Hi,
I am curious about how to interpret the table produced by
anova(ols(...)), from the Design package. I have a multiple linear
regression model, with some interaction, defined by:
ols(formula = log(ksat * 60 * 60) ~ log(sar) * pol(activity,
3) + log(conc) * pol(sand, 3), data = sm.clean, x
Hi all,
Is there anyone know about R function for getting robust estimate of
Buckley-James general liner model for censored data?
tq,
Abdul Kudus
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/m
With my data I get
sc[["(Unknown).{Unknown)"]]
returns NULL
but sc[[1]] returns
DayOfYear Quantity Fraction Category SubCategory
1 1 82 3.903927e-05 (Unknown) (Unknown)
2 2 78 3.713492e-05 (Unknown) (Unknown)
3 3 112 5.332193e-05 (Unkn
You don't have a closing parens on the 'sum'
On Mon, Jul 14, 2008 at 11:25 PM, <[EMAIL PROTECTED]> wrote:
> One more question? I am trying to iterate through this array
>
> I have:
>
> sc <- split(x, list(x$Category, x$SubCategory), drop=TRUE)
>
> I think I understand 'length(sc)' It would be the
One more question? I am trying to iterate through this array
I have:
sc <- split(x, list(x$Category, x$SubCategory), drop=TRUE)
I think I understand 'length(sc)' It would be the total number of non empty
category and sub category pairs (in this case 2415).
I don't seems to be able to iterate t
Jim Lindsey's repeated package has the function gnlmm which will fit
beta regressions with a random intercept and one level of nesting. I
don't know of any other options.
Cheers,
Simon.
Not sure On Mon, 2008-07-14 at 19:16 -0700, Daniel Malter wrote:
> I have a connector-question to that. Is be
I have a connector-question to that. Is beta-regression available for
repeated measures or panel data and if so is it available in R?
thx,
Daniel
Kevin J Emerson wrote:
>
> R-devotees,
>
> I have a question about modeling in the case where the response variable
> is
> binary.
>
> I have a c
Dear R Help,
At the risk of sounding naïve, is there a way for obtaining something
resembling an r^2 value for the fit obtained using SSasymp?
I have read ?SSasymp and looked through the archives but to no avail.
Many thanks,
Robert Terwilliger
__
R-
Wait, are the proportions (probabilities) based on discrete data, or are
they truly continuous? If the latter, then beta regression might be more
appropriate (e.g. package betareg). If the former, include the sample
size for each proportion in the call to glm using the weights= argument.
Or set the
Hi Kevin, you mean an s-shaped relationship of a variable with your response?
So you have a response that is strictly constrained to the interval 0,1 or,
and these limits are not due to truncation or censoring (i.e. your response
variable is truly a proportion).
This sounds like a good applicatio
Hello,
I am trying to select the following headers from a data frame but when I try
and run the command it executes halfway through and give me an error at V188
and V359.
Temp <- data.frame(V4, V5, V6, V7, V8, V9, V10, V11, V12, V13, V14, V15,
V16, V17, V18, V19, V20, V21, V22, V23, V24, V25, V2
I'm having a problem with rgl.snapshot. If I run the following code *once* and
*once only*, all is well.
data(trees)
attach(trees)
plot3d(Height, Girth, Volume, type='s')
rgl.snapshot("/home/user/pic.png", fmt="png", top=TRUE )
And the pic.png looks right. But if I try to rerun rgl.snapshot a
I have a table of data, and columns 6-8 of row 1 each contain an array. How
do I extract those arrays and plot them against eachother? Here is my
current code, which only plots the first point of the array.
circle <- as.matrix(AllData[1,6:8])
triax.points(
circle,
pch = 20,
col.symbols = "Blac
R-devotees,
I have a question about modeling in the case where the response variable is
binary.
I have a case where I have a response variable that is the probability of
success, and four descriptor variables, The response has a sigmoid response
with one of the variables. I would like to test for
Is this what you want:
> x <- data.frame(a=1:10,b=1:10)
> x
a b
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 8 8
9 9 9
10 10 10
> t(x)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
a12345678910
b12345
gregexpr("\\(", frg)
Hua Li wrote:
Hi All,
I'm wondering whether there is a quick way to know the position of a
specific charcater in a long character:
for example
frg="((D:41.04,I:41.04):45.05,(((E:2.32,((G:0.67,J:0.67):0.44,H:1.11):1.21)"
and I would like to know that the 1st, 2nd, 26th,
Well, actually, what I really want is a way to quickly find the fragment that
contained equal number of "(" and ")", starting from the end of the character.
In the case of
frg = ="((D:41.04,I:41.04):45.05,(((E:2.32,((G:0.67,J:0.67):0.44,H:1.11):1.21)"
that means finding the part: "(E:2.32,((G:0
Hi All,
I'm wondering whether there is a quick way to know the position of a specific
charcater in a long character:
for example
frg="((D:41.04,I:41.04):45.05,(((E:2.32,((G:0.67,J:0.67):0.44,H:1.11):1.21)"
and I would like to know that the 1st, 2nd, 26th, 27th, 28th ... character is
"(", is t
The sqldf R package can do that using SQL syntax. See:
http://sqldf.googlecode.com
The R merge command can do a join but not using SQL syntax.
See ?merge
On Mon, Jul 14, 2008 at 5:29 PM, Willa Wei <[EMAIL PROTECTED]> wrote:
>
> Greetings,
>
> I am new to R and have some background knowledge abou
Thank you to all for your ideas. They all solved my problem nicely.
And I learned about paste() too!
--Chris
Christopher W. Ryan, MD
SUNY Upstate Medical University Clinical Campus at Binghamton
40 Arch Street, Johnson City, NY 13790
cryanatbinghamtondotedu
PGP public keys available at http://ho
why don't you give us a snippett of your data and then we may be able to
help (read the posting guide)
On Mon, Jul 14, 2008 at 6:04 PM, Zroutik Zroutik <[EMAIL PROTECTED]> wrote:
> Nope, this assigns the 26th, 28th, 30th anf 32nd point of the total
> nrow(matrix) number of points. This does not a
--- Begin Message ---
Hi R Guru:
Thank you for your answer!
No, Dr. Velez, it is a closed source and propietary software,
that several companies here in Medellin are starting to use
to face actuarial problems.
Kenneth
El lun, 14-07-2008 a las 17:59 -0400, Jorge Ivan Velez escribió:
> Dear Kenn
Nope, this assigns the 26th, 28th, 30th anf 32nd point of the total
nrow(matrix) number of points. This does not assign the points (or places of
the x-axis) with such a rowname.
row number <-> row name
1:nrow(matrix) <-> rowname(matrix[1]) .. rowname(matrix[nrow(matrix)])
I try to rephrase my qu
May be this helps:
jaw<-data.frame(vara=1:20,age=runif(20))
for (i in 2:6) {
jaw[paste("agepow",i, sep="")] <- jaw["age"]^i
}
jaw<-round(jaw,2) # It is just to help the check task
Cheers,
Miltinho Astronauta
Brazil
On 7/14/08, Christopher W. Ryan <[EMAIL PROTECTED]> wrote:
>
> R
Hi R users:
I will like to know if somebody works on insurance statistics
(actuarial problems) and had use TRICAST, and can tell me
if with all the R tools it can be build a solution
like TRICAST or similar.
In a word:
Do you think that R has all the statistical tools
(I mean modeling tools) to
[EMAIL PROTECTED] wrote:
In Julian Faraway's text on pgs 117-119, he gives a very nice, pretty
simple description of how a glm can be thought of as linear model
with non constant variance. I just didn't understand one of his
statements on the top of 118. To quote :
"We can use a similar idea
Hi. I'm a new R user and think what I'm trying to do is pretty basic, but
haven't been able to find the answer in the mailing list archives. How do I
write a function that creates new variables in a given data.frame with a
prefix attached to the variable names? I do a lot of repetitive logging a
Greetings,
I have a data set look like this:
> print(MyResult)
259 10
1113
mean6.108394 3.272211e+01 3.951703e+02 2.086217e+03
79.838867 2.592272e+04
median 1.29 2.743511e+00 4.061491e+00 4.943710e+01
-2.821667
Christopher W. Ryan wrote:
R 2.5.1 on WinXP
I'm trying to create new variables in a dataframe called jaw, as powers
of jaw$age up to the sixth power, and name them
jaw$age.(--the digit corresponding to the power--)
Obviously none of these work (silly for me to try, I suppose):
for (i in 2:6)
Greetings,
I am new to R and have some background knowledge about SQL. I'd like to
know whether there is a way to manipulate the R datasets (or data
frames) using SQL statements. For example, I have two data frames and
both of them have a column called "id", then I want to join this two
data fr
[EMAIL PROTECTED] wrote:
In Julian Faraway's text on pgs 117-119, he gives a very nice, pretty
simple description of how a glm can be thought of as linear model
with non constant variance. I just didn't understand one of his
statements on the top of 118. To quote :
"We can use a similar idea
R 2.5.1 on WinXP
I'm trying to create new variables in a dataframe called jaw, as powers
of jaw$age up to the sixth power, and name them
jaw$age.(--the digit corresponding to the power--)
Obviously none of these work (silly for me to try, I suppose):
for (i in 2:6) {
jaw$age.'i' <- jaw$age^i
For the binomial the standard link function is the logit:
g(y) = log( y/(1-y) )
In the binomial glm model the observed y values are 0, or 1 which give g(0) =
-Inf and g(1) = Inf. Switching to g(mu) with 0 < mu < 1 results in finite
values which are much easier for the computer to work with.
Windows XP Pro
-Original Message-
From: John Kane [mailto:[EMAIL PROTECTED]
Sent: Monday, July 14, 2008 2:57 PM
To: r-help@r-project.org; Darin Brooks
Subject: Re: [R] R Commander question
What OS?
--- On Mon, 7/14/08, Darin Brooks <[EMAIL PROTECTED]> wrote:
> From: Darin Brooks <[EMA
For those people who feel the need for a p-value to test normality on large
sample sizes, I propose the following test/function:
SnowsPenultimateNormalityTest <- function(x){
# the following function works for current implementations of R
# to my knowledge, eventually it may need
What OS?
--- On Mon, 7/14/08, Darin Brooks <[EMAIL PROTECTED]> wrote:
> From: Darin Brooks <[EMAIL PROTECTED]>
> Subject: [R] R Commander question
> To: r-help@r-project.org
> Received: Monday, July 14, 2008, 1:56 PM
> Good afternoon
>
> New to R ... new to the list.
>
> I have installed R C
In Julian Faraway's text on pgs 117-119, he gives a very nice, pretty
simple description of how a glm can be thought of as linear model
with non constant variance. I just didn't understand one of his
statements on the top of 118. To quote :
"We can use a similar idea to fit a GLM. Roughly spea
I accomplish this a little differently.
On the mac, in your home directory (e.g. /Users/susanamrose) there is/could
be a hidden file called .Rprofile You can edit it with vi for instance by
getting a terminal window and >vi .Rprofile It will be created it if it
doesn't exist.
I keep all my loca
On 15/07/2008, at 4:07 AM, R P Herrold wrote:
On Mon, 14 Jul 2008, Peter Dalgaard wrote:
Rolf Turner wrote:
Can anyone give me a hint as to why it doesn't work when the the
executable
script is invoked directly (from tcsh)? Other than the fact that
I'm
using a Mac :-) .
[Invoking unna
Hi Eli
I have "inherited" a cDNA macroarray dataset that is structured as follows.
Three different stressors were tested. For each stressor, there are two
treatments (control and stressed). For each treatment, two biological
replicates exist, and these are paired (i.e., there is a stressed arr
icebreake web.de> writes:
>
>
> Thank's for the answer Marcelo. You are right, I'm working on microarrays
> (Affymetrix).
> I know the two packages you mentioned, but I have absolute know idea which
> function I could use for my problem.
Dear Icebreake,
I'm a biologists and I didn't have kno
On 15/07/2008, at 6:34 AM, Susan Amrose wrote:
I'm trying to source a file automatically every time I start R. I
tried
adding the following .First function in a file Rprofile.site in my
$R_HOME/etc/ directory (verified $R_HOME by Sys.getenv()) as well
as in a
file .Rprofile in my $HOME dire
Write a function that opens a new graphics window (quartz in your case) then
changes the parameters to what you want as the default. Then any time you
want/need a new graphics window, call your function.
Also look at ?options and scroll down to 'device' and ?Startup
Hope this helps,
--
Gregor
On 14/07/2008, at 7:15 PM, Peter Dalgaard wrote:
Odd... That _does_ actually work for me, so I guess the issue is
Mac-specific. One thought:
Tcl is like R; a core language where you load packages in order to
do any real work and wish is just Tcl with the Tk package
preloaded. I.
I'm trying to source a file automatically every time I start R. I tried
adding the following .First function in a file Rprofile.site in my
$R_HOME/etc/ directory (verified $R_HOME by Sys.getenv()) as well as in a
file .Rprofile in my $HOME directory and .Rprofile in the working directory:
.First <-
Thank you very much for all messages and answers, it's ok now :)
Charles C. Berry wrote:
>
> On Sun, 13 Jul 2008, Fran100681 wrote:
>
>>
>> Thank you, but this is not what i want exactly.. i would want to launch
>> function "myfun" with this script:
>>
>>> table <- sample(LETTERS[1:5], 20,TRU
on 07/14/2008 01:39 PM Stephan Kolassa wrote:
Phil:
thanks a lot! However, my original problem[1] remains: in fact, I am
trying to \Sexpr the result for Sweave, and for some reason Sweave
regurgitates the \Sexpr command unchanged:
foo.Rnw contains:
\Sexpr{"DV~IV"}
\Sexpr{cat("DV~IV")}
\Sexpr
Duncan,
thank you, that works perfectly! And a big thank you also to all others
who helped! I'm back to Sweaving!
Best wishes
Stephan
Duncan Murdoch schrieb:
On 7/14/2008 2:39 PM, Stephan Kolassa wrote:
Phil:
thanks a lot! However, my original problem[1] remains: in fact, I am
trying to
This question has already been answered on R-devel.
In any case, the correct procedure is discussed in the 'R Installation and
Administration Manual', and does not involve 'cygwin'.
On Mon, 14 Jul 2008, stephenb wrote:
I am trying to load a dll compiled with the latest cygwin into R 2.7.0
R
On 7/14/2008 2:39 PM, Stephan Kolassa wrote:
Phil:
thanks a lot! However, my original problem[1] remains: in fact, I am
trying to \Sexpr the result for Sweave, and for some reason Sweave
regurgitates the \Sexpr command unchanged:
foo.Rnw contains:
\Sexpr{"DV~IV"}
\Sexpr{cat("DV~IV")}
\Sexpr{
Greetings,
I have "inherited" a cDNA macroarray dataset that is structured as follows.
Three different stressors were tested. For each stressor, there are two
treatments (control and stressed). For each treatment, two biological
replicates exist, and these are paired (i.e., there is a stressed a
Phil:
thanks a lot! However, my original problem[1] remains: in fact, I am
trying to \Sexpr the result for Sweave, and for some reason Sweave
regurgitates the \Sexpr command unchanged:
foo.Rnw contains:
\Sexpr{"DV~IV"}
\Sexpr{cat("DV~IV")}
\Sexpr{sub('~','$sim$',"DV~IV")}
\Sexpr{cat(sub('
on 07/14/2008 01:09 PM Stephan Kolassa wrote:
Dear guRus,
I am trying to replace "~" by "$\sim$" for TeX. However, I can't get the
backslash to work. I would like to turn "DV~IV" into "DV$\sim$IV".
sub("~","$\sim$","DV~IV") => "DV$sim$IV"
sub("~","$\\sim$","DV~IV") => "DV$sim$IV"
sub("~","$\\
Dear guRus,
I am trying to replace "~" by "$\sim$" for TeX. However, I can't get the
backslash to work. I would like to turn "DV~IV" into "DV$\sim$IV".
sub("~","$\sim$","DV~IV") => "DV$sim$IV"
sub("~","$\\sim$","DV~IV") => "DV$sim$IV"
sub("~","$\\\sim$","DV~IV") => "DV$sim$IV"
sub("~","$si
I am trying to load a dll compiled with the latest cygwin into R 2.7.0
R version 2.7.0 (2008-04-22)
Copyright (C) 2008 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
I have:
Sys.getenv("Path")
Good afternoon
New to R ... new to the list.
I have installed R Commander 1.2-9 and it functions perfectly. I would,
however, like to upgrade my Rcmdr to version 1.3-15 ... but I can't seem to
shake the 1.2 version. Do you have any tips on how to upgrade?
Thanks for your time and considerat
Your x values look to be multiples of two, therefore you might
be interested in using log values for the x axis.
x = c(1,2,4,8)
y = c(1,2,3,4)
plot(x, y, log="x", xaxt='n')
axis(1, at=x, labels=x)
__
R-help@r-project.org mailing list
https://stat.ethz
can you provide commented, minimal, self-contained, reproducible
code. Exactly what statements are you using. Normally I would use
'split' to partition the data and then 'lapply' to work on each of the
groups. An idea of what type of processing you have tried and what
your expectation of output
How about:
plot.default(factor(x), y, xaxt='n', xlab='x')
axis(1, at=factor(x), labels=x)
- Original Message -
From: "Jason Liao" <[EMAIL PROTECTED]>
To:
Sent: Monday, July 14, 2008 10:00 AM
Subject: [R] A question about using function plot
Hello, everyone! I have spent two hou
What about making x a factor?
Sarah
On Mon, Jul 14, 2008 at 11:00 AM, Jason Liao <[EMAIL PROTECTED]> wrote:
> Hello, everyone! I have spent two hours to get what I wanted in a simple
> situation and have not been successful. The set up is extremely simple
>
> x = c(1,2,4,8)
> y = c(1,2,3,4)
> pl
Thank's for the answer Marcelo. You are right, I'm working on microarrays
(Affymetrix).
I know the two packages you mentioned, but I have absolute know idea which
function I could use for my problem.
--
View this message in context:
http://www.nabble.com/Tissue-specific-genes-by-ANOVA--tp184412
On 7/14/2008 11:00 AM, Jason Liao wrote:
Hello, everyone! I have spent two hours to get what I wanted in a simple
situation and have not been successful. The set up is extremely simple
x = c(1,2,4,8)
y = c(1,2,3,4)
plot(x, y)
What I need, however, is for the 4 points of 1,2,4,8 to be spaced ev
Try:
library(lattice)
xyplot(y ~ factor(x))
or
plot(1:length(x), y, xaxt='n')
axis(1, at=1:length(x), labels=x)
On Mon, Jul 14, 2008 at 12:00 PM, Jason Liao <[EMAIL PROTECTED]> wrote:
> Hello, everyone! I have spent two hours to get what I wanted in a simple
> situation and have not been suc
On Mon, 14 Jul 2008, Peter Dalgaard wrote:
Rolf Turner wrote:
Can anyone give me a hint as to why it doesn't work when the the executable
script is invoked directly (from tcsh)? Other than the fact that I'm
using a Mac :-) .
[Invoking unname -a gives ``Darwin TURNER-ROLF.local 8.11.1 Darwin
Hello, everyone! I have spent two hours to get what I wanted in a simple
situation and have not been successful. The set up is extremely simple
x = c(1,2,4,8)
y = c(1,2,3,4)
plot(x, y)
What I need, however, is for the 4 points of 1,2,4,8 to be spaced evenly, not
by their numerical scale. Also,
Dear R-help list,
for a package that uses Rmpi calls and a namespace,
I need to define a .Last() function to call mpi.finalize() - directly
before R quits.
The hooks like on .onUnload are not useful in this case, because they
are not executed when the R session is finished.
The defined .L
?unlist
On Mon, Jul 14, 2008 at 11:49 AM, <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I am writing a very long program that deals with multiple large databases
> which
> often have missing data cells and other quirks. I've been testing it on
> small
> samples of the data so it doesn't take too long, but
> From: Marc Schwartz [mailto:[EMAIL PROTECTED]
> Sent: Friday, July 11, 2008 4:54 PM
>
> on 07/11/2008 02:02 PM Woolner, Keith wrote:
> >> From: Marc Schwartz [mailto:[EMAIL PROTECTED]
> >> Sent: Friday, July 11, 2008 12:14 PM
> >>
> >> on 07/11/2008 10:50 AM Woolner, Keith wrote:
> >>> Hi everyo
try at=seq(from=26, to=32, by=2) in the axis statment
hope this helps
Stephen
On Mon, Jul 14, 2008 at 11:21 AM, Zroutik Zroutik <[EMAIL PROTECTED]> wrote:
> Dear R-users,
>
> I'm tackling with a problem which causing me a head-ache for a long time. I
> would like to create a nice x-axis to my pl
Hi Caroline,
>> is.na(strptime("19810329012000",format="%Y%m%d%H%M%S"))
>> [1] TRUE
>> The problem was to do with daylight saving time. I need to specify a
>> time zone as this time >> doesn't exist in my operating system's current
>> time zone. I still think this is odd behaviour >> though!
Hi,
I am writing a very long program that deals with multiple large databases which
often have missing data cells and other quirks. I've been testing it on small
samples of the data so it doesn't take too long, but when I tried it using the
real files, it ran for about a minute and then gave me th
Dear R-users,
I'm tackling with a problem which causing me a head-ache for a long time. I
would like to create a nice x-axis to my plots, but I do not know how to
implement the method.
Imagine a matrix where you have rownames real numbers -- these rownames
should be written in the x-axis nicely.
icebreake web.de> writes:
> Hello,
>
> unfortunately I have I big problem I can't solve.
(...)
> Is there any possibility to do that? ANOVA? Duncan? Constrasts?
>
Hi Icebreake,
Did you try maanova or limma package from www.bioconductor.org?
I suppose that you are on microarray experiment. If
On 7/14/2008 11:03 AM, Erin Hodgess wrote:
Hi R People:
Could someone point out a reference for creating macros in R, please?
Thanks,
Erin
See Programmer’s Niche: Macros in R by Thomas Lumley:
http://www.r-project.org/doc/Rnews/Rnews_2001-3.pdf
--
Chuck Cleland, Ph.D.
NDRI, Inc. (www.ndri
Hi R People:
Could someone point out a reference for creating macros in R, please?
Thanks,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: [EMAIL PROTECTED]
__
R-help
Hello,
Looking over the swap_tail macro in pnorm.c, the comment inside the code
indicates it is to swap the values of '*cum' and '*ccum'. According to the
code below that is taken from the source file, the swap functionality is
dependent on what values are provided for 'x' and 'lower'.
#define s
Hi,
I have a matrix that is indexed by groups and looks something like this:
1 1 2
1 2 1
.
.
.
2 1 1
2 1 2
2 NA 1
.
.
.
3 1 NA
3 2 NA
etc.
The first column is the group variable and I would like to apply categorical
data imputation functions to the other two columns, doing so by groups. I have
Dear users,
is anybody able to reproduce the following behaviour?
I have two copies of my project with the .rhistory files in a directory with
and without ampersand
a) d:\calc\foo1 & foo2 foo3\
b) d:\calc\foo1 - foo2 foo3\
I'm running "rscript .rhistory" in the dir. At the case a) I'm getting a
f
I'm trying to compile R, not getting pass the configure step.
R version is 2.7.1 from source tarball.
Machine is HPUX 11.31, 32-CPUs of Itanium II.
Compilers are HPUX C, C++ and Fortran.
(B3910B A.06.15, May 2007)
My config string:
./configure --prefix=/apps/INTEL/R --enable-R-shlib CC=/opt/aC
Hi
[EMAIL PROTECTED] napsal dne 14.07.2008 14:37:36:
> Many thanks for the recent answers to my question about pattern
recognition.
> The hint to "grep" and the "pattern" argument brought me a big deal
forward.
>
> # Now, I have a data frame:
>
> DATE <- c("1930-01-01", "1930-01-01", "1930-02
Try this:
aggregate(ex$FREQ, list(Year = substr(ex$DATE, 1, 4)), sum)
or
aggregate(ex$FREQ, list(Year = format(as.Date(ex$DATE), "%Y")), sum)
On Mon, Jul 14, 2008 at 9:37 AM, Oehler, Friderike (AGPP)
<[EMAIL PROTECTED]> wrote:
> Many thanks for the recent answers to my question about pattern
Many thanks for the recent answers to my question about pattern recognition.
The hint to "grep" and the "pattern" argument brought me a big deal forward.
# Now, I have a data frame:
DATE <- c("1930-01-01", "1930-01-01", "1930-02-01", "1931-01-01",
"1931-02-01", "1931-03-01") # almost all months u
On 7/14/2008 7:18 AM, Surendra champanerkar wrote:
Dear users
i m not able to find stats4 package to calculate mle
plz give me link for sttats4 package if possible
library(stats4) should do it. It's a base package, i.e. part of R.
Duncan Murdoch
Try this:
pk[rep(1, 4), ]
On Mon, Jul 14, 2008 at 4:33 AM, fernanda lopez <[EMAIL PROTECTED]> wrote:
> Dear all,
> I want to
> write ck<-cbind(rep(pk[1,1],4),rep(pk[1,2],4),rep(pk[1,3],4)) state ment
> in a loop . How can I write it ?
>
>
>
>
>
>> pk
> [,1] [,2]
You can copy the code below to your R-code editor. For Yule's Q, the data is
expected in two vectors. For cramer's phi, the data is expected in separate
columns of a matrix or dataframe.
##Run this code
yule.Q=function(x,y){(table(x,y)[1,1]*table(x,y)[2,2]-table(x,y)[1,2]*table(x,y)[2,1])/(table(
Hi Henrique,
Many thanks. That seems to work for individual numbers passed in as the
first argument.
Would anyone know of a way to get things to print this way by default all
the time in the GUI ? Setting a big.mark option through options does not
seem to do it.
> prettyNum("100.00", big
You can use this for print:
prettyNum("100.00", big.mark=",")
Perhaps you can use a print method for this.
On Mon, Jul 14, 2008 at 8:50 AM, <[EMAIL PROTECTED]> wrote:
> Dear R Users,
>
> I would like the R GUI to display numbers with comma separators, i.e.
>
> 1,000,000.00
>
> instead of
>
Dear R Users,
I would like the R GUI to display numbers with comma separators, i.e.
1,000,000.00
instead of
100
Is there a flag in option() for this behaviour. A google/RSiteSearch has
not revealed anything but perhaps I have missed it.
Thanks in advance,
Tolga
Generally, this communica
On Jul 14, 2008, at 3:03 AM, Megh Dal wrote:
Hi,
following data is taken from http://www.economagic.com/em-cgi/
data.exe/var/west-texas-crude-long. Problem with this data is when
you copy it from this site you would get something like that :
1946 063331.27
1946 079991.27
1946 087771.52
Dear users
i m not able to find stats4 package to calculate mle
plz give me link for sttats4 package if possible
--
Shubham
Surendra
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Hello,
I have an error since I run my R code. This error is :
Fehler in eval.with.vis(expr, envir, enclos) :
>
My code is:
#CONDITION1 : check if the right logfile is chosen
c1=log(z[,3],n)
if (c1==FALSE) {
plot(0:1,0:1,type = "n", axes=FALSE,xlab="",ylab="")
text(0.4,0.8,adj=0,paste("FEHLER"
Hello,
unfortunately I have I big problem I can't solve.
I have to analyse if a gene is tissue specific. For example for the gene xyz
I have following expression values:
HeartLiver Brain
8.998497 10.013561 12.277407
9.743556 10.137574 11.033957
For every tissue I have
Hello everybody,
I've a multi-level structured data.frame.
I = Individual Level (factor)
G = Group Level (factor)
SW= Software (factor)
I G SW
1 1 A
2 1 A
3 2 B
Im looking for a frequncy-talbe of the used software on the group level.
In this case the result has to be:
A = 0.5
B = 0.5
Is ther
Dear R-Users,
I need functions to calculate Yule's Y or Cramérs Index, in order to
correlate variables that are nominally scaled?
Am I wrong? Are such functions existing?
Sincerely,
Timo
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>>> Ted Harding <[EMAIL PROTECTED]> 14/07/2008 00:16 >>>
>said:
>What constitutes "reasonable doubt" can become a very interesting
>question, but there are some crimes for which it has a definite
>statistical interpretation
Warning for potential courtgoers: "reasonable doubt" NEVER has a direct
Hi all,
I've been trying to fit a mixed effects model and I've been having problems.
=>My aim:
to know whether states atributes, political parties and individual atributes
affect the electoral results of men and women candidates.
I use candidates as replications for states and for politi
On Mon, 14 Jul 2008, Gabor Csardi wrote:
Maybe there is a simpler way, but this works fine:
l1 <- 1
l2 <-2
m <-10
ls()
[1] "l1" "l2" "m"
rm(list=grep("^l.*", ls(), value=TRUE))
ls()
[1] "m"
You can supply a regular expression to grep.
And as the 'pattern' argument of ls(). So followi
Hi, do you mean this?
pk<-matrix(rnorm(12),nrow=4,ncol=3)
ck1<-cbind(rep(pk[,1],4),rep(pk[,2],4),rep(pk[,3],4)) ## or
ck2<-cbind(rep(pk[,1],each=4),rep(pk[,2],each=4),rep(pk[,3],each=4))
pk
ck1
ck2
best,
Daniel
fernanda lopez wrote:
>
> Dear all,
> I want to
> write ck<-cbi
is.na(strptime("19810329012000",format="%Y%m%d%H%M%S"))
[1] TRUE
The problem was to do with daylight saving time. I need to specify a time zone
as this time doesn't exist in my operating system's current time zone. I still
think this is odd behaviour though! When you look at the missing obje
Maybe there is a simpler way, but this works fine:
> l1 <- 1
> l2 <-2
> m <-10
> ls()
[1] "l1" "l2" "m"
> rm(list=grep("^l.*", ls(), value=TRUE))
> ls()
[1] "m"
>
You can supply a regular expression to grep.
Gabor
On Mon, Jul 14, 2008 at 10:45:13AM +0200, Oehler, Friderike (AGPP) wrote:
> Dea
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