Re: [R] Changing font in boxplots

2007-08-08 Thread Robert Baer
 Hi John,

 Thanks so much for such a quick reply.
 I have tried to set all to Times font running

 par(font.lab=6) (not 4, maybe this is a local setting on my machine?)

 but now the boxplot shown has the x and y labels in Times New Roman and 
 the
 x and y axis still in Arial. Any idea why R is not setting those in Times?

Try
par(font.axis=6)


 Thanks a lot for your advice,
 Graziella

 --On 08 August 2007 09:16 -0400 John Kane [EMAIL PROTECTED] wrote:

 I don't know if boxplot will accept a font argument.m
 From ?boxplot it is not clear.
 You may need to set the par() command before the
 boxplot

 Example:
 par(font.lab=4)
 boxplot(mass ~ family, data=mydata, ylab=mass %,
 xlab=family,las=1, cex.axis=1)

 --- G Iossa, School Biological Sciences
 [EMAIL PROTECTED] wrote:

 Hi all,

 I am very new to R and this might be a simple
 question but I have looked
 everywhere you suggest before writing to you.

 I am trying to change font type from san-serif to a
 serif (Times New
 Romans) on all labels and axis of my boxplot. I have
 used this function in
 other plots before, e.g.:

 plot(residuals~lnlifespan, data=mydata, pch=psymb,
 font=6, xlab=ln
 reproductive lifespan, ylab=residuals ln mass,
 font.lab=6, cex=1.5,
 cex.axis=1.5, cex.lab=1.5)

 and found that font.lab or font.axis=6 gives Times
 font. However, when I
 try for boxplot:

 boxplot(mass ~ family, data=mydata, ylab=mass %,
 xlab=family,
 font.axis=6,  font=6, par(las=1), cex.axis=1)

 it does not work (R does not give any warning
 messages). I have also tried
 family=Times but without success. Any idea of why
 is not doing it and
 what I can do to get Times font on my boxplot?
 I run R on Windows.

 Thanks a lot,
 Graziella


 *
 Dr. Graziella Iossa

 Mammal Research Unit
 School Biological Sciences
 University of Bristol
 Woodland Road
 Bristol BS8 1UG, UK

 E-mail: [EMAIL PROTECTED]
 Tel 0044 (0)117 9288918
 Fax 0044 (0)117 3317985
 http://www.bio.bris.ac.uk/research/mammal/index.html
 http://www.bio.bris.ac.uk/people/Iossa.htm

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 *
 Dr. Graziella Iossa

 Mammal Research Unit
 School Biological Sciences
 University of Bristol
 Woodland Road
 Bristol BS8 1UG, UK

 E-mail: [EMAIL PROTECTED]
 Tel 0044 (0)117 9288918
 Fax 0044 (0)117 3317985
 http://www.bio.bris.ac.uk/research/mammal/index.html
 http://www.bio.bris.ac.uk/people/Iossa.htm

 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide 
 http://www.R-project.org/posting-guide.html
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 --- Scanned by M+ Guardian Messaging Firewall ---

HTH
Rob

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[R] dataframe layout

2007-03-14 Thread Robert Baer
Can someone remind me how to change the columns in df.a into a two column 
df.b that contains one column of data and another column of the original 
column headings as levels.

Example:
a=1:3
b=4:6
c=7:9
df.a=data.frame(a,b,c)

Should become in df.b:
dat   lev
1  a
2  a
3  a
4  b
5  b
6  b
7  c
8  c
9  c

Thanks.

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Re: [R] R for copying and pasting selected image files?

2007-03-12 Thread Robert Baer

You don't tell us your OS, but the system() command should let you use your 
OS to copy/move files on most OSs.

see ?system
Other commands that might be helpful for this job are:
?setwd
?getwd
?dir

- Original Message - 
From: Kim Milferstedt [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Monday, March 12, 2007 8:33 AM
Subject: [R] R for copying and pasting selected image files?


 Hello,

 I would like to use R to process a list of text strings. The text
 strings are filenames, encoding experimental settings. Based on the
 information in there I'd like to select certain files and copy only
 the selected files to another directory.

 The files are images and there is no need, actually no desire, to
 open them. Is there a way to use R for copying and pasting files with
 any file extension to another directory without opening them?

 Thanks already for any help,

 Kim

 __

 Kim Milferstedt
 University of Illinois at Urbana-Champaign
 Department of Civil and Environmental Engineering
 4125 Newmark Civil Engineering Laboratory
 205 North Mathews Avenue MC-250
 Urbana, IL 61801
 USA
 phone: (001) 217 333-9663
 fax: (001) 217 333-6968
 email: [EMAIL PROTECTED]
 http://cee.uiuc.edu/research/morgenroth

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Re: [R] Examples on how to READ/WRITE to database using R-Project

2007-02-21 Thread Robert Baer

 We are working on a project on forecast modeling and would like to know if
 there are any examples on how to READ/WRITE to a database (e.g. 
 PostgreSQL)
 using R-Project. I do have a sample R Script which takes input as files 
 from
 a directory and writes back output files to a directory.  I would like to
 convert this script to read from a database and write back the output to 
 a
 database.
 - Venkatesh Mantha
You probably want to install the RODBC package.  Then,
library(RODBC)
?RODBC

Also, the the relational database section of the R Data Import/Export Manual 
is worth reading:
http://cran.r-project.org/doc/manuals/R-data.html

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Re: [R] simpleR or usingR package by Verzani

2007-02-15 Thread Robert Baer
I think the name of what you want is simple.freqpoly()
Try:
library(UsingR)
?simple.freqpoly

HTH,
Rob Baer

I am a new R user and so I thought I could start with Using R for
Introductory statistics by Verzani.
In order to use some of the functions and datasets I have to install the
simpleR package which is is now inside the UsingR package. I did so
using
install.packages(UsingR). However, the functions such as
simple.freqpoly.R do not work.
I also tried to install the Simple_0.4.zip available by the author, but
I get the following message:

Error in gzfile(file, r) : unable to open connection
In addition: Warning messages:
1: error 1 in extracting from zip file
2: cannot open compressed file 'Simple/DESCRIPTION'

I should also note that I use windows XP and the latest R version.

Any suggestions?

Thank you!!

__
Danmarks Fiskeriundersøgelser er den 1.1.07 fusioneret med Danmarks Tekniske 
Universitet (DTU), Forskningscenter Risø, Danmarks Fødevareforskning, 
Danmarks Rumcenter og Danmarks Transportforskning. DTU er den fortsættende 
enhed. Læs mere om fusionen på www.detnyedtu.dk

As of 1 January 2007 The Danish Institute for Fisheries Research has merged 
with the Technical University of Denmark, Risø National Laboratory, the 
Danish Institute for Food and Veterinary Research, the Danish National Space 
Centre and the Danish Transport Research Institute. The Technical University 
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Re: [R] axis command and excel time format

2006-11-10 Thread Robert Baer
You should look at
?plot.default
?times
You need to supply an x vector (time) along with a y value vector of the 
same length. Your error message tells you that you vectors were not of equal 
length.  You are repeatedly supplying different length vectors.  times() 
takes a text vector and translates it into an object of class time. Try the 
following to see if it unscrambles things for you.  You can go on to 
suppress axis labels, etc. by using other arguments as necessary .

xrange=times(c(2:25:00,3:14:00))
yrange=c(0,10)
plot(xrange,yrange,type='n')


- Original Message - 
From: Carmen Meier [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Friday, November 10, 2006 5:13 AM
Subject: Re: [R] axis command and excel time format


 

 Carmen,

 Gabor has already given you the detail you ask for, but might try the
 following plot to see what is going wrong:

 plot(times(tt), x, type='l')

 This does not give you the EXACT control of the axis you asked for,
 but this simple plot command gives you a fairly nice result. It
 illustrates that your code is failing becasue you are plotting x
 against the index of x rather than plotting x against time. At least
 this is what I think the misunderstanding is.

 Thank you Robert,
 this is nearly what I need,  but my problem is that I need an empty
 window with axes.
 I am able to build this with any data but not with the time axis.
 I need for a special issue only horizontal lines with different colors
 and arrows, which will be inserted from a function.

 Maybe you could explain me what's the difference between
 library(zoo)
 library(chron)
 time -
 c(2:25:00,2:26:00,2:27:00,2:28:00,2:29:00,2:30:00,2:31:00,

 2:32:00,2:33:00,2:34:00,2:35:00,2:36:00,2:37:00,2:38:00,

 2:39:00,2:40:00,2:41:00,2:42:00,2:43:00,2:44:00,2:45:00,

 2:46:00,2:47:00,2:48:00,2:49:00,2:50:00,2:51:00,2:52:00,

 2:53:00,2:54:00,2:55:00,2:56:00,2:57:00,2:58:00,2:59:00,

 3:00:00,3:01:00,3:02:00,3:03:00,3:04:00,3:05:00,3:06:00,

 3:07:00,3:08:00,3:09:00,3:10:00,3:11:00,3:12:00,3:13:00,
  3:14:00)
 y - c(0,10)
 plot(times(time), y, type='n')
 -   error  in xy.coords(x, y, xlabel, ylabel, log) :'x' and 'y'
 lengths differ
 -   the error is only with type='n' and not with type='l'

 and the example without errors

 library(zoo)
 library(chron)
 time -
 c(2:25:00,2:26:00,2:27:00,2:28:00,2:29:00,2:30:00,2:31:00,

 2:32:00,2:33:00,2:34:00,2:35:00,2:36:00,2:37:00,2:38:00,

 2:39:00,2:40:00,2:41:00,2:42:00,2:43:00,2:44:00,2:45:00,

 2:46:00,2:47:00,2:48:00,2:49:00,2:50:00,2:51:00,2:52:00,

 2:53:00,2:54:00,2:55:00,2:56:00,2:57:00,2:58:00,2:59:00,

 3:00:00,3:01:00,3:02:00,3:03:00,3:04:00,3:05:00,3:06:00,

 3:07:00,3:08:00,3:09:00,3:10:00,3:11:00,3:12:00,3:13:00,
  3:14:00)
 y - c(0,10)
 plot(times(time), y, type='l')


 It is not the type='n' every other combination of data types which i
 tried  was working with the no plot option
 y - c(0,10)
 z - c(0,10)
 plot(z, y, type='n')

 Carmen

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Re: [R] axis command and excel time format

2006-11-08 Thread Robert Baer

Carmen,

Gabor has already given you the detail you ask for, but might try the 
following plot to see what is going wrong:

plot(times(tt), x, type='l')

This does not give you the EXACT control of the axis you asked for, but this 
simple plot command gives you a fairly nice result.  It illustrates that 
your code is failing becasue you are plotting x against the index of x 
rather than plotting x against time.  At least this is what I think the 
misunderstanding is.

HTH

 Please provide a complete self contained example.  I can't follow the
 partial code below; however, its likely you are plotting one thing
 and creating axes using another so there is no reason it should
 come out right.

 On 11/8/06, Carmen Meier [EMAIL PROTECTED] wrote:
 Gabor Grothendieck schrieb:
  Your code plots x which has nothing to do with xt.
 
 The same result if you change xt to x: 02:25 at the origin nothing else
 - I do not know why
 #-- your suggestion 
  mn - times(min_time)
  mx - times(max_time)
  n - 12

  t - times(seq(mn, mx, length = n))

  t - times(unique(sub(..$, 00, t)))

  axis(1, x, sub(:00$, , x)) # works only with plot data before



 Regards Carmen


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[R] unexpected read.csv behavior

2006-10-23 Thread Robert Baer
Example data in a recent post was:
   LandFill Ruminants
United States (USA) .214280   5528.16
France   200.527083   1299.87
Australia185.878368   2448.17
Russian Federation  1752.833400   2024.29
Argentina283.987320   2567.02
Brazil  1048.422480   8839.61
Colombia 265.125000   1307.61
Mexico   981.023351   1814.89
Ethiopia   9.380204   1237.49
Sudan 16.018494   1796.67
India553.425328  12360.30
Pakistan  47.159393   2346.71
China455.680191   8041.79

In trying to play with this posting, I saved the data as a .csv file where the 
first row had only two entries and read:
 , LandFill, Reminants  

I tried reading the data in with 
read.csv(ex.csv, header=TRUE) 

To my surprise this created a 3 column dataframe with the first column labeled 
as X.  X was a factor.  According to the help file:

If there is a header and the first row contains one fewer field than the 
number of columns, the first column in the input is used for the row names. 
Otherwise if row.names is missing, the rows are numbered. 

I got the expected default behavior of having the unlabeled column become row 
labels when I tried,
read.csv(ex.csv, header=TRUE, row.names=1)

Is this unexpected behavior somehow related to how I designed a .csv file or is 
there something I'm misinterpreting in the documentation?

Thanks,
Rob Baer

 version
   _   
platform   i386-pc-mingw32 
arch   i386
os mingw32 
system i386, mingw32   
status 
major  2   
minor  4.0 
year   2006
month  10  
day03  
svn rev39566   
language   R   
version.string R version 2.4.0 (2006-10-03)


 

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Re: [R] Labels for Points- 2 character labels?

2006-10-18 Thread Robert Baer
I think you want to use text()

 x=1:15; y=16:30
 plot(x,y)
 text(x,y, as.character(x), pos=4)

- Original Message - 
From: ableape [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Wednesday, October 18, 2006 4:59 PM
Subject: [R] Labels for Points- 2 character labels?


I would like to be able to label each point in a scatter plot with
 the numeric order of that point. for example, I create the following plot:

 plot(e,n)
 #
 # now I go back and create my labels
 #
 for(i in 1:length(e)) {# lets say e 
 10
pc - as.character(e_order[i])   # e_order has an
 integer array 1,4,3,2... which is the order of e
points(e15[i], n15[i],pch=pc)# this will plot
 0-9 as data labels

  }
 #
 The above works for single characters.  Now if the length of e[] is
 greater than 9, the character string converted by as.character will
 yield more than one digit. At this point, my labeling trick fails:-(

 1) is there a way to create two character labels for a plot?
 2) Or even better, is there a way to do what I am trying to
 do.  Label each point by its numeric order?

 Thanks for your help.

 Renaldo

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Re: [R] frequency table

2006-09-21 Thread Robert Baer



Robert W. Baer, Ph.D.
Associate Professor
Department of Physiology
A. T. Still University of Health Science
800 W. Jefferson St. 
Kirksville, MO 63501-1497 USA


- Original Message -   

 z = rep(c(M,F),c(50,60))
How can I get the following frequency table:
 
  Sex counts %
  F60   54.5
  M   50   45.5

How about:
data.frame(table(z),percent=100*as.vector(table(z)/sum(table(z

HTH,
Rob

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Re: [R] how to construct stripchart with coincident points centered

2006-09-08 Thread Robert Baer
You may be looking for:
 stripchart(y, method=jitter, pch=o, vertical=TRUE,jitter=.5)

Rob

Robert W. Baer, Ph.D.
Associate Professor
Department of Physiology
A. T. Still University of Health Science
800 W. Jefferson St.
Kirksville, MO 63501-1497 USA

- Original Message - 
From: Berta [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Friday, September 08, 2006 3:53 AM
Subject: [R] how to construct stripchart with coincident points centered


 Hi R-users,
 I am using stripchart with coincident points,

 y -rbinom(100, 3, 0.5)
 stripchart(y, method=stack, pch=o, vertical=TRUE)

 But the result is not centered in the sense that if a value (say value 0)
is
 repeated 7 times, the first point is ploted in the middle and the rest at
 its right side, in stead of ploting 3 at its right and 3 at its left.  Can
 anybody help?

 Thanks,
 Berta.

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Re: [R] I am surprised (and a little irritated)

2006-04-19 Thread Robert Baer
Detlef Steuer [EMAIL PROTECTED]write:
  I`m the one to blame for the readme :-) and for providing the rpms.

 If you encounter such big problems my readme sucks. But I'm open for
 critisism and will improve on the current situation for the
 release of R-2.3.0 next monday.

Even as you sit under seige, let me take this opportunity to thank you (and
ALL the other countless developers, maintainers and list contributers) for
the TIRELESS work you do to make R one of the most robust open source
projects I know about.  We the 'end users'. are indebted for your energies
and talents in ways most of us can never even contemplate.

Thanks to you all for your service to the greater good, and for being
willing to step up and incrementally strive for excellence.

Rob



Robert W. Baer, Ph.D.
Associate Professor
Department of Physiology
A. T. Still University of Health Science
800 W. Jefferson St.
Kirksville, MO 63501-1497 USA

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Re: [R] R, RMysql, and MySQL 5 Decimal Type Support

2006-03-28 Thread Robert Baer
 Whenever I have a MySQL query that returns a Decimal result to R I get
 the following warning in R:
 I get this for a simple query like SELECT 2, 2.5 !!
 Warning message:
 RS-DBI driver warning: (unrecognized MySQL field type 246 in column 1)

If I understand  what you are saying, I think you have garbled SQL syntax.
The select statement synatx is:

SELECT list_of_columns FROM tables[s] [WHERE search_conditions]

My guess is that you don't have columns named 2 or 2.5

Rob

Robert W. Baer, Ph.D.
Associate Professor
Department of Physiology
A. T. Still University of Health Science
800 W. Jefferson St.
Kirksville, MO 63501-1497 USA
- Original Message - 
From: Jason Trimble [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Tuesday, March 28, 2006 12:08 PM
Subject: [R] R, RMysql, and MySQL 5 Decimal Type Support


 Hi,












 I am using:

 R ver 2.1.1

 RMySQL ver. 0.5-7

 DBI ver. 0.1-10

 MySQL  Ver 14.12 Distrib 5.0.18.



 Please Help!

 Jason



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Re: [R] How to use: library lattice: barchart

2006-03-21 Thread Robert Baer
 it looks totaly different and I get the error message:
 x should be numeric in: bwplot.formula(x = ayield ~ avariety | asite,
 data = list(ayield = c(2,

---

 cbind(some numeric and not numeric columns)

 gives you all columns to be character and when you make data.frame
 from it it is converted to factors.

 so

  abarley = data.frame(ayield,avariety,ayear,asite)

 brings you close but than you need ayear to be factor. Either convert
 it in data frame or on fly
Actually, as the warning suggests, you have all factors in the dataframe but
you need to convert ayield to a numeric.  Something like,

abarley = data.frame(cbind(as.numeric(ayield),avariety,ayear,asite))


 barchart(ayield ~ avariety | asite, data = abarley, groups =
 factor(ayear), layout = c(1,5) )

 HTH
 Petr



Robert W. Baer, Ph.D.
Associate Professor
Department of Physiology
A. T. Still University of Health Science
800 W. Jefferson St.
Kirksville, MO 63501-1497 USA

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Re: [R] Surv object in data.frame

2006-03-15 Thread Robert Baer
This does work:
coxph(survobj~group, data=df.test[[1]]) # this works like your original

To get insight compare:
str(survobj)
str(df.test)
str(df.test[[1]])

Then note the 2nd sentence of the  following from ?coxph
Arguments:

 formula: a formula object, with the response on the left of a '~'
  operator, and the terms on the right.  The response must be a
  survival object as returned by the 'Surv' function.



Robert W. Baer, Ph.D.
Associate Professor
Department of Physiology
A. T. Still University of Health Science
800 W. Jefferson St.
Kirksville, MO 63501-1497 USA


Dear All,

a Surv object I put in a data frame behaves somehow unexpected (see
example).
If I do a Cox regression on the original Surv object it works. If I put it
in a data.frame and do the regression on the data frame it does not work.
Seemingly it has to do with the class attribute, because if I change the
class attribute to let Surv appeare first, again it works.
Is this known? Should I have found information on it?
Any comments?

Thanks

Heinz Tüchler

## example data
starttime - rep(0,5)
stoptime  - 1:5
event - c(1,0,1,1,1)
group - c(1,1,1,2,2)
## Surv object
survobj   - Surv(starttime, stoptime, event)
## Cox-regression
coxph(survobj~group) # this works
## put Surv object in data.frame
df.test - data.frame(survobj=I(survobj), group)
## Cox-regression on data.frame
coxph(survobj~group, data=df.test) # this does not work
attr(df.test$survobj, 'class') # survobject has class AsIs, Surv
attr(df.test$survobj, 'class') - c('Surv', 'AsIs') # put Surv first
attr(df.test$survobj, 'class') # survobject has class Surv, AsIs
coxph(survobj~group, data=df.test) # now it works

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Re: [R] 2 barplots in the same graph

2006-02-22 Thread Robert Baer
See ?barplot

If I understand what you want, try:

barplot(x1,border=red,density=0)
par(new=TRUE)
barplot(x2,border=green,density=0)

Rob

Robert W. Baer, Ph.D.
Associate Professor
Department of Physiology
A. T. Still University of Health Science
800 W. Jefferson St.
Kirksville, MO 63501-1497 USA
- Original Message - 
From: jia ding [EMAIL PROTECTED]
To: R-help r-help@stat.math.ethz.ch
Sent: Wednesday, February 22, 2006 7:31 AM
Subject: [R] 2 barplots in the same graph


 Hello,

 I have a very simple question about 2 barplots in the same graph.

 It seems quite easy, but I searched google for long time, haven't find
 solution.

 For example, I want one graph like:
 x1=seq(0,2,by=0.3)
 x2=seq(3,0,by=-0.1)
 barplot(x1,col=red)
 barplot(x2,col=green)

 It means if it's on the same graph, some bars are overlaped.
 So if the bars are hollow, instead of filled with color, it will be
better.

 Actually, I think it's something similar with matlab's hold on command.

 Thanks!

 Nina

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Re: [R] print formula on plot

2006-02-08 Thread Robert Baer
Only put the expression inside the expression?

 plot(x,z[3]*x^2+z[2]*x+z[3],type=l, main=My nice plot)
 text(-0.9,5,paste(zs[1],' ',expression(x^3)))  # should work


Robert W. Baer, Ph.D.
Associate Professor
Department of Physiology
A. T. Still University of Health Science
800 W. Jefferson St.
Kirksville, MO 63501-1497 USA
- Original Message - 
From: Thomas Steiner [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Wednesday, February 08, 2006 7:47 AM
Subject: [R] print formula on plot


 I estimate some parameters and I want to print them (pretty) on my plot:

 # somehow estimated parameters
 z-c(1.543523e+00, 1.23453e+00, 3.454000e+00)

 x-seq(-1,1,length=100)
 plot(x,z[3]*x^2+z[2]*x+z[3],type=l, main=My nice plot of the
 estimated function)
 zs-format(z,digits=4,scientific=FALSE,trim=TRUE)

 text(-0.9,7,expression(1.54*x^2))   # is what I want,
 but DYNAMIC
 text(-0.9,6,expression(paste(zf[1],x^3))) # not really
 text(-0.9,5,expression(paste(toSting(zf[1]),x^3)))  # not really
 #using z (double) instead of zf (string) does not help.

 So my question is:
 How do I evaluate zf[1] from the variable to it's (String) value? Here
 it is used within an expression: this makes all the trouble.

 Thanks for help, I tried 2 hrs now...
 Thomas

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Re: [R] Newbie question on using friedman.test()

2006-01-17 Thread Robert Baer
?friedman.test

Says:
Description:
Performs a Friedman rank sum test with unreplicated blocked data.

Usage:
friedman.test(y, ...)

  y: either a numeric vector of data values, or a data matrix.

So assuming your data, d, is unreplicated blocked data, perhaps
 d=as.matrix(d)
 friedman.test(d)
or simply,
friedman.test(as.matrix(d))
???




Robert W. Baer, Ph.D.
Associate Professor
Department of Physiology
A. T. Still University of Health Science
800 W. Jefferson St.
Kirksville, MO 63501-1497 USA
- Original Message - 
From: Bill Kules [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Tuesday, January 17, 2006 12:16 PM
Subject: [R] Newbie question on using friedman.test()


 I am trying to use the friedman.test() on a data frame, d, but
 I am receiving the following error message:


  d
AW HS IAC WA
 1   6  8   3  5
 2   2  2   3  6
 3   7  7   8  3
 4   8  5   4  5
 
 20  2  5   2  7
 21  7  7   6  7
 22  7  8   6  8
 23  6  8   4  5
 24  5  7   5  2
  friedman.test(d)
 Error in any(is.na(groups)) : argument groups is missing, with no
default

 I think I just need to convert the data frame to a matrix, and then
 friedman.test() will get the roups and blocks automatically.

 Question 1) Is my understanding correct?

 Question 2) What R function will convert the data frame to the matrix I
 need?
 I'm still figuring out the matrix functions, and I would appreciate any
 pointers or examples.  The help() section is sometimes a bit terse...

 Thanks in advance from a Newbie,
 Bill

 PS - to recreate the above data frame d:
 d -
 structure(list(AW = as.integer(c(6, 2, 7, 8, 8, 8, 7, 5, 3, 6,
 8, 7, 6, 4, 8, 7, 8, 7, 7, 2, 7, 7, 6, 5)), HS = as.integer(c(8,
 2, 7, 5, 4, 5, 7, 7, 2, 8, 4, 7, 8, 7, 6, 7, 5, 8, 8, 5, 7, 8,
 8, 7)), IAC = as.integer(c(3, 3, 8, 4, 7, 5, 8, 3, 4, 3, 7, 4,
 6, 5, 6, 7, 8, 6, 8, 2, 6, 6, 4, 5)), WA = as.integer(c(5, 6,
 3, 5, 3, 6, 7, 8, 3, 5, 6, 7, 7, 7, 7, 6, 4, 7, 8, 7, 7, 8, 5,
 2))), .Names = c(AW, HS, IAC, WA), row.names = c(1,
 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,
 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24
 ), class = data.frame)

 
 Bill Kules
 Principal, Takoma Software, Inc., Takoma Park, MD
   www.takomasoftware.com
 Ph.D. Candidate, University of Maryland Human-Computer Interaction Lab
   www.cs.umd.edu/hcil

 [EMAIL PROTECTED]
 (301) 405-2725 voice
 (301) 891-7271 voice + voicemail
 (301) 891-7273 fax
 (301) 755-7982 mobile
 

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Re: [R] ylim problem in barplot

2006-01-05 Thread Robert Baer
 PaulB When I use barplot but select a ylim value greater
 PaulB than zero, the graph is distorted.  The bars extend
 PaulB below the bottom of the graph.
 PaulB For instance the command produces a problematic graph.
 PaulB barplot(c(200,300,250,350),ylim=c(150,400))

 Well, my question would be if that is not a feature :-)
 Many people would consider barplots that do not start at 0 as
  Cheating with Graphics  (in the vein of Lying with Statistics).

Well, consider this example:
barplot(c(-200,300,-250,350),ylim=c(-99,400))

It seems that barplot uses ylim and pretty to decide things about the axis
but does some slightly unexpected things with the bars themselves that are
not just at the 'zero' end of the bar.

Rob


Robert W. Baer, Ph.D.
Associate Professor
Department of Physiology
A. T. Still University of Health Science
800 W. Jefferson St.
Kirksville, MO 63501-1497 USA

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Re: [R] forrest plot

2005-10-20 Thread Robert Baer
Try:
metaplot(mn=c(-0.28174,-0.71444,-0.12375,-0.12426,-0.30011,-0.45058,-0.07324
),se=c(0.20766,0.42691,0.26366,0.30357,0.31819,0.28636,0.37758),xlab=HR and
95%CI,logeffect=T,xaxt=n,xlim=c(-1.7,1.7))

axis(side=1,at=c(0.2,0.3,0.6,1.0,1.4,2.0,3.0,5.4),labels=c(0.2,0.3,0.6,1.0,1
4,2.0,3.0,5.4))

Rob

Robert W. Baer, Ph.D.
Associate Professor
Department of Physiology
A. T. Still University of Health Science
800 W. Jefferson St.
Kirksville, MO 63501-1497 USA
- Original Message - 
From: Michela Ballardini [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Thursday, October 20, 2005 8:55 AM
Subject: [R] forrest plot


Hello
I'm trying to plot hazard risk values using the function metaplot with
 the specifications:

metaplot(mn=c(-0.28174,-0.71444,-0.12375,-0.12426,-0.30011,-0.45058,-0.07324
),se=c(0.20766,0.42691,0.26366,0.30357,0.31819,0.28636,0.37758),xlab=HR and
95%CI,logeffect=T,xaxt=n)

axis(side=1,at=c(0,0.2,0.4,0.6,0.8,1.0,1.2,1.4,1.6,1.8,2.0),labels=c(0,0.2,0
4,0.6,0.8,1.0,1.2,1.4,1.6,1.8,2.0))



However, in the plot the x axis is on a log scale and tends to overextend
the left end of the axis. How can I transform the x-scale on a linear scale
with equidistant points?


Thank you very much for your attenction

Mic

**
Dr.ssa Michela Ballardini
Unità di Biostatistica e Sperimentazioni Cliniche
c/o Osp. Morgagni-Pierantoni - Pad. Valsalva
Via Forlanini, 34
47100 Forlì
Tel 0543-731836
Tel/Fax 0543-731612
**


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Re: [R] Dunn's post hoc test

2005-10-17 Thread Robert Baer
I think Martin told you the basic approach to the indexing:

averank-sort(sample(1:100,25,replace=TRUE))
averank[-1] - averank[-length(averank)]
 [1]  1  1  6  3  4 14  1  1  8  1  2  6  5  4 10  0  3  2  1 11  1  1  2  0
averank
 [1]  4  5  6 12 15 19 33 34 35 43 44 46 52 57 61 71 71 74 76 77 88 89 90 92
92


- Original Message - 
From: IAIN GALLAGHER [EMAIL PROTECTED]
To: Martin Henry H. Stevens [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Sent: Monday, October 17, 2005 2:29 PM
Subject: Re: [R] Dunn's post hoc test


 Thanks for your reply Hank. It's not really what I'm
 after (though it's good to know).

 For the test ( as described in Statistics for the
 Biosciences by W. Gardiner. Prentice Hall, 1997) I
 have to rank my groups, calculate the average rank,
 then subtratc each average rank from every other. Any
 value greater than the test statistic is significant.

 eg average rank difference table:

2 5 89
 ---|-- 
 2  - 3 67
|
 5  - - 34
|
 8  - - -1
|
 9  - - --
|

 I can't get my head around writing an algorithm for
 this if I have a vector of average ranks eg averank-
 c(2,5,8,9).

 I know I can address the vector by index and that this
 is probably the correct route but I can't get the
 indexing algorithm right!

 I'm sure someone will point out somethng simple and
 I'll kick myself but the help would be appreciated.

 Thanks again.

 Iain Gallagher

 --- Martin Henry H. Stevens [EMAIL PROTECTED]
 wrote:

  I don't know Dunn's rank test, but the following
  substracts each of
  the sums of averanks from the next rank.
 
  cumsum(averank)[-length(averank)] - averank[-1]
 
  Hank
 
  On Oct 17, 2005, at 4:30 AM, Iain Gallagher wrote:
 
   Hi Everyone.
  
   I am rather new to R and I've been trying to
  implement a function to
   carry out the above test. For a couple of days now
  I've been stuck on
   how to generate average rank differences.
  
   Say I have a vector of average ranks:
  
   averank- c(2,5,9,12)
  
   I would like to subtract averank[1] from
  averank[2], averank[1] and
   averank[2] from averank[3] and averank[1],
  averank[2] and averank[3]
   from averank[4] etc (I know the syntax is wrong
  here... it's just for
   illustration) but I can't work out how to do it.
  
   Ideally I would like to generate an array showing
  the differences
   between the average ranks so I could tell at a
  glance which ones were
   greater than my critical value
  
   I've been looking at loops etc but it's a little
  beyond me at the
   moment. Thanks for any suggestions.
  
   Iain Gallagher
   IIIR
   Edinburgh University
  
   __
   R-help@stat.math.ethz.ch mailing list
   https://stat.ethz.ch/mailman/listinfo/r-help
   PLEASE do read the posting guide!
  http://www.R-project.org/posting-
   guide.html
  
 
  Dr. Martin Henry H. Stevens, Assistant Professor
  338 Pearson Hall
  Botany Department
  Miami University
  Oxford, OH 45056
 
  Office: (513) 529-4206
  Lab: (513) 529-4262
  FAX: (513) 529-4243
  http://www.cas.muohio.edu/~stevenmh/
  http://www.muohio.edu/ecology/
  http://www.muohio.edu/botany/
  E Pluribus Unum
 
 

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Re: [R] Dunn's post hoc test

2005-10-17 Thread Robert Baer
I think I misunderstood your follow-up question.  Try this:

 averank-sort(sample(1:100,10,replace=TRUE))
 x=matrix(nrow=length(averank),ncol=length(averank))
 for (i in 1:length(averank)){
+ for (j in 1:length(averank)){
+ x[i,j] - averank[i] - averank[j]
+ }}
 x
  [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
 [1,]0  -14  -15  -16  -16  -23  -37  -51  -67   -79
 [2,]   140   -1   -2   -2   -9  -23  -37  -53   -65
 [3,]   1510   -1   -1   -8  -22  -36  -52   -64
 [4,]   162100   -7  -21  -35  -51   -63
 [5,]   162100   -7  -21  -35  -51   -63
 [6,]   2398770  -14  -28  -44   -56
 [7,]   37   23   22   21   21   140  -14  -30   -42
 [8,]   51   37   36   35   35   28   140  -16   -28
 [9,]   67   53   52   51   51   44   30   160   -12
[10,]   79   65   64   63   63   56   42   28   12 0
 averank
 [1] 15 29 30 31 31 38 52 66 82 94



- Original Message - 
From: IAIN GALLAGHER [EMAIL PROTECTED]
To: Martin Henry H. Stevens [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Sent: Monday, October 17, 2005 2:29 PM
Subject: Re: [R] Dunn's post hoc test


 Thanks for your reply Hank. It's not really what I'm
 after (though it's good to know).

 For the test ( as described in Statistics for the
 Biosciences by W. Gardiner. Prentice Hall, 1997) I
 have to rank my groups, calculate the average rank,
 then subtratc each average rank from every other. Any
 value greater than the test statistic is significant.

 eg average rank difference table:

2 5 89
 ---|-- 
 2  - 3 67
|
 5  - - 34
|
 8  - - -1
|
 9  - - --
|

 I can't get my head around writing an algorithm for
 this if I have a vector of average ranks eg averank-
 c(2,5,8,9).

 I know I can address the vector by index and that this
 is probably the correct route but I can't get the
 indexing algorithm right!

 I'm sure someone will point out somethng simple and
 I'll kick myself but the help would be appreciated.

 Thanks again.

 Iain Gallagher

 --- Martin Henry H. Stevens [EMAIL PROTECTED]
 wrote:

  I don't know Dunn's rank test, but the following
  substracts each of
  the sums of averanks from the next rank.
 
  cumsum(averank)[-length(averank)] - averank[-1]
 
  Hank
 
  On Oct 17, 2005, at 4:30 AM, Iain Gallagher wrote:
 
   Hi Everyone.
  
   I am rather new to R and I've been trying to
  implement a function to
   carry out the above test. For a couple of days now
  I've been stuck on
   how to generate average rank differences.
  
   Say I have a vector of average ranks:
  
   averank- c(2,5,9,12)
  
   I would like to subtract averank[1] from
  averank[2], averank[1] and
   averank[2] from averank[3] and averank[1],
  averank[2] and averank[3]
   from averank[4] etc (I know the syntax is wrong
  here... it's just for
   illustration) but I can't work out how to do it.
  
   Ideally I would like to generate an array showing
  the differences
   between the average ranks so I could tell at a
  glance which ones were
   greater than my critical value
  
   I've been looking at loops etc but it's a little
  beyond me at the
   moment. Thanks for any suggestions.
  
   Iain Gallagher
   IIIR
   Edinburgh University
  
   __
   R-help@stat.math.ethz.ch mailing list
   https://stat.ethz.ch/mailman/listinfo/r-help
   PLEASE do read the posting guide!
  http://www.R-project.org/posting-
   guide.html
  
 
  Dr. Martin Henry H. Stevens, Assistant Professor
  338 Pearson Hall
  Botany Department
  Miami University
  Oxford, OH 45056
 
  Office: (513) 529-4206
  Lab: (513) 529-4262
  FAX: (513) 529-4243
  http://www.cas.muohio.edu/~stevenmh/
  http://www.muohio.edu/ecology/
  http://www.muohio.edu/botany/
  E Pluribus Unum
 
 

 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html


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Re: [R] Getting eps into Word documents.

2005-10-04 Thread Robert Baer
  On 03-Oct-05 Marc Schwartz (via MN) wrote:
   On Mon, 2005-10-03 at 16:31 -0300, Rolf Turner wrote:
   A student in one of my courses has asked me about getting R graphics
   output (under Linux) into a Word document.  I.e. she wants to do her
   R thing under Linux, but then do her word processing using Word.
--snip--
   So use something like the following:
  
  
   postscript(RPlot.eps, height = 4, width = 4,
  horizontal = FALSE, onefile = FALSE,
  paper = special)
  
   plot(1:5)
  
   dev.off()
  
  
   You can then import the .eps file into Word or most other such
   applications that can import encapsulated postscript files.

-snip
   More information is available from MS here:
  
   http://support.microsoft.com/?kbid=290362
  
   HTH,
  
   Marc Schwartz
--snip---
  b) It won't work anyway if printed to a non-PostScript printer.

 True, which is the case irrespective of Word/Windows. If you don't have
 a PS printer locally or accessible via network, you can always install a
 PS printer driver and print to a file, which can then be printed by a
 third party if required.

Well, as a lowly Windows and Office user, I most often right click on R
grahics, cut to clipboard, and paste into Word.   So one possiblility is for
the student to install R on her own machine (Windows or Mac?).

But I just tried Marc's suggestion, and it looks VERY VIABLE to me.  I
generated the graph from his code snippet and used Insert picture from
file in Word 2003 to place the graphic in a Word document.  I then tried
printing on both an HP 4100 TN laserjet and an HP 960c deskjet.  The image
printed perfectly on both printers with crisp lines and text that apprear to
be vector-based not degraded bitmapped representations.  Certainly worth the
student trying.

HTH,
Rob Baer

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Re: [R] Problem with tick marks in lines.survfit (package survival)

2005-09-19 Thread Robert Baer
How about:
plot(fit, mark.time=TRUE, xscale=365.24,xlab='Years',
ylab='Survival',lwd=c(2,1))

Rob
- Original Message - 
From: Rachel Pearce [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Monday, September 19, 2005 4:59 AM
Subject: [R] Problem with tick marks in lines.survfit (package survival)


 I have attempted to follow posting guidelines but I have failed to find
out
 what I am doing wrong here.
  I am trying to use lines.survfit to plot a second curve onto a survival
 curve produced by plot.survfit. In my case this is to be a progression
free
 survival curve superimposed upon an overall survival curve, but I will
 illustrate my problem using the example given in the help for
lines.survfit.
  I would like to have tick marks for censored data points on both curves,
so
 I would like to do something like:
   fit - survfit(Surv(time, status) ~ sex, pbc,subset=1:312)
  plot(fit[2], mark.time=TRUE, xscale=365.24,
 + xlab='Years', ylab='Survival')
  lines(fit[1], lwd=2, xscale=365.24, mark.time=TRUE)
  but when I do this, no tick marks appear on the second curve. Can anyone
 see what I am doing wrong here?
  Rachel

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Re: [R] The Perils of PowerPoint

2005-09-02 Thread Robert Baer
The R relevance here might be that all the statistics in the world wrongly
applied to data will only bury its information content...R and
Powerpoint (and Matlab and Perl and...) are all terrific tools for turning
data into knowledge, but tools DO NOT relieve us of the necessity of
thinking about and analyzing the meaning of the data with our intellect as
well.  It is wrong to blame ANY tool for our own shortcomings!

My two cents,
Rob

- Original Message - 
From: Marc Schwartz [EMAIL PROTECTED]
To: R-Help r-help@stat.math.ethz.ch
Sent: Friday, September 02, 2005 8:18 AM
Subject: [R] The Perils of PowerPoint


 Hi all,

 Below is a URL for an editorial published today in our local newspaper,
 the Minneapolis StarTribune. It was originally published in the
 Washington Post a couple of days ago:


http://www.washingtonpost.com/wp-dyn/content/article/2005/08/29/AR2005082901444.html

 but that site requires registration. The 'Strib site seems to be open
 for the moment:

 http://www.startribune.com/stories/1519/5591930.html


 I thought folks might find it interesting.

 Best regards,

 Marc Schwartz

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