the variable mortality_probability so
> that this variable exhibits the same pattern as the observed distribution?
One simple way is to use the uniform random output of randomLHS as input to the
quantile function for your desired distribution(s).
For exa
13 matches
Mail list logo
