On 4 Jan 2005 at 10:25, Derek Margetts wrote:
I have a question about changing rownames. In the
following function I am plotting the regression
coeficients with their corresponding mean. Right now,
the labels on the plot and the rownames in the
dataframe are x1,x2,x3...etc. Is there a
Hi
has anyone coded up integer factorization? I want, for example,
R factorize(60)
$primes
[1] 2 3 5
$powers
[1] 2 1 1
(actually, I want the divisor function $\sigma_a(n)$, but coding this
up requires integer factorization).
--
Robin Hankin
Uncertainty Analyst
Southampton Oceanography Centre
It is sometimes important to know that row.names accesses the row names of
a data frame, and rownames the first dimname of an array (including a
matrix). They are almost equivalent, but not quite (e.g. row.names is
generic and has methods for matrices and arrays).
So the example would
When I try to unlist a very large list, R is killed
without any other warning:
A-as.list(as.data.frame(matrix(1:21639744,nrow=3578,ncol=6048)))
with
AA-unlist(A)
or
AA-c(A,recursive=TRUE)
I get a
R terminado (killed)
and the end of the session.
I think I'll need to get more RAM (now 1Gb, any
Hello!
I am producing a set of images and I would like them to be sorted by names
I give. I was able to produce my names and add integer to them. That is
easy. But my problem lies in sort of file from this process:
figure_10.png
figure_11.png
figure_12.png
...
figure_1.png
figure_20.png
...
So
Hello,
Am Mittwoch, den 05.01.2005, 06:47 + schrieb Prof Brian Ripley:
On Wed, 5 Jan 2005, thomas wrote:
But there is no package `pkg1' on CRAN. Try a real name like
install.packages(tree)
Well, of cause I didn't want install pkg1 or pkg2,
the precise commandline was:
1. R
Hi Manuel,
try the following:
A - data.frame(matrix(1:21639744, nrow=3578, ncol=6048))
AA - unlist(A, use.names=FALSE)
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35,
Hi,
I made a package on a linux box. All worked fine.
The package contains only R code (no C). I then wanted to make a zip
file so that I could test the package on a windows machine. I have tried
all the obvious ways to do this (and even some that are not!), but to no
avail. The only
Hi Gregor,
There still exist simple functions to achive that goal:
Look at:
x=1:111
formatC(format=d,x,flag=0,width=ceiling(log10(max(x
[1] 001 002 003 004 005 006 007 008 009 010 011
012 013 014 015 016 017 018 019 020
[21] 021 022 023 024 025 026 027 028 029 030 031
032 033 034 035
On 05-Jan-05 Gregor GORJANC wrote:
Hello!
I am producing a set of images and I would like them to be
sorted by names I give.
[...]
So I would like to convert integers to something like 01 if
upper limit for this conert is 10 or 001 for 100.
[...]
Hi Gregor,
'formatC' provides access to
? formatC
Else (if you are linux) the following shell command can be useful...
rename file_name_ file_name_0 file_name_[0-9]_bleh_*.png
Where file_name_[0-9]_bleh_*.png is supposed to match all those files
with a single digit, and the rename command adds a zero before that digit.
rename can
Hello,
I have an unbalanced mixed model design with two fixed effects
site (2 levels) and timeOfDay (4 levels) and two random effects
day (3 consecutive days) and trap (6 unique traps, 3 per site).
The dependent variable is the body length (BL) of insect larvae from 7
to 29 individuals per trap
Hi Robin
There is a function factorize() in the package conf.design
library(conf.design)
factorize(60)
[1] 2 2 3 5
Hope this helps
Christoph
--
Christoph Buser [EMAIL PROTECTED]
Seminar fuer Statistik, LEO C11
ETH (Federal Inst. Technology) 8092 Zurich SWITZERLAND
phone: x-41-1-632-5414
Eric == Eric Lecoutre [EMAIL PROTECTED]
on Wed, 05 Jan 2005 10:29:48 +0100 writes:
Eric Hi Gregor,
Eric There still exist simple functions to achive that goal:
Eric Look at:
x=1:111
formatC(format=d,x,flag=0,width=ceiling(log10(max(x
Eric [1] 001 002 003 004
Gregor GORJANC wrote:
Hello!
I am producing a set of images and I would like them to be sorted by
names I give. I was able to produce my names and add integer to them.
That is easy. But my problem lies in sort of file from this process:
figure_10.png
figure_11.png
figure_12.png
...
figure_1.png
thomas wrote:
Hello,
Am Mittwoch, den 05.01.2005, 06:47 + schrieb Prof Brian Ripley:
On Wed, 5 Jan 2005, thomas wrote:
But there is no package `pkg1' on CRAN. Try a real name like
install.packages(tree)
Well, of cause I didn't want install pkg1 or pkg2,
the precise commandline was:
1. R CMD
Hi Thomas,
random=~1 works if your data frame is in groupedData format, check
this:
# Orthodont is in groupedData format
fm1 - lme(distance~age+Sex, data=Orthodont, random=~1)
#
dat - as.data.frame(Orthodont)
fm2.1 - lme(distance~age+Sex, data=dat, random=~1)
`dat' is an ordinary data.frame
Thanks to all for valuable suggestions!
--
Lep pozdrav / With regards,
Gregor GORJANC
---
University of Ljubljana
Biotechnical Faculty URI: http://www.bfro.uni-lj.si
Zootechnical Departmentmail: gregor.gorjanc at
Hi
How do I get the output from table() in matrix form?
If I have
R table(c(1,1,1,1,2,20))
1 2 20
4 1 1
I want
[,1] [,2] [,3]
[1,]12 20
[2,]411
The problem is that names(table) is a vector of characters and I need
the numeric values.
I am using
R
Dimitris Rizopoulos wrote:
Hi Thomas,
random=~1 works if your data frame is in groupedData format, check
this:
# Orthodont is in groupedData format
fm1 - lme(distance~age+Sex, data=Orthodont, random=~1)
#
dat - as.data.frame(Orthodont)
fm2.1 - lme(distance~age+Sex, data=dat, random=~1)
`dat'
Hi Robin,
does this help:
x - table(c(1,1,1,1,2,20))
matrix(c(as.numeric(names(x)), x), ncol=length(x), byrow=TRUE,
dimnames=NULL)
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven,
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Thomas Petzoldt wrote:
Hello,
I have an unbalanced mixed model design with two fixed effects
site (2 levels) and timeOfDay (4 levels) and two random effects
day (3 consecutive days) and trap (6 unique traps, 3 per site).
The dependent variable is the body length (BL) of insect larvae from 7
to 29
You probably want something like:
t1 - table( x )
t1
x
1 2 20
3 2 1
t2 - rbind( as.numeric( names( t1 ) ), t1 )
t2
1 2 20
1 2 20
t1 3 2 1
dimnames( t2 ) - NULL
t2
[,1] [,2] [,3]
[1,]12 20
[2,]321
Bendix
--
Bendix Carstensen
Hi Thomas,
I don't quite understand what you want to do. If you use random=~1
(or random=~1|Subject if you don't have a groupedData), then you
just fit a random-intercepts model. ranef(fm1) gives you the
Empirical Bayes estimates (i.e., posterior means) for the
random-effects which you can use
table works the way it does because it applies to *factors*, so the names
are the factor levels of the argument after conversion. So if anything is
wasteful, that is.
How about using the guts of factor and table, via
xx - unique(x)
rbind(vals=xx, cnts=tabulate(match(x, xx)))
?
On Wed, 5 Jan
On Wed, 05 Jan 2005 13:31:51 +0100, Thomas Petzoldt
[EMAIL PROTECTED] wrote:
Dimitris Rizopoulos wrote:
Hi Thomas,
random=~1 works if your data frame is in groupedData format, check
this:
# Orthodont is in groupedData format
fm1 - lme(distance~age+Sex, data=Orthodont, random=~1)
On Jan 5, 2005, at 12:59 pm, Prof Brian Ripley wrote:
table works the way it does because it applies to *factors*, so the
names are the factor levels of the argument after conversion. So if
anything is wasteful, that is.
How about using the guts of factor and table, via
xx - unique(x)
Hi all,
I'm plotting a histogram of dates and would like to shade the bars, e.g.
hist(.leap.seconds,years,col='gray',freq=T)
-but the axis color also changes, how do I prevent that?
thx in advance
Janus
[[alternative HTML version deleted]]
On Wed, 2005-01-05 at 09:34 +, Halldor Bjrnsson wrote:
Hi,
I made a package on a linux box. All worked fine.
The package contains only R code (no C). I then wanted to make a zip
file so that I could test the package on a windows machine. I have tried
all the obvious ways to do this (and
OK, dumb question, and it is probably in the docs somewhere, but after
12 months working with R and quite a while looking at the docs, I still
don't know (or have forgotten) how to replace all NA values in a matrix
at once with some other value. I can do it column by column using
is.na(), but I
Hi
I do it for the geoR package following the excelent and detailed
instructions provided by Yan and Rossini:
Building Microsoft Windows Versions of R and R packages under Intel
Linux
available at the Contributed Documentation in the R web-site
Notice that not having C/Fortran/C++ code make
You have it already
h - matrix(rnorm(100),nrow=20)
h
[,1] [,2][,3] [,4] [,5]
[1,] 0.4669801 0.3349176 -1.60686041 0.981491440 0.1627222
[2,] -0.2580262 -0.2620413 0.53852801 1.294129626 -0.1632906
[3,] 0.9654591 1.0077212 -0.45603772
x - matrix(rnorm(100),ncol=10)
x[x0] - NA
x[is.na(x)] - 1000
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
michael watson (IAH-C)
Sent: Wednesday, January 05, 2005 9:26 AM
To: R-help@stat.math.ethz.ch
Subject: [R] Replacing all NA values in a
Doh, replace() does the job just fine.
Sheesh, I'm not coping well with work post christmas ;-)
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of michael watson
(IAH-C)
Sent: 05 January 2005 14:26
To: R-help@stat.math.ethz.ch
Subject: [R] Replacing all NA
From: Marc Schwartz
On Wed, 2005-01-05 at 09:34 +, Halldor Björnsson wrote:
Hi,
I made a package on a linux box. All worked fine.
The package contains only R code (no C). I then wanted to
make a zip
file so that I could test the package on a windows machine.
I have tried
Replacing the NA´s with eg. 1 :
a
1 2 3 4
1 20 50 10 80
2 NA 19 NA 49
3 NA 32 NA 61
4 45 101 44 190
a[try(is.na(a)) == TRUE] - 1
a
1 2 3 4
1 20 50 10 80
2 1 19 1 49
3 1 32 1 61
4 45 101 44 190
I this can help you,
Matthias
-Ursprüngliche Nachricht-
What you should keep in mind is that a matrix in R is nothing more than a
vector (formed by stacking the columns of the matrix) with the dim
attribute. Thus you can do what you want to do by treating the matrix as a
vector; e.g.,
mymat[is.na(mymat)] - myFavoriteValue
HTH,
Andy
From: michael
Hi Michael,
try this:
mat - matrix(1:25, 5, 5)
mat[sample(25, 10)] - NA
mat
#
mat[is.na(mat)] - 100
mat
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
See also primeFactors() and allFactors():
http://www.maths.lth.se/help/R/.R/library/R.basic/html/primeFactors.html
http://www.maths.lth.se/help/R/.R/library/R.basic/html/allFactors.html
in the R.basic package part of the R.classes bundle:
http://www.maths.lth.se/help/R/R.classes/
Cheers
michael watson (IAH-C) wrote:
OK, dumb question, and it is probably in the docs somewhere, but after
12 months working with R and quite a while looking at the docs, I still
don't know (or have forgotten) how to replace all NA values in a matrix
at once with some other value. I can do it column by
Dear Michael,
A[is.na(A)] - value
I hope this helps,
John
John Fox
Department of Sociology
McMaster University
Hamilton, Ontario
Canada L8S 4M4
905-525-9140x23604
http://socserv.mcmaster.ca/jfox
-Original Message-
Janus Larsen wrote:
Hi all,
I'm plotting a histogram of dates and would like to shade the bars, e.g.
hist(.leap.seconds,years,col='gray',freq=T)
-but the axis color also changes, how do I prevent that?
Looks like that's not very easy with the current implementation, see the
code in
Michael --
is.na works on the full matrix. The commands below construct a matrix,
insert some NA's, and then convert them all to 0.
temp1 - matrix(runif(25), 5, 5)
temp1[temp1 0.1] - NA
temp1[is.na(temp1)] - 0
Hope this helps.
Regards,
Matt Wiener
-Original Message-
From:
Douglas Bates wrote:
I'm not sure what model you want to fit here. To specify a random
effect in lme you need both a grouping factor and a model matrix. The
error message indicates that lme is unable to determine a grouping
factor. It would be correct syntax if you added a single level
Hi,
I 've a matrix n*1 (thus a column) and I would like to
count the number of negative element inside.
Can you help me?
Thanks!
eg:
res[,1]= 1
-3
-1
How obtain the number 2 (number of negative-element)?
__
Hello, and please excuse this off-topic question, but I have not been
able to find an answer elsewhere. Consider a value Z that is calculated
using the product (or ratio) of two means X_mean and Y_mean:
Z=X_mean*Y_mean. More generally, Z=f(X_mean, Y_mean). The standard
error of Z will be a
Hi all -
R2.0.1, OS X
Perhaps while there is some discussion of lme going on.
I am trying to execute a glmm using glmmPQL from the MASS libray, using
the example data set from McCullagh and Nelder's (1989, p442) table
14.4 (it happens to be the glmm example for GENSTAT as well). The data
Thomas Petzoldt wrote:
Douglas Bates wrote:
I'm not sure what model you want to fit here. To specify a random
effect in lme you need both a grouping factor and a model matrix. The
error message indicates that lme is unable to determine a grouping
factor. It would be correct syntax if you
ChrBu == Christoph Buser [EMAIL PROTECTED]
on Wed, 5 Jan 2005 11:13:30 +0100 writes:
ChrBu Hi Robin
ChrBu There is a function factorize() in the package conf.design
ChrBu library(conf.design)
ChrBu factorize(60)
ChrBu [1] 2 2 3 5
yes, as quick search on Jonathan
Hi,
I'm trying to use the version of dchip combined with R to analyze my data.
I need R version 1.6 which fits for dchip as dchip manual said. So I
would appreciate a lot if someone could tell me where I could find this
version and download? I'm using Windows 2000.
Thanks, Jean
Qun(Jean) Shi
How about this?
length(res[res 0, 1])
HTH, Andy
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Frederic renaud
Sent: Wednesday, January 05, 2005 11:08 AM
To: r-help@stat.math.ethz.ch
Subject: [R] count element in column
Hi,
I 've a matrix n*1
How about
sum(res0)?
--
[EMAIL PROTECTED]
The computer should be doing the hard work. That's what it's paid to
do, after all.
-- Larry Wall in [EMAIL PROTECTED]
It is rumored that on Wed, 5 Jan 2005 08:07:34 -0800 (PST)
Frederic renaud [EMAIL PROTECTED] wrote:
Hi,
I
How about
sum(res0)?
--
[EMAIL PROTECTED]
The computer should be doing the hard work. That's what it's paid to
do, after all.
-- Larry Wall in [EMAIL PROTECTED]
It is rumored that on Wed, 5 Jan 2005 08:07:34 -0800 (PST)
Frederic renaud [EMAIL PROTECTED] wrote:
Hi,
I
You could try googling for delta method. I believe MASS even has code for
that...
Andy
From: Bill Shipley
Hello, and please excuse this off-topic question, but I have not been
able to find an answer elsewhere. Consider a value Z that is
calculated
using the product (or ratio) of two
How about this?
length(res[res 0, 1])
HTH, Andy
or simply sum(res0) .
For R beginners: This works because the logical res0 vector is
automatically coerced to a numeric vector of 0's and 1's by sum().
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
Dear R-List,
here my problem:
a - c(gio,gao,geo,NA,1,alpha)
b - 1:6
data.frame(V1=a,V2=b) - c
c
V1 V2
1 gio 1
2 gao 2
3 geo 3
4 NA 4
5 1 5
6 alpha 6
rownames(c) %in% grep(a,as.character(c$V1))
[1] FALSE TRUE FALSE FALSE FALSE TRUE
while I would like to obtain
[1]
x-c(-3. -4.7, -.005, 1, 9)
length(x[x0])
[1] 3
Anne
- Original Message -
From: Frederic renaud [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Wednesday, January 05, 2005 5:07 PM
Subject: [R] count element in column
Hi,
I 've a matrix n*1 (thus a column) and I would like to
I know two standard ways to approach this. The traditional
approximation is called the delta method; it uses a Taylor series
approximation, usually of first order but could be higher. Googling for
delta method produced several useful hits just now. The second method
is Monte Carlo.
I have an S4 class which extends array and has other slots, and in upgrading
to 2.0.1 (from 1.9.1) I have encountered a change in behaviour. This causes
me some difficulties if I want to allow 2-dimensional arrays in the slot
(which I do).
The following (in 2.0.1) illustrates the point:
Dear all
I'm using Jim Lindsay's glnr to perform some maximum likelihood fitting.
What I would like to know is to what does the shape parameter returned
correspond to when using the Laplace distribution and the Cauchy
distribution?
In particular, given a Laplace distrib
1/(2*s)
On Wed, 5 Jan 2005 [EMAIL PROTECTED] wrote:
Dear list,
I spent about two hours searching on the message archive, with no avail.
I have a list of people that have to pass an on-line test, but only a fraction
of them do it. Moreover, as they input their names, the resulting string do not
always
Hi, Andy:
MASS4 has section 5.7 Bootstrap and Permutation Methods. Is
this what you are suggesting? It certainly is relevant to the question
(but not to the delta method, except as a means of checking on it).
Thanks,
spencer graves
Liaw, Andy wrote:
You could try googling
Hi
Say I have some data, two columns in a table being a binary outcome plus
a predictor and I want to plot a graph that shows the percentage
positives of the binary outcome within bands of the predictor, e.g.
Outcome predictor
0 1
1 2
1
It sounds like what you want is a rudimentary spell-checker whose word
is the input name, and whose dictionary is an array of your database
names. Spell checking rules are designed to find missing repeats,
transposed letters, extra letters... precisely the reasons you're not
matching your names to
What I have in mind is the discussion on pp. 167-172 of `S Programming'.
Cheers,
Andy
From: Spencer Graves
Hi, Andy:
MASS4 has section 5.7 Bootstrap and Permutation Methods. Is
this what you are suggesting? It certainly is relevant to
the question
(but not to the delta
This is a rather complex problem. I'm not aware of an R function /
package that can do something like this, but in case you need to build
it from scratch read
http://support.sas.com/documentation/periodicals/obs/obswww15/index.html
If you're familiar with SAS you could translate the code to R.
hello,
i have just started exploring R as an alternative to matlab for data analysis.
so
+far everything is _very_ promising. i have a question though regarding
parameter
+estimation. i have some data which, from a histogram plot, appears to arise
from
+a gamma distribution. i gather that you
Giovanni Malerba bobo at medgen.univr.it writes:
:
: Dear R-List,
: here my problem:
: a - c(gio,gao,geo,NA,1,alpha)
: b - 1:6
: data.frame(V1=a,V2=b) - c
: c
: V1 V2
: 1 gio 1
: 2 gao 2
: 3 geo 3
: 4 NA 4
: 5 1 5
: 6 alpha 6
:
: rownames(c) %in%
Hi,
I am trying to use fSeries library form CRAN, especialy the armaFit
function.
armaFit(formula = x ~ arima(2, 0, 1), method = c(CSS-ML, ML, CSS,
yw, burg, ols, mle), include.mean = TRUE, fixed = NULL, fracdiff.M
= 100, fracdiff.h = -1, title = , description = , ...)
The problem that I
Happy New Year!
XLSolutions Corporation (www.xlsolutions-corp.com) is proud to
announce 2-day R/S-plus Fundamentals and Programming
Techniques.
Chicago, IL January 13th-14th, 2005
San Francisco, CA -- January 13th-14th, 2005
You want:
tapply( Outcome, predictor, mean )
Bendix Carstensen
--
Bendix Carstensen
Senior Statistician
Steno Diabetes Center
Niels Steensens Vej 2
DK-2820 Gentofte
Denmark
tel: +45 44 43 87 38
mob: +45 30 75 87 38
fax: +45 44 43 07 06
[EMAIL PROTECTED]
www.biostat.ku.dk/~bxc
Sorry for joining late, but I wanted to see if my search page
could help. (I don't know which search archive you looked at.)
I entered
fuzzy string match*
and got a few things that look relevant, including the agrep
function.
As for the second part of the question, that seems to be a coding
Hi Andy,
Thanks a lot for your promptly response. I searched the whole web site, I
found the source code for version 1.6.X. Since I'm not a computer person,
I don't how to compile it, but what I want is binary file for Windows 2000
so that I could continue to work on my data. Could you or someone
You can use 'cut' to create the breaks.. Actually there are 8 in the 3-4
range:
Outcome predictor
10 1
21 2
31 2
40 3
50 3
60 2
71 3
81 4
91 4
10
Dear list,
I ran into a strange behaviour of the pnbinom function - or maybe I
just made a stupid mistake.
First thing is that pnbinom seems to be very slow. The other - more
interesting one - is that I get two different curves when I plot the
estimated density and the density given by pnbinom.
Hello. I want to estimate the predicted values and standard errors of
Y=f(t) and its first derivative at each unique value of t using the
smooth.spline function. However, the data (plant growth as a function
of time) show substantial heterogeneity of variance since the variance
of plant mass
Andrew Collier [EMAIL PROTECTED] writes:
hello,
i have just started exploring R as an alternative to matlab for data
analysis. so
+far everything is _very_ promising. i have a question though regarding
parameter
+estimation. i have some data which, from a histogram plot, appears to
You want:
library(MASS)
?fitdist
cheers
Brett
Brett Melbourne, Postdoctoral Fellow
Biological Invasions IGERT www.cpb.ucdavis.edu/bioinv
Center for Population Biology
University of California Davis CA 95616
- Original Message -
From: Andrew Collier [EMAIL PROTECTED]
To:
First, you want to fit the data not the histogram counts (binned data).
Second, glm does not do a very principled fit of a gamma (it is a moment
estimator). Something like
d - rgamma(1000, 20, scale = 2)
summary(glm(d ~ 1, Gamma))
Coefficients:
Estimate Std. Error t value Pr(|t|)
Actually, that should be:
library(MASS)
?fitdistr
You want:
library(MASS)
?fitdist
cheers
Brett
Brett Melbourne, Postdoctoral Fellow
Biological Invasions IGERT www.cpb.ucdavis.edu/bioinv
Center for Population Biology
University of California Davis CA 95616
- Original Message -
From:
Qun Shi [EMAIL PROTECTED] writes:
Hi Andy,
Thanks a lot for your promptly response. I searched the whole web site, I
found the source code for version 1.6.X. Since I'm not a computer person,
I don't how to compile it, but what I want is binary file for Windows 2000
so that I could continue
On Wed, 5 Jan 2005, dax42 wrote:
Dear list,
I ran into a strange behaviour of the pnbinom function - or maybe I just made
a stupid mistake.
First thing is that pnbinom seems to be very slow. The other - more
interesting one - is that I get two different curves when I plot the
estimated density
On Wed, 5 Jan 2005, Prof Brian Ripley wrote:
First, you want to fit the data not the histogram counts (binned data).
Second, glm does not do a very principled fit of a gamma (it is a moment
estimator). Something like
d - rgamma(1000, 20, scale = 2)
summary(glm(d ~ 1, Gamma))
Coefficients:
Googling for `rw1061.exe' turned up:
http://www.cipic.ucdavis.edu/~dmrocke/Class/EAD289D/R/rw1061.exe
Andy
From: Qun Shi
Hi Andy,
Thanks a lot for your promptly response. I searched the whole
web site, I
found the source code for version 1.6.X. Since I'm not a
computer person,
I
dax42 [EMAIL PROTECTED] writes:
Dear list,
I ran into a strange behaviour of the pnbinom function - or maybe I
just made a stupid mistake.
First thing is that pnbinom seems to be very slow. The other - more
interesting one - is that I get two different curves when I plot the
estimated
dax42 Dax42 at web.de writes:
:
: Dear list,
:
: I ran into a strange behaviour of the pnbinom function - or maybe I
: just made a stupid mistake.
: First thing is that pnbinom seems to be very slow. The other - more
: interesting one - is that I get two different curves when I plot the
:
R-help,
I'm the primary developer for an increasingly large R package with over
three thousand lines of code. Unfortunately, do the complexity of the
code, I sometimes am required to change several interoperating parts of
the package before testing for bugs and performance. And sometimes
unnoticed
Dear R Help Members,
I have some R functions that plot semiconductor data. I would like to
automate these plots for individuals in our group such that they don't
have to know R. I have read the R help manuals and postings but have
not found this problem.
I am using R version 2.0.1 under a
Hi,
I am sorry for the incorrect terms. Of course I mean the cumulative
distribution functions.
As I just got two answers telling me my problem does not exist, I tried
again on a different setup - and indeed it works.
So, my first run was done on a Mac OS X 10.3.7 running R 2.0.0, GUI
version.
Andrew Collier wrote:
hello,
i have just started exploring R as an alternative to matlab for data analysis.
so
+far everything is _very_ promising. i have a question though regarding
parameter
+estimation. i have some data which, from a histogram plot, appears to arise
from
+a gamma
Andrew Collier wrote:
hello,
i have just started exploring R as an alternative to matlab for data analysis.
so
+far everything is _very_ promising. i have a question though regarding
parameter
+estimation. i have some data which, from a histogram plot, appears to arise
from
+a gamma
The help page of ?dgamma says:
The mean and variance are E(X) = a*s and Var(X) = a*s^2.
So, to estimate the parameters in your example, try the following:
d - rgamma(10, 20, scale = 2)
var(d)/mean(d)
[1] 1.992091
mean(d)^2/var(d)
[1] 20.09559
... you may use these as start values for
Liaw, Andy wrote:
You could try googling for delta method. I believe MASS even has code for
that...
Andy
If you have the original data you can bootstrap --- else you need the
standar errors and
correlation between the means, and can use the delta methos as above.
You could even
use D or
I had a quick question on the RSvgDevice package if there are any users
out there. I perused the archives and the docs and didn't see anything,
maybe I missed? I have created simple boxplots via devSVG() in both
Windoze and Linux and it seems that there is an attribute difference in
that the
Hi!
For the person who asked about the armaFit from the fSeries
library, here is an example:
library(its)
x1 - priceIts(instrument = c(^ftse), start = 1998-01-01,
+ quote = Close)
fit - armaFit(x1 ~ arima(1,1,1))
fit
Call:
armaFit(formula = x1 ~ arima(1, 1,
McGehee, Robert Robert.McGehee at geodecapital.com writes:
:
: R-help,
: I'm the primary developer for an increasingly large R package with over
: three thousand lines of code. Unfortunately, do the complexity of the
: code, I sometimes am required to change several interoperating parts of
: the
Dear John,
I belive your problem has to do with the sequence of startup. I think that
.Rprofile is called before the required libraries are attached.
You might like to try putting your code into a .First() function and run it
that way.
Cheers,
Andreas
Dr Andreas Kiermeier
Statistician
SARDI
I tried it on Windows XP under both 2.0.1 and 2.1.0 and it seemed to
work both times. I issued the command:
set.seed(1)
before both runs to ensure that I was actually using the same
random numbers. I suggest you try that in all your runs too just
in case its a problem with some sets of
Hi, there:
I made a function to do k-fold cross-validation as
below. Basically whenever I call cv(test) for example,
an error message like:
20Fold 1
Error in model.frame(formula, rownames, variables,
varnames, extras, extranames, :
variable lengths differ
please help.
My test
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