I am trying to write a function that will simulate observed counts X and
Y as below, I want the function to be able give many replicates, can
somebody advise?
thanks
Oarabile
#alpha,n sdx and sdy are constant, and N is a vector of length n.
unstructured<-function(n,N,alpha,sdx,sdy){
Vx<-rnorm(n
Dear R users,
Can anyone please tell me how to generate a large number of samples in R, given
certain distribution and size.
For example, if I want to generate 1000 samples of size n=100, with a N(0,1)
distribution, how should I proceed?
(Since I dont want to do "rnorm(100,0,1)" in R for 10
hey guys, i've been following this discussion about the simulation, and being a
beginner myself, im really unsure of the best method.
I hve the same problem as the initial one, except i need 1000 samples of size
15, and my distribution is Exp(1). I've adjusted some of the loop formulas for
sday, October 16, 2007 11:47 AM
Subject: [R] simulation
>I am trying to write a function that will simulate observed counts X and
> Y as below, I want the function to be able give many replicates, can
> somebody advise?
> thanks
> Oarabile
>
> #alpha,n sdx and sdy are con
On Wed, May 13, 2009 at 5:13 PM, Debbie Zhang wrote:
>
>
> Dear R users,
>
> Can anyone please tell me how to generate a large number of samples in R,
> given certain distribution and size.
>
> For example, if I want to generate 1000 samples of size n=100, with a N(0,1)
> distribution, how shoul
what about putting in a matrix, e.g.,
matrix(rnorm(1000*100), 1000, 100)
I hope it helps.
Best,
Dimitris
Debbie Zhang wrote:
Dear R users,
Can anyone please tell me how to generate a large number of samples in R, given
certain distribution and size.
For example, if I want to generate 10
If you want k samples of size n, why generate k*n samples and put them
in a k-by-n matrix where you can do what you want to each sample:
k = 10
n = 100
x=matrix(rnorm(k*n),k,n)
rowMeans(x)
If you need to do more complex things to each sample and if k is large
enough that you don't want the matrix
On Wed, May 13, 2009 at 4:26 PM, Gábor Csárdi wrote:
> On Wed, May 13, 2009 at 5:13 PM, Debbie Zhang wrote:
>>
>>
>> Dear R users,
>>
>> Can anyone please tell me how to generate a large number of samples in R,
>> given certain distribution and size.
>>
>> For example, if I want to generate 1000
Dear Debbie,
Here are two options:
# Parameters
N <- 1000
n <- 100
# Option 1
mys <- replicate(N, rnorm(n))
mys
# Option 2
mys2 <- matrix(rnorm(N*n),ncol=N)
mys2
HTH,
Jorge
On Wed, May 13, 2009 at 11:13 AM, Debbie Zhang wrote:
>
>
> Dear R users,
>
> Can anyone please tell me how to generat
Does every 100 numbers in rnorm(100 * 1000, 0, 1) have the N(0,1) distribution?
On Wed, May 13, 2009 at 11:13 PM, Debbie Zhang wrote:
>
>
> Dear R users,
>
> Can anyone please tell me how to generate a large number of samples in R,
> given certain distribution and size.
>
> For example, if I wan
Barry Rowlingson wrote:
> On Wed, May 13, 2009 at 4:26 PM, Gábor Csárdi wrote:
>
>> On Wed, May 13, 2009 at 5:13 PM, Debbie Zhang wrote:
>>
>>> Dear R users,
>>>
>>> Can anyone please tell me how to generate a large number of samples in R,
>>> given certain distribution and size.
>>>
>>>
On Wed, May 13, 2009 at 5:36 PM, Wacek Kusnierczyk
wrote:
> Barry Rowlingson wrote:
>> Soln - "for" loop:
>>
>> > z=list()
>> > for(i in 1:1000){z[[i]]=rnorm(100,0,1)}
>>
>> now inspect the individual bits:
>>
>> > hist(z[[1]])
>> > hist(z[[545]])
>>
>> If that's the problem, then I suggest
Barry Rowlingson wrote:
> On Wed, May 13, 2009 at 5:36 PM, Wacek Kusnierczyk
> wrote:
>> Barry Rowlingson wrote:
>
>>> Soln - "for" loop:
>>>
>>> > z=list()
>>> > for(i in 1:1000){z[[i]]=rnorm(100,0,1)}
>>>
>>> now inspect the individual bits:
>>>
>>> > hist(z[[1]])
>>> > hist(z[[545]])
>>>
>
So far nobody seems to have warned the OP about seeding.
Presumably Debbie wants 1000 different sets of samples, but as we all
know there are ways to get the same sequence (initial seed) every time.
If there's a starting seed for one of the "generate a single giant
matrix" methods proposed, t
On Wed, May 13, 2009 at 9:56 PM, Wacek Kusnierczyk
wrote:
> Barry Rowlingson wrote:
> n = 1000
> benchmark(columns=c('test', 'elapsed'), order=NULL,
> 'for'={ l = list(); for (i in 1:n) l[[i]] = rnorm(i, 0, 1) },
> lapply=lapply(1:n, rnorm, 0, 1) )
> # test elapsed
> #
On 14/05/2009, at 10:04 AM, Carl Witthoft wrote:
So far nobody seems to have warned the OP about seeding.
Presumably Debbie wants 1000 different sets of samples, but as we all
know there are ways to get the same sequence (initial seed) every
time.
If there's a starting seed for one of the
Wacek Kusnierczyk wrote:
Barry Rowlingson wrote:
On Wed, May 13, 2009 at 5:36 PM, Wacek Kusnierczyk
wrote:
Barry Rowlingson wrote:
Soln - "for" loop:
> z=list()
> for(i in 1:1000){z[[i]]=rnorm(100,0,1)}
now inspect the individual bits:
> hist(z[[1]])
> hist(z[[545]])
If that's the pr
Barry Rowlingson wrote:
> On Wed, May 13, 2009 at 9:56 PM, Wacek Kusnierczyk
> wrote:
>
>> Barry Rowlingson wrote:
>>
>
>
>>n = 1000
>>benchmark(columns=c('test', 'elapsed'), order=NULL,
>> 'for'={ l = list(); for (i in 1:n) l[[i]] = rnorm(i, 0, 1) },
>> lapply=lappl
Barry Rowlingson wrote:
[...]
>>> Yes, you can probably vectorize this with lapply or something, but I
>>> prefer clarity over concision when dealing with beginners...
>>>
>> but where's the preferred clarity in the for loop solution?
>>
>
> Seriously? You think:
>
> lapply(1:n, rno
Dimitris Rizopoulos wrote:
> Wacek Kusnierczyk wrote:
>> Barry Rowlingson wrote:
>>> On Wed, May 13, 2009 at 5:36 PM, Wacek Kusnierczyk
>>> wrote:
Barry Rowlingson wrote:
> Soln - "for" loop:
>
> > z=list()
> > for(i in 1:1000){z[[i]]=rnorm(100,0,1)}
>
> now inspect
Wacek Kusnierczyk wrote
>> Seriously? You think:
>>
>> lapply(1:n, rnorm, 0, 1)
>>
>> is 'clearer' than:
>>
>> x=list()
>> for(i in 1:n){
>> x[[i]]=rnorm(i,0,1)
>> }
>>
>> for beginners?
>>
>> Firstly, using 'lapply' introduces a function (lapply) that doesn't
>> have an intuitive name. Also,
Peter Flom wrote:
>
>>> Seriously? You think:
>>>
>>> lapply(1:n, rnorm, 0, 1)
>>>
>>> is 'clearer' than:
>>>
>>> x=list()
>>> for(i in 1:n){
>>> x[[i]]=rnorm(i,0,1)
>>> }
>>>
>>> for beginners?
>>>
>>> Firstly, using 'lapply' introduces a function (lapply) that doesn't
>>> have an intuitive n
I wrote
As a beginner, I agree the for loop is much clearer to me.
Wacek Kusnierczyk replied
>
>well, that's quite likely. especially given that typical courses in
>programming, afaik, include for looping but not necessarily functional
>stuff -- are you an r beginner, or a programming
> As a beginner, I agree the for loop is much clearer to me.
>
[Warning: Contains mostly philosophy]
To me, the world and how I interact with it is procedural. When I want
to break six eggs I do 'get six eggs, repeat "break egg" until all
eggs broken'. I don't apply an instance of the break
Peter Flom wrote:
> As a beginner, I agree the for loop is much clearer to me.
>
>
>> well, that's quite likely. especially given that typical courses in
>> programming, afaik, include for looping but not necessarily functional
>> stuff -- are you an r beginner, or a programming beginne
Barry Rowlingson wrote:
>> As a beginner, I agree the for loop is much clearer to me.
>>
>>
>
> [Warning: Contains mostly philosophy]
>
maybe quasi ;)
> To me, the world and how I interact with it is procedural. When I want
> to break six eggs I do 'get six eggs, repeat "break egg"
On 14-May-09 11:28:17, Wacek Kusnierczyk wrote:
> Barry Rowlingson wrote:
>>> As a beginner, I agree the for loop is much clearer to me.
>>
>> [Warning: Contains mostly philosophy]
>>
> maybe quasi ;)
>
>> To me, the world and how I interact with it is procedural. When I want
>> to break
Barry Rowlingson wrote:
>
> Computer scientists will write their beautiful manuscripts, but how
> many people who come to R because they want to do a t-test or fit a
> GLM will read them? That's the R-help audience now.
>
don't forget that r seems to take, maybe undeserved, the pride of being
ierc...@idi.ntnu.no
> To: b.rowling...@lancaster.ac.uk
> CC: r-help@r-project.org
> Subject: Re: [R] Simulation
>
> Barry Rowlingson wrote:
> > On Wed, May 13, 2009 at 9:56 PM, Wacek Kusnierczyk
> > wrote:
> >
> >> Barry Rowlingson wrote:
> >>
>
Debbie Zhang schrieb:
> Now, I am trying to obtain the sample variance (S^2) of the 1000 samples that
> I have generated before.
>
> I am wondering what command I should use in order to get the sample variance
> for all the 1000 samples.
>
>
>
> What I am capable of doing now is just typing in
Since you got the most suitable way to get x, why can't you get the
variances in the same way? Just like:
v = vector()
for (i in 1:length(x)) v[i] = var(x[[i]])
BTW, it is much better to use lapply, like this:
lapply(x, var)
On Thu, May 14, 2009 at 8:26 PM, Debbie Zhang wrote:
>
> Thanks f
Stefan Grosse wrote:
> Debbie Zhang schrieb:
>
>> Now, I am trying to obtain the sample variance (S^2) of the 1000 samples
>> that I have generated before.
>>
>> I am wondering what command I should use in order to get the sample variance
>> for all the 1000 samples.
>>
>>
>>
>> What I am ca
rmulas for
my n=15, but im unsure how to proceed in the quickest way.
Can someone please help?
Much appreciated :)
> From: r.tur...@auckland.ac.nz
> Date: Thu, 14 May 2009 10:26:38 +1200
> To: c...@witthoft.com
> CC: r-help@r-project.org
> Subject: Re: [R] Simulation
>
On Fri, 15 May 2009 19:17:37 +1000 Kon Knafelman
wrote:
KK> I hve the same problem as the initial one, except i need 1000
KK> samples of size 15, and my distribution is Exp(1). I've adjusted
KK> some of the loop formulas for my n=15, but im unsure how to proceed
KK> in the quickest way.
KK> Can
On Fri, 15 May 2009 19:17:37 +1000 Kon Knafelman
wrote:
KK> I hve the same problem as the initial one, except i need 1000
KK> samples of size 15, and my distribution is Exp(1). I've adjusted
KK> some of the loop formulas for my n=15, but im unsure how to proceed
KK> in the quickest way.
KK> Ca
n...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Ben Bolker
> Sent: Friday, May 15, 2009 6:37 AM
> To: r-help@r-project.org
> Subject: Re: [R] Simulation
>
>
>
> On Fri, 15 May 2009 19:17:37 +1000 Kon Knafelman
> wrote:
>
> KK> I hve the
Greg Snow wrote:
> Another possibility (maybe more readable, gives the option of a list,
> probably not faster):
>
> Replicate(1000, rexp(15,1) )
>
I think that should be "replicate"
The matrix form is quite a bit faster, but don't know if that will
matter -- times below are for doing this
(Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: Ben Bolker [mailto:bol...@ufl.edu]
> Sent: Friday, May 15, 2009 10:19 AM
> To: Greg Snow
> Cc: r-help@r-project.org
> Subject: Re: [R] Simula
Greg Snow wrote:
> Another possibility (maybe more readable, gives the option of a list,
> probably not faster):
>
> Replicate(1000, rexp(15,1) )
>
>
provided that simplify=FALSE:
is(replicate(10, rexp(15, 1)))
# "matrix" ...
is(replicate(10, rexp(15, 1), simplify=FALSE))
# "
Hey Guys,
i need 1000 samples of size 15, for distributions of Exp(1) and Norm(0,1). here
is what i have done so far.For the exponential
z<-list()
for(i in 1:1000){z[[i]]=rexp(15,1)}
For the Normal
y<-list()
for(i in 1:1000){y[[i]]=rnorm(15,0,1)}
Is this correct?
after this i need to compu
first
place, your input is highly appreciated
using your method, im getting a bit confused.
> Date: Fri, 15 May 2009 12:16:45 +0100
> From: s.elli...@lgc.co.uk
> To: konk2...@hotmail.com
> Subject: Re: [R] Simulation
>
> The tidiest way of doing something 'simple' with a se
gt; Date: Thu, 14 May 2009 12:05:30 +0100
> From: b.rowling...@lancaster.ac.uk
> To: peterflomconsult...@mindspring.com
> CC: waclaw.marcin.kusnierc...@idi.ntnu.no; r-help@r-project.org
> Subject: Re: [R] Simulation
>
> > As a beginner, I agree the for loop is much clea
Hi listers,
I've been looking for a procedure, but I am not succeding...
I have a function that give multiple results...
Then, I would like to simulate this function n times, so I need to save/keep
the n multiple results, in order to calculate my desired statistics...
I have already tried with the
deer friends i am working in simulation with using R
i know how fonction reconst work but my query is how to simulate using PCA
with R??
THANK YOU
-
[[alternative HTML version deleted]]
___
Hi listers,
I am working on a program of statistical analysis of simulated data and I've
been searching the error at the program, but I didn't find it!
It is something about the WHILE procedure, the error says: Error in while
(ecart >= d) { : missing value where TRUE/FALSE needed
Thanks in advance
Hi there, how can i do simulations(100,000) using ranks. I mean i subtract main
effects from the observations to form residuals, then i ranked the residuals.
Next i allocate the ranks in place of the observations and do a nonparametric
analysis on the ranks. The original data (observation) is ge
Dear List,
I have some problem with my simulation code. Here is output from R:
> sim.sp <- function(data,CM,n,N)
+ {
+ C <- matrix(rep(NA,N),ncol=1)
+ for(i in 1:N)
+ {
+ j <- n
+ xx <- which(colSums(CM[j,])==1)
+ V <- names(xx)
+ V <- paste(V, collapse="+")
+ V <- paste("SBA~", V)
+ rd <- round
You need to store your results in a list and then access the
information in the list to get the values:
> boot<-function(a,b,c){
+ media<-(a+b+c)/3
+ var_app<-var(a)
+ list(media=media,var_app=var_app)
+ }
> boot(2,4,10)
$media
[1] 5.33
$var_app
[1] NA
>
> simul<-function(S){
+
Hi there
has anybody experimented with the implementation in R of linear lists,
as described e.g. in books on Pascal (Jensen and Wirth), or Wirth,
"Algorithms and data structures" ? See
\url{www.dbnet.ece.ntua.gr/~adamo/csbooksonline/AD.pdf }
Pointers would be welcome.
Christian
--
--
Chri
Here I am in a simulation study where I want to find different values
of x and y such that f(x,y)=c (some known constant) w.r.t. x, y >0,
y<=x and x<=c1 (another known constant). Can anyone please tell me how
to do it efficiently in R. One way I thought that I will draw
different random numbers fro
Dear All,
I am running a Monte Carlo simulation study and have some questions on how
to manage data storage efficiently at the end of each 1000 replication loop.
I have three conditions coded using the FOR {} loops and a FOR loop that
generates data for each condition, performs analysis, and compu
2008/10/21 Marcioestat <[EMAIL PROTECTED]>:
>
> Hi listers,
> I am working on a program of statistical analysis of simulated data and I've
> been searching the error at the program, but I didn't find it!
> It is something about the WHILE procedure, the error says: Error in while
> (ecart >= d) { :
Hi Barry,
As you explained I find out my mistake... I wasn't supposed to use a
recursive formula!
So, what I needed, it's just to use the proportion by the function mean to
reach my results...
Thanks,
Márcio
Barry Rowlingson wrote:
>
> 2008/10/21 Marcioestat <[EMAIL PROTECTED]>:
>>
>> Hi lister
Dear R users,
I would like to simulate, for 2 replications, an autoregressive process:
y(t)=0.8*y(t-1)+e(t) where e(t) is i.i.d.(0,sigma*sigma),
Thank you in advance
Ãcoutez gratuitement le nouveau single de Noir Désir et découvrez d'au
after obtaining the result, the result is displayed below:
individual RS_number Phy_Posi LOH_intensity
1 1718890 rs1496555 2.2241110.8121
2 1668776 rs2376495 3.0849860.786 <---
3 1723597 rs4648462 3.1551270.784 <---
4 1728870 rs10492940 3.1876070.7831
Dear List Members,
I am running a random intercept and random slope (2 cross-level
interactions) MLM using package lme4 (individuals nested in countries,
number of 2nd level units is 21). My DV (member of a trade union or not) is
dichotomous, and thus it is a logistic multilevel model. The model c
This message is one of a number of identical copies, also cross-posted to
R-sig-geo, and in fact with an obvious solution that the author gives at the
end. The thread will be continued on R-sig-geo.
Cross posting is advised against expressly in e.g.
http://en.wikipedia.org/wiki/Crossposting, and
Hi listers,
I am working on a simulation... But I am having some troubles...
Suppose I have a function A which produces two results (mean and
variance)...
Then I would like to simulate this function A with a function B many times
using the results from function A
For example:
#Function A
boot<-fu
Are the pairs (x,y) belong to some lattice or can
change continuously?
Does f assume some discrete values (or is constant on
sets of positive measure)? If not then it will be hard
to randomly select x and y which satisfy the exact
equality (this still can happen since there are
finitely many comput
x, y are cont. variable, and f also have to be cont.. And your second
suggestion is correct of course, it actually should be |f(x,y) - c| <
epsilon
Thanks
On Tue, Apr 29, 2008 at 12:34 PM, Moshe Olshansky <[EMAIL PROTECTED]> wrote:
> Are the pairs (x,y) belong to some lattice or can
> change cont
At 02:40 AM 4/29/2008, Arun Kumar Saha wrote:
Here I am in a simulation study where I want to find different values
of x and y such that f(x,y)=c (some known constant) w.r.t. x, y >0,
y<=x and x<=c1 (another known constant). Can anyone please tell me how
to do it efficiently in R. One way I thoug
I am looking for a way to simulate genotypes of cases and control at a
disease locus in R. I am supposed to set the allele frequency as
control/cases. for each of the column below simulate 200 snp dataset.
I am looking at treesim function from popgen to stimulate the genotypes in
R.
Here is what
What is the format of the data you are storing (single value,
multivalued vector, matrix, dataframe, ...)? This will help formulate
a solution. What do you plan to do with the data? Are you going to
do further analysis, write it to flat files, store it in a data base,
etc.? How big are the data
Thanks for your reply. For each condition, I will have a matrix or data
frames of 1000 rows and 4 columns. I also have a total of 64 conditions for
now. So, in total, I will have 64 matrices or data frames of 1000 rows and 4
columns. The format of data I would like to store would be data frames or
One of the things you might take a look at is the 'filehash' package.
It is an easy way of storing/retrieving R objects. I have an
application where my objects are matrices of about the same size and I
can quickly store the data and then come back later with a different
script to do further analys
Davood Tofighi wrote:
> Thanks for your reply. For each condition, I will have a matrix or data
> frames of 1000 rows and 4 columns. I also have a total of 64 conditions for
> now. So, in total, I will have 64 matrices or data frames of 1000 rows and 4
> columns. The format of data I would like to
Thanks to All,
The comments were very helpful; however, the the simulation is running very
slow. I reduced the number of loops (conditions) so I have 36 loops, and the
data-generation occurs 1000 times within each loop. At the end of each 1000
reps, I saved the summary (e.g., mean) of the reps to
?arima.sim
On Wed, 19 Nov 2008, [EMAIL PROTECTED] wrote:
Dear R users,
I would like to simulate, for 2 replications, an autoregressive process:
y(t)=0.8*y(t-1)+e(t) where e(t) is i.i.d.(0,sigma*sigma),
Thank you in advance
??coutez gra
On 20/11/2008, at 10:54 AM, [EMAIL PROTECTED] wrote:
Dear R users,
I would like to simulate, for 2 replications, an autoregressive
process: y(t)=0.8*y(t-1)+e(t) where e(t) is i.i.d.(0,sigma*sigma),
Thank you in advance
?arima.sim
Note to R-core: This is actually rather hard for a neo
Use something like sprintf if you want trailing zeros.
> x
[1] 0.78
> sprintf("%.4f", x)
[1] "0.7800"
>
On Fri, Nov 21, 2008 at 5:31 AM, Abelian <[EMAIL PROTECTED]> wrote:
> after obtaining the result, the result is displayed below:
>
> individual RS_number Phy_Posi LOH_intensity
> 1 1718
in order to return more multiple variables, you can put them in a list
and then return this list.
e.g.
#Function A
boot<-function(a,b,c){
mean_boot<-(a+b)/2
var_boot<-c
list(mean_boot = mean_boot, var_boot = var_boot)
}
out <- boot(1,2,3)
out
$mean_boot
[1] 1.5
$var_boot
[1] 3
On Fri, Aug 1
Ok, the LIST function I understood...
What I would like now is to simulate this Function A many times (S) in order
to get S results for the MEAN and for the VARIANCE...
Zhiliang Ma wrote:
>
> in order to return more multiple variables, you can put them in a list
> and then return this list.
>
On Aug 17, 2009, at 1:40 PM, MarcioRibeiro wrote:
Ok, the LIST function I understood...
I didn't see how you got a "random" input to that function. Would seem
to need one of the r functions as input.
What I would like now is to simulate this Function A many times (S)
in order
to get S
I must to create an array with dimensions 120x8x500. Better I have to make 500
simulations of 8 series of return from a multivariate
normal distribution. there's the command "mvrnorm" but how I can do this
repeating the simulation 500 times?"
[[alternative HTML version deleted]]
___
Hello,
are there simulation packages to simulate population dynamics,
for example for epidemiology?
Or are there (open source) tools that can do that task and can be
used from within R (or at least can read the resulting data files)?
Oliver
__
R-help
March 2008 11:06
To: r-help@r-project.org
Subject: [R] simulation study using R
Dear All,
I am running a Monte Carlo simulation study and have some questions on how to
manage data storage efficiently at the end of each 1000 replication loop. I
have three conditions coded using the FOR {} loo
Check out the help page for replicate().
Andy
From: barbara.r...@uniroma1.it
>
> I must to create an array with dimensions 120x8x500. Better I
> have to make 500 simulations of 8 series of return from a multivariate
> normal distribution. there's the command "mvrnorm" but how I
> can do this
barbara.r...@uniroma1.it wrote:
I must to create an array with dimensions 120x8x500. Better I have to make 500
simulations of 8 series of return from a multivariate
normal distribution. there's the command "mvrnorm" but how I can do this repeating
the simulation 500 times?"
?replicate
Uwe
Liaw, Andy wrote:
Check out the help page for replicate().
Andy
Or the 'n' argument to mvrnorm (or mvtnorm::rmvnorm for that matter)...
From: barbara.r...@uniroma1.it
I must to create an array with dimensions 120x8x500. Better I
have to make 500 simulations of 8 series of return from a mul
ss to Linux due to
professionnal reasons ... so I must work with windows XP.
Do somebody know a way to unleash the calculation power of all four cores
under windows XP ?
Thanks for your help !!
Quant15
--
View this message in context:
http://www.nabble.com/Desperatly-seeking-Parallel-R-simul
Dear R users
I would like to simulate underdispersed Poisson and binomial
distributions somehow.
I know you can do this for overdispersed counterparts - using
rnbinom() for Poisson and rbetabinom() for binomial.
Could anyone share functions to do this? Or please share some tips for
modi
your help !!
>
> Quant15
> --
> View this message in context:
> http://www.nabble.com/Desperatly-seeking-Parallel-R-simulation-with-Windows-XP-tp19737748p19737748.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
a cluster of all your four cores.
>
> As I don't have a dual or quadcore running windows, I can't confirm
> that it is actually using all cores (it does under Linux), but the
> cluster is created.
>
> If you have further questions, just ask - I am learning t
t confirm
> that it is actually using all cores (it does under Linux), but the
> cluster is created.
>
> If you have further questions, just ask - I am learning tio use snow as
> well.
>
> Rainer
>
>
>
>>
>> Thanks for your
On Wed, 15 Jul 2009, Shinichi Nakagawa wrote:
Dear R users
I would like to simulate underdispersed Poisson and binomial distributions
somehow.
I know you can do this for overdispersed counterparts - using rnbinom() for
Poisson and rbetabinom() for binomial.
Could anyone share functions to
Hi all,
I am looking for toolboxes that can handle simulation and/or solution of
SDE with jumps, say levy processes, jump diffusion processes, etc.
Could anybody give me some pointers to really good ones in R, (or Maple or
Matlab)? I did google search myself and haven't found much...I would
also l
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