> 1. First of all, more general question. Standard anova() function for lm()
> or aov() models in R implements Type I sum of squares (sequential), which
> is not well suited for unbalanced ANOVA. Therefore it is better to use
> Anova() function from car package, which was programmed by John Fox to
[EMAIL PROTECTED] a écrit :
> Hi everyone,
> I recently ran a simulation on a computer using R that was hooked up to a
> UPS. There was one time when the power was out for length and the computer
> shut down. I was worried that I had lost the simulation, but upon booting
> the machine up, I h
David Kaplan a écrit :
> Greetings,
> I made changes to my gui preferences and saved them. When I close and
> then open R, it reverts back to default preferences. How do I
> permanently change gui preferences?
one way is using options()
and also using your Rprofile.site file (in ~/etc).
edit
Thanks!
I see, that do.call-function is used often in R-algorithms... Looking over
some extra do.call-examples seems useful. This tail-function is also new for
me.
Is there some reason to use seq(along = VECTOR) instead of 1:length(VECTOR)?
Atte
> You can replace the for with lapply like thi
If you have undergraduates who have trouble writing a decent term paper, or
postgraduates who are having problems with their theses or publications, we may
be able to help you. We offer a personalised, online, tuition service designed
to help students and academics produce solid, well argued tex
Hi everyone,
I recently ran a simulation on a computer using R that was hooked up to a UPS.
There was one time when the power was out for length and the computer shut
down. I was worried that I had lost the simulation, but upon booting the
machine up, I heard the processor kick in. It sounde
Hi, Jim,
It is you again. I couldn't remember how many times you answered my
silly questions. ^_^
I am not sure assign() is what I want. Say, if I want to create 1000
linear model objects with names lm1, lm2lm1000, it seems assign
can't solve it.
But your second solution is close to what I a
Yes there is with statements like:
assign(paste('m', i, sep='', value)
But I would suggest that you put the values in a list to make it
easier to access since all the data is in a single object. You could
do it in a loop:
result <- list()
for(i in 1:100){
..computation.
result[[
Dear Lister,
Is there a way to create many objects with sequencial names, say lm1,
lm2...lm100?
Thanks.
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Thanks for the reply Sego, Landon H. The vector is created as you
say. However, only the first iteration of integrate is captured.
The other twenty vector values remain at 0. Integrate output includes
several attributes and I only want the first one which is under $names
and called "value"
On 8/26/06, Patrick Connolly <[EMAIL PROTECTED]> wrote:
> On Sat, 26-Aug-2006 at 09:57AM +0100, Patrick Burns wrote:
>
> |> > sub.m <- lapply(m, function(x)x[x>2])
> |> > sub.m
> |> [[1]]
> |> [1] 3 4
> |>
> |> [[2]]
> |> [1] 4 5
> |>
> |> [[3]]
> |> [1] 4
> |>
> |> > sub.m[unlist(lapply(sub.m,
On Sat, 26-Aug-2006 at 09:57AM +0100, Patrick Burns wrote:
|> > sub.m <- lapply(m, function(x)x[x>2])
|> > sub.m
|> [[1]]
|> [1] 3 4
|>
|> [[2]]
|> [1] 4 5
|>
|> [[3]]
|> [1] 4
|>
|> > sub.m[unlist(lapply(sub.m, function(x) length(x) == 2))]
|> [[1]]
|> [1] 3 4
|>
|> [[2]]
|> [1] 4 5
|>
|>
Here's one way to do this:
Before you begin the loop, define a vector to capture the output. Then
each iteration of the loop will be assigned to the corresponding element
in that vector:
capture <- double(n) # creates a vector of length n, all with values of
0
for (i in 1:n) {
_ some code _
Dear R users,
I am trying to get data from the clipboard into R on MacOSX. I tried
the following, but got an error message:
read.delim("clipboard")
Error in file(file, "r") : unable to open connection
In addition: Warning message:
unable to contact X11 display
Obviously, I'm not running R usin
Greetings,
I made changes to my gui preferences and saved them. When I close and
then open R, it reverts back to default preferences. How do I
permanently change gui preferences?
Thanks in advance.
David
--
David Kap
Hello,
I am a novice R user and am having difficulty retrieving the values
from 21 iterations of the R function integrate.
The only way I have found is to do a write.table and then a read.table
as shown in the code below. I would rather capture the 21 values inside
the braces ( sapply might
Hello everybody,
I have some questions on ANOVA in general and on ANOVA in R particularly.
I am not Statistician, therefore I would be very appreciated if you answer
it in a simple way.
1. First of all, more general question. Standard anova() function for lm()
or aov() models in R implements Type
I'm running R 2.3.1 on Windows.
When calling boxplot(), shouldn't the "axes" and "frame.plot" arguments
get passed down to bxp()?
A specific example where the plot is not framed (but seems like it
should be):
X <- data.frame(x=as.factor(rep(c(1,2,3), 10)), y=rnorm(30))
boxplot(y~x, data=X, fr
Whoops, I forgot to add, many thanks to all who replied publicly and
privately. I am very appreciative of all comments and suggestions.
Mark Leeds was especially kind in terms of clarifying why my
description of the situation was so confusing.
Cheers,
Michael
On Aug 26, 2006, at 2:51 PM, Mi
Implementing the EM algorithm will be easy if you know what the algorithm is
for your particular problem. This will be very specific to your problem. The
trick is to augment your data to get something for which there is an easy ML
estimate. I do not believe there is a unique recipe to perform the E
Success! The line I needed was:
gcoeffs <-nls(y~(a/b)*exp(-(x-c)^2/(2*b^2)),start=list
(a=0.4,b=2,c=-10), trace=TRUE)
I also needed to provide good guesses for a, b and c. The attached
PNG should explain what I was going after, which is the line in the
center of that curve. I am sorry for
Yes.
On 8/26/06, Paul Smith <[EMAIL PROTECTED]> wrote:
> Dear All
>
> Can R compute the expected value of a random variable?
>
> Thanks in advance,
>
> Paul
>
> __
> R-help@stat.math.ethz.ch mailing list
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> P
Dear All
Can R compute the expected value of a random variable?
Thanks in advance,
Paul
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Thank you so much Marc.
Your solution is exactly what I am looking for.
Have a nice weekend.
wensui
On 8/26/06, Marc Schwartz <[EMAIL PROTECTED]> wrote:
> On Sat, 2006-08-26 at 13:06 -0400, Wensui Liu wrote:
> > Dear Lister,
> >
> > If I have a list of number, say x<-c(0.1, 0.5, 0.6...), how to
let us know what you want to do because the beauty of R is that, in many
cases, you may not have to loop.
- Original Message -
From: "Wensui Liu" <[EMAIL PROTECTED]>
To:
Sent: Saturday, August 26, 2006 1:06 PM
Subject: [R] for() loop question
> Dear Lister,
>
> If I have a list of nu
On Sat, 2006-08-26 at 13:06 -0400, Wensui Liu wrote:
> Dear Lister,
>
> If I have a list of number, say x<-c(0.1, 0.5, 0.6...), how to use a for()
> to loop through each number in x one by one?
>
> Thank you so much!
>
> wensui
Two options:
x <- c(0.1, 0.5, 0.6)
> for (i in x) {print (i)}
[1]
--- Friedrich Leisch <[EMAIL PROTECTED]> wrote:
> > On Sat, 26 Aug 2006 05:38:58 -0700 (PDT),
> > Thomas Harte (TH) wrote:
>
>
> > hallo, friedrich, and thanks for your reply.
>
> > if i Stangle your code i get:
>
> > sp<- make.ar.1(alpha=.5, n=800)
> > plot(s
Dear Lister,
If I have a list of number, say x<-c(0.1, 0.5, 0.6...), how to use a for()
to loop through each number in x one by one?
Thank you so much!
wensui
[[alternative HTML version deleted]]
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https
I think it's a scoping problem. Your function NLL() looks for "new" in
the environment in which NLL() was defined, but you generate your
simulated datasets in a different environment (local to sim.estim()).
There are a number of ways to deal with this:
- pass the dataset as a parameter to NLL()
Thank you!
On 8/26/06, Patrick Burns <[EMAIL PROTECTED]> wrote:
> > sub.m <- lapply(m, function(x)x[x>2])
> > sub.m
> [[1]]
> [1] 3 4
>
> [[2]]
> [1] 4 5
>
> [[3]]
> [1] 4
>
> > sub.m[unlist(lapply(sub.m, function(x) length(x) == 2))]
> [[1]]
> [1] 3 4
>
> [[2]]
> [1] 4 5
>
> > sub4.m <- lappl
Hi All,
I need some help in how one can implement maximumlikelihood estimation for
models with discrete hidden variables in EM in R.
Regards
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lapply(m,function(x)x[x>2])
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
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and provide commented
> On Sat, 26 Aug 2006 05:38:58 -0700 (PDT),
> Thomas Harte (TH) wrote:
> hallo, friedrich, and thanks for your reply.
> if i Stangle your code i get:
> sp<- make.ar.1(alpha=.5, n=800)
> plot(sp, type="l", col="blue")
> whereas if i Stangle my code, i get:
>
?legend
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and provide commented, minimal, self-contained, reproducible code.
Dear all R users,
Suppose,
> x = rnorm(1000)
> y = rt(1000,3)
> plot(range(1:1000),range(x,y),type="n",xlab=NA,ylab=NA)
> lines(x,col="red")
> lines(y,col="blue")
Now I want to put a footnote in the plot window to tell that RED lines
represents the random numbers from normal-dist and blue line rep
Dear all,
I am trying to evaluate the optimisation behaviour of a function. Originally
I have optimised a model with real data and got a set of parameters. Now I
am creating simulated data sets based on these estimates. With these
simulations I am estimating the parameters again to see how vari
--- Friedrich Leisch <[EMAIL PROTECTED]> wrote:
> > On Fri, 25 Aug 2006 11:05:48 -0700 (PDT),
> > Thomas Harte (TH) wrote:
> > i can get pretty close to this in linux by writing a function to save the
> > plot to a pdf device:
> > <>=
> > # no savePlot in Linux ... so write my ow
You can replace the for with lapply like this:
VECTOR <- c(0, 3, 6, 3, 11, 2, 11, 4, 3, 4, 7, 7, 6, 4, 8)
f <- function(i) unique(tail(VECTOR, length(VECTOR)-i+1))[1:5]
out <- do.call(rbind, lapply(seq(along = VECTOR), f))
na.omit(rbind(rep(NA, 5), out))
Note that a matrix with zero rows is retu
try thie:
r <- rnorm(100,4,1)
hist(r, freq=FALSE)
r.f <- function(x)dnorm(x, mean(r), sd(r))
curve(r.f, from=min(r), to=max(r), add=TRUE, col='red')
On 8/26/06, stat stat <[EMAIL PROTECTED]> wrote:
> Dear all,
>
> May be question seems trivial for most of the R
> users, but really at least for
> On Fri, 25 Aug 2006 11:05:48 -0700 (PDT),
> Thomas Harte (TH) wrote:
> --- Prof Brian Ripley <[EMAIL PROTECTED]> wrote:
>> savePlot is just an internal version of dev.copy, part of the support for
>> the menus on the windows() graphics device.
>>
>> It is described in `An Int
> sub.m <- lapply(m, function(x)x[x>2])
> sub.m
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
> sub.m[unlist(lapply(sub.m, function(x) length(x) == 2))]
[[1]]
[1] 3 4
[[2]]
[1] 4 5
> sub4.m <- lapply(m, function(x)x[x>4])
> sub4.m[unlist(lapply(sub4.m, function(x) length(x) > 0))]
[[1]]
[1] 5
On Sat, 26 Aug 2006, Klaus Thul wrote:
> Dear all,
>
> I have the following problem:
>
> - I have written a program in R which runs out of physical memory
> on my MacBook Pro with 1.5 GB RAM
How does R know about physical memory on a virtual-memory OS? I presume
the symptom is swapping by
On Fri, 25 Aug 2006, Maria Montez wrote:
> Thank you for your answers yesterday. I now have another question!
>
> Suppose that instead of creating a formula for a regression model I
> simply wanted to add the variables. I believe I cannot use the
> as.formula anymore. Again I tried to use expre
Dear all,
May be question seems trivial for most of the R
users, but really at least for me, this comes out to
be very problematic.
Suppose I have the following data:
> r
[1] -0.0008179960 -0.0277968529 -0.0105731583
-0.0254050262 0.0321847131 0.0328170674
[7] 0.0431894392 -0.0217614918
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