Re: [R] "ungültige Versionsspezifikation"

2007-08-16 Thread Mag. Ferri Leberl
Thank you. Hereby I send you the return to sessionInfo(). I have
meanwhile updated to 2.5.1.

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> library(cairo)
Fehler in package_version(vers) : ungültige Versionsspezifikation
> sessionInfo()
R version 2.5.1 (2007-06-27)
i486-pc-linux-gnu

locale:
LC_CTYPE=de_DE.UTF-8;LC_NUMERIC=C;LC_TIME=de_DE.UTF-8;LC_COLLATE=de_DE.UTF-8;LC_MONETARY=de_DE.UTF-8;LC_MESSAGES=de_DE.UTF-8;LC_PAPER=de_DE.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=de_DE.UTF-8;LC_IDENTIFICATION=C

attached base packages:
[1] "stats" "graphics"  "grDevices" "utils" "datasets"
"methods"
[7] "base"
>




Am Mittwoch, den 15.08.2007, 15:26 +0200 schrieb Martin Maechler:
> >>>>> "JK" == John Kane <[EMAIL PROTECTED]>
> >>>>> on Wed, 15 Aug 2007 08:55:59 -0400 (EDT) writes:
> 
> JK> I think we need more information about your system.
> JK> Please run
> JK> sessionInfo()
> JK> and include the information in another posting.
> 
> Yes, indeed.
> However R version 2.3.1 seems a bit too old to fit to a current
> version of cairo (rather 'cairoDevice' ??).
> 
> And BTW: It is a *package*, not a library!!!
> 
> Martin Maechler, ETH Zurich
> 
> JK> --- "Mag. Ferri Leberl" <[EMAIL PROTECTED]> wrote:
> 
> >> Dear everybody,
> >> excuse me if this question ist trivial, however, I
> >> have now looked for
> >> an answer for quite a while and therefore dare
> >> placing it here.
> >> I want to export .svg-files and got here the advice
> >> to employ the
> >> cairo-library.
> >> I downloaded the *current*-version here and expanded
> >> it
> >> to /usr/local/lib/R/site-library.
> >> library(cairo) returned ungültige
> >> Versionsspezifikation (INVALID VERSION
> >> SPECIFICATION).
> >> I got some answer here (EPM39), but, stupid enough,
> >> I could not employ
> >> it as I don't know where to look for the variable
> >> named there.
> >> The R-version I employ is 2.3.1.
> >> Can anybody solve the (probably simple) problem?
> >> Thank you in advance.
> >> Yours, Ferri
> 
> 
> 
> JK> 
> 
> JK> __
> JK> R-help@stat.math.ethz.ch mailing list
> JK> https://stat.ethz.ch/mailman/listinfo/r-help
> JK> PLEASE do read the posting guide 
> http://www.R-project.org/posting-guide.html
> JK> and provide commented, minimal, self-contained, reproducible code.

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[R] "ungültige Versionsspezifikation"

2007-08-15 Thread Mag. Ferri Leberl
Dear everybody,
excuse me if this question ist trivial, however, I have now looked for
an answer for quite a while and therefore dare placing it here.
I want to export .svg-files and got here the advice to employ the
cairo-library.
I downloaded the *current*-version here and expanded it
to /usr/local/lib/R/site-library.
library(cairo) returned ungültige Versionsspezifikation (INVALID VERSION
SPECIFICATION).
I got some answer here (EPM39), but, stupid enough, I could not employ
it as I don't know where to look for the variable named there.
The R-version I employ is 2.3.1.
Can anybody solve the (probably simple) problem?
Thank you in advance.
Yours, Ferri

[[alternative HTML version deleted]]

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[R] making groups

2007-07-09 Thread Mag. Ferri Leberl
Dear everybody!
If I have an array of numbers e.g. the points my students got at an
examination, and a  key to group the numbers, e.g. the key which
interval corresponds with which mark (two arrays of the same length or
one 2x(number of marks)), how can I get the array of absolute
frequencies of marks?
I hope I have expressed my problem clearly.
Thank you in advance.
Mag. Ferri Leberl

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Re: [R] histogram with absolute figures

2007-07-09 Thread Mag. Ferri Leberl
Meanwhile I have recognized, that the breaks-option enforces density as
the default. But if I try to force frequencies (freq=TRUE) I get the
following feedback:

Warning message:
the AREAS in the plot are wrong -- rather use freq=FALSE in:
plot.histogram(r, freq = freq, col = col, border = border, angle =
angle,

And the machine hasn't promised too much: the result IS wrong.
Yours,
Mag. Ferri Leberl



Am Freitag, den 06.07.2007, 16:17 -0400 schrieb Sarah Goslee:
> The default of hist() is counts rather than percentages.
> 
> Sarah
> 
> On 7/6/07, Mag. Ferri Leberl <[EMAIL PROTECTED]> wrote:
> > Dear everybody!
> > Is ist easily possible to make up a histogram with absolute numbers
> > instead of percentages?
> > Thank you in advance!
> > Yours, Mag. Ferri Leberl
> >
> > ___

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[R] histogram with absolute figures

2007-07-06 Thread Mag. Ferri Leberl
Dear everybody!
Is ist easily possible to make up a histogram with absolute numbers
instead of percentages?
Thank you in advance!
Yours, Mag. Ferri Leberl

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[R] inequality with NA

2006-09-22 Thread Mag. Ferri Leberl
Dear everybody!
take a<-c(5,3,NA,6).

if(a[1]!=NA){b<-7}
if(a[3]!=5){b<-7}
if(a[3]!=NA){b<-7}
if(a[3]==NA){b<-7}

will alltogeather return

Fehler in if (a[1] != NA) { : Fehlender Wert, wo TRUE/FALSE nötig ist

(or simularly). Somehow this is logical. But how else should I get out,
whether a certain vector-component has an existing value?
Thank you in advance!
Yours,
Mag. Ferri Leberl

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[R] changing the index of a histogram

2006-09-21 Thread Mag. Ferri Leberl
Dear everybody!
I am trying to make up a histogram of marks of a recent test I have
corrified.
Say I have a vector "points" and a vektor "breaks" to segregate the
marks. Finally there is a vector "marks".

hist(points, breaks=breaks) leaves me with two problems:
At first the width of the bars of the histogram corresponds with the
width of the intervall. This will look confusing and not very elaborated
when in the end the index shall not be "breaks", but "marks".
At second, you guessed it, I have the problem to exchange "breaks" for
"marks" on the y-axis, at least surfacially.
A glance into old mailing list brought me 
plot(table(cut(points,breaks=breaks,right=FALSE,include.lowest=TRUE)))
as a good solution to the first matter (except that right=FALSE seems
contradictory ;)
I would be very glad if you could give me a hint to solve the second
problem, most preferably such, that the first would be circumvaded by
transforming the variable intervals of "breaks" into the uniform ones of
"marks" and then employing hist().
Thank you in advance.
Yours,
Mag. Ferri Leberl

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[R] Export to LaTeX

2006-07-14 Thread Mag. Ferri Leberl
Dear Everybody!
I want to export data to LaTeX. As I want to employ the data as freely as 
possible I want to avoid the xtable-command and instead generate some List 
like

\MyOwnPrettyCommand{Adam}{Auer}{17}
\MyOwnPrettyCommand{Bertram}{Bauer}{14}
\MyOwnPrettyCommand{Christoph}{Huber}{75}
\MyOwnPrettyCommand{Damian}{Dorfer}{69}
\MyOwnPrettyCommand{Emanuel}{Eder}{43}

with \MyOwnPrittyCommand defined elsewhere.
As a pitty, if I make up about such a table in r, lets call it "A", and use 
the commands

sink("tabelle.tex")
A
sink("anderedatei")

tabelle.tex will look like this:

 [,1]  [,2][,3] [,4] [,5] [,6] [,7]
[1,] "MyOwnPrettyCommand{" "Adam"  "}{" "Auer"   "}{" "17" "}" 
[2,] "MyOwnPrettyCommand{" "Bertram"   "}{" "Bauer"  "}{" "14" "}" 
[3,] "MyOwnPrettyCommand{" "Christoph" "}{" "Huber"  "}{" "75" "}" 
[4,] "MyOwnPrettyCommand{" "Damian""}{" "Dorfer" "}{" "69" "}" 
[5,] "MyOwnPrettyCommand{" "Emanuel"   "}{" "Eder"   "}{" "43" "}" 

So my question is how to export the data properly, without line-indices, 
without quotes but WITH backslashes.
Thank you in advance.
Yours,
Mag. Ferri Leberl

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[R] xtable

2005-08-30 Thread Mag. Ferri Leberl
I have installed package xtable with

su -c 'R CMD INSTALL xtable'

and got this promising feedback:

* Installing *source* package 'xtable' ...
** R
** data
** help
 >>> Building/Updating help pages for package 'xtable'
 Formats: text html latex example
* DONE (xtable)

Despite that, R returns:

Error: couldn't find function "print.xtable"
Execution halted

when I call that function.
What have I done wrong? Do I have to activate the package everytime when I 
start R?
Thank you in advance!

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[R] tiing an array to a data frame

2005-08-29 Thread Mag. Ferri Leberl
Dear colleagues!
I am afraid this is an easy question but as a pitty I did not find out on my 
own, so please be patient with my question:
If I have a data frame X and an array Y. Which is the command to make Y become 
an additional row of X?
Thank you in advance.
Yours, Mag. Ferri Leberl

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[R] Contingency-Coefficient, Variance

2005-07-17 Thread Mag. Ferri Leberl
Dear Everybody!
Excuse me for this newbie-questions, but I have not found answers to these 
Question:

- Is there a command calculating the variance if the whole group is known 
(thus dividing by n instead of n-1)?

- Which is the command to calculate the contingency-coefficient?

Thank you in advance.
Mag. Ferri Leberl

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[R] visualizing frequencies

2005-06-26 Thread Mag. Ferri Leberl
Dear everybody,

In our game-theory lesson we have run several classroom-experiments where the 
students had to decide for a natural number between one and seven. I have 
troubles now to visualize the results: be a the vector of answers.

hist(a) will not assume natural numbers as answers, but rational. It will make 
the brakes exactly at the natural numbers,  which is difficult to interpret, 
as only natural numbers may be employed.

barplot(a) or barplot(a,1:7) will not aggregate the answers. If three students 
returned the number seven, it will show three bars to the size of seven 
instead one bar to the size of three on index seven.

barplot(1:7,a), in this case, will show the bar at index seven, wut it will be 
to the hight of seven and to the width of three.

I also wanted to show the results of the different versions of the experiment 
in ONE plot. As the number of participants varied I googled around in the 
R-Archives and got to recognize plot.edf. As a pitty, this function seems to 
set the index the wrong way round: the function starts with the number of 
students deciding for seven, but indexes them with one.

Can anybody help me with these two problems?
Thank you in advance!

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[R] empty pdf

2005-04-21 Thread Mag. Ferri Leberl
How can I produce an empty pdf-page (thus with or without frame, but anyway 
without axes)?
I want to add a textpage to my sheets, as in 
<http://www.r-project.org/nocvs/mail/r-devel/2002/0987.html>
Thank you in advance.
Yours, sincerely
Mag. Ferri Leberl

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[R] Histogram

2005-04-20 Thread Mag. Ferri Leberl
Dear everybody!
I am analysing data from an enquette. The answers are either A or B. How can I 
draw a histogram without transforming the data from characters to numbers? If 
the data are saved in a list M, hist(M[,1]) returns:

Error in hist.default(M[, 1]) : `x' must be numeric
Execution halted

Thank you in advance!

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[R] Error in hist.default(A) : `x' must be numeric

2005-04-06 Thread Mag. Ferri Leberl
Dear everybody!
I have load a list A of numbers and want a histogram to be drawn.
on
hist(Y)
the Machine returns:
Error in hist.default(A) : `x' must be numeric
I found out, that the list is of type data.frame.
Y<-as.numeric(Y)
returns
Error in as.double.default(A) : (list) object cannot be coerced to double
What schould I do?
Than you in advance!

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[R] Warning: number of items to replace is not a multiple of replacement length

2004-09-29 Thread Mag. Ferri Leberl
What does this warning mean precisely?
Is there any reason to care about it?
Can I Avoid it by another way of programming?
Thank you in advance.

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[R] There were 50 or more warnings (use warnings() to see the first 50)

2004-09-16 Thread Mag. Ferri Leberl
I employ R in the Slave-Mode.
The slave returns me the following feedback:

There were 50 or more warnings (use warnings() to see the first 50)

I have found no way so far to get the warnings viewed. Which command would be 
appropriate? warnings() (without an argument) returns NULL.
Thank you in advance.

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[R] adress the current index

2004-05-24 Thread Mag. Ferri Leberl
How can I adress the current index of a vector?

I want to work with time series and therefore give the n-th element of a 
vector some value dependent on the value of the n-1th element.

Thank you in advance.

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[R] (kein Betreff)

2004-03-31 Thread Mag. Ferri Leberl
Dear colleagues!

How can I calculate the mean of every line of "feld" without using the command 
"for"?

Thank You in advance


feld<-array(,c(100,10))
mittel<-array(,c(100,1))
feld[,]<-rnorm(1000)
for(a in 1:100){mittel[a]<-mean(feld[a,])}

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