Doing a web search on
R CRAN GJR GARCH
brought up the rugarch package. The models you mentioned are discussed in
the documentation to that package
https://cran.r-project.org/web/packages/rugarch/vignettes/Introduction_to_the_rugarch_package.pdf
On Mon, Mar 25, 2019 at 2:06 PM Amon kiregu wrot
Hello,
Right. Missed that one.
Rui Barradas
Enviado a partir do meu smartphone Samsung Galaxy. Mensagem original
De: Eric Berger Data: 30/01/2018 10:12
(GMT+00:00) Para: Rui Barradas Cc: Daniel Nordlund
, smart hendsome ,
r-help@r-project.org Assunto: Re: [R] Simulation
Or a shorter version of Rui's approach:
set.seed(2511)# Make the results reproducible
fun <- function(n){
f <- function(){
c(mean(runif(5,1,10)),mean(runif(5,10,20)))
}
replicate(n, f())
}
fun(10)
On Tue, Jan 30, 2018 at 12:03 PM, Rui Barradas wrote:
> Hello,
>
> Another way would
Hello,
Another way would be to use ?replicate and ?colMeans.
set.seed(2511)# Make the results reproducible
fun <- function(n){
f <- function(){
a <- runif(5, 1, 10)
b <- runif(5, 10, 20)
colMeans(cbind(a, b))
}
replicate(n, f())
}
fun(10)
Hope this hel
On 1/29/2018 9:03 PM, smart hendsome via R-help wrote:
Hello everyone,
I have a question regarding simulating based on runif. Let say I have
generated matrix A and B based on runif. Then I find mean for each matrix A and
matrix B. I want this process to be done let say 10 times. Anyone can he
I don't think we have enough information to help you with this.
Do you intent the simulated values for growthrate to be selected from the
values in daT$growthrate? Or do these values define a distribution of values
(perhaps ranging between 0 and 1) and the simulation should use that empirical
Before you post on Rhelp you should first read the Posting Guide. It tells you
that requests for statistical advice might be answered but are not really
on-topic. Furthermore requests for tutorials or extended worked examples should
probably be accompanied by evidence of searching using Google o
PLZ mke sure the package installed which contains "mvrnorm" function.
--
PO SU
mail: desolato...@163.com
Majored in Statistics from SJTU
At 2014-09-14 09:56:34, "thanoon younis" wrote:
>Dear R members
>I want to simulate data depending on SEM and when i applied the code below
>i found
Adding back the list address:
On Sep 14, 2014, at 9:53 AM, thanoon younis wrote:
> thank you for your help but i still have error after putting semicolon
> "Error: unexpected symbol in:
> "Ro<-matrix(data=c(7.0,2.1,2.1,7.0), ncol=2)
> yo<-matrix(data=NA,nrow=N,ncol=P) p""
The error message sho
On Sep 14, 2014, at 8:48 AM, Don McKenzie wrote:
ccing to list, as requested in the posting guide, so that others
may be able to help you.
On Sep 14, 2014, at 8:45 AM, thanoon younis > wrote:
Thank you very much for your reply
the output is
#Do simulation for 100 replications
N<-1000;
cc�ing to list, as requested in the posting guide, so that others may be able
to help you.
On Sep 14, 2014, at 8:45 AM, thanoon younis wrote:
> Thank you very much for your reply
>
> the output is
>
> > #Do simulation for 100 replications
> > N<-1000; P<-10
> >
> > phi<-matrix(data=c(1.0,0.3
What errors? What is your output? What output did you expect?
On Sep 14, 2014, at 6:56 AM, thanoon younis wrote:
> Dear R members
> I want to simulate data depending on SEM and when i applied the code below
> i found some errors and i still cannot run it.
> many thanks in advance
>
>
> Thano
Dear Thanoon,
You might look at the various item simulation functions in the psych package.
In particular, for your problem:
R1 <- sim.irt(10,1000,a=3,low = -2, high=2)
R2 <- sim.irt(10,1000,a=3,low = -2, high=2)
R12 <- data.frame(R1$items,R2$items)
#this gives you 20 items, grouped with high c
Please remember the 'reply all' for the r-help page.
First Question: How can i use Pearson correlation with dichotomous data? i
want to use a correlation between dichotomous variables like spearman
correlation in ordered categorical variables?
cor(variable1, variable2, *method = "pearson"*)
Seco
Thanoon,
You should still send the question to the R help list even when I helped
you with the code you are currently using. I will not always know the best
way or even how to proceed with some questions. As for to your question
with the code below.
Firstly, there is no 'phi' method for cor in
Thanoon,
My reply to your previous post should be more than enough for you to
accomplish your goal. Please look over that script again:
ords <- seq(4)
p <- 10
N <- 1000
percent_change <- 0.9
R <- as.data.frame(replicate(p, sample(ords, N, replace = T)))
or alternatively as Mr. Barradas suggest
Hello,
At an R prompt type
?rbinom
?replicate
Hope this helps,
Rui Barradas
Em 10-04-2014 02:28, thanoon younis escreveu:
hi
i want to simulate multivariate dichotomous data matrix with categories
(0,1) and n=1000 and p=10.
thanks alot in advance
[[alternative HTML version deleted
On 10/04/14 09:28, thanoon younis wrote:
hi
i want to simulate multivariate dichotomous data matrix with categories
(0,1) and n=1000 and p=10.
Nobody's stopping you! :-)
cheers,
Rolf Turner
__
R-help@r-project.org mailing list
https://stat.ethz.ch
Thanoon,
Firstly, please remember to reply to the R help list as well so that other
may benefit from your questions as well.
Regarding your second request, I have written the following as a very naive
way of inducing correlations. Hopefully this makes it perfectly clear what
you change for diffe
Hi Thanoon,
How about this?
# replicate p=10 times random sampling n=1000 from a vector containing your
ordinal categories (1,2,3,4)
R <- replicate(10, sample(as.vector(seq(4)), 1000, replace = T))
Cheers,
Charles
On Fri, Apr 4, 2014 at 7:10 AM, thanoon younis
wrote:
> dear sir
> i want to si
So can the same student be associated with multiple instructors? only
within the same school? or more general? by repeated student ID's do
you mean that a student in school A and a student in school B can both
have ID 1, but they are different students? or do you mean that
instructor #1 and Instr
On 04/08/2013 09:30, Rui Barradas wrote:
Hello,
See the help page for ?sample.
X <- sample(0:1, 1, replace = TRUE, prob = c(0.25, 0.75))
Hope this helps,
?rbinom would have been a better answer since simpler, faster,
algorithms are available in that case.
Or even
as.integer(runif(100
Thank you very much guys
On Sun, Aug 4, 2013 at 3:51 PM, Prof Brian Ripley wrote:
> On 04/08/2013 09:30, Rui Barradas wrote:
>
>> Hello,
>>
>> See the help page for ?sample.
>>
>> X <- sample(0:1, 1, replace = TRUE, prob = c(0.25, 0.75))
>>
>> Hope this helps,
>>
>
> ?rbinom would have been
So you looked at some unspecified help pages online and tried some unspecified
stuff? Try being more specific next time you post. For example, try reading the
footer of any R-help email. Note that it says read the Posting Guide, and
provide a reproducible example (at least of what you tried that
Hello,
See the help page for ?sample.
X <- sample(0:1, 1, replace = TRUE, prob = c(0.25, 0.75))
Hope this helps,
Rui Barradas
Em 04-08-2013 08:51, Preetam Pal escreveu:
Hi All,
I want to simulate a random variable X which takes values 1 and 0 with
probabilities 75% and 25% respectively
I am not aware of any such command so, I think, you may have to write one
yourself:
invert the CDF and use uniform random variable (runif) to sample
Mikhail
On Tuesday, June 11, 2013 16:18:59 cassie jones wrote:
> Hello R-users,
>
> I am trying to simulate from truncated skew normal distributi
Tobias,
I'm not sure if this is what you're after, but perhaps it will help.
# create a list of 5 vectors
n <- 5
subsets <- lapply(1:n, function(x) rnorm(5, mean=80, sd=1))
# create another list that takes 2 bootstrap samples from each of the 5
vectors and puts them in a matrix
nbootstrap <- 2
t
On 03-01-2013, at 17:40, Simone Gogna wrote:
> Dear R users,
> suppose we have a random walk such as:
>
> v_t+1 = v_t + e_t+1
>
> where e_t is a normal IID noise pocess with mean = m and standard deviation =
> sd and v_t is the fundamental value of a stock.
>
> Now suppose I want a trading s
On 03-01-2013, at 17:40, Simone Gogna wrote:
> Dear R users,
> suppose we have a random walk such as:
>
> v_t+1 = v_t + e_t+1
>
> where e_t is a normal IID noise pocess with mean = m and standard deviation =
> sd and v_t is the fundamental value of a stock.
>
> Now suppose I want a trading s
look at functions replicate and mvrnorm functions (the later in the MASS
package).
On Sat, Dec 1, 2012 at 12:02 PM, mboricgs wrote:
> Hello!
>
> How can I do 100 simulations of length 17 from bivariate bivariate normal
> distribution, if I know all 5 parameters?
>
>
>
> --
> View this message
On 13.11.2012 15:45, Christopher Desjardins wrote:
Hi,
I am running the following code based on the cpm vignette's code. I believe
the code is syntactically correct but it just seems to hang R. I can get
this to run if I set the sims to 100 but with 2000 it just hangs. Any ideas
why?
No: Work
On Mon, May 28, 2012 at 12:14 PM, Özgür Asar wrote:
> Dear Dila,
>
> Try the following:
>
> library(Rcmdr)
Or avoid the unncessary overhead of Rcmdr and use
library(car)
to provide levenTest instead.
> asim <- 1000
> pv<-NULL
It's also many orders of magnitude more efficient to preallocate "
Dear Dila,
Try the following:
library(Rcmdr)
asim <- 1000
pv<-NULL
for(i in 1:asim)
{
print(i)
set.seed(i)
g1 <- rnorm(20,0,2)
g2 <- rnorm(20,0,2)
g3 <- rnorm(20,0,2)
x <- c(g1,g2,g3)
group<-as.factor(c(rep(1,20),rep(2,20),rep(3,20)))
pv<-c(pv,leveneTest(x,group)$"Pr(>F)"[1])
}
Best
Ozgur
On Apr 5, 2012, at 10:57 PM, Christopher Kelvin wrote:
Hello,
i need to simulate 100 times, n=40 ,
the distribution has 90% from X~N(0,1) + 10% from X~N(20,10)
Is my loop below correct?
Thank you
n=40
for(i in 1:100){
x<-rnorm(40,0,1) # 90% of n
You are overwriting x and y and at the end o
Suggestions? -- Yes.
1) Wrong list.. Post on R-sig-mixed-models, not here.
2) Follow the posting guide and provide the modelformula, which may
well be the source of the difficulties (overfitting).
-- Bert
On Fri, Dec 16, 2011 at 1:56 PM, Scott Raynaud wrote:
> I'm using an R program (which I d
Perhaps you might want to abstract your code a bit and try something like:
X = rnorm(500) # Some Data
replicate(1e4, mean(sample(X, 500, replace = T)))
Obviously you can set up a loop over your data sets as needed.
Michael
On Sat, Nov 12, 2011 at 6:46 PM, Francesca wrote:
> Dear Contributors,
Why don't you use sample;
> sample(1:10,10,replace=TRUE)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of BSanders
Sent: 26 October 2011 08:49
To: r-help@r-project.org
Subject: Re: [R] Simulation from discrete uniform
If yo
If you wanted a discrete uniform from 1-10 use: ceiling(10*runif(1))
if you wanted from 0-12, use: ceiling(13*runif(1))-1
--
View this message in context:
http://r.789695.n4.nabble.com/Simulation-from-discrete-uniform-tp3434980p3939694.html
Sent from the R help mailing list archive at Nabble.com.
On Jul 21, 2011, at 1:42 PM, Benjamin Caldwell wrote:
That is, run all possible combinations of the two vectors through the
equation.
*Ben *
For "all combinations" the usual route is data preparation with either
expand.grid() or outer()
On Thu, Jul 21, 2011 at 10:04 AM, Benjamin Caldwel
That is, run all possible combinations of the two vectors through the
equation.
*Ben *
On Thu, Jul 21, 2011 at 10:04 AM, Benjamin Caldwell wrote:
> Hi,
> I'm trying to run a basic simulation and sensitivity test by running an
> equation with two variables and then plotting the results against
On Mon, Jun 06, 2011 at 04:50:57PM +1000, Stat Consult wrote:
> Dear ALL
> I want to simulate data from Multivariate normal distribution.
> GE.N<-mvrnorm(25,mu,S)
> S <-matrix(rep(0,1),nrow=100)
> for( i in 1:100){sigma<-runif(100,0.1,10);S
> [i,i]=sigma[i];mu<-runif(100,0,10)}
> for (i in 1:20
Hi Serdar,
Take a look at the following:
> sample(0:9, 100, replace = FALSE)
Error in sample(0:9, 100, replace = FALSE) :
cannot take a sample larger than the population when 'replace = FALSE'
> sample(0:9, 100, replace = TRUE)
[1] 5 6 5 7 3 0 8 4 8 2 2 4 7 6 0 7 0 0 0 7 5 6 3 6 0 9 6 1 2 6 9
Hi
Also , same problem to create discrete uniform Distribution ,
But sample () and runif() not useful to generate discrete uniform .
Ex:
> u<-round(runif(10*10,min=1,max=10),0)
> table(u)
u
1 2 3 4 5 6 7 8 9 10
6 10 9 10 14 6 11 14 12 8
Not useful for large number
OR
It is truncated from left.
On Mon, May 16, 2011 at 6:33 PM, Greg Snow wrote:
> Which direction is it truncated? (only values less than a allowed or only
> greater?).
>
> One simple approach is rejection sampling, just generate from a regular
> poisson distribution, then throw away any values in
uniform between that value and 1, then feed that
> uniform into the qpois function.
>
>
>
> *From:* cassie jones [mailto:cassiejone...@gmail.com]
> *Sent:* Monday, May 16, 2011 7:46 PM
> *To:* Greg Snow
> *Cc:* r-help@r-project.org
> *Subject:* Re: [R] simulation from tr
l.com]
Sent: Monday, May 16, 2011 7:46 PM
To: Greg Snow
Cc: r-help@r-project.org
Subject: Re: [R] simulation from truncated poisson
It is truncated from left.
On Mon, May 16, 2011 at 6:33 PM, Greg Snow
mailto:greg.s...@imail.org>> wrote:
Which direction is it truncated? (only values
Which direction is it truncated? (only values less than a allowed or only
greater?).
One simple approach is rejection sampling, just generate from a regular poisson
distribution, then throw away any values in the truncated region. Another
approach if the legal values are those from 0 to a, so
Hi Shane,
it sounds to me as though you have a fairly well-defined problem. You
want to generate random numbers with a specific mean, variance, and
correlation with another random varaible. I would reverse-enginerr
the fuinctions for simple linear regression to get a result like
y = beta_0 + be
?sample
-Oprindelig meddelelse-
Fra: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] På
vegne af cassie jones
Sendt: 8. april 2011 03:16
Til: r-help@r-project.org
Emne: [R] Simulation from discrete uniform
Dear all,
I am trying to simulate from discrete uniform distrib
Hi:
Suppose X has a discrete uniform distribution on the sample space S = {0, 1,
2, ..., 9}. Then a random sample of size 100 from this distribution, for
example, would be
dus <- sample(0:9, 100, replace = TRUE)
# Checks:
table(dus)
lattice::barchart( ~ table(dus), xlim = c(0, 20))
The sample sp
> Date: Mon, 28 Feb 2011 19:18:18 -0800
> From: kadodamb...@hotmail.com
> To: r-help@r-project.org
> Subject: [R] Simulation
>
> I tried looking for help but I couldn't locate the exact solution.
> I have data that has several variables. I want to do se
Well, knowing how your data looks like would definitely help!
Say your data object is called "mydata", just paste the output from
dput(mydata) into the email you want to send to the list.
Ivan
Le 3/1/2011 04:18, bwaxxlo a écrit :
I tried looking for help but I couldn't locate the exact solutio
Dear Kjetil,
Thank you so much for your advice on my question.
Best Regards,
Wonsang
2011/2/10 Kjetil Halvorsen
> What you can do to find out is to type into your R session
> RSiteSearch("multivariate fractional gaussian")
>
> That seems to give some usefull results.
>
> Kjetil
>
> On Tue, F
What you can do to find out is to type into your R session
RSiteSearch("multivariate fractional gaussian")
That seems to give some usefull results.
Kjetil
On Tue, Feb 8, 2011 at 1:51 PM, Wonsang You wrote:
>
> Dear R Helpers,
>
> I have searched for any R package or code for simulating multivar
On 05/01/2011 17:40, Bert Gunter wrote:
My hypothesis was specified before I did my experiment. Whilst far from
perfect, I've tried to do the best I can to assess rise in resistance,
without going into genetics as it's not possible. (Although may be at the
next institution I've applied for MSc).
On 05/01/2011 16:37, Mike Marchywka wrote:
Date: Wed, 5 Jan 2011 15:48:46 +
From: benjamin.w...@bathspa.org
To: r-help@r-project.org
Subject: [R] Simulation - Natrual Selection
Hi,
I've been modelling some data over the past few days, of my work,
repeatedly challenging microbes to a ce
> Date: Wed, 5 Jan 2011 15:48:46 +
> From: benjamin.w...@bathspa.org
> To: r-help@r-project.org
> Subject: [R] Simulation - Natrual Selection
>
> Hi,
>
> I've been modelling some data over the past few days, of my work,
> repeatedly challenging microbes to a certain concentration of cleane
For a Pareto distribution, even a truncated one, the inverse CDF method
should be straightforward to implement.
Giovanni Petris
On Tue, 2010-11-09 at 10:50 -0600, cassie jones wrote:
> Dear all,
>
> I am trying to simulate from truncated Pareto distribution. I know there is
> a package called P
Hi:
library(sos)
findFn('truncated Pareto')
On my system, it scared up 17 matches. It looks like the VGAM package would
be a reasonable place to start looking.
HTH,
Dennis
On Tue, Nov 9, 2010 at 8:50 AM, cassie jones wrote:
> Dear all,
>
> I am trying to simulate from truncated Pareto distribu
Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Jim Silverton
> Sent: Friday, September 10, 2010 1:03 AM
> To: r-help@r-project.org
> Subject:
I have two questions:
(1) How do you 'create' an 2 x 2 table in R using say an Odd ratio of 3 or
even 0.5
(2) If I have several 2 x 2 tables, how can I 'implement' dependence in the
tables with say 25 of the Tables having an odds ratio of 1 and 75 of the
tables having an odds ratio of 4?
Jim
This looks like homework. If it is, you should really tell us along with what
your teacher's policy is on getting help over the internet is (and note that
many teachers monitor this list and can see if you are getting help).
You have done the first part yourself, much better than some who have
Yes I looked into "dse" package. Here I have implemented two approach for
simulation like following :
library(dse)
A1 <- matrix(rnorm(16),4)
A2 <- matrix(rnorm(16),4)
mu <- rnorm(4)
sigma <- matrix(c(0.006594712,
0.006467731,
-0.000254914,
0.005939934,
0.006467731,
0.006654184,
-0.000384097,
0.00
Dear Ron,
have you had a look at the package dse? Here, ARMA models can be
specified and simulated. The only exercise left for you, is to transform
the VECM coefficients into their level-VAR values.
Best,
Bernhard
|> -Original Message-
|> From: r-help-boun...@r-project.org
|> [m
Check out the help page of the lrm function in the rms library. To
show how lrm is used,
the examples simulate data for logistic regression. This may give you
some ideas.
On Wed, Jan 20, 2010 at 10:41 AM, omar kairan wrote:
> Hi,
>
> could someone help me with dilemma on the simulation of logisti
On 21/01/2010, at 4:41 AM, omar kairan wrote:
Hi,
could someone help me with dilemma on the simulation of logistic
regressiondata with multicollinearity effect and high leverage point..
If that is the clearest way in which you can phrase your question then
I doubt that *anyone* can help you.
Try this:
dat <- data.frame(x=11:14, pa=1:4/10, pb=4:1/10)
f <- function(numreps, data){
pmat <- as.matrix(data[-1])
x <- data[,1]
result <- matrix(0, nrow=numreps, ncol=ncol(pmat))
colnames(result) <- c("A", "B")
for(i in seq_len(numreps)){
result[i,] <- apply(pmat, 2, function(p)
If the trials are not connected then I would consider melting the table
using melt() from the reshape package.
And then using lapply() with the function
random.function <- function(my.prob, number.of.observations = 10)
{
sum(rbinom(number.of.observations, 1, my.prob))
}
in case the trials are con
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){
+
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
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.
>
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
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
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
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
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
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
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
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
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))
# "
(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) )
>
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
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
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
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
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
>
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
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
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
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:
> >>
>
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
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:
>> 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"
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
> 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
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
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
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,
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
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
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