Peter Tillmann peter.tillm...@t-online.de writes:
can anyone give an example how to use cross-validation in the plsr package.
There are examples in the references cited on
http://mevik.net/work/software/pls.html
I miss to find the number of factors proposed by cross-validation as
optimum.
Dear all,
does anyone know if it is possible to connect a Windows RExcel
instance to a linux R instance?
Within Rexcel, I find the option Remote Server Address, but I
wonder what the installation procedure on my linux (ubuntu) R looks
like (if possible at all)?
Thanks
Philipp
This solution such as data-na.omit(original database) before you
run step() or stepAIC() has some limitations, I think. I reduced the
number of data lines, and it enhance R square value.
If you have some tips or advices for another solution, I welcome.
Kum
Urban and Regional Planning, GRI
On
Kjaere Bjørn-Helge,
can anyone give an example how to use cross-validation in the plsr
package.
There are examples in the references cited on
http://mevik.net/work/software/pls.html
I miss to find the number of factors proposed by cross-validation as
optimum.
The
Hi
r-help-boun...@r-project.org napsal dne 22.02.2010 08:49:01:
hi everyone,
i am new to R project can anyone please help me by providing
documents
If you already succeeded to install R it has installed into some
directory. Use means of your operating system (maybe Windows?) to find
chinna wrote:
I connected to database and i am accessing the tables but i dont know how
to generate graphs from the database tables.
can anyone please help me
i am new to R project
Hi,
Take a look at the plot() command. Or you can have a look at the lattice
or ggplot2 packages. The
Hi all,
I'm announcing this version of plotrix because I have made a couple of
changes that will affect users. Although the function that used to be
known as hierobarp is fairly new, there may be some users who are
surprised when they find that it has changed its name to barNest. I will
How do I wait for a single character input without terminating Enter?
Following Brian Ripley on
http://markmail.org/message/ptmbkhdfnpnf5zcd
In general R does not have a character-by-character interface with a
keyboard.
Dieter
__
I'm running R 2.10.1 on a GNU/Linux Debian Squeeze (testing)
I type :
help.start()
I get :
starting httpd help server ... Terminé.
Si 'xdg-open' est déjà lancé, il ne sera pas relancé, et vous devez
passer à sa fenêtre.
Sinon, soyez patient...
I think the translation is :
If
Hello,
Having the matrix d
d
value value2 class
1 1 1 x
2 2 2 x
3 3 3 x
4 4 2 x
5 5 1 y
6 11 3 y
7 12 4 z
8 13 5 z
9 14 6 z
1015 7 z
I want to calculate the
Hi Jim,
I just went thorough the demo(plotrix) , what amazing work you have done
there, thank you!
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) |
Hi:
Here are three ways, but there are others:
# (1)
# package doBy:
library(doBy)
summaryBy(value + value2 ~ class, data = df)
class value.mean value2.mean
1 x2.5 2.0
2 y8.0 2.0
3 z 13.5 5.5
# (2) aggregate():
with(df,
Hi,
On R 2.10.1 for Windows, when I do the following to duplicate the
structure of a large numeric matrix called matrix1:
matrix2 - matrix(0,nrow=nrow(matrix1),ncol=ncol(matrix1))
and then
rownames(matrix2) - rownames(matrix1)
I get a cannot allocate vector of size xxMb error
but if I
assigning rownames (after the object is created) triggers a copy of
the object... if you assign the rownames at creation time, no extra
copies...
b
On Mon, Feb 22, 2010 at 9:46 AM, Larson, TR t...@york.ac.uk wrote:
Hi,
On R 2.10.1 for Windows, when I do the following to duplicate the
Hi Wartan,
Not really an answer to your specific problem, but you could try to use
pdf instead of eps, in combination ofcourse with pdflatex. I've never
had problems with pdf nad Sweave. If you need to use eps, than my reply
is of no help to you :).
cheers,
Paul
Wartan Hachaturow wrote:
Thanks,
These functions solve 1 problem ( the mean).
I tried to use them also for the cov matrix but I didn't succed.
Is there any way to calculate the cov matrix for groups?
--
View this message in context:
MY REQUIREMENT:
I have a database and i want to connect to the database through R Project
and using the tables from the database i need to generate the reports.
can anyone plese tell me whether this is possible with R project or not.
Thanks in advance
Chinna.
--
View this message in context:
List Requirements:
PLEASE do read the posting guide [1]http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
With that said.. We need a *bit* more information What database are you
connecting to? MySQL? MSSQL? Are you
we r using teradata database and through R Project we have to access tables
from teradata database and we need to generate graphs using that
dataand also we need forecasted results of that data
...can please tell me the possible ways.
Thanks in advance
Regards
Chinna
--
View this
Hi,
I got the error message:
Error in cmdscale(1 - similarity, k = 1) :
'k' must be in {1, 2, .. n - 1}
It turned out that similarity was just a scalar and therefore n-1 = 0 :(
Maybe the error message could be improved.
Bset Sigbert
__
Once I suggested to BMDP to introduce a module-statement that would
direct the syntax to the specified module (1L, 2L, ...), so that all
syntax could reside in one job, but they did not like that idea.
Heinz
At 14:55 19.02.2010, Terry Therneau wrote:
I used both BMDP and SAS in my earlier
Hello David,
I wanted to thank you for pointing to this code for performing posthoc
friedman test. I just used this example to create a function to perform this
test (and also added some illustrative flexible plots to the mix), here:
Try this:
lapply(list(mean, cov), by, data = d[,1:2], INDICES = d[,'class'])
On Mon, Feb 22, 2010 at 6:13 AM, mirauta bmira...@yahoo.com wrote:
Hello,
Having the matrix d
d
value value2 class
1 1 1 x
2 2 2 x
3 3 3 x
4 4 2 x
5
Dear R team,
As this is a university email address with very small inbox limit, can I
please change my R email subscription to another email address instead of
this one? My new email address is wendy2.q...@gmail.com. I have change my
email in my profile, but seems that does not change my
Dear all,
I intend to do some statistics with my raster table, like correlations or
regression analysis. For that, I need that each pixel be in one line, that
is, there is no count 1. So, all columns should have the same number of
lines, when I compare (combine) rasters.
I'd like to know if
I don't think this information can be found in the documentation, but you can
always just check the actual influence.measures() and print.infl() code to find
out. Most importantly, influence.measures() incldues the following code:
function (model)
{
is.influential - function(infmat, n) {
chinna wrote:
hi everyone,
i am new to R project can anyone please help me by providing documents
my goal is using R i have to connect to the database and i have to generate
reports.
Thanks in advance
chinna.
R is a complex program. If you can't work out how to find the documentation, I
On Feb 21, 2010, at 10:48 PM, Wartan Hachaturow wrote:
Hello.
I'm writing some simple text using sweave, and faced a strange problem
with eps files produced for my plots (one example attached).
Individual eps files are interpreted by ghostscript just fine, and
show up without errors. But
I sent an email to current maintainer Vaidotas Zemlys 2 weeks ago but
no response yet.
Here is a mod, it's an unfinished version. Please have a look, thanks.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read
I wonder if someone can give some pointers on alternatives to linear
regression (e.g. Loess) when dealing with multiple variables.
Taking any simple table with three variables, you can very easily get the
intercept and coefficients with:
summary(lm(read_table))
For obvious reasons, the
Hi!
You should first go to the home page of the R project and read the
manuals available there (and there are a lot).
When you'll understand how R works, read the posting guide and ask
specific questions.
I don't think you'll get answers if you have such imprecise questions.
Ivan
Le 2/22/2010
One vote goes to Dave's answer!
You made (saved) my day
detlef
On Mon, 22 Feb 2010 13:07:23 +
Dr. David Kirkby david.kir...@onetel.net wrote:
chinna wrote:
hi everyone,
i am new to R project can anyone please help me by providing documents
my goal is using R i have to connect
The devel version of odfWeave has some multicolumn capabilities. these
were added by Patrick Lenon and Zekai Otles recently, but I have not
had a chance to test it out.
You can go to:
https://r-forge.r-project.org/projects/odfweave/
and download the version there if you want to evaluate the
The cross-validation in the pls package does not propose a number of
factors as optimum, you have to select this yourself. (The reason for
this is that there is AFAIK no theoretically founded and widely accepted
way of doing this automatically. I'd be happy to learn otherwise.)
The caret
Thanks for the pointer, Detlef, added to the devel-version on R-Forge.
Z
On Mon, 22 Feb 2010, Detlef Steuer wrote:
One vote goes to Dave's answer!
You made (saved) my day
detlef
On Mon, 22 Feb 2010 13:07:23 +
Dr. David Kirkby david.kir...@onetel.net wrote:
chinna wrote:
hi everyone,
Chinna,
I found that there are ODBC drivers for the teradatabase available, so with
the use of RODBC you should be able to connect to your database.
This will allow you to extract your data into an R dataframe, and make
reports of it.
But you noticed in another thread that you are an R
Hello,
I need to have a kaufmaennisches Runden function.
Is there already something like that?
It means: rounding up the 5, instead of rounding it down.
So, 245.455 would give 245.46
I found no option for this.
Maybe there is a package for it?
Oliver
round(245.455, 2) ?
On Mon, Feb 22, 2010 at 11:06 AM, Oliver oli...@first.in-berlin.de wrote:
Hello,
I need to have a kaufmaennisches Runden function.
Is there already something like that?
It means: rounding up the 5, instead of rounding it down.
So, 245.455 would give 245.46
I found
The function prediction() returns this:
Formal class 'performance' [package ROCR] with 6 slots
..@ x.name : chr Cutoff
..@ y.name : chr Accuracy
..@ alpha.name : chr none
..@ x.values:List of 1
.. ..$ : Named num [1:89933] Inf 2.23 2.22 2.17 2.16 ...
.. .. ..- attr(*,
On Feb 22, 2010, at 6:40 AM, Simone R. Freitas wrote:
Dear all,
I intend to do some statistics with my raster table, like
correlations or
regression analysis. For that, I need that each pixel be in one
line, that
is, there is no count 1. So, all columns should have the same
number of
On Feb 21, 2010, at 1:37 PM, wendy wrote:
Dear all,
I am analysising a set of gene expression data using multiplicative
Gaussian
Kernel function. I calculated the kernels between two gene sets with
2
genes for each set. The kernels are less than 1, therefore the
product of
the
Hi,
I would like to test the normality of errors in an
split-plot design using R.
I used the following program:
anava = aov(ganhos ~ Blocos + Trat*Supl +
Error(Blocos/Trat))
summary(anava)
bartlett.test(ganhos, Trat)
bartlett.test(ganhos, Supl)
How can I test the normality of the errors?
Try this:
names(o...@x.values)
On Mon, Feb 22, 2010 at 11:05 AM, Jay josip.2...@gmail.com wrote:
The function prediction() returns this:
Formal class 'performance' [package ROCR] with 6 slots
..@ x.name : chr Cutoff
..@ y.name : chr Accuracy
..@ alpha.name : chr none
..@
Oh strange...
round(108.275 , 2)
[1] 108.28
round(208.275 , 2)
[1] 208.28
round(308.275 , 2)
[1] 308.27
looks not like what one should expect...
R version 2.9.2 (2009-08-24)
Ciao,
Oliver
__
R-help@r-project.org mailing list
On Feb 22, 2010, at 8:05 AM, Jay wrote:
The function prediction() returns this:
Formal class 'performance' [package ROCR] with 6 slots
..@ x.name : chr Cutoff
..@ y.name : chr Accuracy
..@ alpha.name : chr none
..@ x.values:List of 1
.. ..$ : Named num [1:89933] Inf 2.23
On 22/02/2010 9:06 AM, Oliver wrote:
Hello,
I need to have a kaufmaennisches Runden function.
Is there already something like that?
It means: rounding up the 5, instead of rounding it down.
So, 245.455 would give 245.46
I found no option for this.
Maybe there is a package for it?
This is
same result with R2.10.1 on Windows...
Le 2/22/2010 15:34, Oliver a écrit :
Oh strange...
round(108.275 , 2)
[1] 108.28
round(208.275 , 2)
[1] 208.28
round(308.275 , 2)
[1] 308.27
looks not like what one should expect...
R version 2.9.2
Hi,
On Mon, Feb 22, 2010 at 7:27 AM, Wendy Qiao wendy.q...@utoronto.ca wrote:
Dear R team,
As this is a university email address with very small inbox limit, can I
please change my R email subscription to another email address instead of
this one? My new email address is
Hello,
which packages are you talking about?
And... are thoise packages using integer-based calculations?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Hi Amy,
On Sat, Feb 20, 2010 at 9:29 PM, Amy Hessen amy_4_5...@hotmail.com wrote:
Hi ,
Could you please help me in this question:?
After trying this code:
library(e1071)
mydata - as.matrix(read.delim(iris.txt))
train.x - mydata[,-1]
train.y - mydata[,1]
mymodel - svm(train.x,
Hi all,
I take a simple dataset like this:
-1.3
1
-1.5
-1
1.5
-2.5
3
-0.5
Now I want to count how many values there are between -2 and 2.
The answer here would be 6.
Can anyone do this with R?
Thank you in advance.
Gr. Bosken
--
View this message in context:
Hello Sarah, thanks for answering
For example if I have the following example
test - as.data.frame(matrix(c(1,2,3,4, 11,12,13,14, a,b,b,c), nrow
= 3, ncol=3,dimnames = list(c(r1,r2,r3,r4),NULL))
V1 V2 V3
r1 1 11 a
r2 2 12 b
r3 3 13 b
r4 4 14 c
it is easy to select test -
Dear all,
I want to run the VECM BEKK model, but I cannot find the corresponding
package to run this model. Anybody can help?
Thanks a lot
Ted
--
View this message in context:
http://n4.nabble.com/How-to-run-the-VECM-BEKK-model-in-R-tp1564555p1564555.html
Sent from the R help mailing list
Dieter Menne wrote:
Peter Dalgaard wrote:
d - data.frame(f=c(rare, medium,well-done))
#To get the cast in order of appearance, this can be used:
d$f - factor(d$f, levels=unique(d$f))
d$f
[1] rare mediumwell-done
Levels: rare medium well-done
.. which caused some
Thank you, Wolfgang! Now that I know what the function does I can at
least search some literature to learn about those criteria.
Thanks,
Frank Tamborello
On Feb 22, 2010, at 7:00 AM, Viechtbauer Wolfgang (STAT) wrote:
I don't think this information can be found in the documentation,
but
On Feb 22, 2010, at 8:45 AM, statquant wrote:
Hello Sarah, thanks for answering
For example if I have the following example
test - as.data.frame(matrix(c(1,2,3,4, 11,12,13,14,
a,b,b,c), nrow
= 3, ncol=3,dimnames = list(c(r1,r2,r3,r4),NULL))
This is a malformed example. Should be:
test
Your attachment did not get though. Here is what did:
https://stat.ethz.ch/pipermail/r-help/2010-February/229132.html
On Mon, Feb 22, 2010 at 8:11 AM, zhuojun chen uifid...@gmail.com wrote:
I sent an email to current maintainer Vaidotas Zemlys 2 weeks ago but
no response yet.
Here is a mod,
Hi,
ab-read.table(header=F,file=textConnection(-1.3
1
-1.5
-1
1.5
-2.5
3
-0.5))
length(which(ab =-2 ab=2))
Bosken schrieb:
Hi all,
I take a simple dataset like this:
-1.3
1
-1.5
-1
1.5
-2.5
3
-0.5
Now I want to count how many values there are between -2 and 2.
The answer here would be 6.
On 22/02/2010 9:57 AM, Oliver wrote:
Hello,
which packages are you talking about?
And... are thoise packages using integer-based calculations?
I don't use this, but the ones I'd look in would be:
MASS (function rational())
rcdd
Rmpfr
You can use
RSiteSearch(rational)
to look for more.
thank you for reply. I just type: hist(x) from SSH terminal, expecting a
histogram to pop up like what i got under windows.instead I got the
following error msg:
Error in X11(d$display, d$width, d$height, d$pointsize, d$gamma,
d$colortype, :
unable to start device X11cairo
In addition:
Hi!
22.02.2010 17:45, xin wei wrote:
thank you for reply. I just type: hist(x) from SSH terminal, expecting a
histogram to pop up like what i got under windows.instead I got the
following error msg:
Error in X11(d$display, d$width, d$height, d$pointsize, d$gamma,
d$colortype, :
xin wei wrote:
thank you for reply. I just type: hist(x) from SSH terminal, expecting a
histogram to pop up like what i got under windows.instead I got the
following error msg:
Error in X11(d$display, d$width, d$height, d$pointsize, d$gamma,
d$colortype, :
unable to start device
Uwe, you're right, it turns out that cairo hadn't been previously installed
on the system.
Andrew
2010/2/19 Uwe Ligges lig...@statistik.tu-dortmund.de
On 19.02.2010 17:44, Andrew Yee wrote:
I was wondering if someone could help with the antialias option in tiff().
I'm running R 2.9.2 on
On 20.02.2010 08:51, Dan Edgcumbe wrote:
I'm trying to do some confirmatory factor analysis on some data. My SEM
model solves in 22 iterations, but when I try to look at the modification
indices, using mod.indices, I get the following error message:
Error in solve.default(hessian) :
system
Duncan Murdoch murdoch at stats.uwo.ca writes:
[...]
MASS (function rational())
rcdd
Rmpfr
[...]
OK, I will look for those packages.
Maybe they will help.
Thank you.
__
R-help@r-project.org mailing list
Hi,
Here is a suggeston:
x - c(-1.3, 1, -1.5, -1, 1.5, -2.5, 3, -0.5)
sum(x =-2 x=2)
[1] 6
HTH,
Jorge
On Mon, Feb 22, 2010 at 9:49 AM, Bosken wrote:
Hi all,
I take a simple dataset like this:
-1.3
1
-1.5
-1
1.5
-2.5
3
-0.5
Now I want to count how many values there are
See the footer of your message.
Uwe Ligges
On 22.02.2010 13:27, Wendy Qiao wrote:
Dear R team,
As this is a university email address with very small inbox limit, can I
please change my R email subscription to another email address instead of
this one? My new email address is
On 20.02.2010 19:10, Nancy Adam wrote:
Hi all
I receive this error message: “Error in mymodel$MSE : $ operator is invalid
for atomic vectors”
When I use KNN.
This is the code:
library(e1071)
train- rbind(iris3[,,1], iris3[,,2], iris3[,,3])
cl- factor(c(rep(s,50), rep(c,50), rep(v,50)))
Dear list,
I have two vectors:
x - c(one,two)
y - paste(rep(x,2),blah)
I want to replace all occurrences of each element of x in y with
something else, so that y looks like this:
y
[1] something else blah something else blah something else blah
[4] something else blah
I can do this using a
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of wendy
Sent: Saturday, February 20, 2010 6:58 PM
To: r-help@r-project.org
Subject: [R] replicate matrix
Hi all,
I have a matrix, for example
[,1] [,2]
[1,] 13
gsub(one|two, something else, y)
?
On Mon, Feb 22, 2010 at 4:25 PM, Marianne Promberger
marianne.promber...@kcl.ac.uk wrote:
Dear list,
I have two vectors:
x - c(one,two)
y - paste(rep(x,2),blah)
I want to replace all occurrences of each element of x in y with
something else, so that y
On Feb 22, 2010, at 10:25 AM, Marianne Promberger wrote:
Dear list,
I have two vectors:
x - c(one,two)
y - paste(rep(x,2),blah)
I want to replace all occurrences of each element of x in y with
something else, so that y looks like this:
y
[1] something else blah something else
And as long as the interval is symmetric about zero:
sum(abs(x) = 2)
[1] 6
-Peter Ehlers
On 2010-02-22 9:18, Jorge Ivan Velez wrote:
Hi,
Here is a suggeston:
x- c(-1.3, 1, -1.5, -1, 1.5, -2.5, 3, -0.5)
sum(x=-2 x=2)
[1] 6
HTH,
Jorge
On Mon, Feb 22, 2010 at 9:49 AM, Bosken wrote:
gsub(paste(x, collapse = |), something else, y)
[1] something else blah something else blah something else blah
[4] something else blah
Many thanks! I didn't know about collapse. Should have thought about
reading up in ?paste ...
Thanks
Marianne
--
Marianne Promberger PhD, King's
I have often found this to happen if the scale of one variable is
orders of magnitude different than the scale of other variables. Have
you tried inspecting the covariance matrix and log transforming any
such variables?
On Feb 22, 2010, at 8:14 AM, Uwe Ligges wrote:
On 20.02.2010
I'm sure there is a clever way to do the following, but I've been unable to
find it on this forum or by writing my own functions. I have 8 years worth
of weekly data but would like to restrict the labels on the x-axis to months
only. I've included the first year's worth of data below.
My line
Hello,
I am trying to plot selected data points to a preexisting persp plot
that satisfies a condition. I used the following statement -
text(coords[,1], coords[,2], names(act[which(act 8.75)]), cex=0.7)
But I get all the points labeled, instead of the points that satisfy
the condition
OKKK
Thanks a lot for letting me know the subset function
Cheers
--
View this message in context:
http://n4.nabble.com/Accessing-values-of-a-matrix-tp1561932p1564724.html
Sent from the R help mailing list archive at Nabble.com.
__
Tahnk you. But, when I try the command you both suggested I get a NULL
as the results.
names(object1 @ x.values)
NULL
Where did I go wrong?
On Feb 22, 4:34 pm, David Winsemius dwinsem...@comcast.net wrote:
On Feb 22, 2010, at 8:05 AM, Jay wrote:
The function prediction() returns this:
Hello,
I have a very long (~50,000) sequence of repeating numbers. The first
100 are:
[1] 0 0 0 0 0 0 0 0 0 0 0 429
[13] 429 429 429 429 429 429 429 858 858 858 858
858
[25] 858 1287 1287 1287 1287 1287 1716
Try this:
rle(x)$length
On Mon, Feb 22, 2010 at 2:27 PM, Larson, TR t...@york.ac.uk wrote:
Hello,
I have a very long (~50,000) sequence of repeating numbers. The first 100
are:
[1] 0 0 0 0 0 0 0 0 0 0 0 429
[13] 429 429 429 429 429
Hi Mohamed,
Please read the posting guides for both, the R-help and BioC mailing
lists, so you learn the basics of how to ask questions on these
lists (and software related mailing lists in general). Following
those guides will greatly increase the chance that someone will
actually be willing
Preeti,
Could you please post a sample of your data and all the code? It is hard to
answer your question without knowing what coords and act are
-
Try http://prettygraph.com Pretty Graph , the easiest way to make R-powered
graphs on the web.
--
View this message in context:
Dear Users,
Consider a multivariate time series model:
a_1*y(t)-...-a_k*y(t-k)=b+[c_1*z(t)-...-c_j*z(t-j)]
i.e., a simple multivariate time series model with one exogenous variable.
I would like to know what package can I use to do the following, using R:
1) Select k and j jointly;
2) Estimate
Hello Hrishi,
The command you suggested plotted the years on the x-axis only. It was the
same exact plot as the one I included in the original post.
Respectfully,
Eric
--
View this message in context:
http://n4.nabble.com/adding-infrequent-date-labels-to-x-axis-tp1564804p1564875.html
Sent
Hi emorway,
It seems to me that all you need is one command:
plot(as.Date(cropped.cast1$date,%Y-%m-%d),cropped.cast1$Frac_ET_Satsfd_mean)
-
Try http://prettygraph.com Pretty Graph , the easiest way to make R-powered
graphs on the web.
--
View this message in context:
hi, Kevin and K.Elo:
thank you for the suggestion. Can you be more specific on these? (like how
exactly get into x-switch or man ssh). I am totally ignorant about linux and
SSH:( Memory limitation forces me to switch from windows to Linux
cluster.
Xin
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View this message in
Dear Jarret, Uwe, and Dan,
Sorry -- I missed the initial question. What's a bit odd here is that the
singularity occurs only in the computation of the modification indices. It
might help to look at the conditioning of the covariance matrix of the
parameter estimates (i.e., the eigenvalues or
Marianne,
The function substring2 from the Hmisc packages works nicely for me to
do this (without explicit gsub, though):
x- c(one,two)
y- paste(rep(x,2),blah)
y
[1] one blah two blah one blah two blah
substring2(y,x)- something else
y
[1] something else blah something else blah something
Dne Po 22. února 2010 18:53:55 xin wei napsal(a):
hi, Kevin and K.Elo:
thank you for the suggestion. Can you be more specific on these? (like how
exactly get into x-switch or man ssh). I am totally ignorant about linux
Hello,
in Linux You can plot as in Windows, for example using hist(x). But
Hi!
22.02.2010 19:53, xin wei wrote:
hi, Kevin and K.Elo:
thank you for the suggestion. Can you be more specific on these? (like how
exactly get into x-switch or man ssh). I am totally ignorant about linux and
SSH:( Memory limitation forces me to switch from windows to Linux
You can try the locfit package, which I believe can handle up to 5
variables. E.g.,
R library(locfit)
Loading required package: akima
Loading required package: lattice
locfit 1.5-6 2010-01-20
R x - matrix(runif(1000 * 3), 1000, 3)
R y - rnorm(1000)
R mydata - data.frame(x, y)
R str(mydata)
K. Elo wrote:
Hi!
22.02.2010 19:53, xin wei wrote:
hi, Kevin and K.Elo:
thank you for the suggestion. Can you be more specific on these? (like how
exactly get into x-switch or man ssh). I am totally ignorant about linux and
SSH:( Memory limitation forces me to switch from windows to
Hi emorway,
You are right. I tried the command only with one year's data, in which case
it plots month names as labels. But for multiple years, the labels become
years. I found this old post useful -
http://n4.nabble.com/Month-tick-marks-on-a-plot-td879121.html#a879121.
Picking the code from the
Hello Hrishi,
That worked great, and in the process I learned some new ways of going about
writing R code. Thank you very much for helping me out!
Eric
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View this message in context:
http://n4.nabble.com/adding-infrequent-date-labels-to-x-axis-tp1564804p1564943.html
Sent from the R help
Here are a few possibilities using gsubfn in the gsubfn package
x - c(one,two)
y - paste(rep(x,2),blah)
library(gsubfn)
# 1
gsubfn(\\w+, w ~ if (w %in% x) something else else w, y)
# 2
gsubfn(\\w+, list(one = something else, two = something else), y)
# 3
L - sapply(x, function(...) something
Hi,
You just need to install X11 windows software on windows, such as
xming, and start xming before ssh -X use...@remotehost
Hope this helps.
On 2/22/10, xin wei xin...@stat.psu.edu wrote:
thank you for reply. I just type: hist(x) from SSH terminal, expecting a
histogram to pop up like
Hello,
I have a large matrix (foo), I have to split by weeks and in which measures
start at different points of time.
Now I am looking for the number of the week, in which a specific value
appears. I do not need the date or any other information from a row, but
(due to structure of result
Dear all,
I am trying to test goodness of fit. I assume that a data follow Poisson or
Negative binomial distribution. I can test the goodness of fit in case of no
truncated data. However, I could not find any good function or packages when
a data is truncated.
For example, a frequency table
Dear R-help members,
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for the R User Conference, useR! 2010, is one week away:
Submission deadline: Monday, March 1, 2010
The deadline for early registration is also Monday, March 1, 2010.
I encourage you all to
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