A radical solution is to edit the file .../etc/Rdevga and to change
the definition of the first set of fonts, e.g.
# TT Arial : plain
# TT Arial : bold
# TT Arial : italic
# TT Arial : bolditalic
TT Times New Roman : plain
TT Times New Roman : bold
TT Times New Roman : italic
TT Times New Roman
On Fri, 30 Jun 2006, Renaud Lancelot wrote:
A radical solution is to edit the file .../etc/Rdevga and to change
the definition of the first set of fonts, e.g.
# TT Arial : plain
# TT Arial : bold
# TT Arial : italic
# TT Arial : bolditalic
TT Times New Roman : plain
TT Times New Roman :
Thanks at all for your carefully help. I decided to used both version of your
proposals and to combine with a request of the operating system.
All the best - J?rn Schulz.
--
View this message in context:
http://www.nabble.com/cat-and-positioning-of-the-output-tf1869521.html#a5115737
Sent from
[EMAIL PROTECTED] writes:
Good day everyone,
I want to assess the error when fitting a Gram-Charlier
CDF to some data 'ws', that is, I want to calculate:
Err = |ecdf(ws) - GCh_ser(ws)|
The problem is, I cannot get the F(x) values from the
ecdf.
'Summary(ecdf())' returns some
Hello Dennis,
have you considered the function bh6lrtest() in the package urca?
To my knowledge, there is no other package available that offers VECM
functionalities.
Best,
Bernhard
ps: As you migth be interested in VAR and SVAR too: I am currently working on
such a package which should be
There is the Rserve package which you can look at here:
http://stats.math.uni-augsburg.de/Rserve/down.shtml
JS
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R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide!
Steven,
I cannot vouch for the behaviour of the function smooth.spline(), but
the theoretical answer to your question is yes. If g = Sy is the transformation
from data vector y to spline vector g, the equivalent degrees of freedom are
usually defined as EDF = trace(S), where S is the n x n
Mike, may also want to check out Rpad. Its mainly used to delivery R
programs to other users via a web browser, so the other users don't need R
installed on their machine. If you make a nice HTML gui, the other users
don't even have to know R. Each user gets his own environment, so one
person's
Dear mailing list I have a data that have 20,000 rows and 20 columns. Io
wonted to extract the 10th row only. Example the 10th, 20th, 30th 40th
..2
th. can you please help me how do I do that.Than kyou.
Example is below.
Inpute:
AG GG GG AG
CC CC CC CC
CT CC CT CT
GG GG GG GG
CC CC CC CC
GG GG
This should work:
data[seq(1,nrow(data),10),]
Andris
On Piektdiena, 30. Jūnijs 2006 14:45, yohannes alazar wrote:
Dear mailing list I have a data that have 20,000 rows and 20 columns. Io
wonted to extract the 10th row only. Example the 10th, 20th, 30th
40th�..2 th. can you please help
Just a note to say what I did. I think that the results were OK but I have
yet to hear from the journal.
1. I saved the Word document under another name.
2. I deleted all the contents of the document except the target graphic.
3. I printed to file yielding a .prn file.
4. I changed the extension
mydf[seq(10,2,10),]
yohannes alazar wrote:
Dear mailing list I have a data that have 20,000 rows and 20 columns. Io
wonted to extract the 10th row only. Example the 10th, 20th, 30th 40th…..2
th. can you please help me how do I do that.Than kyou.
Example is below.
Inpute:
AG GG GG AG
you need something like:
new.data - data[seq(10, nrow(data), 10), ]
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax:
Horace,
Last year I found another Jonckheere-Terpstra function at
http://www.biostat.wustl.edu/archives/html/s-news/2000-10/msg00126.html
Using that and the version in SAGx and wanting a few other options, I
created the version below.
Hope it is useful,
Chris
--
Christopher Andrews, PhD
SUNY
Hi,
I would really appreciate any suggestions on this (rather trivial) problem..
Say I have two vectors:
v1=seq(1:10)
v2=seq(1:15)
For a set of common breaks I need to divide the density of v2 over v1. This
means that I want to avoid having 0 counts for any v1 breakpoint. But
(unsuprisingly)
I think the following part of lme.formula
if (numIter controlvals$maxIter) {
stop(Maximum number of iterations reached without convergence.)
}
should be something like
if (numIter controlvals$maxIter) {
if (controlvals$returnObject) {
warning(Maximum number of iterations
For those who can find no documention on the @ operator, ?@
says it is used to Extract tbe contents of a slot in a object with a
formal class structure. ?slot provides more information, including
some critical references. The primary reference is Chambers (1998)
Programming with
Does any one have solution for R not running? I am getting the error
message. Please Advice.
Thanks
1. gunzip R-patched.tar.gz
2. tar -xvf R-patched.tar
3. changed the directory to the newly created directory R-patched
4. Typed ./configure
5. Typed make
6. Make check
7.
The Toolkit for Weighting and Analysis of Nonequivalent Groups (twang
1.0) has been released to CRAN. The package collects functions useful
for computing propensity score weights for treatment effect estimation,
developing nonresponse weights, and diagnosing the quality of those
weights. The
Dear all,
gmp package is now available on cran with version 0.4
Aim of gmp is to provide a lot a function to manipulate big integers, big
rationals and last but not least big integers in Z/nZ (modulo n)
We add in last version support for matrix computation (standards operators, + *
- / %*%
Dear R-users
I need to read some text files produced by some other software.
I used readLines (with n = -1 ) command and then tried to find some numbers
I liked to extract or some line numbers I like to know.
The problem is that there is no empty line at the end of the text files.
R gave me
I want to calculate chi square test of goodness of fit to test,
Sample coming from Poisson distribution.
please copy this script in R run the script
The R script is as follows
## start
#
No_of_Frouds-
Have you checked and see if there is any error in steps 5 through 8? If you
can't start R, I doubt steps 6 and 7 ran fine.
Andy
From: Pramod Anugu
Does any one have solution for R not running? I am getting
the error message. Please Advice.
Thanks
1. gunzip R-patched.tar.gz
On Fri, Jun 30, 2006 at 07:56:41AM -0500, Pramod Anugu wrote:
Does any one have solution for R not running? I am getting the error
message. Please Advice.
Yes, you have emailing the group fairly regularly about your troubles.
And you have been told just about each and every time to *check the
Hi there:
I'd thought these two versions of noncentral t-distribution are essentially
the same:
qqplot(rt(1000,df=20,ncp=3),qt(runif(1000),df=20,ncp=3))
But, the scales of the x-axis and the y-axis are quite different according to
the QQ-plot.
Did I make any mistakes
[Taka Matzmoto]
Is there any way to prevent [this] warning message.
Hi, Taka. The easiest might be using the suppressWarnings wrapper.
See ?suppressWarnings for more information.
--
François Pinard http://pinard.progiciels-bpi.ca
__
chisq.test(counts, p=Expected/sum(Expected), simulate.p.value =FALSE,
correct = FALSE)
Chi-squared test for given probabilities
data: counts
X-squared = 40.5207, df = 13, p-value = 0.0001139
Warning message:
l'approximation du Chi-2 est peut-être incorrecte in: chisq.test(counts,
I have a text file that I have imported into R. It contains 3 columns and
316940 rows. The first column is vegetation plot ID, the second species
names and the third is a cover value (numeric). I imported using the
read.table function.
My problem is this. I need to reformat the information
?as.matrix
On Friday 30 June 2006 09:11, Wade Wall wrote:
I have a text file that I have imported into R. It contains 3 columns
and 316940 rows. The first column is vegetation plot ID, the second
species names and the third is a cover value (numeric). I imported using
the read.table
On 6/30/2006 10:11 AM, Wade Wall wrote:
I have a text file that I have imported into R. It contains 3 columns and
316940 rows. The first column is vegetation plot ID, the second species
names and the third is a cover value (numeric). I imported using the
read.table function.
My problem
On Fri, Jun 30, 2006 at 10:11:15AM -0400, Wade Wall wrote:
I have a text file that I have imported into R. It contains 3 columns and
316940 rows. The first column is vegetation plot ID, the second species
names and the third is a cover value (numeric). I imported using the
read.table
Hi Wade,
On 6/30/06, Wade Wall [EMAIL PROTECTED] wrote:
I have a text file that I have imported into R. It contains 3 columns and
316940 rows. The first column is vegetation plot ID, the second species
names and the third is a cover value (numeric). I imported using the
read.table
Hi there
I want to pass arguments (i.e. the response variable and the subset
argument) in a self-made function to glm.
Here is one way I can do this:
f.myglm - function(y,subfact,subval) {
glm(d.mydata[,y]~d.mydata[,'x1'],family=binomial,subset=d.mydata[,subfact]==subval)
}
str(d.mydata)
Dear R-user
I have 100 lists.
Each list has several components.
For example,
data1
$a
[1] 1 2
$b
[1] 3 4
$c
[1] 5
There are data1, data2,, data100. All lists have the same number and the
same name of components.
Is there any function I can use for applying to only a specific component
f.myglm - function(y=y, subset=x2 == 'yes', data=d.d.mydata)
eval(parse(text=glm(,
deparse(substitute(y)), ~ x1, family=binomial, data=,
deparse(substitute(data)), , subset =,
subset, )))
f.myglm()
---
Jacques VESLOT
CNRS
i forgot a paste():
f.myglm - function(y=y, subset=x2 == 'yes', data=d.d.mydata)
eval(parse(text=paste(glm(,
deparse(substitute(y)), ~ x1, family=binomial, data=,
deparse(substitute(data)), , subset =,
subset, ), sep=)))
---
Jacques
Doese anyone know a R function to find the median of a gamma distribution?
[[alternative HTML version deleted]]
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PLEASE do read the posting guide!
Philip He [EMAIL PROTECTED] writes:
Doese anyone know a R function to find the median of a gamma distribution?
qgamma(0.5, )
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of
On Fri, 30 Jun 2006, Philip He wrote:
Doese anyone know a R function to find the median of a gamma distribution?
It's not clear what you mean. If you know the parameters of a gamma
distribution then qgamma() will give you any quantile.
If you have data and want to estimate the median then
From: Philip He [EMAIL PROTECTED]
Date: Fri Jun 30 10:30:28 CDT 2006
To: R help list r-help@stat.math.ethz.ch
Subject: [R] median of gamma distribution
someone might know a more elegant way but one way
is to use the distribution ( i think it's dnorm for
the normal but i get confused because
On 30-Jun-06 Philip He wrote:
Doese anyone know a R function to find the median of a gamma
distribution?
qgamma will do it. Test:
-log(0.5)
[1] 0.6931472
qgamma(0.5,1)
[1] 0.6931472
Ted.
E-Mail: (Ted Harding) [EMAIL
smooth.matrix = function(x, df){
n = length(x);
A = matrix(0, n, n);
for(i in 1:n){
y = rep(0, n); y[i]=1;
yi = predict(smooth.spline(x, y, df=df),x)$y;
A[,i]= yi;
}
(A+t(A))/2;
}
- Simon Wood, Mathematical Sciences, University of Bath, Bath BA2 7AY
-
On Fri, 30 Jun 2006, Long Qu wrote:
Hi there:
I'd thought these two versions of noncentral t-distribution are essentially
the same:
qqplot(rt(1000,df=20,ncp=3),qt(runif(1000),df=20,ncp=3))
But, the scales of the x-axis and the y-axis are quite different according
to the QQ-plot.
Perhaps this could be developed into a spline smooth method
for model.matrix and included in R.
On 6/30/06, Simon Wood [EMAIL PROTECTED] wrote:
smooth.matrix = function(x, df){
n = length(x);
A = matrix(0, n, n);
for(i in 1:n){
y = rep(0, n); y[i]=1;
yi =
This is the same as glm except that
- formula may be of class character in which case its
regarded as the name of the response variable and
the formula defaults to resp ~ Species for that response
- the data frame defaults to iris
- modify as appropriate for your case
myglm - function
It looks like in the call to lme
fm1 - lme(distance ~ age, data = Orthodont,
+control=lmeControl(msMaxIter=1))
you did not specify any random effects. Why not try:
fm1 - lme(distance ~ age, random= ~1| groupID, data = Orthodont,
+control=lmeControl(msMaxIter=1))
where
Thank you very much for your kind reply. It solved the problem of rt( ). :D
But it seems that the qt( ) also have problems:
I modified the rt( ) function as you suggested,
rt - function (n, df, ncp = 0)
{
if (ncp == 0)
.Internal(rt(n, df))
else rnorm(n, ncp)/sqrt(rchisq(n, df)/df)
}
In the old version of lme, one could construct a grouped data object and
this would alleviate the need to specify the random portion of the
model. So, Spencer's call is equivalent to
fm1 - lme(distance ~ age, random= ~age| Subject, data = Orthodont)
This condition does not hold under lmer,
On Sat, 1 Jul 2006, Long Qu wrote:
Thank you very much for your kind reply. It solved the problem of rt( ). :D
But it seems that the qt( ) also have problems:
Yes, there does seem to be a problem near zero. A clearer version is
curve(qt(x,df=20,ncp=3),from=0,to=0.004)
Just for the record. (One of the) problem was that configure picked up
ATLAS, but had problem with it at link time for whatever reason.
(This is on some version of Redhat, x86_64. I should think there are people
who have similar setup and got both ATLAS and readline to work.)
Andy
From:
In the below code fragment, print(arg) always prints the
last element of rekeningen$rekening.
Is this because of lazy evaluation? I.e. arg is evaluated at
the time the button is pressed?
And, if so, how can I avoid this?
I tried function() {force(arg); print(arg)} but that didn't work either.
Harold is correct. The help page for 'Orthodont' includes the
following example:
formula(Orthodont)
distance ~ age | Subject
If 'random' is not specified, 'lme' sets random = formula(data). If
that's NULL, the 'lme' help page says it Defaults to a formula
consisting of
Maybe this helps
( data1 = list(a=c(1,2), b=c(3,4), c=c(5,6,7)) )
( data2 = list(a=c(10,11), b=c(30,40), c=c(70,80)) )
cc - NULL
for(data in ls(pattern=^data[0-9]+$)) {
cc - c(cc, with(get(data), c))
}
mean(cc)
JeeBee.
On Fri, 30 Jun 2006 09:50:51 -0500, Taka Matzmoto wrote:
Dear R-user
It has to be a simple thing, but I could not figure it out:
How do I send the text output from object x to the printer?
As a shell user I would expect a pipe to the printer... |kprinter or
|lpr -Pmyprinter somehow. And yes, I'm on Linux.
Thanks!
__
I am trying to use lme to fit a mixed effects model to get the same
results as when using the following SAS code:
proc mixed;
class refseqid probeid probeno end;
model expression=end logpgc / ddfm=satterth;
random probeno probeid / subject=refseqid type=cs;
lsmeans end / diff cl; run;
There are
I am trying to use lme to fit a mixed effects model to get the same
results as when using the following SAS code:
proc mixed;
class refseqid probeid probeno end;
model expression=end logpgc / ddfm=satterth;
random probeno probeid / subject=refseqid type=cs;
lsmeans end / diff cl; run;
There are
type count
0 20
1 15
0 10
1 35
0 28
I would like to create two vectors from the data above. For example,
type1=c(15, 35) and type0 = c(20, 10, 28). Can any one help
Raphael
__
R-help@stat.math.ethz.ch mailing list
JeeBee [EMAIL PROTECTED] writes:
In the below code fragment, print(arg) always prints the
last element of rekeningen$rekening.
Is this because of lazy evaluation? I.e. arg is evaluated at
the time the button is pressed?
No and yes. Lazy evaluation has nothing to do with it, but the
function
Dear useRs,
the useR! 2006 conference took place in Vienna two weeks ago: it was an
exciting and interesting meeting with about 400 useRs from all over the
world and more than 150 presentations.
Especially for those of you who could not make it to the conference, we
have made 4up PDF versions of
Hi all.
I have a factor variable distributed over time. I am looking for an elegant
way to code duration of a state. Suppose,
rainfall.shocks - factor(sample(c(1,2,3), size = 15, replace = TRUE, prob
= unit.p),
+ label = c(Drought, Normal, High))
rainfall.shocks
[1] Normal
-- Forwarded message --
From: Alexander Nervedi [EMAIL PROTECTED]
Date: Jun 30, 2006 4:56 PM
Subject: time series patterns
To: [EMAIL PROTECTED]
Hi all.
I have a factor variable distributed over time. I am looking for an elegant
way to code duration of a state. Suppose,
I am out of the office and will be back on Monday 31 July.
Your mail will not be forwarded automatically.
Your mail regarding {Spam?} {Virus?} Message could not be delivered will be
read when I return
In urgent cases please contact Mr Kholdoun Torki
mailto:[EMAIL PROTECTED]
Hi!
In the example below why is x[[1]] == 0.2237724 false?
Alexander Nervedi
x - runif(10)
x[[1]]
[1] 0.2237724
x
[1] 0.2237724 0.2678944 0.9375811 0.5963889 0.6180519 0.6449580 0.7308510
[8] 0.7347386 0.4837286 0.1416100
x[[1]] == 0.2237724
FALSE
From: Alexander Nervedi [EMAIL
Does this do what you want? It creates a 'list' with the vectors:
x - 'type count
+ 0 20
+ 1 15
+0 10
+ 1 35
+ 0 28
+ '
x - read.table(textConnection(x), header=TRUE)
x
type count
1020
2115
3010
4135
5028
type -
This is FAQ 7.31:
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
Also please do not piggy back on other threads since it makes the
archives less useful.
On 6/30/06, Alexander Nervedi [EMAIL PROTECTED] wrote:
Hi!
In the example below why is
i created a postscipt file in R and then i downloaded a free version
of ghostview to view it. unfortunately, i get the message
fata error : dynamic memory exhausted
when i try to view it.
when i do a dir on windows xp, the file size is 149,034,475
and i know there about 17,000 graphs. is there
a
Hi:
I have two vectors of data, x and y and I want to get the polynomial
expansion of (x+y)^p with any integer power p in R. Suppose p=2, then I want a
matrix of five vectors, namely, x y x^2 y^2 x*y. The coefficient of the
polynomial is not needed. I can write it manully if p is small. But
Dear all,
I would like to generate bi-variate normal data given that the first column
of the data is known. for example:
I first generate a set of data using the command,
x - rmvnorm(10, c(0, 0), matrix(c(1, 0, 0, 1), 2))
then I would like to sum up the two columns of x:
x.sum - apply(x, 1,
From: Shin, David [EMAIL PROTECTED]
Date: Sat Jul 01 00:15:21 CDT 2006
To: 'r-help@stat.math.ethz.ch' r-help@stat.math.ethz.ch
Subject: [R] generate bi-variate normal data
it's an interesting question. someone else
on this list can answer more explicitly but
i think you have to use the result for
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