You can use regular expressions:
as.Date(sub('(\\d+)m(\\d+)','\\1-\\2-01',dts,perl=TRUE))
but as.Date isn't as inflexible as you think:
as.Date(paste(dts,'m01',sep=''),'%Ym%mm%d')
- Phil Spector
Statistical Computing
Dear Cristopher,
I have a question and hope you may help me...I designed a model with a
data set, the model
Call:
lm(formula = outcom ~ vari1 + vari2 + dummy1 + dummy2)
Residuals:
Min 1Q Median 3Q Max
-0.7090 -0.2648 0.1466 0.2167 0.7513
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(In
First, package the new values of your predictor variables into a data frame:
data.frame(VAR1=8, VAR2=64)
Then, use the predict() method to apply the model to the predictor values:
predict(model, newdata=data.frame(VAR1=8, VAR2=64))
Finally, tell the predict() method that these are
hi: i tried regular expressions and , as usual, failed. but below does
do the job uglily using strsplit. i'd still be curious and appreciate if
someone could do the regular expression method. thanks.
dts <- c("1990m12", "1992m8") #March 1990 and Aug 1992
#SPLIT IT USING m
temp <- strsplit(dts,
Hello,
If I have a character string like
d <- c("1990m3", "1992m8") #March 1990 and Aug 1992
what is the easiest way to convert it into any standard date form; for
example,
d <- c("01/03/1990", "01/08/1992")
I looked at as.Date but it doesn't seem to address my problem as I have an
"m" stuck
Your question has already been answered but as an aside
note that the zoo package has a yearqtr class that can
be helpful when dealing with quarterly dates.
On Wed, Dec 17, 2008 at 10:25 AM, Boriss wrote:
> Dear all,
>
> I have an array of strings
>
> a <- c("2008q3","2005q1","2004q3")
>
> I woul
Remove the comma in the line with the error.
On Wed, Dec 17, 2008 at 11:24 AM, wrote:
> Dear R Users,
>
> I am trying to do something quite simple: replace the elements of a zoo
> object. For some reason, the following code does not seem to work. How can
> I replace the value for the 14th of Dec
Its not clear what the result would be but you may be able to use read.table.
Try this:
Lines <- "19 c:/data/WF-100/2008/20080911/trk/20080911.013115.007.17.txt
10 s name of program that wrote this file trkplt name of program that
wrote this file
10 GORDON machine that generated this file
On 17/12/2008 9:47 PM, Duncan Murdoch wrote:
On 17/12/2008 8:56 PM, Peter Dalgaard wrote:
Ben Bolker wrote:
I can get the errors to happen on Ubuntu 8.10 with R --vanilla (*without*
valgrind) -- but
editing momfit.r line 742 so that plot.progress=FALSE seems to make the
problem go away. (Thi
On 17/12/2008 8:56 PM, Peter Dalgaard wrote:
Ben Bolker wrote:
I can get the errors to happen on Ubuntu 8.10 with R --vanilla (*without*
valgrind) -- but
editing momfit.r line 742 so that plot.progress=FALSE seems to make the
problem go away. (This was a lucky guess, it looked like there was
Ben Bolker wrote:
I can get the errors to happen on Ubuntu 8.10 with R --vanilla (*without*
valgrind) -- but
editing momfit.r line 742 so that plot.progress=FALSE seems to make the
problem go away. (This was a lucky guess, it looked like there was
something
odd going on with the plots.)
Ho
I can get the errors to happen on Ubuntu 8.10 with R --vanilla (*without*
valgrind) -- but
editing momfit.r line 742 so that plot.progress=FALSE seems to make the
problem go away. (This was a lucky guess, it looked like there was
something
odd going on with the plots.)
Hope that helps someo
Gundala Viswanath wrote:
>
> Dear Ivar,
>
> How can I extend the limit of "n" size?
>
> When I tried this function with n>= 15, it fails:
>
>> f <- function(bases, n){apply(expand.grid(rep(list(bases),n)), 1, paste,
>> collapse="")}
>> f(c("A", "T", "C", "G"), 15)
> Error in rep.int(rep.int
glenn roberts wrote:
I have got a general problem when applying a function to a dataframe using
the function;
Apply(df,1,myfunct)
Do you mean "apply"? Or does the "Aaply" function come from a separate
package?
Where myfunct has an IF statement in it along the lines of;
If (z == 0)
X
I have got a general problem when applying a function to a dataframe using
the function;
Apply(df,1,myfunct)
Where myfunct has an IF statement in it along the lines of;
If (z == 0)
X = 1
If (z ==0)
Z = 1
I.e. Two If statements
Is there something I am missing or have a just formed the if statem
Dear Ivar,
How can I extend the limit of "n" size?
When I tried this function with n>= 15, it fails:
> f <- function(bases, n){apply(expand.grid(rep(list(bases),n)), 1, paste,
> collapse="")}
> f(c("A", "T", "C", "G"), 15)
Error in rep.int(rep.int(seq_len(nx), rep.int(rep.fac, nx)), orep) :
Peter Dalgaard wrote:
Rolf Turner wrote:
On 18/12/2008, at 11:34 AM, Richard E. Chandler wrote:
Dear R-help,
Like several other subscribers, I have recently encountered a problem
whereby R will execute code apparently correctly and without error,
but any subsequent command will yield "Error:
On 18/12/2008, at 1:09 PM, Peter Dalgaard wrote:
Rolf Turner wrote:
I can confirm that the error occurs.
I downloaded the files and sourced ``ErrorDemo.R''. Doing x <-
rnorm(10)
after doing so triggered the error. Subsequently attempting
traceback()
(or anything else) simply
On 17/12/2008 5:34 PM, Richard E. Chandler wrote:
Dear R-help,
Like several other subscribers, I have recently encountered a problem
whereby R will execute code apparently correctly and without error,
but any subsequent command will yield "Error: bad value" so that R
has to be killed and rest
Rolf Turner wrote:
On 18/12/2008, at 11:34 AM, Richard E. Chandler wrote:
Dear R-help,
Like several other subscribers, I have recently encountered a problem
whereby R will execute code apparently correctly and without error,
but any subsequent command will yield "Error: bad value" so that R
h
On 18/12/2008, at 11:34 AM, Richard E. Chandler wrote:
Dear R-help,
Like several other subscribers, I have recently encountered a problem
whereby R will execute code apparently correctly and without error,
but any subsequent command will yield "Error: bad value" so that R
has to be killed and
In addition, your model statement is odd. Note that within-S factor
Type is tested with both the type I and the type II residuals, whereas
only the latter should be used. Try this model instead:
aov.errs.ae <- aov(TrainErrs ~ idio*Type + Error(Subject/
Type),data=learnDat.ae)
or, for mor
This is what I ended up using:
Data.subset<-data.frame(PlotFinal,YearFinal,BioFinal,ritFinal) ###Subset
of main dataset
Data.length<-sapply(split(Data.subset,PlotFinal),nrow) ### The number
of data points in each plot
Data.sub<-split(Data.subset,PlotFinal) ##Split Data.subset
Dear R-help,
Like several other subscribers, I have recently encountered a problem
whereby R will execute code apparently correctly and without error,
but any subsequent command will yield "Error: bad value" so that R
has to be killed and restarted. We have checked this out with a few
differe
I'm not sure what you are trying to do: modulo arithmetic isn't exactly
splitting a matrix into 2. Or even remotely.
Regardless of the goal, when numbers are shown in quotes, it's a sure
sign that R thinks they are strings, and thus non-number, just as the error
message (not a warning) states.
To
hi i have a huge matrix and want to split it into 2. with the command %%2.
but i get this warning message: *"Error in mmat%%2 : non-numeric argument to
binary operator*"
here's the bit from my matrix:
V1 V2
[1,] "Affymetrix:CompositeSequence:ATH1-
> Referees demand that the line in the KM-curve should be changed to
> dotted at the point where standarerror is <= 10 %. I don't think it's
> a good habit but I urgently need to implement such a thing in R with
> survfit, survplot or another program. They also want numbers at risk
> below the cu
NB: Not reply needed (Ben was extremely helpful!)
I've just started using R last week and am still scratching my head.
I have a data set and want to run a separate regression across each level of
a factor (treating each one separately). The data right now is arranged such
that the value of the f
Dear Hao,
It works for me. Here is my sessionInfo():
> sessionInfo()
R version 2.8.0 Patched (2008-11-08 r46864)
i386-pc-mingw32
locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252
On Wed, 17 Dec 2008, Titan8883 wrote:
Hi all,
I am a new graduate student who is also new to R. I am ok with the basics,
but the problem I am having right now seems beyond what I can do..so I am
looking for advice.
Advice? OK. Here goes.
I would suggest you pull one of the data files into
Could any one help ? I start to learn the glmnet package. I tried with
the example in the manual.
x=matrix(rnorm(100*20),100,20)
y=rnorm(100)
fit1=glmnet(x,y)
When I tried to fit the model, I received the error message:
Error in validObject(.Object) :
invalid class "dgCMatrix" object: row in
On Wed, Dec 17, 2008 at 8:11 AM, Mark Heckmann wrote:
> Dear R-experts,
>
> I have two problems:
>
> PROBLEM (1)
> ---
>
> I want to produce a very small png file (35 x 18 px) that contains a
> histogram without a figure region or margins, only the pure heights.
> In the base graphic syste
On Wed, Dec 17, 2008 at 3:06 AM, Martyn Plummer wrote:
> The artefacts that you see are a normal result of using bitmap graphics
> devices. I have tried to explain these below:
Thanks very much for your explanations, MP; they were quite informative!!
I recognize that others may feel differently
On Wed, 17 Dec 2008 20:07:08 +0100, GOUACHE David
wrote:
argument which I will pass on to subset() somewhere inside my function.
I would use the example of .() function from plyr package in this case:
.<-function (...){
structure(as.list(match.call()[-1]), class = "quoted")
}
myfuncti
The output I would be looking for would be one row for each data file with
columns for each variable, so using a .csv example with a few variables
would be:
-
File_name,date_written,program_ver,data_file_ver,bin_width
20080911
It would be helpful if you could show what the output would be for the
example given. Exactly what are 'values' and what would be the
'headings'. As mentioned before, you can use readLines and then parse
the data you want, but something like Perl might be easier, but it is
hard to tell from the m
This thread may help?
https://stat.ethz.ch/pipermail/r-help/2007-November/145345.html
On Wed, 17 Dec 2008 20:07:08 +0100, "GOUACHE David"
wrote:
> Hello R-helpers,
>
> I'm writing a long function in which I manipulate a certain number of
> datasets. I want the arguments of said function to allo
I usually use Unix tools to process really data beforehand (sed, awk), but
if you want a pure R solution it is usually possible to kludge
something together
with scan() working line by line.
# read a line
# if it contains stuff you aren't interested in, go on to the next line
# if it contains one
I wrote a dirty hack last time I faced this problem, I'll be curious
to see what is the proper way of dealing with the scoping and
evaluation rules.
library(datasets)
myfunction<-function(table, extraction) {
table2<-subset(table,extraction)
return(table2)
}
condition1 <- quote(iris$S
Hi all,
I am a new graduate student who is also new to R. I am ok with the basics,
but the problem I am having right now seems beyond what I can do..so I am
looking for advice. I am trying to pull data from flat ASCII files, but they
do not have a "nice" structure so a simple "read.table" doesn't
Available free for the typing are the functions for the default and
the dataframe methods of subset:
> subset.default
function (x, subset, ...)
{
if (!is.logical(subset))
stop("'subset' must be logical")
x[subset & !is.na(subset)]
}
> subset.data.frame
function (x, subset, se
Many thanks to all responders. It turns out that there is a winbugs
update available, which defines a new version of inprod, inprod2.
Using inprod2 the vectorized code runs in about the same time as the
"scalar" version.
On the print problem: the issue here turns out to be that the arm
pack
Hello R-helpers,
I'm writing a long function in which I manipulate a certain number of datasets.
I want the arguments of said function to allow me to adapt the way I do this.
Among other things, I want my function to have an argument which I will pass on
to subset() somewhere inside my functio
On 18/12/2008, at 4:52 AM, zack holden wrote:
Dear list,
I would like to plot 2 series of numbers with very different ranges/
scales as lines on the same plot. I assumed this is commonly done
and easy, but I have not found any help files (e.g. axis() or
matplot() that show how. I've sear
> [R] Yet another set of codes to optimize
> Daren Tan daren76 at hotmail.com
> Fri Dec 5 03:41:23 CET 2008
>
> I have problems converting my dataset from long to wide format.
Previous
> attempts using reshape package and aggregate function were
unsuccessful
> as they took too long. Apparently, my
zackfire wrote:
>
>
> Dear list,
>
> I would like to plot 2 series of numbers with very different ranges/scales
> as lines on the same plot. I assumed this is commonly done and easy, but I
> have not found any help files (e.g. axis() or matplot() that show how.
> I've searched many old posts
Try this:
x[index(x) == as.Date("2008-12-14")] <- 1
x
On Wed, Dec 17, 2008 at 2:24 PM, wrote:
> Dear R Users,
>
> I am trying to do something quite simple: replace the elements of a zoo
> object. For some reason, the following code does not seem to work. How can
> I replace the value for the
I'm also getting this error with ggplot2_0.8.1 on winXP with R 2.8.0 and R
2.7.1.
hadley wrote:
>
> Hi David,
>
> I inadvertently introduced a bug in ggplot in the last release. I
> uploaded a fix to CRAN this morning and it should be available in the
> near future. Sorry for the inconveni
Dear R-experts,
I have two problems:
PROBLEM (1)
---
I want to produce a very small png file (35 x 18 px) that contains a
histogram without a figure region or margins, only the pure heights.
In the base graphic system this is simple:
png(filename = "hist.png", res = 72, width=35, hei
Dear list,
I would like to plot 2 series of numbers with very different ranges/scales as
lines on the same plot. I assumed this is commonly done and easy, but I have
not found any help files (e.g. axis() or matplot() that show how. I've searched
many old posts to no avail.
I'll be very gra
Gavin:
I think its important to point out two probably obvious things
(1) the dataset is very imbalanced...we have an overabundance of
'analogs==FALSE' roughly 94% of the data. If we think of a purely
non-parametric test of equality of the underlying CDF's then we have
alot of confidence in F0
On 12/17/08, Javier PB wrote:
>
> Dear R-users,
>
> Sorry if someone came out with a similar question but after one day of
> searching I am giving up:
>
> Does anyone know how to plot the original points used in a lm model and the
> set of resulting regression lines generated by the model?
>
On 12/17/2008 12:27 PM, Kirk Wythers wrote:
I have what I hope is a ridiculously simple question. I am trying to
follow an example that uses the function win.graph(), but my machine
does not recognize win.graph(). I am running R 2.8.0 on OS X. Is there
some OS X specific function that replac
I have what I hope is a ridiculously simple question. I am trying to
follow an example that uses the function win.graph(), but my machine
does not recognize win.graph(). I am running R 2.8.0 on OS X. Is there
some OS X specific function that replaces win.graph or am I missing
some package?
If this were copy and paste-able then I could probably give you a
solution, but I would have a look at
?coredata
On Wed, Dec 17, 2008 at 11:24 AM, wrote:
> Dear R Users,
>
> I am trying to do something quite simple: replace the elements of a zoo
> object. For some reason, the following code do
Dear Bernhard
You were right, pushing my knowledge about R once again to the next level!
Sys.getenv("R_SHELL") indeed returns "". The fix you are proposing works for
me. If I am right, your function dvi.latex2 makes use of shell rather than of
sys in the original function dvi.latex and stores r
Dear R Users,
I am trying to do something quite simple: replace the elements of a zoo
object. For some reason, the following code does not seem to work. How can
I replace the value for the 14th of Dec of 2008 in the zoo object x below
with 1 (it is currently NA).
> x
2008-12-11 2008-12-12 2008
?predict.lm
--
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] On Behalf Of Ricardo Gomez
> Sent: Wednesday, December 17, 2008
Dear Luciano,
The "1" in (1|NestID) indicates only a random intercept. Note that in
most models in R, a "1" on the righthandside of the formula indicates
the intercept, "-1" or "0" indicates no intercept. ~X, which is
equivalent to ~X + 1, indicates a slope along X and an intercept. Hence
a random
Greetings,
I'd be grateful if a good Samaritan helps me to approach this problem
with my data, I've created the following model
lm(formula = OUTCOME ~ VAR1 + VAR2)
summary(model)
Call:
lm(formula = OUTCOME ~ VAR1 + VAR2)
Residuals:
Min 1Q Median 3Q Max
-1.4341 -0.3621 0.1879 0.
Hi Boris,
Yes. Try this:
paste('IMT',a,sep="")
See ?paste.
HTH,
Jorge
On Wed, Dec 17, 2008 at 10:25 AM, Boriss wrote:
> Dear all,
>
> I have an array of strings
>
> a <- c("2008q3","2005q1","2004q3")
>
> I would like to add to each a[i], with i = 1,2,3
>
> the following string "IMT",
>
> s
Dear all,
I have an array of strings
a <- c("2008q3","2005q1","2004q3")
I would like to add to each a[i], with i = 1,2,3
the following string "IMT",
such that in the end I could get
b <- c("IMT2008q3","IMT2005q1","IMT2004q3")
Is it possible to accomplish this without a loop command?
Thank
Hello Marco,
as might not be evident at first sight, but have you set the environment
variable "R_SHELL"? If you spot at the dvi method for latex you will find a
call to sys(), which will call shell() and if the argument shell is unset then
the contents of "R_SHELL" will be used. Hence, what do
The boxcox function in the MASS package computes these intervals. If you want
to see how it computes the interval you can look at the source code. MASS the
book also describes the general process.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcar
Look at rotate.wireframe in the TeachingDemos package.
Hope this helps,
--
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] On
Dear R-experts,
Quite new to R on this end, but learning fast (I hope).
I am running version 2.7.1 on Windows Vista. I have small dataset
which consists of:
# NestID: nest indicator for each chicken. Siblings sharing the same nest have
the same nest indicator.
# Chick: chick indicator consis
Hello,
I was wondering if you could tell me how to calculate 95% confidence
intervals for lambda for a box-cox power transformation.
Best wishes
Eoin
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.et
AllenL wrote:
>
> Much thanks! This helped a lot. Another quick one:
> In using the lmList function in the nlme package, is it possible to subset
> my data according to the number of observations in each level? (ie. I
> obviously want to include only those levels in which the observations are
>
Hello all.
I wish to run k-means with "manhattan" distance.
Since this is not supported by the function "kmeans", I turned to the "pam"
function in the "fpc" package.
Yet, when I tried to have the algorithm run with different starting points,
I found that pam ignores and keep on starting the algori
Dear R-users,
Sorry if someone came out with a similar question but after one day of
searching I am giving up:
Does anyone know how to plot the original points used in a lm model and the
set of resulting regression lines generated by the model?
This is how I do it using the "plot" and "lines" f
> In general I try not to post questions to forums until I've tried my best to
> read about them in the available documentation. I recently undertook a
> project that used odfWeave and have been very pleased with the package.
> But, the R help documentation suggests that there are more sophisticat
On Wed, Dec 17, 2008 at 02:09:57PM +0100, Philipp Pagel wrote:
> or maybe
>
> ggregate(foo$V1, by=list(foo$V1), FUN=length)
Oops- that was supposed to be 'aggegate'...
--
Dr. Philipp Pagel
Lehrstuhl für Genomorientierte Bioinformatik
Technische Universität München
Wissenschaftszentrum Weihens
On Wed, Dec 17, 2008 at 03:36:44AM -0800, RON70 wrote:
>
> I have a dataframe with two columns :
>
> 11600 238'4
> 12000 218'0
[...]
> There are repeatations in 1st column and I want to use this as Level. Next I
> want to report for each level how many data points are there, with
> corresponding
Thanks a lot for every answer I got!
I could solve my problem!
Greg, your proposal seems to be quite useful for me :-) Thank you.
Ciao,
Antje
Antje schrieb:
Hi all,
I'd like to know, if I can solve this with a shorter command:
a <- rnorm(100)
which(a > -0.5 & a < 0.5)
# would give me all
Dear Achim,
Thanks for the script. It works fine except it sometimes yields extreme
wide confidence intervals. That is for a factor level with only a few
replications or a level with all zeros. I noticed that the se for those
predictions was Nan. Therefore I've added two lines (marked with #% at
t
Dear listers,
I am using the pcnm function (spacemakeR) to obtain eigenvectors for a
spatial grid of sampling sites. These pcnm eigenvectors are then used in
multivariate ordination to test where community composition follows local
environment or rather shows spatial autocorrelation, and get suppo
Hi,
I have been testing some of the alternative suggested approaches. The best PC
set may not be the best predictors subset, but is that true that it is not
generally the case? If you have to explore data patterns and (potential)
relationships between a response variables and a large set of can
I have a dataframe with two columns :
11600 238'4
12000 218'0
12200 209'0
12600 192'0
13000 176'4
14000 145'0
15000 119'0
16000 103'0
1800080'0
1900068'3
2 59'0
11600 208'1
12000 189'2
12200 180'3
There are repeatations in 1st column and I want to use this as
Dear Tal,
pam is not in the fpc package but in the cluster package. Look at ?pam and
?pam.object to find out what it does.
As far as I see, the medoids in the output object are the final cluster
medoids, not the initial ones, which presumably explains the observed
behaviour.
Best regards,
Ch
Dear Sara,
You could also use the Subbotools package by Giulio Bottazzi and test the
goodness of fit of your data to this distribution.
See:
http://www.lem.sssup.it/WPLem/files/2004-14.pdf
Also, http://cafim.sssup.it/~giulio/software/subbotools/install_cygwin.html
Regards,
José
Mr José Luis
Dear friends of R,
I want to produce a pdf file with the contents of a matrix. I employ the latex
command in combination with dvi, both contained in the Hmisc package. It seems
to me that the function does not correctly set the directory.
> tbl.loc <- matrix(1:4, nc=2)
> latex.obj <- latex(t
The artefacts that you see are a normal result of using bitmap graphics
devices. I have tried to explain these below:
I have looked at your figures in Eye of Gnome, with anti-aliasing turned
off (Menu Edit/Preferences; Tab "Image View"; option "Smooth images when
zoomed"). I recommend that you d
Hi,
If you can formulate your question it in terms of actual problem you have with data.frame it would be easier to answer.
for the time being check subset() if it is what you want.
SV.
On Tue, 16 Dec 2008 11:09:19 +0100, Antje wrote:
Hi all,
I'd like to know, if I can solve this with a sho
see ?system.time
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
On Wed, 17 Dec 2008, Ingmar Visser wrote:
I see, I was afraid for an answer along these lines as my problem now turns
into a stat problem (-;Any suggestions on how to analyze such data as given
You mean, you were afraid that the help page was correct?
Weights can be specified by a 'weigh
Dear all,
What's the common R idiom to do this task?
For example I want to compute the running time
of one function in R
> mylong_running_func(100)
How can I compute the running time of the function
above?
- Gundala Viswanath
Jakarta - Indonesia
__
R
On Wed, 17 Dec 2008, Ingmar Visser wrote:
I see, I was afraid for an answer along these lines as my problem now turns
into a stat problem (-;Any suggestions on how to analyze such data as given
below with weights AND at the same time
taking into account that there are repeated measurements?
It
I see, I was afraid for an answer along these lines as my problem now
turns into a stat problem (-;
Any suggestions on how to analyze such data as given below with
weights AND at the same time
taking into account that there are repeated measurements?
Best, Ingmar
On 17 Dec 2008, at 11:02, Pro
Dear all,
This is off-topic,
however I hope someone can give me useful suggestion..
Given the regression model
y = b0 + b1*x + e
I am interested in testing for positive coeffs, namely
H0: b0>0 AND b1>0
H1: b0,b1 unconstrained
It is simple to estimate the model under H0 and H1 (there are several
On Tue, 16 Dec 2008 16:25:07 +0100, Mark Heckmann wrote:
Hi R-users,
I want to apply a function to each column of a data frame that is
numeric.
colwise(), numcolwise() and catcolwise() in plyr package turn a function that
operates on vectors into one that operates on columns of data fra
On Wed, 17 Dec 2008, Tom Cattaert wrote:
Hi,
I am using the packages tree and rpart to build a classification tree to
predict a 0/1 outcome. The package rpart has the advantage that the function
plotcp gives a visual representation of the cross-validation results with a
horizontal line indicati
Weights are not supported: multistratum aov is designed for balanced
designs and uses projection for which weighting is inappropriate.
On Wed, 17 Dec 2008, Ingmar Visser wrote:
Dear R-help,
I'm facing a problem with defining a repeated measures anova with
weighted data.
Here's the code to rep
To add on Robin Hankin's solution, if you want to generate the strings
you can try:
f <- function(bases, n){apply(expand.grid(rep(list(bases),n)), 1,
paste, collapse="")}
f(c("A", "T", "C", "G"), 2)
f(c("A", "T", "C", "G"), 4)
best
Ivar
Robin Hankin wrote:
Gundala
f <- function(n){expand.
Hi,
I am using the packages tree and rpart to build a classification tree to
predict a 0/1 outcome. The package rpart has the advantage that the function
plotcp gives a visual representation of the cross-validation results with a
horizontal line indicating the 1 standard error rule, i.e. the
recom
On Wed, Dec 17, 2008 at 12:54 AM, Prof Brian Ripley
wrote:
> Your PDF problems indicate a broken viewer. How were you viewing PDF?
You are absolutely correct about the PDF files; I've since checked the
PDF files in other viewers and have not been able to reproduce the
problem. There was definite
Dear R-help,
I'm facing a problem with defining a repeated measures anova with
weighted data.
Here's the code to reproduce the problem:
# generate some data
seed=11
rtrep <- data.frame(rt=rnorm(100),ti=rep(1:5,20),subj=gl
(20,5,100),we=runif(100))
# model with within factor for subjects/repea
Gundala
f <- function(n){expand.grid(rep(list(seq_len(4)),n))}
HTH
Robin
Gundala Viswanath wrote:
Dear all,
Is there an efficient way in R to construct all strings from 4 bases (ATCG).
If we want a length L string, there are 4 ^ L possible strings of such.
e . g with L = 2 we have AA
Stefan Gengenbach schrieb:
Hello all,
I`m a computer science student from Frankfurt/main (germany).
Our student team (10 students) are working on a R/Interface project (Java
and R) and we have a problem with R and JRI.
We are working on 10 computers with Windows XP SP2, R 2.8.0, jdk1.6.0_11
Dear all,
Is there an efficient way in R to construct all strings from 4 bases (ATCG).
If we want a length L string, there are 4 ^ L possible strings of such.
e . g with L = 2 we have AA, AT, AC, AG, .. GC, GA, GT, GG as many as
4 ^ 2 = 16 strings,
with L = 3 we have as many as 4 ^ 3 = 64 str
Your PDF problems indicate a broken viewer. How were you viewing PDF?
You have also not told us how you view PNG, but you would expect
anti-aliased output to be blurry when viewed at 100% (or more). You need
to be careful not to have anti-aliasing turned on in the viewer as well as
in the fi
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