Hello,
I am struggling with computing nodes in Unix.
I have the use of a Unix server that has 30 nodes and I would like to
batch scripts.
Here is an R example that results in 72 repeated tasks based on the 2
loops. If I wanted to send these out to the different nodes, each node
has 1 task and th
Hi List,
I am trying to create a spatial representation of some wind data.
I have the season, frequency, strength and direction of the wind from 10
different locations, the coverage of the area that I am interested in is
not 100% there are small gaps in my coverage due to the location of the
we
Thank you for your help
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Give us more information to work with. What does
str(pred$posterior)
show so that we can see the structure of the data. Is it a matrix, if
so then you would do
pred$posterior[, 'o']
On Wed, Sep 29, 2010 at 10:41 PM, Gundala Viswanath wrote:
> I have a variable that looks like this:
>
>> prin
I have a variable that looks like this:
> print(pred$posterior)
ox
1 2.356964e-03 9.976430e-01
2 8.988153e-01 1.011847e-01
3 9.466137e-01 5.338627e-02
4 2.731429e-11 1.00e+00
Now what I want to do is to access "o" and "x"
How come this approach fa
Try this:
trunc(x / 10 ^ nchar(x) * 100)
On Wed, Sep 29, 2010 at 11:18 PM, Christian Schoder
wrote:
> hi R-users!
>
> does anyone know how I can access/print only the first two digits of a
> number? if i have the number 23732, i would like to get 23. if i have
> 355 i would like to get 35. if i
On Wed, Sep 29, 2010 at 7:18 PM, Christian Schoder
wrote:
> hi R-users!
>
> does anyone know how I can access/print only the first two digits of a
> number? if i have the number 23732, i would like to get 23. if i have
> 355 i would like to get 35. if i have 4 i would like to get 40.
It's a stran
hi R-users!
does anyone know how I can access/print only the first two digits of a
number? if i have the number 23732, i would like to get 23. if i have
355 i would like to get 35. if i have 4 i would like to get 40.
thanks for your help!
christian
_
On Wed, Sep 29, 2010 at 1:27 PM, Jyotasana Gulati wrote:
> Hi,
>
> I am have a data set of around 43000 probes(rows), and have to calculate
> correlation matrix. When I run cor function in R, its throwing an error
> message of RAM shortage which was obvious for such huge number of rows. I am
>
Hello
I am trying to use the correlog function to estimate a spatial correlogram
for the residuals of a logistic regression and I have run accross the
following error.
> summary(binom1 <- glm(Use~X20mslop+X20mdem+soilsst, family=binomial,
+ data=M60m2000NE_1.df))
> correlog1.1 <- correlog(M60
Hi
Thanks for the wonderful package! I have a question on plotting.
How do I control the dodging (like the position_dodge in ggplot2) of
the points after creating a plot with ezPlot? when plotting a 2-way
ANOVA the error bars cover each other...
Thanks!
shai
On Aug 31, 2:53 pm, Mike Lawrence
Here's my cooked up example:
# Faked data
x <- sample(1:100, 300, replace = TRUE)
# y = a + bx + cx^2 + noise, where a, b, c differ in each group
y <- rep(c(2, 5, 3), each = 100) + rep(c(-0.5, 0.5, 1), each = 100) * x +
rep(c(0.01, -0.01, 0.02), each = 100) * x^2 + rnorm(300, 0, 10)
g <- rep
sum(foo=="o")/length(foo)
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Hello,
My apologies, it was the hotmail writer which ate my post (hopefully this will
get there, intact).
dd<-rbind(data.frame(rbind(c("V1","A",0.3),c("V2","A",0.5),c("V3","A",0.2))),
data.frame(rbind(c("V1","B",0.3),c("V2","B",0.4),c("V3","B",0.8))),
data.frame(rbind(c("V1","C",0.9)
Try this:
prop.table(table(foo))
On Wed, Sep 29, 2010 at 10:43 PM, Gundala Viswanath wrote:
> I have a vector that looks like this:
>
> > foo
> [1] "o" "o" "o" "x" "o" "o" "o" "o" "o" "x" "x" "o" "x
>
> How can we find the percentage of "o" and "x" in
> that vector in R?
>
> - G.V
>
> _
Hi,
I am have a data set of around 43000 probes(rows), and have to calculate
correlation matrix. When I run cor function in R, its throwing an error message
of RAM shortage which was obvious for such huge number of rows. I am not
getting a logical way to cut off this huge number of entities,
Dear R Users,
I have model simulated data for 240 days.
The day 1 is Jan 1, 2009, 00:00 hrs and then with 3-hourly interval and so
on.
The time axis is 1,2,3,4..1920; so the total rows in the data are
1920.
How to convert above time axis in "year" "month" "day" "hour" format
Great Thank
On Wed, Sep 29, 2010 at 6:43 PM, Gundala Viswanath wrote:
> I have a vector that looks like this:
>
>> foo
> [1] "o" "o" "o" "x" "o" "o" "o" "o" "o" "x" "x" "o" "x
>
> How can we find the percentage of "o" and "x" in
> that vector in R?
table(foo)/length(foo)
Peter
I have a vector that looks like this:
> foo
[1] "o" "o" "o" "x" "o" "o" "o" "o" "o" "x" "x" "o" "x
How can we find the percentage of "o" and "x" in
that vector in R?
- G.V
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Perhaps use lgamma?
> lgamma(220)
[1] 964.8206
Jonathan
On Wed, Sep 29, 2010 at 3:22 PM, song song wrote:
> for example, when I am calculating a posterior density, I need to calculate
> gamma(75*3+5)=gamma(220) which is out of the bound of gamma function. what
> shall I do for this condition>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Michael Bedward
> Sent: Wednesday, September 29, 2010 5:34 PM
> To: Michael Larkin; Rhelp
> Subject: Re: [R] repeat a function
>
> On 30 September 2010 02:48, Michael Larkin
> w
On 30 September 2010 02:48, Michael Larkin wrote:
> >
> > testdat <- replicate( 50, growth[ sample(nrow(growth), 8, rep=TRUE) ] )
>
> I can't seem to get it to work. I keep getting the error message of
> "undefined columns selected"
>
> Any advice?
I'd need to know the dimensions of your matrix
On Wed, Sep 29, 2010 at 6:55 AM, hairryharry wrote:
> Hi,
>
> Fairly new to R - have done basic plots but now faced with plotting a
> matrix/table of results -I know what I want but cannot find out how to do
> it.
>
> Basically have individual questions ( x) to which an organization can rate
> the
THanks for all the replies.
I guess I should have been clearer from the beginning.
When I said I don't want to write my code, I meant I don't want to CREATE a new
function.
Not because I can't, but because I don't want to.
The cmh.test in the {lawstat} package still doesn't look like the cocrahn
Hadley,
I'm not sure this will solve the issue because if I move the script, I would
still have to go into the script and edit the "/path/to/my/script.r", or do I
misunderstand your workaround?
I'm looking for something like:
file.path.is.here("myscript.r")
and which would return something li
On Thu, 30 Sep 2010, Matthew Finkbeiner wrote:
I don't have enough RAM for this problem, so I need a work around. This is
what I want to do:
y<- sample(2^32, 10, replace=FALSE)
y <- trunc(runif( 10, 1, 2^32+1))
while( any( dup.y <-duplicated(y) ) ) y[dup.y] <-
trunc(
I don't have enough RAM for this problem, so I need a work around. This is
what I want to do:
y<- sample(2^32, 10, replace=FALSE)
but my machine won't let me do that. so I now do this:
x<- seq(1,2^32, by=100)
y<- sample(x, 10, replace=FALSE)
this works fine, but by selecting every 100
Can you send your code and data as separate files so we can get it into R
easily?
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University
Ph. (410) 5
Hi Dennis,
Sorry for not being considerate. I should have at least mentioned I was
using MCMCpack.
I have no idea about traceback() though.
I appreciate your suggestion.
Best,
Zhongyi
On Wed, Sep 29, 2010 at 5:00 PM, Dennis Murphy wrote:
> Hi:
>
> It might be helpful to inform the list whic
Ali -
A reproducible example would be very helpful, but I'll
try to guess what you mean.
mydat = data.frame(year=rep(2008:2010,each=5),var=1:15)
sdat = split(mydat$var,mydat$year)
do.call(cbind,sdat)
2008 2009 2010
[1,]16 11
[2,]27 12
[3,]38 13
[4,]4
Hi,
I've been reading quite a bit about the proper way of analyzing repeated
measured data and understand the advantages/pitfalls of doing it using
either a MANCOVA or linear mixed model approach. But I was wondering, for
the sake of really understanding, if anyone has some data to show how a
stan
Thanks much for all the help, R-helpers. Ended up getting the counts of the
categories of the matching variable in both x and y and then limiting the
sample from there. No longer really random, but I think it's fine for my
purposes.
Thanks again.
LB
On 28 September 2010 18:40, Michael Bedward wro
I used split() to split a variable by 3 years, but am wondering how to call up
that split data and use it in further analyses. Can I make separate columns for
the 3 resulting year groups?
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>
> Forgive me if this question has been addressed, but I was unable to find
> anything in the r-help list or in cyberspace. My question is this: is there a
> function, or set of functions, that will enable a script to detect its own
> path? I have tried file.path() but that was not what I was l
for example, when I am calculating a posterior density, I need to calculate
gamma(75*3+5)=gamma(220) which is out of the bound of gamma function. what
shall I do for this condition>
Thank you
[[alternative HTML version deleted]]
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Dear R users,
I am leaning MCMC sampling, and have a problem while trying to sample
exponential r.v.'s via the following code:
samp <- MCMCmetrop1R(dexp, theta.init=1, rate=2,
mcmc=5000, burnin=500,
thin=10, verbose=500, logfun=FALSE)
I tried o
Hi:
There's no way you could produce a loess plot based on the data supplied
below. V1, your purported x-variable, is a factor; moreover, you have one
point per V1 * V2 factor combination. (BTW, you might also consider using
the carriage return when demarcating individual lines of code.)
The reas
Use par(oma=c(1,1,1,1)) # oma = "outer margin area"
Is this what your looking for?
On Wed, Sep 29, 2010 at 5:03 PM, Mohsen Jafarikia wrote:
> Hello All,
>
> I am drawing a graph having 18 small graphs inside using par(mfrow =
> c(6,3))
> command. My problem is how to specify the margins of the
Hello All,
I am drawing a graph having 18 small graphs inside using par(mfrow = c(6,3))
command. My problem is how to specify the margins of the whole 18 graphs. I
used par(mar=c(6.5, 6.5, 1.5, 1.5)) for each graph separately already but it
does not left any margins for the 'mtext()' for the margi
HI:
I've seen a few threads about this topic but
still can't find a straightforward way on this.
Is there a package that can control R within an access form. For example,
I want to send a query to R, perform some statistics in R and send the output or
summary back to Access and display it on a f
Try this:
growthBoot <- replicate(3, growth[sample(9,12,replace=T),], simplify =
FALSE)
lapply(growthBoot, nls, formula = Length ~ Linf * (1 - exp(-K * (Age -
to))), start = par)
On Wed, Sep 29, 2010 at 4:56 PM, Michael Larkin wrote:
> I apologize if this comes across as confusing. I will try t
Mike,
Without completely knowing your end game with these questions and this
procedure it does seem like you are "re-inventing the wheel" here. If that is
true and given the nls() fit that you are using I would suggest that you look
at boot.case() in the alr3 package or nlsBoot() in the nlstoo
Hi:
The deal with replicate() is that its second argument is a *function*; more
specifically, a function *call*. That's why Henrique's solution worked and
your attempt didn't. Inside replicate(), if the function has arguments, they
need to be supplied. This works:
testdat <- function(df, n) df[sa
Try this:
PATH <- dirname(sys.frame(1)$ofile)
On Wed, Sep 29, 2010 at 5:00 PM, Stu Field wrote:
> Hi all,
>
> Forgive me if this question has been addressed, but I was unable to find
> anything in the r-help list or in cyberspace. My question is this: is there
> a function, or set of functions,
Hi all,
Forgive me if this question has been addressed, but I was unable to find
anything in the r-help list or in cyberspace. My question is this: is there a
function, or set of functions, that will enable a script to detect its own
path? I have tried file.path() but that was not what I was lo
Hello,
I have been struggling to do a plot in ggplot(2) that's of lattice equivalent.
The following code shows the lattice plot.
dd<-rbind(data.frame(rbind(c("V1","A",0.3),c("V2","A",0.5),c("V3","A",0.2))),data.frame(rbind(c("V1","B",0.3),c("V2","B",0.4),c("V3","B",0.8))),data.frame(rbind(c("V1"
I apologize if this comes across as confusing. I will try to explain my
situation as best I can.
I have R bootstrapping my growth data for fish. It's resampling my database
of age and length data and then produces several new datasets for me. In
this case, it's resampling my data to create
Dear All,
I am given a time series such at, at every time t_i, I am given a set
of data (every element of the set is just an integer number).
What I need is an injective function able to map every set into a
number (possibly an integer number, but that is not engraved in the
stone). Does anybody kn
#I am trying to understand how R fits models for contrasts in a
#simple one-way anova. This is an example, I am not stupid enough to want
#to simultaneously apply all of these contrasts to real data. With a few
#exceptions, the tests that I would compute by hand (or by other software)
#will give
Hi
On 29/09/2010 8:17 p.m., Tal Galili wrote:
My honor.
A short question: if there is something in the device that is sensitive
to the overlapping of the text, then is it possible to add a warning
massage output when the length of the text is longer then the device
dimensions?
The graphics en
Mathieu -
First of all, you can combine as many conditions as
you want in an if statement, using && (and) and || (or).
So to say
coef$st[i-1] < obs[t] < coef$st[i]
use
coef$st[i-1] < obs[t] && obs[t] < coef$st[i]
So following your logic, you could use
x = numeric(length(obs))
for(t
Dear R-helpers,
I'm trying to associate linear coefficients (intercept and slope) to tens of
thousands of observations based on a table with benchmark values.
#Example - Value table and their corresponding coefficients (intercept and
slope)
coef = data.frame(cbind(st=c(1:5),b = runif(5,0.3,5
Thanks,
Thats great just what I was trying to do.
HH
Thomas Stewart wrote:
HH-
I'm not familiar with the plots you mention, but the following is a
quick attempt to create the plot you describe.
data<-data.frame(
org=1:10,
q1=sample(1:10,replace=T),
q2=sample(1:10,replace=T),
q3=sa
Hi, everyone.
GAD package analyses complex ANOVA models with any combination of
orthogonal/nested and fixed/random factors, as described by Underwood
(1997). There are two restrictions: (i) data must be balanced; (ii)
fixed nested factors are not allowed. Homogeneity of variances is
checked using
> paste( '(', paste( "'", rep(letters[1:3],2), "'", sep="", collapse=','), ')',
> sep="" )
[1] "('a','b','c','a','b','c')"
If you need the space after the comma then just change ',' to ', '.
The outer paste can be replaced with sprintf (and that may be more readable).
--
Gregory (Greg) L. Sno
George Coyle wrote:
>
> I am trying to turn several lines of information into a variable. I used
> the filx function to input my file then the readlines to qualify what I
> want. Essentially I have data in a file every 10 minutes through a day
> for
> several years down a column:
>
> date tim
Hmmm. Maybe a documentation typo in ?spplot.
If you follow the documentation through to ?levelplot, you find
that
cuts: number of levels the range of ‘z’ would be divided into
(no mention of actual breakpoints) but:
at: numeric vector giving breakpoints along the range of ‘z’.
I think that you want:
replicate(5, growth[sample(9,12,replace=T),], simplify = FALSE)
On Wed, Sep 29, 2010 at 3:19 PM, Michael Larkin wrote:
> I am trying to get R to resample my dataset of two columns of age and
> length
> data for fish. I got it to work, but it is not resampling every repli
Mishra, Srikanta battelle.org> writes:
>
> Is there an R function for evaluating the exponential integral
>
> Ei(x) = INT(-inf,x) exp(t)/t dt
Try the gsl package.
Also library(sos); findFn("{exponential integral}")
although admittedly it doesn't find the Expint page
in the gsl package ...
I am trying to get R to resample my dataset of two columns of age and length
data for fish. I got it to work, but it is not resampling every replicate.
Instead, it resamples my data once and then repeated it 5 times.
Here is my dataset of 9 fish samples with an age and length for each one:
A
Is there an R function for evaluating the exponential integral
Ei(x) = INT(-inf,x) exp(t)/t dt
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PLEASE do read the posting gu
Hello
On Tue, Sep 28, 2010 at 1:39 PM, Soumen Pal wrote:
> I need your kind help regarding the following:
>
> I wish to know is there any way to use R in Visual Basic environment. I want
> to develop a VB application where R can be embedded (R will work as a back
> end statistical engine). If ava
for (j in 1:n)
{
if (j%%2==0)
{
iRange = c(n:1)
} else
iRange = c(1:n)
for (i in iRange)
{
your code
}
}
Peter
On Wed, Sep 29, 2010 at 10:40 AM, cassie jones wrote:
> Dear All,
>
>
> I am trying to define a loop for a m*n matrix, where i=1:n and j=1:m. Un
I am trying to turn several lines of information into a variable. I used
the filx function to input my file then the readlines to qualify what I
want. Essentially I have data in a file every 10 minutes through a day for
several years down a column:
date time value
9/28/10 02:00 13
9/28/10 0
Dear All,
I am trying to define a loop for a m*n matrix, where i=1:n and j=1:m. Unlike
the usual for loop, i should go in the following way:
For j=1,
i=1,2,3,n
For j=2,
i=n,n-1,n-2,..,1
For j=3,
i=1,2,3,.n etc.
which means i should go in either increasing or decreasing order
alternat
Thank you very much for your solution but it works only in a dataframe
object. If I am using an ftable object, it doesn't run.
I use, as a workaround, to fill with blank spaces the left of each number,
so when I print the table, it appears aligned to right.
But, obviously, this doesn't work for t
Thanks for the help.
Sharad
On Mon, Sep 27, 2010 at 9:12 PM, Remko Duursma [via R] <
ml-node+2716469-935075351-6...@n4.nabble.com
> wrote:
> Try something like this:
>
>
>
> dfr <- read.table(textConnection("plate.id well.id Group HYB
> rlt1
> 1 P1 A1 Control SKOV3hyb 0.
Dear useRs,
I am currently fitting an advanced failure time model using Göran
Broström's excellent "eha" library with the "aftreg" command.
My question: How do I interpret the "Overall p-value", that is
reported at the very bottom of the output? I already figured out it
must be a chi-square test,
On Tue, 28 Sep 2010, L Brown wrote:
Hi, everyone. I have what I hope will be a simple coding question. It seems
this is a common job, but so far I've had trouble finding the answer in
searches.
I have two matrices (x and y) with a different number of observations in
each. I need to draw a rando
Ted is well aware of how to change his list email. He was advising
people on the list who who have his old email address in their address
books to remove it.
On 09/29/2010 12:16 PM, Vojtěch Zeisek wrote:
Hello,
go to https://stat.ethz.ch/mailman/options/r-help/y...@email and if You do not
remem
On 2010-09-28 18:39, Marlin Keith Cox wrote:
When using a wireframe, I need to move the colorkey from the "right"
position (default0 towards the plot. I have also needed to adjust the
height and used the code
colorkey=list(T,space='right',height=.5)
I have looked at documents (within levelplot
Dear All;
I have searched to see if any code in R was written to calculate mean and
variance of a Ranked Set Sample, but did not find any. Is there any package
for RSS, or kindly can somebody share a code he/she wrote, I am very
grateful and willing to acknowledge that in my work. Thanks much
Ahm
It all depends on the ultimate use of the results.
Frank
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
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On Sep 29, 2010, at 18:21 , Berwin A Turlach wrote:
> G'day Tobias,
>
> On Wed, 29 Sep 2010 14:01:10 +
> "Keller Tobias (kelleto0)" wrote:
>
>> Für unseren Statistik Unterricht brauchen wir das R-Programm.
>> Ich habe das Programm heruntergeladen, deutsch gewählt und
>> installiert. Das Pr
Hello.
How can I write this all in one line?
mydata is a zoo series, limit is a numeric vector of the same size
tmp <- ave(coredata(mydata),as.Date(index(mydata)),FUN = function(x) (
(cummax(x)-x )) )
tmp <- (tmp < limit)
final <- ave(coredata(tmp),as.Date(index(mydata)),FUN = function(x) cumpr
Thats great thanks very much for your help
On 29 Sep 2010, at 17:30, Ben Bolker wrote:
[I'm a little confused: are you "Sam Smith" or "Chris Mcowen" ... ?]
This is admittedly a bit confusing, but the best scale on which to
compute standard errors is the link scale.
It turns out (I hadn't r
[I'm a little confused: are you "Sam Smith" or "Chris Mcowen" ... ?]
This is admittedly a bit confusing, but the best scale on which to
compute standard errors is the link scale.
It turns out (I hadn't realized this) that predict.glm does give
you not-crazy answers when you ask for
se.fit=T
Dear List and Ben
( I apologise if this has been sent twice, but it is not showing in my sent
folder and i have been having trouble with my email of late)
Right, that makes sense, thanks
The reason i used type= response was i wanted to convert the predicted
probabilities to the response scale,
G'day Tobias,
On Wed, 29 Sep 2010 14:01:10 +
"Keller Tobias (kelleto0)" wrote:
> Für unseren Statistik Unterricht brauchen wir das R-Programm.
> Ich habe das Programm heruntergeladen, deutsch gewählt und
> installiert. Das Problem ist nach 3mal neu installieren, dass das
> Programm immer auf
Hello,
go to https://stat.ethz.ch/mailman/options/r-help/y...@email and if You do not
remember Your password, use Password reminder (down) to mail it to Your
address. Then login and do all needed changes. You can replace r-hep in link
above with r-sig-phylo, r-announce and so on. And of course r
Hi:
This 'works' on the lapply end:
# Function to read one file from the list:
g <- function(x) read.zoo(file = x, header = TRUE, FUN = as.chron, sep =
",",
colClasses = c("NULL", "NULL", "character",
"numeric"))
# Apply to all files in list:
lapply(filenames, g)
[[1]]
(0
Version 4.63 of the caret package is now on CRAN.
caret can be used to tune the parameters of predictive models using
resampling, estimate variable importance and visualize the results.
There are also various modeling and "helper" functions that can be
useful for training models.
caret has wrappe
Hi:
Someone off-list (Josh Wiley - thank you) mentioned the Error() term in the
OP's ANOVA, which I missed in responding to the post - sorry for the
misinformation. Using the npk example with code that Josh showed me, we
have, for the following model,
npk.aov2 <- aov(yield ~ N*P*K + Error(block/P
Hello all,
I have been meaning to learn R for a while and have just subscribed to this
list. I am planning to give R a shot at one of my live projects. I am looking
to explore graphical features of R on my data below.
Sample Data:
Cat1 - Cat2 - Cat3 - Cat4 - NumPeople - Salary
H - L - H -
sehr geehrte Damen und Herren
Für unseren Statistik Unterricht brauchen wir das R-Programm.
Ich habe das Programm heruntergeladen, deutsch gewählt und installiert. Das
Problem ist nach 3mal neu installieren, dass das Programm immer auf englisch
kommt.
Können Sie mir helfen?
Vg
Tobias Keller
Thanks. I didn't see any obvious way to do it either. It looks like the
caption option has room for additional input that are as yet not designated.
I'll take a look at the latex() possibility.
Best,
cuz
--
View this message in context:
http://r.789695.n4.nabble.com/short-captions-for-xtable-tp2
Apologies for bothering anyone who may not be interested in
this, but some of you will, for instance, have my current
email address in your address books, etc.
As result of a new policy by Manchester University, retired
former staff who no longer contribute actively to research
in their former de
Hi:
You could look into the gamm4 package. Its description is:
Fit generalized additive mixed models via a version of mgcv's gamm function,
using lme4 for estimation via Fabian Scheipl's trick.
HTH,
Dennis
On Wed, Sep 29, 2010 at 7:28 AM, Camarda, Carlo Giovanni <
cama...@demogr.mpg.de> wrote:
Hej,
Calling newuoa (from the minqa package) makes R crash when the package rgl
is loaded first. This however only on certain selected data.
The data used for testing (saved to 'bugs.R'):
xvals =
c(1,2,4,5,7,8,9,10,11,12,14,15,16,18,19,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36)
yval
On Wed, 29 Sep 2010, Duncan Murdoch wrote:
On 29/09/2010 10:51 AM, Gundala Viswanath wrote:
I am refering to a function call like this:
>data(iris)
>x<- svmlight(Species ~ ., data = iris)
I tried to see the content of it by typing:
> Species ~ .
but it gives nothing. How can I see it's co
On 29/09/2010 10:51 AM, Gundala Viswanath wrote:
I am refering to a function call like this:
>data(iris)
>x<- svmlight(Species ~ ., data = iris)
I tried to see the content of it by typing:
> Species ~ .
but it gives nothing. How can I see it's content ?
str(Species ~ .) will tell you tha
I am refering to a function call like this:
>data(iris)
>x <- svmlight(Species ~ ., data = iris)
I tried to see the content of it by typing:
> Species ~ .
but it gives nothing. How can I see it's content ?
- P.Dubois
__
R-help@r-project.org mailing
Ethan-
You need to be more explicit about what you mean by 'background'. Do you
mean:
(a) the entire plot including margins?, or
(b) only the plotting area?, or
(c) a different color for both margins and plotting area?
If you want (a), the solution is par(bg = '#003D79').
If you want (b), the s
Hi,
I'm just learning to write R extensions in C and to embed R in C.
I was trying to get through the example in the help page on calling the .dll
directly (
http://cran.r-project.org/doc/manuals/R-exts.html#Calling-R_002edll-directly
).
When I compile I consistently get the error that getDLLVer
Right, that makes sense, thanks
The reason i used type= response was i wanted to convert the predicted
probabilities to the response scale, as surely this is the scale at which a
95CI value is most useful for?
I.e
>> pp <- predict(model1,se.fit=TRUE, type = "response")
1 0.68
Probability
I am using both nlminb and optim to get MLEs from a likelihood function I have
developed. AFAIK, the model I has not been previously used in this way and so I
am struggling a bit to unit test my code since I don't have another data set to
compare this kind of estimation to.
The likelihood I hav
On Sep 29, 2010, at 8:31 AM, cuz wrote:
>
> Hi,
>
> For my dissertation, I've made copious use of xtable. I've just gotten
> stumped however. I'm a fan of extended captions explaining the table, but
> now I have to assemble a a list of tables and the captions are unwieldy. I
> presume xtable cal
Dear R-users,
Is there any R-function for fitting generalized additive mixed
models for ordinal data? Do they actually make some sense? I can fit a
generalized linear mixed model for ordinal data using the function
clmm(ordinal) and I'm able to cope with generalized additive model for
ordi
On 10-09-29 10:04 AM, Sam wrote:
> Hi Ben and list,
>
> Sorry to be a pain! I have followed your code, and modified it -
>
**You should not use type="response" here.**
The point is that the (symmetric) confidence intervals are computed on
the link/linear predictor
scale, and then inverse-link-
Try this:
par(bg = '#003D79')
On Wed, Sep 29, 2010 at 11:14 AM, Arenson, Ethan wrote:
>
> Hi.
>
> I want to create a plot with Pantone654 as the background. The RGB for this
> color is (0,61,121), which corresponds to a hex of #003D79. How do I specify
> the bg parameter for this?
>
> All Best,
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