On Wed, 2009-12-02 at 16:52 +0800, Zhijiang Wang wrote:
> Dear All,
>1. why did the problem happen?
>2. How to solve it?
>
>--
>
> Best wishes,
> Zhijiang Wang
Well... The algorithm for Mann-whitney test have problem with ties
To solve you can use jitter
a<-1:10
b<-1:10
wilcox.tes
Hi,
Can anyone please provide the formula used to compute ACF(nlme). I believe the
one that is used in R is of the type mentioned on the website. Please correct
me if I am wrong. The normalization of the numerator (Ch) has been done by 'N'
where as I would like to do it by 'N-k'. Is there anywa
Hi,
My earlier email was sent as html so no visible properly. I am now posting it
as text. Sorry for the duplication of the message.
Can anyone please provide the formula used to compute ACF(nlme). I believe the
one that is used in R is of the type mentioned on the website
(http://www.itl.nis
Baloo,
Why don't you use the built-in acf function?
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie & Kwaliteitszorg
Gaverstraat 4
9500 Geraardsbergen
Belgium
Research Institute for
On Mon, 2009-12-21 at 13:20 -0500, John Tillinghast wrote:
> I'm trying to build a model with an overall smooth function for all
> the data, plus an additional smooth function for *some* of the data.
> If "ind" is my indicator variable (0 for some x, 1 for others), I
> imagined I could write it
On Tue, 22 Dec 2009, alok juneja wrote:
Hi,
My earlier email was sent as html so no visible properly. I am now posting it
as text. Sorry for the duplication of the message.
Can anyone please provide the formula used to compute ACF(nlme). I believe the
one that is used in R is of the type me
Dear List,
I work with multilevel data from psychological group experiments and
have frequently encountered a situation for which I haven't found an
elegant solution: I need to assign the value of a specific group
member to all members of the group. For example, I have a group leader
(ide
Dear list,
I have 2 data sets such as:
> head(calib20090730b)
color XR XG XB L1_1 L1_2 L1_3
1 1 87 55 62 116 124.0 100
2 2 164 125 134 204 203.0 153
3 3 118 64 98 157 101.0 139
4 4 65 72 72 102 111.0 135
5 5 142 95 112 176 161.5 133
6 6 89 113 112
In the ACF(nlme) the normalization of the numerator has been done by N and I
want to normalize it by N-k, where N is the observations and k is the lag.
Baloo
--- On Tue, 12/22/09, ONKELINX, Thierry wrote:
From: ONKELINX, Thierry
Subject: RE: [R] Nested For loops
To: "baloo mia" , r-help@r-proj
On 12/22/2009 06:04 AM, teo wrote:
Hi:
Could you please guide me how to plot "weekly.t2". Below are my values and I
am only able to plot "weekly t1" but failing to add a second line
representng "weekly t2".
Thanks.
weekly.t1
[1] 228.5204 326.1224 387.2449 415.4082 443.6735 454.6939 451.5306
one approach is:
Dat <- read.table(textConnection("
personId groupId groupLeader someAttribute leaderAttribute
11 17 0 0.14583 NA
22 17 1 0.21875 NA
33 17 0 0.089743590 NA
4
Try this:
merge(df, subset(df, groupLeader == 1, select = c(groupId,
someAttribute)), by = "groupId")[,-5]
On Tue, Dec 22, 2009 at 8:42 AM, Bertolt Meyer wrote:
> Dear List,
>
> I work with multilevel data from psychological group experiments and have
> frequently encountered a situation for wh
Hi,
if "a" is a constant, and "u" is a vector of random variables of arbitrary
length, each of which is uniformly distributed between 0 and 1, then h(u) is
a random sample of the function f.
So, if you adjust the code to the desired "a" and to the desired length "n,"
then you get a random sample
ggplot2
ggplot2 is a plotting system for R, based on the grammar of graphics,
which tries to take the good parts of base and lattice graphics and
avoid bad parts. It takes care of many of the fiddly details
that make plotting a hassle (l
Thanks to both of you for the comments and suggestions. Over the next couple of
days I plan to work through my simple problem using the help offered in this
forum.
From: David Winsemius
To: Bert Gunter
Sent: Mon, December 21, 2009 2:31:26 PM
Subject: Re: [R
May be please inform where I can find the source code for "nlme:::ACF.gls"
and "nlme:::ACF.lme". Sorry to ask. As I have looked into the R directory but
don't find that.
Best
Alok
--- On Tue, 12/22/09, Prof Brian Ripley wrote:
> From: Prof Brian Ripley
> Subject: Re: [R] ACF normalization.
>
Please find release 3.6.0 of the package "randomSurvivalForest" now
available for download on CRAN.
-
CHANGES TO RELEASE 3.6.0
RELEASE 3.6.0 represents the last and final major upgrade
of this product. Current and f
Greetings Zaki,
You should really post this question on the R-help forum so that
others might benefit from any responses. It's been a while since I've
done this, but if memory serves, the basic process was to download
xpdf and add it to the windows path, thus making it accessable from
within R. Tw
Hi List, I would like to embed some fonts (AFM format) on 324 eps files
produced by a R loop, Im able to do so one-by-one with the following
command:
embedFonts('C:/Users/Rodrigo/Documents/UFPR/Micropaleontologia/Potiguar/Cata
logo/Mapas/EPS/Example.eps',
outfile='C:/Users/Rodrigo/Documents/U
Maybe this (with enough data for a CI) ? :
> Dataset = data.table(Dataset)
> Dataset[,as.list(wilcox.test(ratio,conf.int=TRUE)$conf.int),by="LEAID"]
LEAID V1 V2
[1,] 6307 0.720 0.92
[2,] 8300 0.5678462 0.83
Warning messages:
1: In switch(alternative, two.sided
Just enter nlme:::ACF.gls at the command prompt and hit enter. This works with
most functions. Another option is to download the source code of nlme from CRAN.
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur-
Wouldn't the best approach be to adapt the acf function already
available in R and change it only to suit your needs?
cheers,
Paul
baloo mia wrote:
In the ACF(nlme) the normalization of the numerator has been done by N and I
want to normalize it by N-k, where N is the observations and k is th
Try this. The function f takes a vector of indices and returns the
attribute for the unique groupLeader row among them. We use ave to
apply to each set of indices having a common groupId.
f <- function(ix) with(df[ix,], someAttribute[groupLeader == 1])
transform(df, leaderAttribute = ave(1:nrow(d
Thanks for the suggestions..
I must admit that on the R-help one of my coworker is asking the same question
named "Nested For loops". My question "ACF normalization" and his question both
lead to the same problem. So now I merge both the questions. Sorry for the
inconvenience.
The main problem
1) I want to calculate a regression, but when I enter >
lm(formula=KS~libor+adj.close) I only get the following:
Call:
lm(formula = KS ~ libor + adj.close)
Coefficients:
(Intercept)liboradj.close
-56.38666 55.39709 -0.01836
I don't get the estimated standard deviation, error
On Tue, 2009-12-22 at 15:30 +0100, Gavriel & Esti Zoladz wrote:
> 1) I want to calculate a regression, but when I enter >
> lm(formula=KS~libor+adj.close) I only get the following:
>
> Call:
> lm(formula = KS ~ libor + adj.close)
> Coefficients:
> (Intercept)liboradj.close
> -56.3866
Hi,
I am having problems getting cohen's kappa to work. I have been using the
function:
><-ckappa(x,y)
from the psy package.
I am trying to test for inter-observer reliability, I have 2 observers and 26
categories, however, the two observers might not necessarily have the same
range of cate
dcemriS4: A package for medical image anlaysis
The dcemriS4 package is a major update, and S4 implementation, of the
dcemri package. All previous functionality has been replicated, for
DCE-MRI and DWI, with specific emphasis on cancer imaging
applications. Additional major features include:
o
Here is one other solution. This one uses sqldf and is the SQL
analogue of Henrique's solution. It merges df with a copy of itself
restricted to those rows with groupLeader equal to 1 displaying the
indicated columns.
library(sqldf)
sqldf("select b.personId, b.groupId, b.groupLeader, b.someAttri
Okay, since you did not provide the information I asked for, this is the
best guess of what I think you want to do (see commented code below).
#draw a random u*
u.star=runif(1)
#draw a random w
w=runif(1)
#Interval boundaries as defined by u
u=c(0.1,0.2,0.3,0.6,0.9)
#All Us smaller than u*
min.
Hi All,
This example code
dDF <- structure(list(y = c(4.75587, 4.8451, 5.04139, 4.85733, 5.20412,
5.92428, 5.69897, 4.78958, 4, 4), t = c(0, 48, 144, 192, 240,
312, 360, 0, 48, 144), Batch = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1
), T = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 2), pH = c(4.6, 4.6,
Just confirming it isn't the bug fixed in 2.11.0dev, and giving an even
simpler example:
R version 2.11.0 Under development (unstable) (2009-12-20 r50794)
> expand.grid(data.frame(y=1:10, t=1:10))
Error in `[[<-.data.frame`(`*tmp*`, i, value = c(1L, 2L, 3L, 4L, 5L, 6L, :
replacement has 100 r
1) Look at the help: a data frame may be a list, but do pass a list
such as unclass(dDF) when it says 'list'.
2) You have 7 columns of 10 items, which gives 10 million rows. Is
that really what you want (especially as some of the columns are
constant)? It's an object of ca 500MB.
On Tue, 2
The help page says:
"...: vectors, factors or a list containing these."
So, you can try this:
expand.grid(as.list(dDF))
On Tue, Dec 22, 2009 at 2:19 PM, Keith Jewell wrote:
> Just confirming it isn't the bug fixed in 2.11.0dev, and giving an even
> simpler example:
>
> R version 2.11.0 Under d
Keith Jewell campden.co.uk> writes:
>
> Hi All,
>
> This example code
>
> dDF <- structure(list(y = c(4.75587, 4.8451, 5.04139, 4.85733, 5.20412,
> 5.92428, 5.69897, 4.78958, 4, 4), t = c(0, 48, 144, 192, 240,
> 312, 360, 0, 48, 144), Batch = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1
>
On Dec 22, 2009, at 11:19 AM, Keith Jewell wrote:
Just confirming it isn't the bug fixed in 2.11.0dev, and giving an
even
simpler example:
R version 2.11.0 Under development (unstable) (2009-12-20 r50794)
expand.grid(data.frame(y=1:10, t=1:10))
Error in `[[<-.data.frame`(`*tmp*`, i, value
Thanks, Prof Ripley (and everyone else who's responding while I type this!).
I thought I was being stupid. What I wanted was:
expand.grid(lapply(dDF, unique))
which works fine!
Seasonal greetings to all,
Keith Jewell
"Prof Brian Ripley" wrote in message
news:alpine.lfd.2.00.0912
On 22/12/2009 11:19 AM, Keith Jewell wrote:
Just confirming it isn't the bug fixed in 2.11.0dev, and giving an even
simpler example:
The docs for expand.grid say it works on lists, but they don't mention
dataframes. Although a dataframe is in many ways a list with extra
structure, apparen
Costa wrote:
Hi,
I installed R.10.1 and Graphviz 2.26 - which doesnt contain the
libcdt-4.dll, so I get the same error message 'This application has failed
to start because libcdt-4.dll was not found' when loading in R
library (Rgraphviz)
I cannot find an installation file for Graphviz 2.
David,
Thank you for the insight.
HA1) StructTS() ships with R in package stats.
HA2) I don't see anything strange or unusual in looking for standard errors
of estimated parameters. In structural time series the only unknown
parameters, estimated by StructTS(), happen to be
Hi all,
I want to fit data called "metal" with a polynominal function as dP ~ a.0 +
a.1 * U0 + a.2 * U0^2 + a.3 * U0^3 + a.4 * U0^4
The data set includes, the independant variable U0 and the dependant
variable dP.
I've seen that the combination of lm() and poly() can do that instead of
using th
Hi, if I see it correctly, the nls you run is a linear model. It will
probably give you the same (or virtually identical) result as
lm(dP~U0+I(U0^2)+I(U0^3)+I(U0^4)).
Your lm model, by contrast, creates orthogonal polynomials such that all
orders of the polynomials are uncorrelated with the other
Get some statistical consulting help or read up on these topics -- any good
textbook on regression should contain the necessary material. This has
nothing to do with nonlinear regression, so you are confused about the basic
ideas. It has nothing to do with R.
If you don't understand how the statis
> Ot should be noted that the performance of single split into training + test
> does not perform satisfactorily unless N > 10,000 in many cases.
Agreed. I've had a few cases where the resampled statistics on the
training set look poor but the test set results are much better. In
the end, the trai
On Mon, Dec 21, 2009 at 4:42 AM, Duncan Murdoch wrote:
> I've just posted a demo made with the rgl package to Youtube, visible here:
> http://www.youtube.com/watch?v=prdZWQD7L5c
>
> For future reference, here are the steps I used:
>
> 1. Design a shape to be displayed, and then play with the ani
On Dec 22, 2009, at 11:55 AM, Giovanni Petris wrote:
David,
Thank you for the insight.
Right.
HA1) StructTS() ships with R in package stats.
HA2) I don't see anything strange or unusual in looking for standard
errors
of estimated parameters. In structural time series the only
On 22/12/2009 12:49 PM, Mark Knecht wrote:
On Mon, Dec 21, 2009 at 4:42 AM, Duncan Murdoch wrote:
> I've just posted a demo made with the rgl package to Youtube, visible here:
> http://www.youtube.com/watch?v=prdZWQD7L5c
>
> For future reference, here are the steps I used:
>
> 1. Design a shap
>You can see an implementation of mps for the normal inverse Gaussian in
>fBasics on R-Forge.
>Have a look at the function .nigFit.mps in the file dist-nigFit.R
That really helps me. Thanks!
--
View this message in context:
http://n4.nabble.com/Maximum-spacing-method-tp976335p977224.html
S
Hi,
I would like to create a 3 dimensional barplot of 16 odds ratios that
demonstrate an interaction between two variables (CD14 and CD23).
Here is an example of what the graph I would like to produce
(http://www.xfy.com/manual/dev/developer/1.4/spec/image/chart/type_3d.gif):
Here is are the o
Hi:
I need to do text mining on PDF files. I understand there is a readPDF
command in tm that can be used. Have read the 2008 posts on converting
PDF files to text by Tony Breyal and others.
Wondering if the procedure has been standardized in any tutorial or
otherwise? Being new to R, I wa
Dear R-Team,
Am I right supposing that the "breakpoints()" function in the strucchange
package is an implementation of the pure structural change model proposed
by Bai and Perron (1997, 2003)?
My question relates to a partial structural change model that Bai and
Perron formulate in their 2003 pap
I am trying to get monthly means for a daily data series using zoo(). I have
found an odd problem, that seems to be caused by zoo()'s handling of leap
years.
Here's my R script with 2 methods (freq=365, 366) for aggregating the daily
data to monthly series:
library(zoo)
J_link <- "http://www.ij
Hans Christian:
Am I right supposing that the "breakpoints()" function in the strucchange
package is an implementation of the pure structural change model proposed
by Bai and Perron (1997, 2003)?
Yes, see ?breakpoints. In the references section there is a pointer to our
CSDA 2003 paper, a pre
On Tue, 22 Dec 2009, D Kelly O'Day wrote:
I am trying to get monthly means for a daily data series using zoo(). I have
found an odd problem, that seems to be caused by zoo()'s handling of leap
years.
It's not really zoo's odd handling, but yours ;-) More seriously, do not
use ts() with eithe
Z
Thanks. I new I was missing something!
Kelly
Achim Zeileis wrote:
>
> On Tue, 22 Dec 2009, D Kelly O'Day wrote:
>
>>
>> I am trying to get monthly means for a daily data series using zoo(). I
>> have
>> found an odd problem, that seems to be caused by zoo()'s handling of leap
>> years.
>
Install the fortunes package, then do:
> library(fortunes)
> fortune(197)
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org]
Copied/pasted from my earlier reply:
It's been a while since I've
done this, but if memory serves, the basic process was to download
xpdf and add it to the windows path, thus making it accessable from
within R. Two methods follow:
Method One (easiest) - using the awesome ?system command:
(1) Do
I upgraded to R2.10.1pat and discovered, along with everybody else,
that static HTML pages are no longer the default. Fine; my tastes
would go the other way, but I'm happy to adapt.
However, I'd still like to build static HTML pages (for stable
bookmarking, use when R is not running, etc.).
I'm
Hi, I was wondering how you can create a pdf where the output is vertically
justified with for example a one inch margin on top using pdf().
For example in
pdf(file='test.pdf', paper='letter', pagecentre=F)
### code for plot ###
dev.off()
is there an option where it generate output that starts f
[Environment: Win Xp, Miktex 2.7, R 2.9.2]
In an Sweave document, I'm displaying the results of car:::Anova()
tests, that look like this in the
generated .tex file:
\begin{Soutput}
Type III MANOVA Tests: Pillai test statistic
Df test stat approx F num Df den Df Pr(>F)
(Intercept) 1 0.86 90.38
Hi, I'm new to R, with some experience with Matlab and SPSS. I've
figured out how to run my repeated measures anova and am getting the
right numbers for my effects (comparing with results from other
software), but am having trouble with the model.tables function.
Specifically, using:
model
On Dec 22, 2009, at 4:22 PM, Jon Prince wrote:
Hi, I'm new to R, with some experience with Matlab and SPSS. I've
figured out how to run my repeated measures anova and am getting the
right numbers for my effects (comparing with results from other
software), but am having trouble with the mo
On Tue, 22 Dec 2009, Michael Friendly wrote:
[Environment: Win Xp, Miktex 2.7, R 2.9.2]
If you want to set R output in a TeX font that (as most are) is very
limited in its coverage, try options(useFancyQuotes = FALSE).
We don't have the 'at a minimum' information asked for in the posting
g
Hi David,
You might not be aware that Giovanni Petris is the author of the dlm
package, a world-class package for modelling stochastic / time-varying
systems.
Not that I know much about this stuff.
Cheers
-Felix
2009/12/23 David Winsemius :
> On Dec 22, 2009, at 11:55 AM, Giovanni Petris wrote
Unfortunately, expand.grid doesn't validate the class of its argument, so it
is reporting an internal error rather than something more intelligible.
On Tue, Dec 22, 2009 at 11:19 AM, Keith Jewell wrote:
> Just confirming it isn't the bug fixed in 2.11.0dev, and giving an even
> simpler example:
>
David Winsemius wrote:
On Dec 22, 2009, at 4:22 PM, Jon Prince wrote:
Hi, I'm new to R, with some experience with Matlab and SPSS. I've
figured out how to run my repeated measures anova and am getting the
right numbers for my effects (comparing with results from other
software), but am ha
Prof Brian Ripley wrote:
On Tue, 22 Dec 2009, Michael Friendly wrote:
[Environment: Win Xp, Miktex 2.7, R 2.9.2]
If you want to set R output in a TeX font that (as most are) is very
limited in its coverage, try options(useFancyQuotes = FALSE).
Perfect, and all I need! Thanks so much for this
On Dec 22, 2009, at 5:19 PM, Jon Prince wrote:
David Winsemius wrote:
On Dec 22, 2009, at 4:22 PM, Jon Prince wrote:
Hi, I'm new to R, with some experience with Matlab and SPSS. I've
figured out how to run my repeated measures anova and am getting
the right numbers for my effects (compa
On Tue, 22 Dec 2009, David Winsemius wrote:
On Dec 22, 2009, at 5:19 PM, Jon Prince wrote:
David Winsemius wrote:
>
> On Dec 22, 2009, at 4:22 PM, Jon Prince wrote:
>
> > Hi, I'm new to R, with some experience with Matlab and SPSS. I've
> > figured out how to run my repeated measures anov
On Dec 22, 2009, at 7:16 PM, Charles C. Berry wrote:
On Tue, 22 Dec 2009, David Winsemius wrote:
On Dec 22, 2009, at 5:19 PM, Jon Prince wrote:
David Winsemius wrote:
> > On Dec 22, 2009, at 4:22 PM, Jon Prince wrote:
> > > Hi, I'm new to R, with some experience with Matlab and SPSS.
I'v
Thanks, but that produces what I think is an estimated interval.
I really want to use the above formula. I just can't figure out how to
get it to run by the LEAID.
It does require 9 observations to produce an interval, but I was showing a
sample.
Thanks again.
L.A.
Matthew Dowle-3 wrote:
>
>
Hi - I'm wondering if there is any existing package in R that
facilitates ACD (Autoregressive Conditional Duration) model. I
googled. Apparently there were packages dynamo and fACD. However, they
do not seem to be on any CRAN. I am wondering if anyone could point me
to an existing package (if any).
David Winsemius wrote:
On Dec 22, 2009, at 5:19 PM, Jon Prince wrote:
David Winsemius wrote:
On Dec 22, 2009, at 4:22 PM, Jon Prince wrote:
Hi, I'm new to R, with some experience with Matlab and SPSS. I've
figured out how to run my repeated measures anova and am getting
the right numbe
On 12/23/2009 01:58 AM, Julia Myatt wrote:
Hi,
I am having problems getting cohen's kappa to work. I have been using the
function:
<-ckappa(x,y)
from the psy package.
I am trying to test for inter-observer reliability, I have 2 observers and 26
categories, however, the two observ
Michael Friendly yorku.ca> writes:
>
> > \usepackage[cp1252]{inputenc}
> > \inputencoding{cp1252}
I usually just use
\usepackage[utf8]{inputenc}
and that seems to work fine. Couldn't give you a detailed description
of my TeX setup though (standard installation on Ubuntu 8.10) ...
chee
On Dec 22, 2009, at 8:52 PM, Jon Prince wrote:
David Winsemius wrote:
On Dec 22, 2009, at 5:19 PM, Jon Prince wrote:
David Winsemius wrote:
On Dec 22, 2009, at 4:22 PM, Jon Prince wrote:
Hi, I'm new to R, with some experience with Matlab and SPSS.
I've figured out how to run my repe
Tena korua David and Jon
Without an Error() in the model, you get standard errors for the effects
and standard errors of the difference for the means. With fully
balanced data, as in the example, these are directly comparable (compare
model.tables(npk.aov, se=T)*sqrt(2) with model.tables(npk.aov,
Hi Everybody,
I know that R has snow package which can be used for Parallel Computing.
However, every time I try making a cluster, the only type of cluster I'm
able to make is the "SOCK" (that too because I disabled the firewalls). For
the rest (i.e. MPI, NWS, and PVM), I get error every time I t
Thanks for the suggestions..
I must admit that on the R-help one of my coworker is asking the same question
named "Nested For loops". My question "ACF normalization" and his question both
lead to the same problem. So now I merge both the questions. Sorry for the
inconvenience.
The main problem
Dear all,
I have a data.frame with some NAs that can
happens anywere, and I would like to keep
only rows without NAs. Thanks in advance
for your help, and Merry Xmas.
myvect<-runif(100)
myvect[sample(1:100)[1:5]]<-NA
myDF<-data.frame(matrix(myvect,ncol=5))
myDF
miltinho
[[alternative HTM
try this:
myDF[complete.cases(myDF), ]
Best,
Dimitris
milton ruser wrote:
Dear all,
I have a data.frame with some NAs that can
happens anywere, and I would like to keep
only rows without NAs. Thanks in advance
for your help, and Merry Xmas.
myvect<-runif(100)
myvect[sample(1:100)[1:5]]<-NA
A few weeks ago, I released my package edtdbg, which integrates R's
debug() with your text editor. At the time, I said I'd release a
version with more features in a couple of days. Well, it's taken a
while, as I've experimented with various approaches, but I'm now releasing
the new version, intr
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