Dear Dr. Dalgaard
Sorry for delay reply..
That's exactly what I was looking for - thanks a lot.
Hsin-Ya
Peter Dalgaard wrote:
>
>
>
> Andrew Robinson wrote:
>> In your data, subject is nested within sequence. Was that your
>> intention?
>>
>>
> Presumably yes. This looks like a standar
You do that by right clicking the R icon on the Windows Desktop
and then choosing Run As Administrator .
Alternately grab el.js and Rgui.bat from the batchfiles distribution --
home page is at:
http://batchfiles.googlecode.com
and place them anywhere in your path and then you can just do this
xy.labels is used to control text by the individual points when
plotting scatterplots of one zoo series vs. another -- not the axis
labels. e.g.
library(zoo)
set.seed(1)
z <- zoo(rnorm(10))
plot(z, z, xy.label = letters[1:10])
Assuming a single panel plot this will suppress the x axis label
but s
Dear all,I'm a newbie to Survival Analysis and I have some questions about the
functions in the survival package.First, what's the difference between survreg
and coxph? Say fit1<- survreg(survObj ~ var1 + var2 + var3, dist= 'whatever')
and fit2<- coxph(survObj ~ var1 + var2 + var3), what's the d
Run R as the Administrator. Install the packages. Then run as an ordinary
user.
Charles Annis, P.E.
[EMAIL PROTECTED]
phone: 561-352-9699
eFax: 614-455-3265
http://www.StatisticalEngineering.com
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of nicho
xaxt="n" and yaxt="n" should do the trick if you just want to suppress the
labels. check
?par for further refinement
On Tue, Jul 1, 2008 at 5:25 PM, <[EMAIL PROTECTED]> wrote:
> I would like to know if there is a way to use plot(zoo) and suppress the
> label text to increase the amount of the sc
Hi Mark,
I'm sorry, it should be
temptable[,,1]+ temptable[,,2]+ temptable[,,3]
HTH,
Jorge
On Tue, Jul 1, 2008 at 10:50 PM, Jorge Ivan Velez <[EMAIL PROTECTED]>
wrote:
>
> Dear Mark,
>
> Is this what you want?
>
> print(margintable,dim=c(8,4))
>
>
> HTH,
>
> Jorge
>
>
>
> On Tue, Jul 1, 200
Dear Mark,
Is this what you want?
print(margintable,dim=c(8,4))
HTH,
Jorge
On Tue, Jul 1, 2008 at 10:41 PM, <[EMAIL PROTECTED]> wrote:
> I have a simple table below called temptable and i want to obtain the same
> structure that prop.table creates except get the counts
> rather than the pr
Please disregard my previous question. I realize that I don't need such
a function because I already have what I need ( duuuh ).
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide htt
I have a simple table below called temptable and i want to obtain the
same structure that prop.table creates except get the counts
rather than the proportions. margin.table seems to create one table with
columns and rows whereas I am looking for the three table
type structure that prop.table gi
I can't seem to install packages to R. Each time I get the following
output...for example
bundle 'VR' successfully unpacked and MD5 sums checked
The downloaded packages are in
C:\Users\DarkBlue\AppData\Local\Temp\RtmpQrD7Le\downloaded_packages
updating HTML package descriptions
Warning m
I could not find any of the other answers in my list. So I don't know what
the conversation was. Anyway, with the approach I suggested, it is no
problem to track which score/loading belongs to which sample because the
results were stored in an array. If you want to trace it back to the
bootstrapped
On 1 July 2008 at 19:47, Dirk Eddelbuettel wrote:
|
| On 1 July 2008 at 14:03, Erin Hodgess wrote: | I'm having some trouble with
| mpiexec and Rmpi.
| |
| | I would like to be able to pass in the number of "children" via the mpiexec
| | command (from the command line).
| |
| | this is in SUSE1
I would like to know if there is a way to use plot(zoo) and suppress the
label text to increase the amount of the screen that if for the plot and
not the labels. I tried xy.labels=FALSE, but that had no effect. Thanks
[[alternative HTML version deleted]]
_
no this is not what I want.
I was using "loess" function or smooth.spline.
but For loess I don't know how would I be able to get the integral.
On 7/1/08, stephen sefick <[EMAIL PROTECTED]> wrote:
>
> ?lm
> lm(x[,1]~x[,2])
>
> On Tue, Jul 1, 2008 at 2:28 PM, Shirin Safa <[EMAIL PROTECTED]> wrote:
use c() to make your condition a vector, try to use all() to return a
logical value for the entire condition vector .
On 2008-7-1, at 下午8:40, mysimbaa wrote:
I'm trying to do realize the following:
I have 4 condtions.
If all conditions are satisfied I will paste("PASS")
If any of these is not
On 1 July 2008 at 14:03, Erin Hodgess wrote: | I'm having some trouble with
mpiexec and Rmpi.
|
| I would like to be able to pass in the number of "children" via the mpiexec
| command (from the command line).
|
| this is in SUSE10.1, with R-2.7.1
|
| Here are my files: cat eb.R library(Rmpi) mp
On 01/07/2008 7:16 PM, Lana Schaffer wrote:
Hi,
I am using version R2.7.0 and I have installed
Rtools27.exe.
Upon calling R CMD INSTALL I got this error:
.../src/gnuwin32/MakePkg:1: missing separator
Can someone tell me what is wrong?
It's very likely that your PATH is incorrect, and the wrong
On 1 July 2008 at 18:49, Erin Hodgess wrote:
| I have used
|
| ./R --no-save -q -f e.in >stuff.out
|
| with great success on SUSE10.1 with R-2.7.1.
|
| My question is, please: is there a way to pass in a variable to the
| e.in file? I'm fairly sure that the answer is no, but thought I'd
| dou
on 07/01/2008 06:49 PM Erin Hodgess wrote:
Dear R People:
I have used
./R --no-save -q -f e.in >stuff.out
with great success on SUSE10.1 with R-2.7.1.
My question is, please: is there a way to pass in a variable to the
e.in file? I'm fairly sure that the answer is no, but thought I'd
double
Do you have a reason to treat all 3 levels together and not have a separate
regression for each level?
--- On Tue, 1/7/08, rlearner309 <[EMAIL PROTECTED]> wrote:
> From: rlearner309 <[EMAIL PROTECTED]>
> Subject: [R] A regression problem using dummy variables
> To: r-help@r-project.org
> Recei
Hi Erin,
On Tue, 1 Jul 2008 18:49:02 -0500,
"Erin Hodgess" <[EMAIL PROTECTED]> wrote:
> Dear R People: I have used
> ./R --no-save -q -f e.in >stuff.out
> with great success on SUSE10.1 with R-2.7.1.
> My question is, please: is there a way to pass in a variable to the
> e.in file? I'm fairl
Are you explicitly 'print'ing the lattice plot?
print(xyplot(...))
On Tue, Jul 1, 2008 at 6:20 AM, Michael Hopkins
<[EMAIL PROTECTED]> wrote:
>
>
> Hi R people
>
> I am using a function to create a pdf device, then send a lot of plots
> to it in a loop then a last lattice xyplot (itself within
On 7/1/08, Michael Hopkins <[EMAIL PROTECTED]> wrote:
>
>
> Hi R people
>
> I am using a function to create a pdf device, then send a lot of plots
> to it in a loop then a last lattice xyplot (itself within a function)
> outside the loop and finally call dev.off() to write to the file.
> This
Hi Erin --
"Erin Hodgess" <[EMAIL PROTECTED]> writes:
> Dear R People:
>
> I'm having some trouble with mpiexec and Rmpi.
>
> I would like to be able to pass in the number of "children" via the
> mpiexec command (from the command line).
>
> this is in SUSE10.1, with R-2.7.1
>
> Here are my files:
Michael Hopkins wrote:
Hi R people
I am using a function to create a pdf device, then send a lot of plots
to it in a loop then a last lattice xyplot (itself within a function)
outside the loop and finally call dev.off() to write to the file.
This works well apart from the fact that the la
Dear R People:
I have used
./R --no-save -q -f e.in >stuff.out
with great success on SUSE10.1 with R-2.7.1.
My question is, please: is there a way to pass in a variable to the
e.in file? I'm fairly sure that the answer is no, but thought I'd
double check.
Thanks in advance,
Sincerely,
Erin
Hi,
I am using version R2.7.0 and I have installed
Rtools27.exe.
Upon calling R CMD INSTALL I got this error:
.../src/gnuwin32/MakePkg:1: missing separator
Can someone tell me what is wrong?
Lana Schaffer
Biostatistics/Informatics
The Scripps Research Institute
DNA Array Core Facility
La Jolla, CA
On 2/07/2008, at 10:33 AM, Bert Gunter wrote:
Hoisted by own petard! How appropriate.
1. I hereby invoke the codicil.
2. Check the dictionary. An accepted pronunciation of inveigle is
"inveegle."
Does this allow me to wriggle through the escape clause?
But of course! (I did check
Hi, Dear all R experts,
I could use the fitdistr() function to estimate the parameters in a
univariate weibull distribution,but for the bivariate case it doesn't work.
Is there a function in R to estimate the parameters in a bivariate weibull
distribution, given I have a set of observations
And, oddly enough :-), integrate.xy does pretty much exactly what I
suggested. Thanks for providing that reference
I would be interested in seeing how the original poster's data works
out using integrate.xy as opposed to simply calculating x*y
By the way, since the original data were 'perc
On 2/07/2008, at 10:38 AM, stephen sefick wrote:
I would like to integrate the area under a curve without any
smoothing or
the like- just on the raw numbers. I looked at integrate() but it
requires
a function which I assume means something like x+x^2+x^3
is there a built in function in R
as.table =TRUE was exactly what I needed. Also thanks for the little aside.
Gabor Grothendieck wrote:
>
> Here are some ways of rearranging panels:
>
> library(lattice)
> p <- xyplot(Sepal.Length ~ Sepal.Width | Species, iris)
> p
> p[c(2, 1, 3)]
>
> xyplot(Sepal.Length ~ Sepal.Width | Specie
There is an integrate.xy in sfsmic. Limitations discussed there.
On Tue, Jul 1, 2008 at 6:27 PM, stephen sefick <[EMAIL PROTECTED]> wrote:
> I would like to know the answer to this question now that I know what we are
> getting at. integrate() looks like it is the right thing, but it has to use
On 7/1/08, Sam Albers <[EMAIL PROTECTED]> wrote:
> I have constructed a Trellis style xyplot.
>
> lengthf <- factor(length)
> xyplot(SLI$velocity ~ SLI$width | SLI$lengthf, layout = c(2,7), xlab =
> "Width (cm)", ylab = "Velocity (m/s^2)", col = "black")
As an aside, the recommended incantation
Hoisted by own petard! How appropriate.
1. I hereby invoke the codicil.
2. Check the dictionary. An accepted pronunciation of inveigle is
"inveegle."
Does this allow me to wriggle through the escape clause?
-- Bert
-Original Message-
From: Rolf Turner [mailto:[EMAIL PROTECTED]
Sent:
I would like to integrate the area under a curve without any smoothing or
the like- just on the raw numbers. I looked at integrate() but it requires
a function which I assume means something like x+x^2+x^3
is there a built in function in R for this?
#let's say
x <- seq(1:50)
y <- seq(1:50)
plot(
Here are some ways of rearranging panels:
library(lattice)
p <- xyplot(Sepal.Length ~ Sepal.Width | Species, iris)
p
p[c(2, 1, 3)]
xyplot(Sepal.Length ~ Sepal.Width | Species, iris, as.table = TRUE)
On Tue, Jul 1, 2008 at 6:20 PM, Sam Albers <[EMAIL PROTECTED]> wrote:
> I have constructed a Tre
I would like to know the answer to this question now that I know what we are
getting at. integrate() looks like it is the right thing, but it has to use
a function- I would like to know how to just integrate the area under a
curve with just an input of x and y coordinates.
Stephen
On Tue, Jul 1
On 2/07/2008, at 10:15 AM, Bert Gunter wrote:
Ow ow ow. I can't stand it any longer! "Weird" is one of those weird
exceptions in English to the "i before e except after c" rule and
is spelled
"W-E-I-R-D" .
In case you're interested, the other exceptions are seize, inveigle ,
either, leisure
I have constructed a Trellis style xyplot.
lengthf <- factor(length)
xyplot(SLI$velocity ~ SLI$width | SLI$lengthf, layout = c(2,7), xlab =
"Width (cm)", ylab = "Velocity (m/s^2)", col = "black")
This produces a lovely little plot. However, the grouping factor(lengthf)
isn't in the right order. M
Ow ow ow. I can't stand it any longer! "Weird" is one of those weird
exceptions in English to the "i before e except after c" rule and is spelled
"W-E-I-R-D" .
In case you're interested, the other exceptions are seize, inveigle ,
either, leisure, neither.
:-( -- Bert Gunter
-Original Messa
On Tue, 2008-07-01 at 17:12 +, Mark Lyman wrote:
> Does anybody have any suggestions on how I might simulate from fitted GAM
> model? I am using the gam function in the mgcv package to fit a variable
> coefficient model like the following from the examples. I would like simulate
> based on t
This is FAQ 7.33
-thomas
On Tue, 1 Jul 2008, [EMAIL PROTECTED] wrote:
Can someone please enlighten me as to why the following happens?
-2.7^8.6
[1] -5125.407
p<- -2.7
q<- 8.6
p^q
[1] NaN
R seems perfectly able to calculate -2.7^8.6, but fails when the exact same
values are assig
I think the previous answer (to use lm() ) is not necessarily the best
option.
Since what you want is the definite integral (area under the curve), you
can just use one of the existing definite integration tools (sorry, I
don't recall the names because I don't use them).
If you want to get a
does loess return the values that it produces- take that and then integrate
under the curve.
On Tue, Jul 1, 2008 at 4:28 PM, Shirin Safa <[EMAIL PROTECTED]> wrote:
> no this is not what I want.
> I was using "loess" function or smooth.spline.
> but For loess I don't know how would I be able to ge
?lm
lm(x[,1]~x[,2])
On Tue, Jul 1, 2008 at 2:28 PM, Shirin Safa <[EMAIL PROTECTED]> wrote:
> Hi
>
> I have a set of data like this:
>
>*Time of Day* *Pct of Daily Volume* 9:45 7.50% 10 6.25% 10:15 4.45%
> 10:30 4.80% 10:45 4.45% 11:00 4.20% 11:15 2.50% 11:30 2.30% 11:45 2.25%
> 12:00 2.45% 12
Hi
I have a set of data like this:
*Time of Day* *Pct of Daily Volume* 9:45 7.50% 10 6.25% 10:15 4.45%
10:30 4.80% 10:45 4.45% 11:00 4.20% 11:15 2.50% 11:30 2.30% 11:45 2.25%
12:00 2.45% 12:15 2.60% 12:30 2.00% 12:45 2.05% 13:00 2.40% 13:15 1.90%
13:30 3.10% 13:45 2.90% 14:00 2.80% 14:15 2.50
You can use names(database)[position]
where position is the number of the column that you want to rename!
Leandro Marino
-Mensagem original-
De: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
nome de [EMAIL PROTECTED]
Enviada em: terca-feira, 1 de julho de 2008 16:18
Para: R-help@r-project.
I will see the code! It is similar that i want!
Thanks a lot!
-Mensagem original-
De: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Enviada em: terça-feira, 1 de julho de 2008 16:22
Para: Leandro Marino
Cc: [EMAIL PROTECTED] Org
Assunto: Re: [R] plot window
Look at the code for the graph
Try
require(gregmisc)
rename(data, from="old_name", to="new_name")
A Smile costs Nothing
But Rewards Everything
Happiness is not perfected until it is shared
-Jane Porter
--- On Tue, 7/1/08, [EMAIL PROTECTED] <[EMAI
> x <- data.frame(a=1,b=2,c=3)
> x
a b c
1 1 2 3
> colnames(x)[2] <- "done"
> x
a done c
1 12 3
>
On Tue, Jul 1, 2008 at 3:17 PM, <[EMAIL PROTECTED]> wrote:
> Hi,
>
> Sorry for the simple question. Is there a way to change the name of only one
> column of an existing data frame?
>
> I kn
Look at the code for the graphic on the R home page.
If you click on it you will see the code.
On Tue, Jul 1, 2008 at 3:15 PM, Leandro Marino
<[EMAIL PROTECTED]> wrote:
> Hi list,
>
> I want to know how can i creat a plot window with this configuration:
> ___
> |
Hi,
Sorry for the simple question. Is there a way to change the name of only one
column of an existing data frame?
I know colnames allows you to set the name of all the columns, but only one
column in the middle of my data frame needs a new name.
Thanks,
-Nina
__
Try:
nf <- layout(matrix(c(1,2, 4, 6, 1, 3, 5, 6), nc = 2))
layout.show(nf)
On Tue, Jul 1, 2008 at 4:15 PM, Leandro Marino <[EMAIL PROTECTED]>
wrote:
> Hi list,
>
> I want to know how can i creat a plot window with this configuration:
> ___
> |
Hi list,
I want to know how can i creat a plot window with this configuration:
___
| |
| PLOT 1|
| |
|-|
| | |
| |
Hi,
I am trying to load the library(ks), but I am getting the following error:
Loading required package: KernSmooth
KernSmooth 2.22 installed
Copyright M. P. Wand 1997
Loading required package: mvtnorm
Loading required package: rgl
Loading required package: misc3d
Error in lazyLoadDBfetch(key, da
Dear R People:
I'm having some trouble with mpiexec and Rmpi.
I would like to be able to pass in the number of "children" via the
mpiexec command (from the command line).
this is in SUSE10.1, with R-2.7.1
Here are my files:
cat eb.R
library(Rmpi)
mpi.remote.exec(paste("i am",mpi.comm.rank(),"o
Hi,
the problem is what you want!
With the -2.7^8.6 you are doing -(2.7^8.6) it exists...
But, if you try (-2.7)^8.6 the R gives you NaN. When you define p=-2.7 and
q=8.6 and do p^q you are doing that (-2.7)^8.6.
If you write p=2.7 and q=8.6 and use the -p^q it will work!
Leandro Marino
www.l
poolloopus yahoo.com yahoo.com> writes:
>
> Can someone please enlighten me as to why the following happens?
> > -2.7^8.6
> [1] -5125.407
>
> > p<- -2.7
> > q<- 8.6
> > p^q
> [1] NaN
> R seems perfectly able to calculate -2.7^8.6, but fails when the exact same
values are assigned to
> variabl
Nina,
read.csv() will default fill = TRUE,
or add to your read.table() argument list.
If that does not help, you will need to use col.names argument. see
?read.table
HTH,
Jim Porzak
Responsys, Inc.
San Francisco, CA
http://www.linkedin.com/in/jimporzak
On Mon, Jun 30, 2008 at 11:16 AM, <[EMAIL
on 07/01/2008 01:15 PM [EMAIL PROTECTED] wrote:
Can someone please enlighten me as to why the following happens?
-2.7^8.6
[1] -5125.407
p<- -2.7 q<- 8.6 p^q
[1] NaN
R seems perfectly able to calculate -2.7^8.6, but fails when
the exact same values are assigned to variables and then the
co
Thank you very much for your precise help.
I will try to implement this tomorrow.
It looks like the best solution which could be done.
Greetings,
Adel Tekari
Marc Schwartz wrote:
>
> on 07/01/2008 07:40 AM mysimbaa wrote:
>> I'm trying to do realize the following:
>> I have 4 condtions.
>> I
Hi,
I am interested in using a bayesian network as a predictor (machine
learning); however, I can't get any of the implementations (deal, nblearn)
to learn & predict stuff.
Shouldn't there also be probabilites for each node after the learning phase,
how can I access these?
Cheers,
Stephan
--
Vi
Dear Nina,
If you have a matrix X like this
# Data set
set.seed(123)
X=matrix(rnorm(10*5),ncol=5)
X[1,1]<-NA
X[2,1]<-NA
X[1,5]<-NA
X[5,2]<-NA
X
and you'd like to remove the NA values for a particular row (for example row
1), you can try something like:
X[1,!is.na(X[1,])]
Now, if you have a ve
On Tue, Jul 01, 2008 at 11:15:58AM -0700, [EMAIL PROTECTED] wrote:
> Can someone please enlighten me as to why the following happens?
> > -2.7^8.6
> [1] -5125.407
>
> > p<- -2.7
> > q<- 8.6
> > p^q
> [1] NaN
> R seems perfectly able to calculate -2.7^8.6, but fails when the exact same
> values ar
Can someone please enlighten me as to why the following happens?
> -2.7^8.6
[1] -5125.407
> p<- -2.7
> q<- 8.6
> p^q
[1] NaN
R seems perfectly able to calculate -2.7^8.6, but fails when the exact same
values are assigned to variables and then the computation is repeated.
Thanks in advance for an
It depends what you mean by 'ignore'. Some functions have an na.rm
argument which throws out NAs before computing the statistic.
if the vector 'x' has NAs, then
x <- x[!is.na(x)]
may be what you're looking for. This removes NAs from x and reassigns
the value to x.
[EMAIL PROTECTED] wrote:
Hi,
I am extracting data from a table where the rows have different column lengths,
and empty columns have NA in them. Whenever I extract a row with some empty
columns, the resulting vector carries all the NAs. Is there a way to ignore the
empty columns?
Thanks,
-Nina
___
HI Miriam,
the problem is that you have the Date and some other factors in your
database.
try to use the
lapply(your_base,class)
and see what appears. If you have any kind of factor or date you cannot use
boxplot.
If you verify that only V1, V2 and V3 are numeric you can do the box plot
with t
The advice is to store numeric data as numeric, and not factors. Try
str(yourdata) and you'll probably see that Diatoms is stored as a
'factor'. Are those commas used as decimal points? read.table can deal
with that if so. If not, you do not need them, and should convert those
fields to num
Does anybody have any suggestions on how I might simulate from fitted GAM
model? I am using the gam function in the mgcv package to fit a variable
coefficient model like the following from the examples. I would like simulate
based on the fitted model like the simulate function in the stats packa
options(warn = -1)
or
suppressWarnings(as.numeric("test"))
On Tue, Jul 1, 2008 at 1:11 PM, Hua Li <[EMAIL PROTECTED]> wrote:
> Hi All,
>
> I'm working with R and want to ignore the warning messages given, is there
> a way to stop R from giving out warning messages any more?
>
> an example:
>
>
On Wed, 2008-07-02 at 00:58 +1000, Jason Lee wrote:
> Hi,
>
> Thanks for the reply.
>
> Basically I dont have any label for my data except column 1 which is labeled
> with Sample1, Sample2...etc...(vertical).
Well that is going to cause you problems, not the one you report below,
but it will bit
You should consult "An Introduction to R", section 11, "Writing your own
functions", along with any beginner books on R.
http://www.r-project.org/doc/bib/R-books.html
test <- function(x = 2) {
x^2
}
test() ## will give 4
test(3) ## will give 9
Marco Salvi wrote:
Hi, i'm tryng to build a
Dear Hua,
Try
options(warn=-1)
tt = "test"
as.numeric(tt)
[1] NA
See also ?options for more info.
HTH,
Jorge
On Tue, Jul 1, 2008 at 12:11 PM, Hua Li <[EMAIL PROTECTED]> wrote:
> Hi All,
>
> I'm working with R and want to ignore the warning messages given, is there
> a way to stop R from gi
Hi,
I'm trying to make a boxplot with the data at the end of the message, and when I
try to execute the command
>boxplot(Diatoms) (or for any other field instead of "Diatoms")
I get the following error message:
Error in oldClass(stats) <- cl : adding class "factor" to an invalid object
Any adv
Dear All,
I have found in the poly help this sentence:
The orthogonal polynomial is summarized by the coefficients, which can be
used to evaluate it via the three-term recursion given in Kennedy & Gentle
(1980, pp. 3434), and used in the predict part of the code.
My question: which type of ortho
Hi,
you can use the jpeg(), pdf(), bmp() functions. In this function you can
determinate the size of the graph, background color, quality of the image.
Leandro Marino
-Mensagem original-
De: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
nome de Ptit_Bleu
Enviada em: terca-feira, 1 de jul
Hi All,
I'm working with R and want to ignore the warning messages given, is there a
way to stop R from giving out warning messages any more?
an example:
tt = "test"
as.numeric(tt)
would give me the following message:
[1] NA
Warning message:
NAs introduced by coercion
I decide to ignore the
on 07/01/2008 07:40 AM mysimbaa wrote:
I'm trying to do realize the following:
I have 4 condtions.
If all conditions are satisfied I will paste("PASS")
If any of these is not satisfied I will paste("FAIL"). But I have to paste
the corresponding failure.
ifelse is a good solution but for a 2 cond
Thanks!
I tried already. Still the same error message...
wishes
Gunther
- Original Message -
From: "ONKELINX, Thierry" <[EMAIL PROTECTED]>
To: "Gunther Höning" <[EMAIL PROTECTED]>;
Sent: Tuesday, July 01, 2008 3:08 PM
Subject: RE: [R] Regression and fitting
Try to set your own star
Ok, the mistake was in the pca(x)<-princomp(SampleD[i,j]), should've used
pca(x)<-princomp(SampleD) instead.
Now, is there anyway to keep track of the matrix index, so in the end of all
PCAs, I can tell which score/loading belongs to which sample?
Thanks everyone!
On Mon, Jun 30, 2008 at 9:08 PM,
This is actually more like a Statistics problem:
I have a dataset with two dummy variables controlling three levels. The
problem is, one level does not have many observations compared with other
two levels (a couple of data points compared with 1000+ points on other
levels). When I run the regre
Hello,
I'm trying to produce graphs automatically from data stored in database.
Before saving the graphs, I would like to maximize the size of the graphs.
The best would be to directly open maximized windows with x11() but up to
now I failed doing it.
I tried different widths and heighs but I ne
on 07/01/2008 04:58 AM Francisco Javier Santos Alamillos wrote:
Hello everyone,
I need reshape an array. For example, if we have next array:
a <- c(1,2,3,4,5,6,7,8,9,10,11,12)
dim(a) <- c(2,2,3)
a
, , 1
[,1] [,2]
[1,]13
[2,]24
, , 2
[,1] [,2]
[1,]57
[2,]
Hi all,
I've tried to plot a vector which has two peaks in the density.
This link shows the figure.
http://docs.google.com/View?docid=dcvdrfrh_1dk9r2rc7
The red line is normal curve and green line is gamma curve.
Notice that red line can correctly fit the histogram that has two peaks
(i.e. red c
Hi. With the previous version of R, it was possible to execute the function
'select.list()' in BATCH mode. In this way, I could write my R code in a .R
file and execute that with a double click on a .bat file, that contain the
instruction tu run my R code in batch mode (R CMD BATCH myRcode.R).
Hello everyone,
I need reshape an array. For example, if we have next array:
> a <- c(1,2,3,4,5,6,7,8,9,10,11,12)
> dim(a) <- c(2,2,3)
> a
, , 1
[,1] [,2]
[1,]13
[2,]24
, , 2
[,1] [,2]
[1,]57
[2,]68
, , 3
[,1] [,2]
[1,]9 11
[2,] 10 12
I'm trying to do realize the following:
I have 4 condtions.
If all conditions are satisfied I will paste("PASS")
If any of these is not satisfied I will paste("FAIL"). But I have to paste
the corresponding failure.
ifelse is a good solution but for a 2 conditions. Maybe switch or something
like t
Hi, i'm tryng to build a function that take some input from user and if the
user doesn't provide that inputs the function should set a deafult value. I
have taken as example a function that i found in the package dlm but with
the same code i receive an error (the component is missing...). According
I solved the problem with
plot(0:1,0:1, type = "n", axes=FALSE,xlab="",ylab="")
text(...)
Thanks,
Adel
mysimbaa wrote:
>
> Dear R users,
>
> Is it possible to add comments in a plot window?
> I have 3 plots -> this plots have to be commented on same window
> Is possible to do the following :
So, I used df <- data.frame( x=I(coef(cancerv1(,2:407))),
y=cancerv1[,408])before feeding to PCR.
However, I get the below error.
Error in coef(cancerv1(, 2:407)) : could not find function "cancerv1".
It's because you have the syntax error 'cancerv1(...)', which looks to
R, and me, a functi
Hi,
Thanks for the reply.
Basically I dont have any label for my data except column 1 which is labeled
with Sample1, Sample2...etc...(vertical).
My cancerv1 data is data.frame.
So, I used df <- data.frame( x=I(coef(cancerv1(,2:407))),
y=cancerv1[,408])before feeding to PCR.
However, I get the b
Simply add a definition for this variable to the source for your package,
e.g.;
.Last.make.date <- NULL
And the warning should go away.
-g
On 6/30/08 11:44PM , "Rolf Turner" <[EMAIL PROTECTED]> wrote:
>
>
> I have written a function make.fun(), which I keep in my personal
> ``miscellaneous
Hi,
Currently I have a data set of 2 classes and I am interedted to use emerging
patterns algorithm. I wondered if there is any packages or recommended code
available for this?
Please advise.Thanks.
[[alternative HTML version deleted]]
__
R-h
Dong-hyun Oh wrote:
> Dear UseRs,
>
> I would like to know the way to find classes of each column of
> data.frame().
>
sapply(, class)
vQ
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Nina,
In which way have you obtained your CSV file? Did you export data
from an Excel spreadsheet?
Regards,
Paulo Barata
---
Paulo Barata
Fundacao Oswaldo Cruz
Rua Leopoldo Bulhoes 1480 - 8A
21041-210 Rio de Janeiro - RJ
Bra
many thanks.
On Jul 1, 2008, at 4:00 PM, Henrique Dallazuanna wrote:
sapply(your_data, class)
On Tue, Jul 1, 2008 at 10:50 AM, Dong-hyun Oh <[EMAIL PROTECTED]>
wrote:
Dear UseRs,
I would like to know the way to find classes of each column of
data.frame().
Thank you in advance.
Many thanks.
On Jul 1, 2008, at 3:57 PM, Romain Francois wrote:
sapply( iris, class )
Dong-hyun Oh wrote:
Dear UseRs,
I would like to know the way to find classes of each column of
data.frame().
Thank you in advance.
=
Dong-hyun
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