> it's a >> package << , not a library, please!
>
>
Sorry for using library instead package, but
library() is one command for using packages.
Therefore I (and it seems that i am not the only one) used library instead
package.
Knut
__
R-help@r-
Turn your problem into an optimization one and use the various
optimization abilities of R, for instance have a look at nlm:
?nlm
As an example to solve the rather simple:
x-y=0
x+y=2
We could do:
f <- function(vals) {
x <- vals[1]
y <- vals[2]
sum(c(x-y,x+y-2)^2)
}
nl
Dear:
I have a paired equation below. Can I solve (x,y) using R.
Thanks!
Xin
A=327.727
B=9517.336
p=0.114^10
(1-p)*y*(1-x)/x/(1-x^y)=A
A(1+(1-x)*(1+y)/x-A))=B
[[alternative HTML version deleted]]
__
R-help@r-project.org ma
On Dec 16, 2007 9:44 PM, Spencer Graves <[EMAIL PROTECTED]> wrote:
> How can I see more of the structure than displayed by 'str'?
> Consider the following:
>
>
> tstDF <- data.frame(a=1, row.names='b')
> > str(tstDF)
> 'data.frame': 1 obs. of 1 variable:
> $ a: num 1
>
>
> The object
On Sun, 16 Dec 2007, Spencer Graves wrote:
> How can I see more of the structure than displayed by 'str'?
str is generic.
> methods(str)
[1] str.data.frame* str.default*str.dendrogram* str.logLik* str.POSIXt*
Non-visible functions are asterisked
>
As you see there is a data.fr
How can I see more of the structure than displayed by 'str'?
Consider the following:
tstDF <- data.frame(a=1, row.names='b')
> str(tstDF)
'data.frame': 1 obs. of 1 variable:
$ a: num 1
The object 'tstDF' has row.names, but I have to suspect they are
there -- AND know a
RSiteSearch("AUC")
would lead you to
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/46416.html
On Dec 13, 2007 12:38 PM, Armin Goralczyk <[EMAIL PROTECTED]> wrote:
> Hello
>
> Is there an easy way, i.e. a function in a package, to calculate the
> area under the curve (AUC) for drug serum levels
You can using the default Linux (Xandros, Debian Based) and enable
Debian Etch Repo in eeepc to install R via apt-get.
http://lnxg.ca/?Hardware:Asus_Eeepc_701:EeePC_Tips
I can install R, LaTeX.
Regards,
C
On Dec 17, 2007 4:37 AM, Gabor Csardi <[EMAIL PROTECTED]> wrote:
> You can easily install
ravi wrote:
>
> I have the following code, where we need to solve for mu and sigma, when
> we have mut and sdt. Can you suggest how to use a solve function in R to
> do that? I am new to R and am not sure how to go from defining the
> functions, to solving for them.
>
> Thanks
>
>
> tru
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
David Winsemius wrote:
> On 15 Dec 2007, you wrote in gmane.comp.lang.r.general:
>
>> If we can assume that the abstract is always the 4th paragraph then we
>> can try something like this:
>>
>> library(XML)
>> doc <-
>> xmlTreeParse("http://eutils.
try this:
> dates <- c("January 12, 1988", "March 4, 2006", "1958")
> gsub(".*, *(.*)", "\\1", dates) # notice the it is comma blank asterisk in
> the string
[1] "1988" "2006" "1958"
On Dec 16, 2007 7:34 AM, Anupa Fabian <[EMAIL PROTECTED]> wrote:
> I have some data whose date column consists
Christian Schulz wrote:
> Hello,
>
> i'm trying to replace different target variables in rpart with a
> function. The data.frame getting always the target variable as last column.
> Try below, i get the target variable in the explained variables, too!?
> Have anybody an advice to avoid this
Hello,
I'm trying to learn how to use lattice and levelplot in particular.
There are three elements of customizing the plots I'm stuck with:
a) Is there a way to put borders around each "cell" within a
level-plot. I'm trying to do something like the
colsep/rowsep/sepcolor/sepwidth paramet
"Gabor Grothendieck" <[EMAIL PROTECTED]> wrote in
news:[EMAIL PROTECTED]:
> On Dec 16, 2007 2:53 PM, David Winsemius <[EMAIL PROTECTED]>
> wrote:
>> # Never did find a way to "print" them when internal.
> ?saveXML
And now I understand where that odd "\n " originated before I
changed the
"Gabor Grothendieck" <[EMAIL PROTECTED]> wrote in
news:[EMAIL PROTECTED]:
> On Dec 16, 2007 2:53 PM, David Winsemius <[EMAIL PROTECTED]>
> wrote:
>> # in the debugging phase I needed to set useInternalNodes = TRUE to
>> see the tags. Never did find a way to "print" them when internal.
>
> I ass
You can easily install ubuntu on it (although it might require
an external drive):
http://hup.hu/node/48116
So running R should not be problem.
G.
On Sun, Dec 16, 2007 at 12:56:39PM -0500, Burton Rothberg wrote:
> I'm thinking of getting one of these lightweight linux laptops for
> traveling. Doe
Thanks Jim. I was using R 2.4.0, that must be the problem. After I upgraded
to 2.6.1, aggregate() generated the correct order of levels. Thanks!
On 12/16/07, jim holtman <[EMAIL PROTECTED]> wrote:
>
> What version of R are you using? Here is the output I got with 2.6.1:
>
> > library(chron)
> > d
I'm thinking of getting one of these lightweight linux laptops for
traveling. Does anyone know if / how R runs on it?
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE d
On Dec 16, 2007 2:53 PM, David Winsemius <[EMAIL PROTECTED]> wrote:
> # in the debugging phase I needed to set useInternalNodes = TRUE to see the
> tags. Never did find a way to "print" them when internal.
I assume you mean FALSE. See:
?saveXML
__
R-he
On Sunday 16 December 2007, Adrian Dusa wrote:
> Dear all,
>
> [...]
>
> Given these examples, I have two questions:
> 1. What is the correct syntax to import the R-exported file
> 2. What can I do to prevent these situations from happening?
> (besides whipping the data entry person :), I am referr
Dear all,
Some very wise data entry person gave me about an hour of a headache, trying
to find out why a 2000x500 dataframe won't be read into R.
After much trial and error, I pinpointed the problem to an accidentally
inserted double quote into a string variable (some comments from an open
que
On 15 Dec 2007, you wrote in gmane.comp.lang.r.general:
> If we can assume that the abstract is always the 4th paragraph then we
> can try something like this:
>
> library(XML)
> doc <-
> xmlTreeParse("http://eutils.ncbi.nlm.nih.gov/entrez/eutils/erss.cgi?rss
> _guid=0_JYbpsax0ZAAPnOd7nFAX-29fXDp
Hello,
i'm trying to replace different target variables in rpart with a
function. The data.frame getting always the target variable as last column.
Try below, i get the target variable in the explained variables, too!?
Have anybody an advice to avoid this.
rp1 <- rpart(eval(parse(text=paste(
On Dec 14, 2007 9:26 PM, Mark Difford <[EMAIL PROTECTED]> wrote:
> >> Is there a solution for this problem?
>
> If there is, then Professor Bates (the gentleman who replied to your
> question) will have tried to find it, and fix it, for you.
What I was trying to indicate in my replies is that "th
Yes, for some ways of running a script (and you haven't told us how your
intend to run this). E.g.
% Rscript foo.R
works, where
% cat foo.R
mydata <- read.table(stdin(), header=TRUE, sep="\t", nrow=6)
freqesperado
117.5
147.5
47.5
17.5
27.7
16.3
mydata
But, e.g. you can't use source() (it pa
Milton Cezar Ribeiro yahoo.com.br> writes:
> I would like to test if are there some significant orientation of frequencies
on a polar analysis.
>
Check packages CircStats and circular.
Dieter
__
R-help@r-project.org mailing list
https://stat.ethz.ch/
On 12/16/07, Bob Green <[EMAIL PROTECTED]> wrote:
> Hello,
>
> Below is the code for a basic bar graph. I was seeking advice
> regarding the following:
>
> (a) For each time period there are values from 16 people. How I can
> change the colour value so that each person has a different colour,
> wh
Hi there,
I would like to test if are there some significant orientation of frequencies
on a polar analysis.
my data looks like:
mydata<-"angule,frequency
22.5,6
67.5,3
112.5,2
157.5,2
202.5,3
247.5,3
292.5,6
337.5,5"
mydata.tab<-read.table(textConnection(mydata), header = TRUE,sep=",")
require
Ugly brute-force approach: col=1:16 . Jim Lemon's
approach with Plotrix is much nicer. You might also
want to have a look at RColorBrewer though I am not
sure how easily it can handle 16 different colours.
barplot(dft, beside = TRUE, main= "Risk score by
assessment", xlab = " Score", ylab = "
You can use the following to get the data from that call:
> x <- do.call('rbind', kaste)
> as.matrix(x[,2:3])
[,1] [,2]
[1,] 53.55 8.58
[2,] 53.87 8.7
[3,] 53.87 10.69
On Dec 16, 2007 10:54 AM, Mag. Ferri Leberl <[EMAIL PROTECTED]> wrote:
> Thank you for the attempt.
> On your advice I am
Thank you for the attempt.
On your advice I am working with lists to avoid the numbers (which are
geographical coordinates) becoming strings. The call you suggest does
not take care of that. Now I am trying to extract the coordinates from
the list efficiently. Of course I could make a loop, e.g.
p
Try this:
Lines <- "freqesperado
117.5
147.5
47.5
17.5
"
read.table(textConnection(Lines), header = TRUE)
On Dec 16, 2007 10:47 AM, Milton Cezar Ribeiro
<[EMAIL PROTECTED]> wrote:
> Dear All,
>
> It there a way of I read my data tab-separated on the own script, without
> read from a external fil
Try:
print(x, digits = 3)
On Dec 15, 2007 9:22 PM, tom soyer <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I am constructing a contingency table using xtabs. The function works great:
> mo
> yr Sep Oct Nov Dec
> 1950 -7.164486e-02 3.152674e-02 -1.283389e-
Dear All,
It there a way of I read my data tab-separated on the own script, without read
from a external file and without type the data?
I would like something like
mydata<- read.data(head=T, sep="\t")
freqesperado
117.5
147.5
47.5
17.5
##END OF DATA
Many thanks, miltinho
para armazenamen
Is this what you want as output:
> do.call('rbind', kaste)
[,1] [,2] [,3]
[1,] "Bremerhaven" 53.55 8.58
[2,] "Cuxhaven"53.87 8.7
[3,] "Lübeck" 53.87 10.69
On Dec 16, 2007 8:54 AM, Mag. Ferri Leberl <[EMAIL PROTECTED]> wrote:
> Dear everybody!
> Please find attached a tin
Dear everybody!
Please find attached a tiny R-program. It returns:
[,1] [,2]
[1,] 53.55 NA
[2,] 53.55 NA
[3,] 53.55 NA
How can I manage the first column to show the second component not only
of the first list in küste but of the second component of every list in
küste respectively, such
Hi,
I am constructing a contingency table using xtabs. The function works great:
mo
yr Sep Oct Nov Dec
1950 -7.164486e-02 3.152674e-02 -1.283389e-02 1.570382e-01
1951 3.054293e-02 4.665234e-02 -2.445499e-04 8.720204e-02
1952 3.937034e-0
I have some data whose date column consists of two types of date entries:
(a) year-only entries (eg "1983") and
(b) full date info (eg September 12, 1962).
Here's what the non-standard date info looks like:
> mode(non.standard.dates)
[1] "numeric"
> head(non.standard.dates)
[1] July 15, 1925 F
What version of R are you using? Here is the output I got with 2.6.1:
> library(chron)
> dts=seq.dates("1/1/01","12/31/03")
> rnum=rnorm(1:length(dts))
> df=data.frame(date=dts,obs=rnum)
> agg=aggregate(df[,2],list(year=years(df[,1]),month=months(df[,1])),sum)
> levels(agg$month) # aggregate() au
In fact, even ordinary aggegate works ok with zoo's as.yearmon:
> aggregate(rnum, list(dts = as.yearmon(dts)), sum)
dts x
1 Jan 2001 4.43610085
2 Feb 2001 0.49842227
3 Mar 2001 7.52139932
4 Apr 2001 1.47917343
5 May 2001 10.64459923
6 Jun 2001 -1.22530586
7 Jul 2001 8
This does look strange. Note that aggregate.zoo in the zoo package
would work here:
> library(zoo)
> aggregate(zoo(rnum, dts), as.yearmon, sum)
Jan 2001Feb 2001Mar 2001Apr 2001May 2001Jun 2001
4.43610085 0.49842227 7.52139932 1.47917343 10.64459923 -1.22530586
Jul 20
On 16/12/2007 9:25 AM, [EMAIL PROTECTED] wrote:
> Hello the list,
>
> I am trying to write a "cleanProgramming" function to test the
> procedure I use. For example, I want to be sure that I am not using
> globals variables. The function "findGlobals" detect that.
>
> To list the globals used in
Its a FAQ
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-turn-a-string-into-a-variable_003f
On Dec 16, 2007 9:25 AM, <[EMAIL PROTECTED]> wrote:
> Hello the list,
>
> I am trying to write a "cleanProgramming" function to test the
> procedure I use. For example, I want to be sure that I am
Hello the list,
I am trying to write a "cleanProgramming" function to test the
procedure I use. For example, I want to be sure that I am not using
globals variables. The function "findGlobals" detect that.
To list the globals used in function "fun", the syntax is :
"findGlobals(fun,FALSE)$vari
Hi,
I am using aggregate() to add up groups of data according to year and month.
It seems that the function aggregate() automatically sorts the levels of
factors of the grouping elements, even if the order of the levels of factors
is supplied. I am wondering if this is a bug, or if I missed someth
I have the following code, where we need to solve for mu and sigma, when we
have mut and sdt. Can you suggest how to use a solve function in R to do
that? I am new to R and am not sure how to go from defining the functions,
to solving for them.
Thanks
truncated <- function(x)
{
mu=x[1];
sigma
Try this:
matplot(t(dft), type = "o", xlab = "Time", ylab = "Score")
On Dec 16, 2007 4:14 AM, Bob Green <[EMAIL PROTECTED]> wrote:
> Hello,
>
> Below is the code for a basic bar graph. I was seeking advice
> regarding the following:
>
> (a) For each time period there are values from 16 people.
Thanks Jim.
That works well, thanks. Is there a way I can specify the range as it
seems to adjust automatically. Also I was hoping to plot multiple
datasets on the same chart but with different colours.
In the normal plot() I can do this with col = I was hoping to do
the same with points.co
am using TNORM - rtnorm to simulate from a truncated normal distribution.
However, the current function available allows us to define the mean and SD
of the non-truncated (original) distribution and then run the simulation.
http://rss.acs.unt.edu/Rdoc/library/msm/html/tnorm.html
I would instead l
Bob Green wrote:
> Hello,
>
> Below is the code for a basic bar graph. I was seeking advice
> regarding the following:
>
> (a) For each time period there are values from 16 people. How I can
> change the colour value so that each person has a different colour,
> which recurs across each of th
John Beamer wrote:
> I am trying to draw a polar plot, which is easy enough to do in the
> plotrix package through the polar.plot function.
>
> However I would like to change the origin of the length vector. For
> instance all my length values are between 75 and 85, so instead of
> having the orig
Hello,
Below is the code for a basic bar graph. I was seeking advice
regarding the following:
(a) For each time period there are values from 16 people. How I can
change the colour value so that each person has a different colour,
which recurs across each of the three graphs/tie epriods?
(b)
On Wed, 12-Dec-2007 at 11:35AM +0100, Peter Dalgaard wrote:
|> Philippe Grosjean wrote:
|> > The problem is often a misspecification of the comment.char argument.
|> > For read.table(), it defaults to '#'. This means that everywhere you
|> > have a '#' char in your Excel sheet, the rest of the l
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