Hi, is there a function to calculate fractional ranks?
Thanks
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal,
Thank you for your very fast response.
I just tried to use the zoo package, after having read the vignettes, but I get
this error message:
Warning messages:
1: In x$DAY : $ operator is invalid for atomic vectors, returning NULL
2: In x$EARNINGS :
$ operator is invalid for atomic vectors,
It works perfectly, thank you so much.
Now I will try to put teh results into a suitable form (a data frame like this):
SEC_ID.xSEC_ID.yEARN_COV
Thank you again
Angelo Linardi
-Messaggio originale-
Da: Patrick Burns [mailto:[EMAIL PROTECTED]
Inviato: giovedì 5 giugno
On Fri, 6 Jun 2008, Rick Bilonick wrote:
I'm using Suse Linux Enterprise Desktop 10.2 (SP2) on an HP 2133 (x86)
mini-notebook. (There apparently are a LOT of bugs in 10.1!) I
downloaded R-base from the openSuse 10.2 repository and was (finally)
able to install it (after installing blas and
Hi all,
Suppose I want to read a text file with read.table.
It containt lines to be skipped that begins with ! and ^.
Is there a way to include this two values in the read.table function?
I tried this but doesn't seem to work.
dat - read.table(mydata.txt, comment.char = c(!,^) , na.strings
=
thank you ! :)
2008/6/5 Rolf Turner [EMAIL PROTECTED]:
On 6/06/2008, at 1:08 AM, biologeeks wrote:
dear all,
in the package pwr , there is the fonction power.anova.test which permit
to
obtain the power for a one-way ANOVA...but I'm looking for a way to
compute
the power of a multiway
I would like to generate a graphics text. I have a 67x2 table with
5-character string in col 1 and 2-character string in col 2.
Is it possible to make such a table appear on a graphics or a
message-box pop-up window ?
Thank you so much.
--
Maura E.M
Hi,
I'm not sure why you think glm doesn't provide goodness of fit tests.
Have a look at anova.glm and summary.glm. All the functions you
mention can deal with multiple predictors. multinom deals with
non-binary data. lrm will deal with ordinal data as well as binary.
polr (in the MASS
Hi,
I haven´t much experience on writing functions and would like to
modify the simple tbrm() function from package dplR in order to save
the weights that it produces. I have tried using the superassignment
operator as explained in the R intro, but is this the right way to
save a
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
library(lattice)
## works as expected
xyplot(1~1, key = list(text = list(c(Maenner
## works as expected
xyplot(1~1, key = list(text = list(c(Maenner))), xlab = M\344nner)
## gives an error
xyplot(1~1, key = list(text = list(c(M\344nner
Is
Hello list,
I was trying to select a column of a data frame using the *which* command. I
was actually selecting the rows of the data frame using *which, *and then
displayed a certain column of it. The command that I was using is:
sequence=*mydata*[*which*(human[,3] %in% genes.sam.names),*9*]
In
hadley wickham írta:
2008/5/27 Mihalicza Péter [EMAIL PROTECTED]:
Dear List and Hadley,
I would like to have a boxplot with ggplot2 and have the outlier values
labelled with their name attribute. So I did
library(ggplot2)
dat=data.frame(num=rep(1,20), val=c(runif(18),3,3.5),
Thanks for the quick reply David,
so far this sums up to:
# logistic on binary data
lrm combined with resid(model,'gof')
# logistic on raw binary data
glm with gof using anova.glm
(i think that anova.glm only makes sence on grouped binary data, not on the raw binary data..)
(so what is the
i a have transformed my data to data frame named df with only column names(no
rownames).each column represnt one sample with 3 observations (in deed
nrow(df)=3, and ncol(df)=92).in order to check homoskedasticity of variance
of my 92 samples i do:
bartlett.test(df)
it work and give me a
Dear R experts,
I am currently facing a tricky problem which I have read a lot about in
the various R mailing lists without finding exactly what I need.
I have a big data frame DF (about 2,000,000 rows) with 7 columns being
variables and 1 being a measure (using reshape package nomeclature).
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Bernd Weiss schrieb:
| library(lattice)
|
| ## works as expected
| xyplot(1~1, key = list(text = list(c(Maenner
|
| ## works as expected
| xyplot(1~1, key = list(text = list(c(Maenner))), xlab = M\344nner)
|
| ## gives an error
| xyplot(1~1, key
Eleni Christodoulou elenichri at gmail.com writes:
I was trying to select a column of a data frame using the *which* command. I
was actually selecting the rows of the data frame using *which, *and then
displayed a certain column of it. The command that I was using is:
Karin Lagesen wrote:
I know this is fairly basic, but I must have somehow missed it in the
manuals.
I have two vectors, often of unequal length. I would like to compare
them for identity. Order of elements do not matter, but they should
contain the same.
I.e: I want this kind of comparison:
jpardila wrote:
Dear List,
I am creating a plot and I want to insert the tabular data below the X axis.
I mean for every value of X I want to show the value in Y as a table below
the plot. I think the attached image gives an idea of what I mean by this.
Below is the code i am using now... but
Well, you failed to give the 'at a minimum information' asked for in the
posting guide, and \344 is locale-specific. I see 'MingW32' below, so
will guess this is German-language Windows. We don't know what the error
was, either.
It works correctly for me in CP1252 with R-patched, and gives
Bernd Weiss bernd.weiss at uni-koeln.de writes:
library(lattice)
## gives an error
xyplot(1~1, key = list(text = list(c(M\344nner
Is this a bug?
You forgot to mention your version, assuming 2.7.0 unpatched.
Corrected by Brian Ripley in developer version (and probably also in patched)
Hi All,
Newbie question for you all but i have been looking at the archieves and the
help dtuff to get a rough idea of what i want to do
I would like to merge two dataframes together based on a keyed variable in
one dataframe linking to the other dataframe. Only some of the cases will
match but
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Prof Brian Ripley schrieb:
[...]
| It works correctly for me in CP1252 with R-patched, and gives an error
| in 2.7.0 (and works in 2.6.2). I think it was fixed as side effect of
|
| oRare string width calculations in package grid were not
Hi all!
So far I learned some R but finilizing my plots so they look
publishable seems not to be possible.
I set up some boxplots. Everything works well but when I put more then
two of them in one plot the labels of the axes appear smaller than the
normal font size.
x - rnorm(30)
y -
if I wanted to use a name for a column with two words say Dick Cheney and
George Bush
can I put these in quotes Dick Cheney and George Bush to get them to
read into R using both read.table and read.zoo to recognize this.
thanks
Stephen
--
Let's not spend our time and resources thinking about
according to the helpfile, comment only takes one character, so you'll
have to do some 'magic' :)
i'd suggest to first run mydata through sed, and replace one of the
comment chars with another, then run read.table with the one comment
char that remains.
sed -e 's/^\^/!/' mydata.txt
should work - don't even have to put them in quotes, if your field
separator is not space. why don't you just try it and see what comes out? :)
on 06/06/2008 08:43 AM stephen sefick said the following:
if I wanted to use a name for a column with two words say Dick Cheney and
George Bush
can I
Update your version of zoo to the latest one.
On Fri, Jun 6, 2008 at 3:18 AM, [EMAIL PROTECTED] wrote:
Thank you for your very fast response.
I just tried to use the zoo package, after having read the vignettes, but I
get this error message:
Warning messages:
1: In x$DAY : $ operator is
try this:
FullData - merge(ETC, SURVEY, by.x = ord, by.y = uid, all.x = T,
all.y = F)
on 06/06/2008 07:30 AM Michael Pearmain said the following:
Hi All,
Newbie question for you all but i have been looking at the archieves and the
help dtuff to get a rough idea of what i want to do
I would
Dear R users
I have a very basic question. I tried but could not find the required result.
using
dat - pima
f - table(dat[,9])
f
0 1
500 268
i want to find that class say 0 having maximum frequency i.e 500. I used
which.max(f)
which provide
0
1
How can i get only the 0. Thanks and
On 6/6/2008 9:14 AM, Muhammad Azam wrote:
Dear R users
I have a very basic question. I tried but could not find the required result.
using
dat - pima
f - table(dat[,9])
f
0 1
500 268
i want to find that class say 0 having maximum frequency i.e 500. I used
which.max(f)
which provide
The 0 is the name of the item and the 1 is the index in f of the maximum
class. (since f is a table, and the first element of the table is the
maximum, which.max returns a 1) So, if you just want to know which class
is maximum you can say
names(which.max(f))
Michael Conklin
Chief
names(f)[which.max(f)]
on 06/06/2008 09:14 AM Muhammad Azam said the following:
Dear R users
I have a very basic question. I tried but could not find the required result.
using
dat - pima
f - table(dat[,9])
f
0 1
500 268
i want to find that class say 0 having maximum frequency i.e
Run the sessionInfo() command in R, as the posting guide requests!
Jason Lee wrote:
Hi,
I am not too sure its what you meant :-
Below is the closest data for each session from top
PID USER PR NI VIRT RES SHR S %CPU %MEMTIME+ COMMAND
26792 jason 25 0 283m 199m 2620 R 100
On 6/6/2008 9:18 AM, Chuck Cleland wrote:
On 6/6/2008 9:14 AM, Muhammad Azam wrote:
Dear R users
I have a very basic question. I tried but could not find the
required result. using
dat - pima
f - table(dat[,9])
f
0 1 500 268
i want to find that class say 0 having maximum frequency
Hello R-users,
I have a very simple problem I wanted to solve. I have a large dataset as
such:
Lag X.Symbol Time TickType ReferenceNumber Price Size X.Symbol.1
Time.1 TickType.1 ReferenceNumber.1
1 ES 3:ESZ7.GB 08:30:00B74390987 151075 44
3:ESZ7.GB08:30:00 A
Use aggregate() for aggregation and use indexing or subset() for selection.
Alternately try the sqldf package: http://sqldf.googlecode.com
which allows one to perform SQL operations on data frames.
On Fri, Jun 6, 2008 at 6:12 AM, [EMAIL PROTECTED] wrote:
Dear R experts,
I am currently facing
I want to take the first row of each unique ID value from a data frame.
For instance
ddTable -
data.frame(Id=c(1,1,2,2),name=c(Paul,Joe,Bob,Larry))
I want a dataset that is
Id Name
1 Paul
2 Bob
unique(ddTable)
Will give me all 4 rows, and
unique(ddTable$Id)
Will give me
Dear Maura,
try the function textplot from the package gplots. you can say
textplot(yourmatrix) and get a plot of a character matrix.
On Fri, 6 Jun 2008, Maura E Monville wrote:
I would like to generate a graphics text. I have a 67x2 table with
5-character string in col 1 and 2-character
Risposta automatica dal 06/06/08 fino al 14/06/08
I'm going to have limited access to my email untill the 14th of june 2008
Avrò accesso limitato all'email fino al 14 giugno 2008
[[alternative HTML version deleted]]
__
R-help@r-project.org
On 6/6/2008 9:35 AM, Emslie, Paul [Ctr] wrote:
I want to take the first row of each unique ID value from a data frame.
For instance
ddTable -
data.frame(Id=c(1,1,2,2),name=c(Paul,Joe,Bob,Larry))
I want a dataset that is
Id Name
1 Paul
2 Bob
unique(ddTable)
Will give me all 4
Good point. Thanks
On Fri, Jun 6, 2008 at 9:05 AM, Daniel Folkinshteyn [EMAIL PROTECTED]
wrote:
should work - don't even have to put them in quotes, if your field
separator is not space. why don't you just try it and see what comes out? :)
on 06/06/2008 08:43 AM stephen sefick said the
I don't have R on this machine but will this work.
myrows - unique(ddTable[,1])
unis - ddTable(myrows, ]
--- On Fri, 6/6/08, Emslie, Paul [Ctr] [EMAIL PROTECTED] wrote:
From: Emslie, Paul [Ctr] [EMAIL PROTECTED]
Subject: [R] Subsetting to unique values
To: r-help@r-project.org
Received:
An example is:
symbol=human[which(human[,3] %in% genes.sam.names),8]
The data* human* and *genes.sam.names* are attached. The result of the above
command is:
symbol
[1] CCL18 MARCO SYT13
[4] FOXC1 CDH3
[7] CA12 CELSR1
Emslie, Paul [Ctr] emsliep at atac.mil writes:
I want to take the first row of each unique ID value from a data frame.
For instance
ddTable -
data.frame(Id=c(1,1,2,2),name=c(Paul,Joe,Bob,Larry))
I want a dataset that is
IdName
1 Paul
2 Bob
unique(ddTable)
Will give
Try and make sure that R is in your windows Path variable
I got your message when I first did this, but when I did the about it
then worked...
==
Please access the attached hyperlink for an important electronic
Colleagues,
Several days ago, I wrote to the list about a lengthy delay in startup
of a a script. I will start with a brief summary of that email. I
have a 10,000 line script of which the final 3000 lines constitute a
function. The script contains time-markers (cat(date()) to that I can
I didn't get any attached data, but my suspicion here is that you have somehow
got RefSeq IDs in column 8 of human, as well as the gene symbols. Did you read
this data in from a text file?
Eleni Christodoulou wrote:
An example is:
symbol=human[which(human[,3] %in% genes.sam.names),8]
The
cool. :) yea, the argument names are by.x and by.y, so your by.etc were
ignored in the black hole of arguments passed to other methods
on 06/06/2008 09:11 AM Michael Pearmain said the following:
Thanks
Works perfectly.
Was the problem due to me putting by.survey and by.etc rather than by.y
Hi,
When i do the next line it work fine:
fit.spherical(var, 0, 2.6, 250, type='c', iterations=10,
tolerance=1e-06, echo=FALSE, plot.it=T, weighted=TRUE, delta=0.1,
verbose=TRUE)
But, i use the next and send one error:
fit.variogram(spherical, var, nugget=0, sill=2.6, range=250,
plot.it=TRUE,
Hello,
I have the next function call:
lme(fixed=Error ~ Temperature * Tumour ,random = ~1|ID, data=error_DB)
which returns an lme object. I am interested on carrying out some kind
of lsmeans on the data returned, but I cannot find any function to do
this in R. I'have seen the effect()
Anybody have any thoughts on this? Please? :)
on 06/05/2008 02:09 PM Daniel Folkinshteyn said the following:
Hi everyone!
I have a question about data processing efficiency.
My data are as follows: I have a data set on quarterly institutional
ownership of equities; some of them have had
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
It's too obvious, so I am positive that there is a good reason for not doing
this, but still:
why is it not possible, to have an outlier output in stat_boxplot that can
be used at geom_text()?
Something like this, with upper:
dat=data.frame(num=rep(1,20), val=c(runif(18),3,3.5),
One thing that is likely to speed the code significantly
is if you create 'result' to be its final size and then
subscript into it. Something like:
result[i, ] - bestpeer
(though I'm not sure if 'i' is the proper index).
Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
Try reading the posting guide before posting.
On Fri, Jun 6, 2008 at 11:12 AM, Daniel Folkinshteyn [EMAIL PROTECTED] wrote:
Anybody have any thoughts on this? Please? :)
on 06/05/2008 02:09 PM Daniel Folkinshteyn said the following:
Hi everyone!
I have a question about data processing
Hi list,
Is it possible to save the name of a filename automatically when
reading it using read.table() or some other function?
My aim is to create then an output table with the name of the original
table with a suffix like _out
example:
mydata = read.table(Run224_v2_060308.txt, sep = \t,
Hi all,
Does anyone know where to download the BRugs package? I did not find
it on r-project website. Thanks.
NL
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R-help@r-project.org mailing list
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PLEASE do read the posting guide
i did! what did i miss?
on 06/06/2008 11:45 AM Gabor Grothendieck said the following:
Try reading the posting guide before posting.
On Fri, Jun 6, 2008 at 11:12 AM, Daniel Folkinshteyn [EMAIL PROTECTED] wrote:
Anybody have any thoughts on this? Please? :)
on 06/05/2008 02:09 PM Daniel
Is there a way to set up a regression in R that forces two coefficients
to be equal but opposite in sign?
I'm trying to setup a model where a subject appears in a pair of
environments where a measurement X is made. There are a total of 5
environments, one of which is a baseline. But each
Its summarized in the last line to r-help. Note reproducible and
minimal.
On Fri, Jun 6, 2008 at 12:03 PM, Daniel Folkinshteyn [EMAIL PROTECTED] wrote:
i did! what did i miss?
on 06/06/2008 11:45 AM Gabor Grothendieck said the following:
Try reading the posting guide before posting.
On
That is the last line of every message to r-help.
On Fri, Jun 6, 2008 at 12:05 PM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
Its summarized in the last line to r-help. Note reproducible and
minimal.
On Fri, Jun 6, 2008 at 12:03 PM, Daniel Folkinshteyn [EMAIL PROTECTED]
wrote:
i did!
well, where are you getting the filename in the first place? are you
looping over a list of filenames that comes from somewhere?
generally, for concatenating strings, look at function 'paste':
write.table(myoutput, paste(myfilename,_out.txt, sep=''),sep=\t)
on 06/06/2008 11:51 AM DAVID ARTETA
You can write your own function, something about like this:
read.table2 - function(file, ...)
{
x - read.table(file, ...)
attributes(x)[[file_name]] - file
return(x)
}
mydata - read.table2(Run224_v2_060308.txt, sep = \t, header = TRUE)
myfile - attr(x, file_name)
On Fri, Jun 6, 2008 at 12:51
Hello,
I am trying to perform a fit.contrast() on a lme object with this code:
attach(error_DB)
model_temperature - lme(Error ~ Temperature, data = error_DB,random=~1|ID)
summary(model_temperature)
fit.contrast(model_temperature, Temperature, c(-1,1), conf.int=0.95 )
detach(error_DB)
but I got
Dear NL.
BRugs is available from the CRAN extras repository hosted by Brian Ripley.
install.packages(BRugs)
should install it as before (for R-2.7.x), if you have not changed the
list of default repositories.
Best wishes,
Uwe Ligges
Nanye Long wrote:
Hi all,
Does anyone know where to
Perhaps this was too big a question, so I'll ask something shorter:
I have fit a linear model, and want to use its prediction intervals to
calculate the sum of many individual predictions.
1) Some of the lower prediction intervals are negative, which is
non-sensical. Should I just set all
Hi,
I want to reorder the colors given by rainbow(7) so that the last half
move to the first 4.
For example:
ci=rainbow(7)
ci
[1] #FFFF #FFDB00FF #49FF00FF #00FF92FF #0092
#4900
[7] #FF00DBFF
I would like #FFFF #FFDB00FF #49FF00FF to be at the end of
ci, and the rest to be at
The package has a plot() method for random-effects meta-analyses as well,
either those produced by meta.DSL or meta.summaries.
There are examples on the help page for meta.DSL.
-thomas
On Tue, 27 May 2008, Shi, Jiajun [BSD] - KNP wrote:
Dear all,
I could not draw a forest plot
Hi,
I am new to R and i am looking for a way to extract a subset from a
vector.
I have a vector of number oscillating around zero (a decreasing
autocorrelation function) and i would like to extract only the first
positive part of the function (from zero lag to the lag where the function
On Fri, 6 Jun 2008, Neil Gupta wrote:
Hello R-users,
I have a very simple problem I wanted to solve. I have a large dataset as
such:
Lag X.Symbol Time TickType ReferenceNumber Price Size X.Symbol.1
Time.1 TickType.1 ReferenceNumber.1
1 ES 3:ESZ7.GB 08:30:00B74390987
i thought since the function code (which i provided in full) was pretty
short, it would be reasonably easy to just read the code and see what
it's doing.
but ok, so... i am attaching a zip file, with a small sample of the data
set (tab delimited), and the function code, in a zip file (posting
Dear Dani,
I intend at some point to extend the effects package to linear and
generalized linear mixed-effects models, probably using lmer() rather
than lme(), but as you discovered, it doesn't handle these models now.
It wouldn't be hard, however, to do the computations yourself, using
the
thanks for the tip! i'll try that and see how big of a difference that
makes... if i am not sure what exactly the size will be, am i better off
making it larger, and then later stripping off the blank rows, or making
it smaller, and appending the missing rows?
on 06/06/2008 11:44 AM Patrick
ci = rainbow(7)[c(4:7, 1:3)]
on 06/06/2008 01:02 PM avilella said the following:
Hi,
I want to reorder the colors given by rainbow(7) so that the last half
move to the first 4.
For example:
ci=rainbow(7)
ci
[1] #FFFF #FFDB00FF #49FF00FF #00FF92FF #0092
#4900
[7] #FF00DBFF
I
I think the posting guide may not be clear enough and have suggested that
it be clarified. Hopefully this better communicates what is required and why
in a shorter amount of space:
https://stat.ethz.ch/pipermail/r-devel/2008-June/049891.html
On Fri, Jun 6, 2008 at 1:25 PM, Daniel Folkinshteyn
just in case, uploaded it to the server, you can get the zip file i
mentioned here:
http://astro.temple.edu/~dfolkins/helplistfiles.zip
on 06/06/2008 01:25 PM Daniel Folkinshteyn said the following:
i thought since the function code (which i provided in full) was pretty
short, it would be
One simple way is to do something like:
fit - lm(y ~ I(x1-x2) + x3, data=mydata)
The first coeficient (after the intercept) will be the slope for x1, the slope
for x2 will be the negative of that. This model is nested in the fuller model
with x1 and x2 fit seperately and you can therefore
On Fri, 6 Jun 2008, Nanye Long wrote:
Hi all,
Does anyone know where to download the BRugs package? I did not find
it on r-project website. Thanks.
It is Windows-only, and you download it from 'CRAN (extras)' which is part
of the default repository set on Windows versions of R. So
Ok, sorry about the zip, then. :) Thanks for taking the trouble to clue
me in as to the best posting procedure!
well, here's a dput-ed version of the small data subset you can use for
testing. below that, an updated version of the function, with extra
explanatory comments, and producing an
The interesting thing about R is that there are several ways to skin the
cat; here is yet another solution:
do.call(rbind, by(ddTable, ddTable$Id, function(z) z[1,,drop=FALSE]))
Id name
1 1 Paul
2 2 Bob
On Fri, Jun 6, 2008 at 9:35 AM, Emslie, Paul [Ctr] [EMAIL PROTECTED] wrote:
I want
Thanx Thierry,
Suggestion #1 had no effect.
I have been playing with variants on #2 along the way.
DaveT.
-Original Message-
From: ONKELINX, Thierry [mailto:[EMAIL PROTECTED]
Sent: June 6, 2008 04:02 AM
To: Thompson, David (MNR); hadley wickham
Cc: r-help@r-project.org
Subject: RE: [R]
That is going to be situation dependent, but if you
have a reasonable upper bound, then that will be
much easier and not far from optimal.
If you pick the possibly too small route, then increasing
the size in largish junks is much better than adding
a row at a time.
Pat
Daniel Folkinshteyn
On Fri, 6 Jun 2008, Luca Mortarini wrote:
Hi,
I am new to R and i am looking for a way to extract a subset from a
vector.
I have a vector of number oscillating around zero (a decreasing
autocorrelation function) and i would like to extract only the first
positive part of the function (from
Does the difference have something to do with ggplot() using ranges
derived from the data?
When I modify my original 'test' dataframe with two extra rows as
defined below, I get expected results in both versions.
Order shouldn't matter - and if it's making a difference, that's a
bug. But I'm
Cool, I do have an upper bound, so I'll try it and how much of a
speedboost it gives me. Thanks for the suggestion!
on 06/06/2008 02:03 PM Patrick Burns said the following:
That is going to be situation dependent, but if you
have a reasonable upper bound, then that will be
much easier and not
Please read the help for par(mfrow)! AFAICS this is nothing to do with
boxplot().
In a layout with exactly two rows and columns the base value
of 'cex' is reduced by a factor of 0.83: if there are three
or more of either rows or columns, the reduction factor is
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Patrick Burns
Sent: Friday, June 06, 2008 12:04 PM
To: Daniel Folkinshteyn
Cc: r-help@r-project.org
Subject: Re: [R] Improving data processing efficiency
That is going to be situation dependent, but
On Fri, Jun 6, 2008 at 2:28 PM, Greg Snow [EMAIL PROTECTED] wrote:
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Patrick Burns
Sent: Friday, June 06, 2008 12:04 PM
To: Daniel Folkinshteyn
Cc: r-help@r-project.org
Subject: Re: [R] Improving data
Hello
Is there exists a package for multivariate random forest, namely for
multivariate response data ? It seems to be impossible with the
randomForest function and I did not find any information about this
in the help pages ...
Thank you for your help
Bertrand
Hi,
I have a simple question. If I have a table and I want to have the mean
for each row, how can I do?!
Es:
c1 c2 c3 mean
1 12 13 14 ??
2 15 24 10 ??
...
Thanks,
Marco
__
The path to R/bin is in the Windows PATH variable. Yet I get this
error.
On Jun 6, 10:37 am, Dumblauskas, Jerry [EMAIL PROTECTED]
suisse.com wrote:
Try and make sure that R is in your windows Path variable
I got your message when I first did this, but when I did the about it
then worked...
I have installed R (D)COM on a (windows) machine that is part of a windows
domain. if I run the test file in a local (log into this machine)
administrative account it works fine. If I run the test file on a domain
account with administrative rights it will not connect to the server, even
is I
Hello
Is there exists a package for multivariate random forest, namely for
multivariate response data ? It seems to be impossible with the
randomForest function and I did not find any information about this
in the help pages ...
Thank you for your help
Bertrand
My guess is that number 2 is closest to the mark.
Typing too fast is unfortunately not one of my
habitual attributes.
Gabor Grothendieck wrote:
On Fri, Jun 6, 2008 at 2:28 PM, Greg Snow [EMAIL PROTECTED] wrote:
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On
Hello
Is there exists a package for multivariate random forest, namely for
multivariate response data ? It seems to be impossible with the
randomForest function and I did not find any information about this
in the help pages ...
Thank you for your help
Bertrand
Dear all;
I'm planning to install Linux on my computer to run R (I'm bored of
W..XP). However, I haven't used Linux before and I would appreciate,
if possible, suggestions/comments about what could be the best option
install, say Fedora, Ubuntu or OpenSuse which to my impression are the
most
-Original Message-
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Sent: Friday, June 06, 2008 12:33 PM
To: Greg Snow
Cc: Patrick Burns; Daniel Folkinshteyn; r-help@r-project.org
Subject: Re: [R] Improving data processing efficiency
On Fri, Jun 6, 2008 at 2:28 PM, Greg Snow
TABLE-matrix(data=c(12,13,14,15,24,10),byrow=T,nrow=2,ncol=3)
TABLE
[,1] [,2] [,3]
[1,] 12 13 14
[2,] 15 24 10
apply(TABLE,1,mean)
[1] 13.0 16.3
Chunhao
Quoting Marco Chiapello [EMAIL PROTECTED]:
Hi,
I have a simple question. If I have a table and I want to have
OK,
The original ggplot() construct (below) on the following two dataframes
(test1, test2) generate different outputs, which I have attached. The
output that I expect is that shown in test2.png. My expectations are
that I have set the plotting limits with 'scale_x_continuous(lim = c(0,
360)) +
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