Again my example was't very clear: there were not enough same numbers in the
VECTOR.
What I need is something like this:
VECTOR-c(0,3,6,3,11,2,11,4,3,4,7,7,6,4,8)
MATRIX-c()
for(i in 1:length(VECTOR)){
v-(unique(VECTOR[i:length(VECTOR)])[1:5])
MATRIX-rbind(MATRIX,v)
}
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
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
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 your
sub.m - lapply(m, function(x)x[x2])
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[x4])
sub4.m[unlist(lapply(sub4.m, function(x) length(x) 0))]
[[1]]
[1] 5
Patrick
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 Introduction to R' (the
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 me,
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
--- 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:
label=first.ar.1, results=hide=
# no savePlot in Linux ... so
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
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 represents
?legend
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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:
width- 20;
lapply(m,function(x)x[x2])
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide
Hi All,
I need some help in how one can implement maximumlikelihood estimation for
models with discrete hidden variables in EM in R.
Regards
[[alternative HTML version deleted]]
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R-help@stat.math.ethz.ch mailing list
Thank you!
On 8/26/06, Patrick Burns [EMAIL PROTECTED] wrote:
sub.m - lapply(m, function(x)x[x2])
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[x4])
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()
-
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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R-help@stat.math.ethz.ch mailing list
--- 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(sp, type=l, col=blue)
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] 0.1
[1]
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: r-help@stat.math.ethz.ch
Sent: Saturday, August 26, 2006 1:06 PM
Subject: [R] for() loop question
Dear Lister,
If I
Dear All
Can R compute the expected value of a random variable?
Thanks in advance,
Paul
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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
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PLEASE do read
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
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
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,
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,
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
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
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
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 using
On Sat, 26-Aug-2006 at 09:57AM +0100, Patrick Burns wrote:
| sub.m - lapply(m, function(x)x[x2])
| 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,
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[x2])
| sub.m
| [[1]]
| [1] 3 4
|
| [[2]]
| [1] 4 5
|
| [[3]]
| [1] 4
|
| sub.m[unlist(lapply(sub.m, function(x) length(x) == 2))]
|
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
Dear Lister,
Is there a way to create many objects with sequencial names, say lm1,
lm2...lm100?
Thanks.
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PLEASE do read the posting guide
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.
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
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
If you have undergraduates who have trouble writing a decent term paper, or
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be able to help you. We offer a personalised, online, tuition service designed
to help students and academics produce solid, well argued
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
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 it.
[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 heard
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 use
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