Re: [R] simplification of code using stamp?

[Rainer M Krug]

hadley wickham wrote:
 I get the following, which is not what I am looking for.

   test[1:10,]
 expertxx seeds runvalue
 1  BW x001025   1 rsqs, slope, d.slope, intercept, d.intercept
 2  BW x001025   2 rsqs, slope, d.slope, intercept, d.intercept
 3  BW x001025   3 rsqs, slope, d.slope, intercept, d.intercept
 4  BW x001025   4 rsqs, slope, d.slope, intercept, d.intercept
 5  BW x001025   5 rsqs, slope, d.slope, intercept, d.intercept
 6  BW x001028   1 rsqs, slope, d.slope, intercept, d.intercept
 7  BW x001028   2 rsqs, slope, d.slope, intercept, d.intercept
 8  BW x001028   3 rsqs, slope, d.slope, intercept, d.intercept
 9  BW x001028   4 rsqs, slope, d.slope, intercept, d.intercept
 10 BW x001028   5 rsqs, slope, d.slope, intercept, d.intercept
 
 That's because stamp doesn't produce great output at the moment (have
 a look at the str(test) to see that you have a data frame containing a
 list of vectors)
 
 The following code should put it a more reasonable form:
 
 tidy - function(x) {
  bind - function(i) data.frame(x[i, -ncol(x),drop=FALSE], t(a$value[[i]]))
  l - lapply(1:nrow(x), bind)
  do.call(rbind.fill, l)
 }
 
 tidy(test)
Thanks Hadley - I'll look at the code.


 
 (but unless you provide a reproducible example I can't be sure that it
 works with your data)
 
 Regards,
 
 Hadley


-- 
Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation
Biology (UCT)

Department of Conservation Ecology and Entomology
University of Stellenbosch
Matieland 7602
South Africa

Tel:+27 - (0)72 808 2975 (w)
Fax:+27 - (0)21 808 3304
Cell:   +27 - (0)83 9479 042

email:  [EMAIL PROTECTED]
[EMAIL PROTECTED]

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[R] Error bars in ggplot?

[Rainer M Krug]

Hi

Is it possible to put error bars in ggplot? I couldn't find anything so far.

Rainer

-- 
Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation
Biology (UCT)

Department of Conservation Ecology and Entomology
University of Stellenbosch
Matieland 7602
South Africa

Tel:+27 - (0)72 808 2975 (w)
Fax:+27 - (0)21 808 3304
Cell:   +27 - (0)83 9479 042

email:  [EMAIL PROTECTED]
[EMAIL PROTECTED]

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[R] Installation of RGrace

[Rainer M Krug]

Hi

I try to install RGrace, but always get the following error below.

Might it have to do that RGtk is not installed and where can I find it 
to install it?

I have GTK 1.2.10 installed.

My R version and platform:

  version
_
platform   i686-pc-linux-gnu
arch   i686
os linux-gnu
system i686, linux-gnu
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)


09:26:46 (1.32 MB/s) - 
`/tmp/RtmpDCW8km/downloaded_packages/RGrace_0.5-5.tar.gz' saved 
[164119/164119]

* Installing *source* package 'RGrace' ...
** libs
gcc -I/usr/lib/R/include -I/usr/lib/R/include -I/usr/include/gtk-1.2 
-I/usr/include/glib-1.2  -I/usr/lib/glib/include -I/usr/local/include 
  -fpic   -c gdk_draw.c -o gdk_draw.o
gdk_draw.c:3:21: error: gdk/gdk.h: No such file or directory
gdk_draw.c:4:21: error: gtk/gtk.h: No such file or directory
gdk_draw.c: In function ‘draw_rubber_box’:
gdk_draw.c:9: error: ‘GtkWidget’ undeclared (first use in this function)
gdk_draw.c:9: error: (Each undeclared identifier is reported only once
gdk_draw.c:9: error: for each function it appears in.)
gdk_draw.c:9: error: ‘drawing_widget’ undeclared (first use in this 
function)
gdk_draw.c:9: error: syntax error before ‘)’ token
gdk_draw.c:10: error: ‘GdkColor’ undeclared (first use in this function)
gdk_draw.c:11: error: ‘GdkGC’ undeclared (first use in this function)
gdk_draw.c:11: error: ‘gc1’ undeclared (first use in this function)
gdk_draw.c:12: error: ‘Color’ undeclared (first use in this function)
gdk_draw.c:17: error: ‘GDK_XOR’ undeclared (first use in this function)
gdk_draw.c:19: error: ‘GDK_LINE_SOLID’ undeclared (first use in this 
function)
gdk_draw.c:19: error: ‘GDK_CAP_PROJECTING’ undeclared (first use in this 
function)
gdk_draw.c:19: error: ‘GDK_JOIN_MITER’ undeclared (first use in this 
function)
gdk_draw.c:20: error: ‘GDK_SOLID’ undeclared (first use in this function)
make: *** [gdk_draw.o] Error 1
ERROR: compilation failed for package 'RGrace'
** Removing '/usr/lib/R/library/RGrace'





-- 
Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation
Biology (UCT)

Department of Conservation Ecology and Entomology
University of Stellenbosch
Matieland 7602
South Africa

Tel:+27 - (0)72 808 2975 (w)
Fax:+27 - (0)21 808 3304
Cell:   +27 - (0)83 9479 042

email:  [EMAIL PROTECTED]
[EMAIL PROTECTED]

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Re: [R] Read.csv

[Petr Pikal]

Hi

On 26 Oct 2006 at 10:31, Wang, Joshua (EQS) wrote:

Date sent:  Thu, 26 Oct 2006 10:31:00 +0900
From:   Wang, Joshua (EQS) [EMAIL PROTECTED]
To: Murray Pung [EMAIL PROTECTED]
Copies to:  r-help@stat.math.ethz.ch
Subject:Re: [R] Read.csv

 Mr. Pung,
 
 Thank you. It works in terms of loading the data. Except that there is
 a header in the file that I would like to load it to be the names in
 the data.frame.

You probably shall set row.names=NULL parameter in read.csv as stated 
in help page

row.names a vector of row names. This can be a vector giving the 
actual row names, or a single number giving the column of the table 
which contains the row names, or character string giving the name of 
the table column containing the row names. 
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. 
Using row.names = NULL forces row numbering.  

However you probably will always get some row numbering as it is an 
attribute of data.frame.

HTH
Petr



 
 Regards,
 Josh
 
 
 
 
  From: Murray Pung [mailto:[EMAIL PROTECTED] 
  Sent: Thursday, October 26, 2006 10:26 AM
  To: Wang, Joshua (EQS)
  Cc: r-help@stat.math.ethz.ch
  Subject: Re: [R] Read.csv
 
 
  Try:
 
  read.csv(name.csv, header = F)
 
 
  On 26/10/06, Wang, Joshua (EQS) [EMAIL PROTECTED] wrote: 
 
   Are there ways to load a csv file without row.names by
 read.csv? Thanks.
 
   [[alternative HTML version deleted]] 
 
   __
   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
   and provide commented, minimal, self-contained,
 reproducible code.
 
 
 
 
 
  -- 
  Murray Pung
  Statistician, Datapharm Australia Pty Ltd
  0404 273 283 
 
 
  [[alternative HTML version deleted]]
 
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 R-help@stat.math.ethz.ch mailing list
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Petr Pikal
[EMAIL PROTECTED]

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Re: [R] Read.csv

[Philipp Pagel]

On Thu, Oct 26, 2006 at 09:41:25AM +0900, Wang, Joshua (EQS) wrote:
 Are there ways to load a csv file without row.names by read.csv? Thanks.

Have a look at ?read.csv. Somewhere in there you will find an
explanation for the option row.name. In your case, you probably want to
set it to NULL.

cu
Philipp

-- 
Dr. Philipp PagelTel.  +49-8161-71 2131
Dept. of Genome Oriented Bioinformatics  Fax.  +49-8161-71 2186
Technical University of Munich
Science Center Weihenstephan
85350 Freising, Germany

 and

Institute for Bioinformatics / MIPS  Tel.  +49-89-3187 3675
GSF - National Research Center   Fax.  +49-89-3187 3585
  for Environment and Health
Ingolstädter Landstrasse 1
85764 Neuherberg, Germany
http://mips.gsf.de/staff/pagel

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Re: [R] Read.csv

[Peter Dalgaard]

Petr Pikal [EMAIL PROTECTED] writes:

 Hi
 
 On 26 Oct 2006 at 10:31, Wang, Joshua (EQS) wrote:
 
 Date sent:Thu, 26 Oct 2006 10:31:00 +0900
 From: Wang, Joshua (EQS) [EMAIL PROTECTED]
 To:   Murray Pung [EMAIL PROTECTED]
 Copies to:r-help@stat.math.ethz.ch
 Subject:  Re: [R] Read.csv
 
  Mr. Pung,
  
  Thank you. It works in terms of loading the data. Except that there is
  a header in the file that I would like to load it to be the names in
  the data.frame.
 
 You probably shall set row.names=NULL parameter in read.csv as stated 
 in help page
 
...
 However you probably will always get some row numbering as it is an 
 attribute of data.frame.

And JW should probably have avoided the two times probably by giving
a concrete example of an input file and the intended result

-- 
   O__   Peter Dalgaard Øster Farimagsgade 5, Entr.B
  c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
 (*) \(*) -- University of Copenhagen   Denmark  Ph:  (+45) 35327918
~~ - ([EMAIL PROTECTED])  FAX: (+45) 35327907

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Re: [R] how to improve the efficiency of the following lapply codes [Broadcast]

[Liaw, Andy]

Make good use of Rprof():  It has helped me a great deal in pinpointing
bottlenecks where I would not have suspected.

Cheers,
Andy 

From: Weiwei Shi
 object.size(intersect.matrix)
 41314204
 
 but my machine has 4 G memory, so it should be ok since after 
 12 hours, it finishes 16k out of 60k but still slow non-linearly.
 
 I am thinking to chop 60k into multiple 5k data.frames to run 
 the program. but just wondering is there a way around it?
 
  version
_
 platform   i686-pc-linux-gnu
 arch   i686
 os linux-gnu
 system i686, linux-gnu
 status
 major  2
 minor  3.1
 year   2006
 month  06
 day01
 svn rev38247
 language   R
 version.string Version 2.3.1 (2006-06-01)
 
 [EMAIL PROTECTED] ox]$ more /proc/meminfo
 total:used:free:  shared: buffers:  cached:
 Mem:  4189724672 3035549696 11541749760 282836992 2057129984
 Swap: 4293586944 645042176 3648544768
 
 [EMAIL PROTECTED] ox]$ more /proc/cpuinfo
 processor   : 0
 vendor_id   : GenuineIntel
 cpu family  : 15
 model   : 4
 model name  : Intel(R) Xeon(TM) CPU 3.60GHz
 stepping: 3
 cpu MHz : 3591.419
 cache size  : 2048 KB
 
 
 
 thanks.
 
 On 10/25/06, Weiwei Shi [EMAIL PROTECTED] wrote:
  Hi,
  I have a series of lda analysis using the following lapply function:
 
  n - dim(intersect.matrix)[1]
  net1.lda - lapply(1:(n), function(k) i.lda(data.list, 
  intersect.matrix, i=k, w))
 
  i.lda is function to do the real lda analysis.
 
  intersect.matrix is a nx1026 matrix, n can be a really huge number 
  like 60k. The target is perform a random search. Building a n=120k 
  matrix is impossible for my machine. When n=5k, the task 
 can be done 
  in 30 min while n=60k, it is estimated to take 5 days. So I am 
  wondering where my coding problem is, which causes this to be a 
  nonlinearity.
 
  If more info is needed, I will provide.
 
  thanks
 
  --
  Weiwei Shi, Ph.D
  Research Scientist
  GeneGO, Inc.
 
  Did you always know?
  No, I did not. But I believed...
  ---Matrix III
 
 
 
 --
 Weiwei Shi, Ph.D
 Research Scientist
 GeneGO, Inc.
 
 Did you always know?
 No, I did not. But I believed...
 ---Matrix III
 
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--
Notice:  This e-mail message, together with any attachments,...{{dropped}}

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[R] Quantile regression questions

[Brian Gardunia]

I am relatively new to R, but am intrigued by its flexibility.  I am interested 
in quantile regression and quantile estimation as regards to cotton fiber 
length distributions.  The length distribution affects spinning and weaving 
properties, so it is desirable to select for certain distribution types.  The 
AFIS fiber testing machinery outputs a vector for each sample of type c(12, 
235, 355, . . . n) with the number of fibers in n=40 1/16 inch length 
categories.  My question is what would be the best way to convert the raw 
output to quantiles and whether it would be appropriate to use quantile 
regression to look at whether location, variety, replication, etc. modify the 
length distribution.

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Re: [R] Error bars in ggplot?

[hadley wickham]

 Is it possible to put error bars in ggplot? I couldn't find anything so far.

Have a look at ggerrorbar.

Hadley

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Re: [R] Error bars in ggplot?

[Rainer M Krug]

hadley wickham wrote:
 Is it possible to put error bars in ggplot? I couldn't find anything so far.
 
 Have a look at ggerrorbar.
Thanks - exactly what I was looking for. By the way - ggerrorbar is not 
mentioned in ?ggplot.

Rainer

 
 Hadley
 
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-- 
Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation
Biology (UCT)

Department of Conservation Ecology and Entomology
University of Stellenbosch
Matieland 7602
South Africa

Tel:+27 - (0)72 808 2975 (w)
Fax:+27 - (0)21 808 3304
Cell:   +27 - (0)83 9479 042

email:  [EMAIL PROTECTED]
[EMAIL PROTECTED]

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[R] legends in ggplot

[David Barron]

Is it possible to change the legend title in ggplot that is generated
by a colour (other than by changing the name of the variable)?  For
example, this produces a legend title, tip/total_bill.  How can that
title be changed?

ggpoint(ggplot(tips,aesthetics=list(y=tip,x=total_bill,colour=tip/total_bill)) )

Thanks.

-- 
=
David Barron
Said Business School
University of Oxford
Park End Street
Oxford OX1 1HP

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[R] distance between legend title and legend box

[Denis Chabot]

Hi,

I've looked at the parameters available for the legend function and  
cannot find a way to change the distance between the top of the box  
surrounding a legend and the legend's title. I have a math expression  
that raises the height of my title.

If you don't mind the non-sensical title I give to the legend for  
this plot (Figure 3.20 in R Graphics):

with(iris,
   plot(Sepal.Length, Sepal.Width,
pch=as.numeric(Species), cex=1.2))
legend(6.5, 4.2, c(setosa, versicolor, virginica),
   cex=1, pch=1:3, title=expression(kg/km^2))

The result depends on the device, but I think any device will show  
the box needs to be raised a bit (in quartz, the top of the box  
passes in the middle of the 2, in pdf it is acceptable, but just  
(the top of the box lightly touches the top of the 2).

Sincerely,

Denis Chabot

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[R] Measurements of 3000 criminals

[Dietrich Trenkler]

Hallo everyone,

excuse me if this is not a genuine R question but I do not know where to
ask else.

Referring  to e.g.

https://stat.ethz.ch/pipermail/r-help/2004-December/062114.html

I wonder if these measurements of 3000 criminals (raw data) are
available anywhere.  At least I didn't find them in the R datasets
package or by means of Google.  What I did find was a table of
frequencies of the central values for *grouped* classifications (finger
lenghts) in the Handbook of Small Data Sets.

Thank you in advance.

D. Trenkler  

-- 
Dietrich Trenkler c/o Universitaet Osnabrueck 
Rolandstr. 8; D-49069 Osnabrueck, Germany
email: [EMAIL PROTECTED]

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Re: [R] distance between legend title and legend box

[David Barron]

I don't know if there is a nice, easy way of doing this, but this
method will work:

with(iris,
  plot(Sepal.Length, Sepal.Width,
   pch=as.numeric(Species), cex=1.2))

lg-legend(6.5, 4.2, c(setosa, versicolor, virginica),
  cex=1, pch=1:3, title=expression(kg/km^2), bty=n)

x1 - lg$rect$left
y1 - lg$rect$top - lg$rect$h
x2 - x1 + lg$rect$w
y2 - lg$rect$top + .1  # change the .1 to get more or less space

rect(x1,y1,x2,y2)


On 26/10/06, Denis Chabot [EMAIL PROTECTED] wrote:
 Hi,

 I've looked at the parameters available for the legend function and
 cannot find a way to change the distance between the top of the box
 surrounding a legend and the legend's title. I have a math expression
 that raises the height of my title.

 If you don't mind the non-sensical title I give to the legend for
 this plot (Figure 3.20 in R Graphics):

 with(iris,
plot(Sepal.Length, Sepal.Width,
 pch=as.numeric(Species), cex=1.2))
 legend(6.5, 4.2, c(setosa, versicolor, virginica),
cex=1, pch=1:3, title=expression(kg/km^2))

 The result depends on the device, but I think any device will show
 the box needs to be raised a bit (in quartz, the top of the box
 passes in the middle of the 2, in pdf it is acceptable, but just
 (the top of the box lightly touches the top of the 2).

 Sincerely,

 Denis Chabot

 __
 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
 and provide commented, minimal, self-contained, reproducible code.



-- 
=
David Barron
Said Business School
University of Oxford
Park End Street
Oxford OX1 1HP

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Re: [R] legends in ggplot

[hadley wickham]

 Is it possible to change the legend title in ggplot that is generated
 by a colour (other than by changing the name of the variable)?  For
 example, this produces a legend title, tip/total_bill.  How can that
 title be changed?

 ggpoint(ggplot(tips,aesthetics=list(y=tip,x=total_bill,colour=tip/total_bill))
  )

You need to add a colour scale, and specify the name to appear on the
legend (by default it just uses the name of the variable):

p - 
ggpoint(ggplot(tips,aesthetics=list(y=tip,x=total_bill,colour=tip/total_bill))
)
sccolour(p, Tip rate)
sccolour(p, This is a very long label\nsplit over two lines)

you can also use expressions to create mathematical annotations:

sccolour(p, expression(beta * x^2))

Regards,

Hadley

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Re: [R] Measurements of 3000 criminals

[Ben Bolker]

Dietrich Trenkler Dietrich.Trenkler at uni-osnabrueck.de writes:

 
 Hallo everyone,
 
 excuse me if this is not a genuine R question but I do not know where to
 ask else.
 
 Referring  to e.g.
 
 https://stat.ethz.ch/pipermail/r-help/2004-December/062114.html
 
 I wonder if these measurements of 3000 criminals (raw data) are
 available anywhere.  At least I didn't find them in the R datasets
 package or by means of Google.  What I did find was a table of
 frequencies of the central values for *grouped* classifications (finger
 lenghts) in the Handbook of Small Data Sets.
 
 Thank you in advance.
 
 D. Trenkler  
 

  The original paper on JSTOR
(Biometrika Vol I, no 2, 1902, W. R. Macdonell)
gives tables of bivariate distributions (head length
vs head breadth, height vs head breadth, height
vs left middle finger length) for the 3000-criminal
sample, a 1306-criminal subsample, and for a sample
of 1000 Cambridge men.
  It would be an interesting challenge for a
historian of science ...

   Ben Bolker

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Re: [R] How to specify a constant in gnls{nlme}

[Jost Burkardt]

Hi,

It works for me if i encapsulate the formula within a function

 a =1; b=1;
 f1  - function(x,v,w) {a/(b+(x/v)^w)}
 fm2 = gnls( y ~ f1(x,v,w), start=list( v=1, w=1)) 

Regards,
Jost

hong qin [EMAIL PROTECTED] writes:

 Hi All,

 I have question about speficifying a constant in gnls() from package nlme.
 Here is a testing code:

 #
 library(nlme)
  x = exp( rnorm(100))
  y = 1/(1+x) + rnorm(100)/10
  plot( y ~ x)
  fm1 = gnls( y ~ 1/(1+(x/v)^w), start=list( v=1, w=1))

  a =1; b=1;
  fm2 = gnls( y ~ a/(b+(x/v)^w), start=list( v=1, w=1)) #This won't work
 because I don't know to set $a$ and $b$ as constant.

 

 Thanks,

 HQ

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[R] Up- or downsampling time series in R

[Brandt, T. (Tobias)]

Hi
 
I have data that is sampled (in time) with a certain frequency and I would
like to express this time series as a time series of a higher (or lower)
frequency with the newly added time points being filled in with NA, 0, or
perhaps interpolated. My data might be regularly or irregularly spaced. For
example, I might have quarterly data that I would like to handle as a
monthly time series with NAs filled in for the missing months.
 
RSiteSearch(upsample) gave one link to a function in the waveslim
package that I'm not familiar with. It seems to me that this would be a
fairly common time series task and thus am hoping to find something in the
more common time series packages/classes such as ts, zoo, tseries, etc...
 
I will now give some example code.
 
If I am lucky enough that my data is irregularly spaced, then a
combination of zoo and ts already accomplishes this task.
 
 require(zoo)
[1] TRUE
 dt - sample(c(1,3,9), 20, replace=TRUE)
 t - zoo(dt, as.yearmon(Sys.Date()) + cumsum(dt)/12)
 t
Jan 2007 Feb 2007 Nov 2007 Feb 2008 Nov 2008 Dec 2008 Mar 2009 Apr 2009 Jul
2009 Aug 2009 
   31939131
31 
Nov 2009 Feb 2010 Nov 2010 Aug 2011 May 2012 Jun 2012 Jul 2012 Oct 2012 Jul
2013 Aug 2013 
   33999113
91 
 as.ts(t)
 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
2007   3   1  NA  NA  NA  NA  NA  NA  NA  NA   9  NA
2008  NA   3  NA  NA  NA  NA  NA  NA  NA  NA   9   1
2009  NA  NA   3   1  NA  NA   3   1  NA  NA   3  NA
2010  NA   3  NA  NA  NA  NA  NA  NA  NA  NA   9  NA
2011  NA  NA  NA  NA  NA  NA  NA   9  NA  NA  NA  NA
2012  NA  NA  NA  NA   9   1   1  NA  NA   3  NA  NA
2013  NA  NA  NA  NA  NA  NA   9   1
 plot(t)

 
However if the data happens to be regularly spaced, upsampling it isn't
quite as straightforward.
 
 t2 - zoo(sample(1:3, 20, replace=TRUE), as.yearmon(seq(2000, by=0.5,
length=20)))
 t2
Jan 2000 Jul 2000 Jan 2001 Jul 2001 Jan 2002 Jul 2002 Jan 2003 Jul 2003 Jan
2004 Jul 2004 
   33221312
33 
Jan 2005 Jul 2005 Jan 2006 Jul 2006 Jan 2007 Jul 2007 Jan 2008 Jul 2008 Jan
2009 Jul 2009 
   22332332
13 
 (t2.ts - as.ts(t2))
Time Series:
Start = c(2000, 1) 
End = c(2009, 2) 
Frequency = 2 
 [1] 3 3 2 2 1 3 1 2 3 3 2 2 3 3 2 3 3 2 1 3
 plot(t2)

 
I would expect this to be as simple as changing the frequency attribute of
t2.ts to 12 but I didn't seem to be able to find out how to do this or if it
is possible.
 
So far, the only way around this that I have found is doing it manually in
the following way:
 
 t2.monthly - zoo(NA, as.yearmon(seq(from=2000, to=2009.5, by=1/12)))
 window(t2.monthly, as.numeric(time(t2)) ) - as.numeric(t2)#
can this be done using [] indexing?
 t2.monthly
Jan 2000 Feb 2000 Mar 2000 Apr 2000 May 2000 Jun 2000 Jul 2000 Aug 2000 Sep
2000 Oct 2000 
   3   NA   NA   NA   NA   NA3   NA
NA   NA 
Nov 2000 Dec 2000 Jan 2001 Feb 2001 Mar 2001 Apr 2001 May 2001 Jun 2001 Jul
2001 Aug 2001 
  NA   NA2   NA   NA   NA   NA   NA
2   NA 
Sep 2001 Oct 2001 Nov 2001 Dec 2001 Jan 2002 Feb 2002 Mar 2002 Apr 2002 May
2002 Jun 2002 
  NA   NA   NA   NA1   NA   NA   NA
NA   NA 
Jul 2002 Aug 2002 Sep 2002 Oct 2002 Nov 2002 Dec 2002 Jan 2003 Feb 2003 Mar
2003 Apr 2003 
   3   NA   NA   NA   NA   NA1   NA
NA   NA 
May 2003 Jun 2003 Jul 2003 Aug 2003 Sep 2003 Oct 2003 Nov 2003 Dec 2003 Jan
2004 Feb 2004 
  NA   NA2   NA   NA   NA   NA   NA
3   NA 
Mar 2004 Apr 2004 May 2004 Jun 2004 Jul 2004 Aug 2004 Sep 2004 Oct 2004 Nov
2004 Dec 2004 
  NA   NA   NA   NA3   NA   NA   NA
NA   NA 
Jan 2005 Feb 2005 Mar 2005 Apr 2005 May 2005 Jun 2005 Jul 2005 Aug 2005 Sep
2005 Oct 2005 
   2   NA   NA   NA   NA   NA2   NA
NA   NA 
Nov 2005 Dec 2005 Jan 2006 Feb 2006 Mar 2006 Apr 2006 May 2006 Jun 2006 Jul
2006 Aug 2006 
  NA   NA3   NA   NA   NA   NA   NA
3   NA 
Sep 2006 Oct 2006 Nov 2006 Dec 2006 Jan 2007 Feb 2007 Mar 2007 Apr 2007 May
2007 Jun 2007 
  NA   NA   NA   NA2   NA   NA   NA
NA   NA 
Jul 2007 Aug 2007 Sep 2007 Oct 2007 Nov 2007 Dec 2007 Jan 2008 Feb 2008 Mar
2008 Apr 2008 
   3   NA   NA   NA   NA   NA3   NA
NA   NA 
May 2008 Jun 2008 Jul 2008 Aug 2008 Sep 2008 Oct 2008 Nov 2008 Dec 2008 Jan
2009 Feb 2009 
  NA   NA2   NA   NA   NA   NA   NA
1   NA 
Mar 2009 Apr 2009 May 2009 Jun 2009 Jul 2009 
  NA   NA   NA   NA3 
 points(t2.monthly, type=p, 

[R] pairs matchning

[Federico Calboli]

Hi All,

I have two numerical matrices of 2 columns and many rows.

The two coulumns of matrix (1) form a number of 'pairs' of numbers, e.g:

  [,1] [,2]
[1,]10
[2,]34
[3,]34
[4,]58
[5,]10
[6,]10
[7,]67

Matrix (2) contains the *unique* pairs:

  [,1] [,2]
[1,]10
[2,]34
[3,]58
[4,]67


I would like to create a vector matching the pairs in matrix (1) to the unique 
pairs in matrix (2), e.g:

[1] 1 2 2 3 1 1 4

(done by hand)

match() does not seem to be able to handle pairs, and I don't seem to be able 
to 
find an elegant solution...

Cheers,

Federico

-- 
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St Mary's Campus
Norfolk Place, London W2 1PG

Tel  +44 (0)20 7594 1602 Fax (+44) 020 7594 3193

f.calboli [.a.t] imperial.ac.uk
f.calboli [.a.t] gmail.com

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Re: [R] avoiding a loop

[chao gai]

Mark,

This is hardly a one-liner. It will only help you if you call it with large 
objects. If you have 0 in your data it needs even more extending.
Here it is:
myconstruct - function(aa) {
   aa - c(1,1,aa,1)
   bb - cumprod(aa)
   taa - aa1
   difftaa - diff(taa)
   starts - which(c(FALSE,difftaa==1))
   ends - which(c(FALSE,difftaa==-1))
   correction - rep(0,length(aa))
   correction[starts] - 1/bb[starts-2]
   correction[ends] - -1/bb[starts-2]
   correction - cumsum(correction)
   correction[correction==0] - 1
   correction
   bbcor - bb*correction
   bbcor[!taa] - aa[!taa]
   bbcor[c(-1,-2,-length(bbcor))]
}


aa - runif(1,min=.5,max=1.5)
system.time(target - constructLt(aa))
system.time(found - myconstruct(aa))
identical(target,found)
max(abs(target-found))


 system.time(target - constructLt(aa))
[1] 0.059 0.001 0.059 0.000 0.000
 system.time(found - myconstruct(aa))
[1] 0.013 0.000 0.014 0.000 0.000
 identical(target,found)
[1] FALSE
 max(abs(target-found))
[1] 4.440892e-16



Appearently the outcome is different is some low decimal, due to other 
calculation method.

Kees




On Tuesday 24 October 2006 20:36, Leeds, Mark (IED) wrote:
 I think I asked a similar question 3 years ago to the  Splus list and I
 think the answer was no or noone answered so noone should spend more
 than 5 minutes on this
 because it could definitely be a waste of time.

 My question is whether the function below can be rewritten without a for
 loop. apply is fine if it can be done that way but i doubt it. I call it
 a lot and would
 prefer to not loop.

 #---
 --

 constructLt-function(invector) {

 outvector-invector

  for ( i in 2:length(invector) ) {
  if ( invector[i]  1 ) {
   outvector[i]-invector[i]*outvector[i-1]
  }
 }

 return(outvector)

 }

 #---
 -
 

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 analyst/research report.  Unless indicated, these views are the author's
 and may differ from those of Morgan Stanley research or others in the Firm.
  We do not represent this is accurate or complete and we may not update
 this.  Past performance is not indicative of future returns.  For
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Re: [R] pairs matchning

[TobiasBr]

Hi

You could try to find an equivalent representation as a string and try to
match those. 

 (A - cbind(sample(1:2, 10, rep=TRUE), sample(1:2, 10, rep=TRUE)))
  [,1] [,2]
 [1,]12
 [2,]12
 [3,]12
 [4,]22
 [5,]11
 [6,]12
 [7,]12
 [8,]11
 [9,]12
[10,]11
 (B - unique(A))
 [,1] [,2]
[1,]12
[2,]22
[3,]11
 strRep - function(mat) apply(mat, 1, function(x) paste(x, collapse=;))
 strRep(B)
[1] 1;2 2;2 1;1
 (idx - match(strRep(A), strRep(B)))
 [1] 1 1 1 2 3 1 1 3 1 3
 B[idx, ]
  [,1] [,2]
 [1,]12
 [2,]12
 [3,]12
 [4,]22
 [5,]11
 [6,]12
 [7,]12
 [8,]11
 [9,]12
[10,]11
 

HTH


-Original Message-
From: [EMAIL PROTECTED] 
[mailto:[EMAIL PROTECTED] On Behalf Of Federico Calboli
Sent: 26 October 2006 06:48 PM
To: r-help
Subject: [R] pairs matchning

Hi All,

I have two numerical matrices of 2 columns and many rows.

The two coulumns of matrix (1) form a number of 'pairs' of 
numbers, e.g:

  [,1] [,2]
[1,]10
[2,]34
[3,]34
[4,]58
[5,]10
[6,]10
[7,]67

Matrix (2) contains the *unique* pairs:

  [,1] [,2]
[1,]10
[2,]34
[3,]58
[4,]67


I would like to create a vector matching the pairs in matrix 
(1) to the unique pairs in matrix (2), e.g:

[1] 1 2 2 3 1 1 4

(done by hand)

match() does not seem to be able to handle pairs, and I don't 
seem to be able to find an elegant solution...

Cheers,

Federico

--
Federico C. F. Calboli
Department of Epidemiology and Public Health Imperial College, 
St Mary's Campus Norfolk Place, London W2 1PG

Tel  +44 (0)20 7594 1602 Fax (+44) 020 7594 3193

f.calboli [.a.t] imperial.ac.uk
f.calboli [.a.t] gmail.com

__
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Re: [R] pairs matchning

[chao gai]

Federico,

Why not 
1 convert to data frame  dfo - as.data.frame(mat)
2 take unique items with ?unique  df- unique(dfo)
3 add an index to the unique itmes df$unique - 1:nrow(df)
4 merge the two data.frames  combi - merge(dfo,df)
5 extract the index? combi$unique

Kees 

On Thursday 26 October 2006 18:47, Federico Calboli wrote:
 Hi All,

 I have two numerical matrices of 2 columns and many rows.

 The two coulumns of matrix (1) form a number of 'pairs' of numbers, e.g:

   [,1] [,2]
 [1,]10
 [2,]34
 [3,]34
 [4,]58
 [5,]10
 [6,]10
 [7,]67

 Matrix (2) contains the *unique* pairs:

   [,1] [,2]
 [1,]10
 [2,]34
 [3,]58
 [4,]67


 I would like to create a vector matching the pairs in matrix (1) to the
 unique pairs in matrix (2), e.g:

 [1] 1 2 2 3 1 1 4

 (done by hand)

 match() does not seem to be able to handle pairs, and I don't seem to be
 able to find an elegant solution...

 Cheers,

 Federico

__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Quantile regression questions

[roger koenker]

Brian,

It is hard to say at this level of resolution of the question, but it  
would seem that you might
be able to start by considering each sample vector as as repeated  
measurement of the
fiber length -- so 12 obs in the first 1/16th bin, 235 in the next  
and so forth, all associated
with some vector of covariates representing location, variety, etc,  
then the conventional
quantile regression would serve to estimate a conditional quantile  
function for fiber length
for each possible covariate setting --- obviously this would require  
some model for the
way that the covariate effects fit together, linearity,  possible  
interactions, etc etc, and it
would also presume that it made sense to treat the vector of  
responses as independent
measurements.  Building in possible dependence involves some new  
challenges, but
there is some recent experience with inferential methods for  
microarrays that have
incorporated these effects.

I'd be happy to hear more about the data and possible models, but  
this should be
routed privately since the topic is rather too specialized for R-help.


url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax:   217-244-6678Champaign, IL 61820


On Oct 26, 2006, at 7:20 AM, Brian Gardunia wrote:

 I am relatively new to R, but am intrigued by its flexibility.  I  
 am interested in quantile regression and quantile estimation as  
 regards to cotton fiber length distributions.  The length  
 distribution affects spinning and weaving properties, so it is  
 desirable to select for certain distribution types.  The AFIS fiber  
 testing machinery outputs a vector for each sample of type c(12,  
 235, 355, . . . n) with the number of fibers in n=40 1/16 inch  
 length categories.  My question is what would be the best way to  
 convert the raw output to quantiles and whether it would be  
 appropriate to use quantile regression to look at whether location,  
 variety, replication, etc. modify the length distribution.

 __
 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
 and provide commented, minimal, self-contained, reproducible code.

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Re: [R] pairs matchning

[Ben Fairbank]

If all of the numbers are one-digit numbers, as your subset shows, you
can create a third column equal to 10 * col 1 + col 2, and then use
match() on the resulting two-digit numbers, no?  If the numbers are
larger than one-digit, and you know the maximum, use that rather than 10
as the multiplier.

Ben

-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Federico Calboli
Sent: Thursday, October 26, 2006 11:48 AM
To: r-help
Subject: [R] pairs matchning

Hi All,

I have two numerical matrices of 2 columns and many rows.

The two coulumns of matrix (1) form a number of 'pairs' of numbers, e.g:

  [,1] [,2]
[1,]10
[2,]34
[3,]34
[4,]58
[5,]10
[6,]10
[7,]67

Matrix (2) contains the *unique* pairs:

  [,1] [,2]
[1,]10
[2,]34
[3,]58
[4,]67


I would like to create a vector matching the pairs in matrix (1) to the
unique 
pairs in matrix (2), e.g:

[1] 1 2 2 3 1 1 4

(done by hand)

match() does not seem to be able to handle pairs, and I don't seem to be
able to 
find an elegant solution...

Cheers,

Federico

-- 
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St Mary's Campus
Norfolk Place, London W2 1PG

Tel  +44 (0)20 7594 1602 Fax (+44) 020 7594 3193

f.calboli [.a.t] imperial.ac.uk
f.calboli [.a.t] gmail.com

__
R-help@stat.math.ethz.ch mailing list
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and provide commented, minimal, self-contained, reproducible code.

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Re: [R] Up- or downsampling time series in R

[Achim Zeileis]

On Thu, 26 Oct 2006, Brandt, T. (Tobias) wrote:

 I have data that is sampled (in time) with a certain frequency and I would
 like to express this time series as a time series of a higher (or lower)
 frequency with the newly added time points being filled in with NA, 0, or
 perhaps interpolated. My data might be regularly or irregularly spaced. For
 example, I might have quarterly data that I would like to handle as a
 monthly time series with NAs filled in for the missing months.

Both can be done easily with zoo, examples for both are given in the
package vignettes and the man pages of the package.

1. extend to a finer grid (upsampling)
## generate some time series
z - zoo(sample(1:3, 20, replace=TRUE),
 as.yearmon(seq(2000, by=0.5, length=20)))
## generate emtpy series on finer grid
z2 - zoo(,seq(start(z), end(z), by = 1/12))
## merge (returns univariate series, by default filled with NAs)
merge(z, z2)

2. aggregate to a coarser grid (downsampling)
## transform to annual data
as.annual - function(x) floor(as.numeric(x))
## average within years
aggregate(z, as.annual, mean)
## first observation within year
aggregate(z, as.annual, head, 1)

hth,
Z

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Re: [R] cloud() works but wireframe() is blank

[Larry Layne]

--On Wednesday, October 25, 2006 4:46 PM -0700 Deepayan Sarkar 
[EMAIL PROTECTED] wrote:

 On 10/25/06, Larry Layne [EMAIL PROTECTED] wrote:
 Per the message from Alexander Nervedi, 29 April 2006:

  I have to be making a riddiculously silly ommission.
  when I run the fillowing i get the cloud plot ok. But I cant figure
  out what I am missing out when I call wireframe.

  Any help would be appreciated.

  x-runif(100)
  y-rnorm(100)
  z-runif(100)

  temp -data.frame(x,y,z)
  wireframe(x~y*z,temp)
  cloud(x~y*z,temp)

 There is something funny about the way wireframe treats a data frame
 object created using data.frame vs. using expand.grid. For instance,
 this works:

 There's nothing funny about it, ?wireframe clearly says:

   In the case of 'wireframe',
   calculations are based on the assumption that the 'x' and 'y'
   values are evaluated on a rectangular grid defined by their
   unique values.

Sorry about incorrectly interpreting this part of the description about 
wireframe.

 In the original example, the (x, y) values are clearly not on a grid,
 and in your (very strange) example below, they are not on a
 rectangular grid, as can be seen using

 xyplot(u ~ v, reggrid)

One of the nice things about posting questions in a list is that it is so 
conveniently easy to embarrass oneself in public. The corrected code 
follows and works just fine:

library(lattice)
###MAKE-BELIEVE DATA
u - mat.or.vec(100,1)
v - mat.or.vec(100,1)
ij=0
for(i in 1:10)
  {
for(j in 1:10)
  {
ij=ij+1
u[ij]=i
v[ij]=j
  }
  }
z - runif(100)
reggrid - data.frame(u,v,z)
is.data.frame(reggrid)
wireframe(z~u*v,reggrid,scales=list(arrows=FALSE),drape=TRUE,colorkey=TRUE)

 In addition, I would also like to
 do what Alexander is trying to do, which is create a wireframe from
 irregular lattice data.

Although what I would really like to do is plot a 3D Triangulated Irregular 
Network (TIN) surface of the irregularly spaced points, the following sort 
of gets me there in the short-term:

##INTERPOLATE USING AKIMA PACKAGE AND PLOT SURFACES
library(akima)
library(lattice)
  #MAKE-BELIEVE DATA
u - runif(30)
v - runif(30)
z - runif(30)
  #LINEAR INTERPOLATION
reggrid - interp(u,v,z,linear=TRUE,extrap=FALSE)
  #WIREFRAME SURFACE PLOT FROM INTERPOLATED DATA
wireframe(reggrid$z,scales=list(arrows=FALSE),drape=TRUE,colorkey=TRUE)
  #PLOT IMAGE, CONTOUR, AND POINTS
plot(u,v)
image  (reggrid,add=TRUE)
contour(reggrid,add=TRUE)
points (u,v,pch = 3)

Larry Layne

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[R] Header of dataframe

[Benjamin Dickgiesser]

Hi
I am fairly new to R and I would appreciate some help to hopefully a
trivial problem.

I created a function:

summary.aggregate - function(y, ...)
{
temp.mean   - aggregate(y, FUN=mean, ...)
temp.sd - aggregate(y, FUN=sd, ...)
temp.length - aggregate(y, FUN=length, ...)
temp - 
data.frame(cbind(mean=temp.mean$x,stdev=temp.sd$x,n=temp.length$x))
}

this outputs e.g.:

  meanstdev  n
1 645.6125 65.94129 60
2 655.2121 70.64094 60
3 633.3161 80.48620 60
4 650.3897 77.59191 60
5 630.4955 84.9 60
6 656.2608 66.16100 60
7 666.1775 74.39796 60
8 663.1543 71.10769 60

Is there an easy way to add a column name to the first column?

Thank you,
Benjamin

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

[R] function to normalize vectors?

[Daniel Elliott]

Hello all.

I can find no function to compute norms (even the basic two-norm) of a
vector in the online help (within the GUI) or the downloadable
documentation.

Thanks.

- dan elliott

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and provide commented, minimal, self-contained, reproducible code.

[R] Generating tables

[Benjamin Dickgiesser]

I am actually trying to write something above the first column ( in
the example the column with numbers 1 to 8) is this possible at all?
If not, is there another way to make a table which supports this? I am
merely using a data frame as a way to create a table I don't need the
data to be this data type.

Benjamin

On 10/26/06, Leeds, Mark (IED) [EMAIL PROTECTED] wrote:
 I think colnames(DF)-c(meanname,sdname,nname) should work if
 that's what you want.



 -Original Message-
 From: [EMAIL PROTECTED]
 [mailto:[EMAIL PROTECTED] On Behalf Of Benjamin
 Dickgiesser
 Sent: Thursday, October 26, 2006 1:08 PM
 To: r-help@stat.math.ethz.ch
 Subject: [R] Header of dataframe

 Hi
 I am fairly new to R and I would appreciate some help to hopefully a
 trivial problem.

 I created a function:

 summary.aggregate - function(y, ...)
 {
 temp.mean   - aggregate(y, FUN=mean, ...)
 temp.sd - aggregate(y, FUN=sd, ...)
 temp.length - aggregate(y, FUN=length, ...)
 temp -
 data.frame(cbind(mean=temp.mean$x,stdev=temp.sd$x,n=temp.length$x))
 }

 this outputs e.g.:

   meanstdev  n
 1 645.6125 65.94129 60
 2 655.2121 70.64094 60
 3 633.3161 80.48620 60
 4 650.3897 77.59191 60
 5 630.4955 84.9 60
 6 656.2608 66.16100 60
 7 666.1775 74.39796 60
 8 663.1543 71.10769 60

 Is there an easy way to add a column name to the first column?

 Thank you,
 Benjamin


__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] help fo finding R package utils

[Christos Hatzis]

Amina,
 
The best way for entering data in R is by importing an existing data file
(text, Excel etc).  This will reduce time and potential data entry errors.
Please read Section 7 of the Introduction to R manual for a detailed
discussion of available options.  The manuals are available on your system
(Help menu in the R Console for Windows) or through the R web site.
 
Regards,
 
-Christos

  _  

From: amna khan [mailto:[EMAIL PROTECTED] 
Sent: Thursday, October 26, 2006 1:21 PM
To: [EMAIL PROTECTED]
Subject: Re: [R] help fo finding R package utils


Respected sir
I have a great problem for data entry in R. using sessionInfo() utils has
appeared. but while entring data there is a message of no object found. Sir
is there any option for spreadsheet for data entry.
I shall be thankful to you.
 
Amina


 
On 10/25/06, Christos Hatzis [EMAIL PROTECTED] wrote: 

Amina,

utils is a base package and should be already installed in a functional R
system.

You can try
sessionInfo()

to verify that it is loaded.

-Christos

-Original Message-
From: [EMAIL PROTECTED]
[mailto: [EMAIL PROTECTED] On Behalf Of amna khan
Sent: Thursday, October 26, 2006 12:11 AM
To: R-help@stat.math.ethz.ch  mailto:R-help@stat.math.ethz.ch 
Subject: [R] help fo finding R package utils

Sir I have a problem that from which country i can assess R package utils
for data entry for loading it in R
Please help me in this regard. 


--
AMINA SHAHZADI
Department of Statistics
GC University Lahore, Pakistan.
Email:
[EMAIL PROTECTED]
[EMAIL PROTECTED]  mailto:[EMAIL PROTECTED] 
[EMAIL PROTECTED]

   [[alternative HTML version deleted]]

__
R-help@stat.math.ethz.ch  mailto: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
http://www.R-project.org/posting-guide.html 
and provide commented, minimal, self-contained, reproducible code.








-- 
AMINA SHAHZADI
Department of Statistics
GC University Lahore, Pakistan.
Email: 
[EMAIL PROTECTED]
[EMAIL PROTECTED]
[EMAIL PROTECTED] 

[[alternative HTML version deleted]]

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and provide commented, minimal, self-contained, reproducible code.

[R] Query about using table

[lalitha viswanath]

Hi
I have data of the following form
ID  age  member_FLAG
125  Y
236.75   N 
375.5N
.
.

I want to get a histogram of this data showing 
distribution of member_flag in each age-bin i.e. how
many values in each age bin have a member_flag of 'Y'
and how many have 'N'.
I was able to do the same using barplot2.

However I also need similar information in a tabular
form using percentages.
i.e in each age bin, what is the PERCENTAGE of IDs
with a member_flag of 'Y' 

I am trying to work with table for the same, but would
appreciate some guidance regarding the above.

Thanks
Lalitha

__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

[R] Query about using table

[lalitha viswanath]

Hi
I have data of the following form
ID  age  member_FLAG
125  Y
236.75   N 
375.5N
.
.

I want to get a histogram of this data showing 
distribution of member_flag in each age-bin i.e. how
many values in each age bin have a member_flag of 'Y'
and how many have 'N'.
I was able to do the same using barplot2.

However I also need similar information in a tabular
form using percentages.
i.e in each age bin, what is the PERCENTAGE of IDs
with a member_flag of 'Y' 

I am trying to work with table for the same, but would
appreciate some guidance regarding the above.

Thanks
Lalitha

__
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
and provide commented, minimal, self-contained, reproducible code.

Re: [R] help fo finding R package utils

[Christos Hatzis]

Please reply to the list, not to me personally.

When I want to import an Excel file, I usually save it as a tab-delimited
text file and use read.delim to import it:
 
my.dataframe - read.delim(file.name.txt, header=TRUE, sep=\t) 

See ?read.delim for further details.  There are many other options, I'm
sure, but this is simple and flexible and does not require any additional
packages.

Good luck.
-Christos



From: amna khan [mailto:[EMAIL PROTECTED] 
Sent: Thursday, October 26, 2006 2:41 PM
To: [EMAIL PROTECTED]
Subject: Re: [R] help fo finding R package utils


sir i have downloaded  poptools for excel to import data but still feeling
problems. I am not understanding what I should do?


On 10/26/06, Christos Hatzis [EMAIL PROTECTED] wrote: 

Amina,
 
The best way for entering data in R is by importing an existing data
file (text, Excel etc).  This will reduce time and potential data entry
errors. 
Please read Section 7 of the Introduction to R manual for a
detailed discussion of available options.  The manuals are available on your
system (Help menu in the R Console for Windows) or through the R web site. 
 
Regards,
 
-Christos



From: amna khan [mailto:[EMAIL PROTECTED] 
Sent: Thursday, October 26, 2006 1:21 PM 
To: [EMAIL PROTECTED]
Subject: Re: [R] help fo finding R package utils 

 
Respected sir
I have a great problem for data entry in R. using sessionInfo()
utils has appeared. but while entring data there is a message of no object
found. Sir is there any option for spreadsheet for data entry.
I shall be thankful to you.
 
Amina


 
On 10/25/06, Christos Hatzis [EMAIL PROTECTED]
mailto:[EMAIL PROTECTED]  wrote: 

Amina,

utils is a base package and should be already installed in a
functional R
system.

You can try
sessionInfo()

to verify that it is loaded.

-Christos

-Original Message-
From: [EMAIL PROTECTED]
[mailto: [EMAIL PROTECTED] On Behalf Of amna
khan 
Sent: Thursday, October 26, 2006 12:11 AM
To: R-help@stat.math.ethz.ch
mailto:R-help@stat.math.ethz.ch 
Subject: [R] help fo finding R package utils 

Sir I have a problem that from which country i can assess R
package utils
for data entry for loading it in R
Please help me in this regard. 


--
AMINA SHAHZADI
Department of Statistics 
GC University Lahore, Pakistan.
Email:
[EMAIL PROTECTED]
[EMAIL PROTECTED] 
[EMAIL PROTECTED]

   [[alternative HTML version deleted]]

__ 
R-help@stat.math.ethz.ch mailto: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 
and provide commented, minimal, self-contained, reproducible
code.








-- 
AMINA SHAHZADI
Department of Statistics
GC University Lahore, Pakistan.
Email: 
[EMAIL PROTECTED]
[EMAIL PROTECTED]
[EMAIL PROTECTED] 




-- 
AMINA SHAHZADI
Department of Statistics
GC University Lahore, Pakistan.
Email: 
[EMAIL PROTECTED]
[EMAIL PROTECTED] 
[EMAIL PROTECTED]

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[R] latent class model

[karen xie]

Dear List,

I try to implement the latent class model with the unknown number of
classes. I wonder whether someone can provide me some sample codes.

Thank you for your help.

Karen

[[alternative HTML version deleted]]

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Query about using table

[apjaworski]

Lalitha,

Try something like

with(data, table(cut(age, n), member_FLAG))

where data is a data frame with columns named age and member_FLAG and n is
the number of age categories (bins) you want.  This will give you the
number of Ys and Ns in each bin, that you will need to convert to
percentages.

Also, you might want to check the help pages for table and cut.

Cheers,

Andy

__
Andy Jaworski
518-1-01
Process Laboratory
3M Corporate Research Laboratory
-
E-mail: [EMAIL PROTECTED]
Tel:  (651) 733-6092
Fax:  (651) 736-3122


   
 lalitha viswanath 
 lalithaviswanath 
 @yahoo.comTo 
 Sent by:  r-help@stat.math.ethz.ch
 [EMAIL PROTECTED]  cc 
 at.math.ethz.ch   
   Subject 
   [R] Query about using table 
 10/26/2006 01:44  
 PM
   
   
   
   




Hi
I have data of the following form
ID  age  member_FLAG
125  Y
236.75   N
375.5N
.
.

I want to get a histogram of this data showing
distribution of member_flag in each age-bin i.e. how
many values in each age bin have a member_flag of 'Y'
and how many have 'N'.
I was able to do the same using barplot2.

However I also need similar information in a tabular
form using percentages.
i.e in each age bin, what is the PERCENTAGE of IDs
with a member_flag of 'Y'

I am trying to work with table for the same, but would
appreciate some guidance regarding the above.

Thanks
Lalitha

__
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
and provide commented, minimal, self-contained, reproducible code.

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] distance between legend title and legend box

[Paul Murrell]

Hi


Denis Chabot wrote:
 Hi,
 
 I've looked at the parameters available for the legend function and  
 cannot find a way to change the distance between the top of the box  
 surrounding a legend and the legend's title. I have a math expression  
 that raises the height of my title.
 
 If you don't mind the non-sensical title I give to the legend for  
 this plot (Figure 3.20 in R Graphics):
 
 with(iris,
plot(Sepal.Length, Sepal.Width,
 pch=as.numeric(Species), cex=1.2))
 legend(6.5, 4.2, c(setosa, versicolor, virginica),
cex=1, pch=1:3, title=expression(kg/km^2))
 
 The result depends on the device, but I think any device will show  
 the box needs to be raised a bit (in quartz, the top of the box  
 passes in the middle of the 2, in pdf it is acceptable, but just  
 (the top of the box lightly touches the top of the 2).


Looking at the source for legend(), it only expects the title to be zero
or 1 lines high (i.e., either no title or a title that is one line of
normal text high).  It needs a rewrite to allow for the exact height of
the title text.

Paul


 Sincerely,
 
 Denis Chabot
 
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 R-help@stat.math.ethz.ch mailing list
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.

-- 
Dr Paul Murrell
Department of Statistics
The University of Auckland
Private Bag 92019
Auckland
New Zealand
64 9 3737599 x85392
[EMAIL PROTECTED]
http://www.stat.auckland.ac.nz/~paul/

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and provide commented, minimal, self-contained, reproducible code.

[R] Newbie: Better way to do compound conditionals in subset?

[Zembower, Kevin]

There must be a better way to select the rows after 22-Apr-2004 and
before 01-Sep-2004 with a temperature below 65 than this:

 before2sw1 - subset(energy.data, as.Date(start, format=%d-%b-%y) 
as.Date(01-Sep-04, format = %d-%b-%y))
 before2sw2 - subset(before2sw1, as.Date(start, format=%d-%b-%y) =
as.Date(22-Apr-04, format = %d-%b-%y), select=c(therms,temp,days))
 before2sw - subset(before2sw2, temp  65)

Is it also possible to combine in this step:

attach(before2sw)
before2sw.HDD - therms / (65 - temp) * days

My data looks like this:
 head(energy.data)
  start therms   gas KWHs elect temp days
1 10-Jun-98  9 16.84  613 63.80   75   40
2 20-Jul-98  6 15.29  721 74.21   76   29
3 18-Aug-98  7 15.73  597 62.22   76   29
4 16-Sep-98 42 35.81  460 43.98   70   33
5 19-Oct-98105 77.28  314 31.45   57   29
6 17-Nov-98106 77.01  342 33.86   48   30


Thanks for your suggestions and advice. I'm continuing to enjoy learning
R.

-Kevin

Kevin Zembower
Internet Services Group manager
Center for Communication Programs
Bloomberg School of Public Health
Johns Hopkins University
111 Market Place, Suite 310
Baltimore, Maryland  21202
410-659-6139

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and provide commented, minimal, self-contained, reproducible code.

[R] Turn off border on barplot

[Frank McCown]

I was wondering if there was any way to completely remove the borders 
around each bar when using the barplot function.  I understand how to 
change the color of the border (border=white), but I need the border 
to be invisible so I can plot many skinny bars right next to each other 
with only the bar's color visible.

Thanks,
Frank

-- 
Frank McCown
Old Dominion University
http://www.cs.odu.edu/~fmccown/

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Newbie: Better way to do compound conditionals in subset?

[Jeffrey Robert Spies]

I would personally use the following method (example using the iris  
data included with R):

data(iris)
tSelect - (iris$Sepal.Length  6.0  iris$Sepal.Length  6.2  iris 
$Sepal.Width == 3.0)
tSelectedData - iris[tSelect,]

Then  you can simply work with tSelectedData for whatever equation  
you use, i.e.:

tSelectedData$Sepal.Length - tSelectedData$Sepal.Width

Of course you could write all of this on one line, but that doesn't  
read well.

Hope that helps,

Jeff.

On Oct 26, 2006, at 3:55 PM, Zembower, Kevin wrote:

 There must be a better way to select the rows after 22-Apr-2004 and
 before 01-Sep-2004 with a temperature below 65 than this:

 before2sw1 - subset(energy.data, as.Date(start, format=%d-%b-%y) 
 as.Date(01-Sep-04, format = %d-%b-%y))
 before2sw2 - subset(before2sw1, as.Date(start, format=%d-%b-%y) =
 as.Date(22-Apr-04, format = %d-%b-%y), select=c(therms,temp,days))
 before2sw - subset(before2sw2, temp  65)

 Is it also possible to combine in this step:

 attach(before2sw)
 before2sw.HDD - therms / (65 - temp) * days

 My data looks like this:
 head(energy.data)
   start therms   gas KWHs elect temp days
 1 10-Jun-98  9 16.84  613 63.80   75   40
 2 20-Jul-98  6 15.29  721 74.21   76   29
 3 18-Aug-98  7 15.73  597 62.22   76   29
 4 16-Sep-98 42 35.81  460 43.98   70   33
 5 19-Oct-98105 77.28  314 31.45   57   29
 6 17-Nov-98106 77.01  342 33.86   48   30


 Thanks for your suggestions and advice. I'm continuing to enjoy  
 learning
 R.

 -Kevin

 Kevin Zembower
 Internet Services Group manager
 Center for Communication Programs
 Bloomberg School of Public Health
 Johns Hopkins University
 111 Market Place, Suite 310
 Baltimore, Maryland  21202
 410-659-6139

 __
 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
 and provide commented, minimal, self-contained, reproducible code.

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and provide commented, minimal, self-contained, reproducible code.

[R] applying order() in a for loop

[Kim Milferstedt]

Hi,

with the code below I try to organize sections of a data.frame. The 
dataframe has 9 data columns. I'd like to order the columns of the 
data.frame by the sum of the columns' entries. The sections of the 
columns are identified by i. The order of the columns is given by 
the vector z. Each section of the data.frame is ordered 
differently. This can be seen when looking at the individual rows of 
h where all different z vectors are stored for later use.

In a third for loop, e is calculated based on the ordering 
calculated before. e is the re-organized data.frame with now all 
column segments in descending order -- in theory.

Problem:
The code below does not update the z in the third for loop. It 
remains the initial z that I calculate the first time around. I 
tried re-initializing z at various places in the code, hoping to 
get rid of the old values but with no apparent consequences. Also in 
the help I could not find anything that I could relate to my problem.

Does anybody have a suggestion on how to get the appropriate z to 
work when shuffling the data around for e?

Thanks already for suggestions!

Kim

### Code 

for (i in 11:19)
 {
 for (j in 3:ncol(data))
 {
 a - sum(data[,j][data[,1] == i/10])
 b - c(b,a)
 }

 z - order(b, decreasing = TRUE)
 d - length(data[,1][data[,1] == i/10])

 h - rbind(h,z) ### Just to check whether I really get 
different orderings for the different data segments.

 for (l in 1:d)
 {
 e - data[((i-10)*d+l-d),(3:ncol(data))][z] ## This 
z does not get updated but remains the initially calculated z
 f - rbind(f,e)
 }
 b - NULL
 e - NULL
 z - NULL
 }

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Header of dataframe

[Mihai Nica]

names(temp)[1]=a

see: ?names

did you really created that function :-)?
 
Mihai Nica
170 East Griffith St. G5
Jackson, MS 39201
601-914-0361

- Original Message 
From: Benjamin Dickgiesser [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Thursday, October 26, 2006 12:08:08 PM
Subject: [R] Header of dataframe

Hi
I am fairly new to R and I would appreciate some help to hopefully a
trivial problem.

I created a function:

summary.aggregate - function(y, ...)
{
temp.mean - aggregate(y, FUN=mean, ...)
temp.sd  - aggregate(y, FUN=sd, ...)
temp.length - aggregate(y, FUN=length, ...)
temp - data.frame(cbind(mean=temp.mean$x,stdev=temp.sd$x,n=temp.length$x))
}

this outputs e.g.:

  meanstdev  n
1 645.6125 65.94129 60
2 655.2121 70.64094 60
3 633.3161 80.48620 60
4 650.3897 77.59191 60
5 630.4955 84.9 60
6 656.2608 66.16100 60
7 666.1775 74.39796 60
8 663.1543 71.10769 60

Is there an easy way to add a column name to the first column?

Thank you,
Benjamin

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[[alternative HTML version deleted]]

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Turn off border on barplot

[Marc Schwartz]

On Thu, 2006-10-26 at 16:13 -0400, Frank McCown wrote:
 I was wondering if there was any way to completely remove the borders 
 around each bar when using the barplot function.  I understand how to 
 change the color of the border (border=white), but I need the border 
 to be invisible so I can plot many skinny bars right next to each other 
 with only the bar's color visible.
 
 Thanks,
 Frank

You can try:

 # Set line type to blank
 # See ?par
 # First save current value
 lty.o - par(lty)

 par(lty = 0)

 # Also set space to 0, rather
 # than default 0.2, if needed for more room
 barplot(1:10, space = 0)

 # Reset to old value
 par(lty = lty.o)

Depending upon the look you need, setting space to 0 may be sufficient
and still allow you to enable a border.

HTH,

Marc Schwartz

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Newbie: Better way to do compound conditionals in subset?

[Peter Dalgaard]

Jeffrey Robert Spies [EMAIL PROTECTED] writes:

 I would personally use the following method (example using the iris  
 data included with R):
 
 data(iris)
 tSelect - (iris$Sepal.Length  6.0  iris$Sepal.Length  6.2  iris 
 $Sepal.Width == 3.0)
 tSelectedData - iris[tSelect,]
 
 Then  you can simply work with tSelectedData for whatever equation  
 you use, i.e.:
 
 tSelectedData$Sepal.Length - tSelectedData$Sepal.Width
 
 Of course you could write all of this on one line, but that doesn't  
 read well.
 
 Hope that helps,

But why not use subset()? For a neat way around the as.Date business,
consider 

between - function(x,x1,x2) x = x1  x  x2
begin - as.Date(2004-04-22)
end - as.Date(2004-09-01)
before2sw - subset(energy.data, 
   between(as.Date(start, format=%d-%b-%y), begin, end)  temp  65)
 

 Jeff.
 
 On Oct 26, 2006, at 3:55 PM, Zembower, Kevin wrote:
 
  There must be a better way to select the rows after 22-Apr-2004 and
  before 01-Sep-2004 with a temperature below 65 than this:
 
  before2sw1 - subset(energy.data, as.Date(start, format=%d-%b-%y) 
  as.Date(01-Sep-04, format = %d-%b-%y))
  before2sw2 - subset(before2sw1, as.Date(start, format=%d-%b-%y) =
  as.Date(22-Apr-04, format = %d-%b-%y), select=c(therms,temp,days))
  before2sw - subset(before2sw2, temp  65)
 
  Is it also possible to combine in this step:
 
  attach(before2sw)
  before2sw.HDD - therms / (65 - temp) * days
 
  My data looks like this:
  head(energy.data)
start therms   gas KWHs elect temp days
  1 10-Jun-98  9 16.84  613 63.80   75   40
  2 20-Jul-98  6 15.29  721 74.21   76   29
  3 18-Aug-98  7 15.73  597 62.22   76   29
  4 16-Sep-98 42 35.81  460 43.98   70   33
  5 19-Oct-98105 77.28  314 31.45   57   29
  6 17-Nov-98106 77.01  342 33.86   48   30
 
 
  Thanks for your suggestions and advice. I'm continuing to enjoy  
  learning
  R.
 
  -Kevin
 
  Kevin Zembower
  Internet Services Group manager
  Center for Communication Programs
  Bloomberg School of Public Health
  Johns Hopkins University
  111 Market Place, Suite 310
  Baltimore, Maryland  21202
  410-659-6139
 
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-- 
   O__   Peter Dalgaard Øster Farimagsgade 5, Entr.B
  c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
 (*) \(*) -- University of Copenhagen   Denmark  Ph:  (+45) 35327918
~~ - ([EMAIL PROTECTED])  FAX: (+45) 35327907

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[R] Organisation of medium/large projects with multiple analyses

[Mark Wardle]

Dear all,

I'm still new to R, but have a fair experience with general programming.
All of my data is stored in postgresql, and I have a number of R files
that generate tables, results, graphs etc.  These are then available to
be imported into powerpoint/latex etc.

I'm using version control (subversion), and as with most small projects,
now have an ever increasing number of R scripts, each with fairly
specific features. With any enlarging project, there are always issues
regarding interdependencies, shared commonality (eg accessing same data
store), and old scripts stopping working with changes made elsewhere (eg
to data schema). For example, I might have a specific inclusion and
exclusion criteria for patients, and this SQL query may have to be
included in a number of analyses; I'm tempted to factor this out into a
project-specific data access library, but is that over the top?

This is a very long-winded and roundabout way of asking people how they
organise medium-sized projects? Do people create their own libraries
for specific projects for shared functionality, or do people just
liberally use source() for this kind of thing? What about namespaces?
I've got unwieldy sounding functions like ataxia.repeats.plot.alleles()
and often these functions are not particularly generic, and are only
called three or four times, but they do save repetition.

Do you go to the effort of creating a library that solves your
particular problem, or only reserve that for more generic functionality?
Do people keep all of their R scripts for a specific project separate,
or in one big file? I can see advantages (knowing it all works) and
disadvantages (time for it all to run after minor changes) in both
approaches, but it is unclear to me which is better. I do know that
I've set-up a variety of analyses, moved on to other things, only to
find later on that old scripts have stopped working because I've changed
some interdependency. Does anyone go as far as to use test suites to
check for sane output (apart from doing things manually)?  Note I'm not
asking about how to run R on all these scripts, as people have already
suggested makefiles.

I realise these are vague high-level questions, and there won't be any
right or wrong answers, but I'm grateful to hear about different
strategies in organising R analyses/files, and how people solve these
problems? I've not seen this kind of thing covered in any of the
textbooks. Apologies for being so verbose!

Best wishes,

Mark



-- 
Dr. Mark Wardle
Clinical research fellow and Specialist Registrar in Neurology,
C2-B2 link, Cardiff University, Heath Park, CARDIFF, CF14 4XN. UK

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Re: [R] Turn off border on barplot

[Paul Murrell]

Hi


Marc Schwartz wrote:
 On Thu, 2006-10-26 at 16:13 -0400, Frank McCown wrote:
 I was wondering if there was any way to completely remove the borders 
 around each bar when using the barplot function.  I understand how to 
 change the color of the border (border=white), but I need the border 
 to be invisible so I can plot many skinny bars right next to each other 
 with only the bar's color visible.

 Thanks,
 Frank
 
 You can try:
 
  # Set line type to blank
  # See ?par
  # First save current value
  lty.o - par(lty)
 
  par(lty = 0)
 
  # Also set space to 0, rather
  # than default 0.2, if needed for more room
  barplot(1:10, space = 0)
 
  # Reset to old value
  par(lty = lty.o)


or ...

barplot(1:10, border=NA)

Paul


 Depending upon the look you need, setting space to 0 may be sufficient
 and still allow you to enable a border.
 
 HTH,
 
 Marc Schwartz
 
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-- 
Dr Paul Murrell
Department of Statistics
The University of Auckland
Private Bag 92019
Auckland
New Zealand
64 9 3737599 x85392
[EMAIL PROTECTED]
http://www.stat.auckland.ac.nz/~paul/

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Re: [R] Turn off border on barplot

[Frank McCown]

 barplot(1:10, border=NA)

That's it!  Thanks a ton, guys.

Frank

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Re: [R] Header of dataframe

[Benjamin Dickgiesser]

Thx for the help but again, this is not what I am looking for. I need
the  a to be over the first column e.g.

ameanstdev  n
1 645.6125 65.94129 60
2 655.2121 70.64094 60
3 633.3161 80.48620 60
4 650.3897 77.59191 60
5 630.4955 84.9 60
6 656.2608 66.16100 60
7 666.1775 74.39796 60
8 663.1543 71.10769 60



On 10/26/06, Mihai Nica [EMAIL PROTECTED] wrote:

 names(temp)[1]=a

 see: ?names

 did you really created that function :-)?
  Mihai Nica
 170 East Griffith St. G5
 Jackson, MS 39201
 601-914-0361


 - Original Message 
 From: Benjamin Dickgiesser [EMAIL PROTECTED]
 To: r-help@stat.math.ethz.ch
 Sent: Thursday, October 26, 2006 12:08:08 PM
 Subject: [R] Header of dataframe

 Hi
 I am fairly new to R and I would appreciate some help to hopefully a
 trivial problem.

 I created a function:

 summary.aggregate - function(y, ...)
 {
 temp.mean - aggregate(y, FUN=mean, ...)
 temp.sd  - aggregate(y, FUN=sd, ...)
 temp.length - aggregate(y, FUN=length, ...)
 temp -
 data.frame(cbind(mean=temp.mean$x,stdev=temp.sd$x,n=temp.length$x))
 }

 this outputs e.g.:

   meanstdev  n
 1 645.6125 65.94129 60
 2 655.2121 70.64094 60
 3 633.3161 80.48620 60
 4 650.3897 77.59191 60
 5 630.4955 84.9 60
 6 656.2608 66.16100 60
 7 666.1775 74.39796 60
 8 663.1543 71.10769 60

 Is there an easy way to add a column name to the first column?

 Thank you,
 Benjamin

 __
 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
 and provide commented, minimal, self-contained, reproducible code.




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Re: [R] Turn off border on barplot

[Marc Schwartz]

On Fri, 2006-10-27 at 10:00 +1300, Paul Murrell wrote:
 Hi
 
 
 Marc Schwartz wrote:
  On Thu, 2006-10-26 at 16:13 -0400, Frank McCown wrote:
  I was wondering if there was any way to completely remove the borders 
  around each bar when using the barplot function.  I understand how to 
  change the color of the border (border=white), but I need the border 
  to be invisible so I can plot many skinny bars right next to each other 
  with only the bar's color visible.
 
  Thanks,
  Frank
  
  You can try:
  
   # Set line type to blank
   # See ?par
   # First save current value
   lty.o - par(lty)
  
   par(lty = 0)
  
   # Also set space to 0, rather
   # than default 0.2, if needed for more room
   barplot(1:10, space = 0)
  
   # Reset to old value
   par(lty = lty.o)
 
 
 or ...
 
 barplot(1:10, border=NA)
 
 Paul

Yep, that'll do it too...  :-)

Thanks Paul. I had forgotten about that.

The only place that particular behavior seems documented is in ?par in
the Color Specification section. Does it make sense to add that to
relevant See Also's?

Best regards,

Marc

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Re: [R] Header of dataframe

[Mihai Nica]

Oh, I can see now what you mean. That's beyond me, but... why :-)?
 
Mihai Nica
170 East Griffith St. G5
Jackson, MS 39201
601-914-0361

- Original Message 
From: Benjamin Dickgiesser [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Thursday, October 26, 2006 4:35:20 PM
Subject: Re: [R] Header of dataframe

Thx for the help but again, this is not what I am looking for. I need
the  a to be over the first column e.g.

ameanstdev  n
1 645.6125 65.94129 60
2 655.2121 70.64094 60
3 633.3161 80.48620 60
4 650.3897 77.59191 60
5 630.4955 84.9 60
6 656.2608 66.16100 60
7 666.1775 74.39796 60
8 663.1543 71.10769 60



On 10/26/06, Mihai Nica [EMAIL PROTECTED] wrote:

 names(temp)[1]=a

 see: ?names

 did you really created that function :-)?
  Mihai Nica
 170 East Griffith St. G5
 Jackson, MS 39201
 601-914-0361


 - Original Message 
 From: Benjamin Dickgiesser [EMAIL PROTECTED]
 To: r-help@stat.math.ethz.ch
 Sent: Thursday, October 26, 2006 12:08:08 PM
 Subject: [R] Header of dataframe

 Hi
 I am fairly new to R and I would appreciate some help to hopefully a
 trivial problem.

 I created a function:

 summary.aggregate - function(y, ...)
 {
 temp.mean - aggregate(y, FUN=mean, ...)
 temp.sd  - aggregate(y, FUN=sd, ...)
 temp.length - aggregate(y, FUN=length, ...)
 temp -
 data.frame(cbind(mean=temp.mean$x,stdev=temp.sd$x,n=temp.length$x))
 }

 this outputs e.g.:

   meanstdev  n
 1 645.6125 65.94129 60
 2 655.2121 70.64094 60
 3 633.3161 80.48620 60
 4 650.3897 77.59191 60
 5 630.4955 84.9 60
 6 656.2608 66.16100 60
 7 666.1775 74.39796 60
 8 663.1543 71.10769 60

 Is there an easy way to add a column name to the first column?

 Thank you,
 Benjamin

 __
 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
 and provide commented, minimal, self-contained, reproducible code.




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[[alternative HTML version deleted]]

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Re: [R] Newbie: Better way to do compound conditionals in subset?

[Marc Schwartz]

On Thu, 2006-10-26 at 15:55 -0400, Zembower, Kevin wrote:
 There must be a better way to select the rows after 22-Apr-2004 and
 before 01-Sep-2004 with a temperature below 65 than this:
 
  before2sw1 - subset(energy.data, as.Date(start, format=%d-%b-%y) 
 as.Date(01-Sep-04, format = %d-%b-%y))
  before2sw2 - subset(before2sw1, as.Date(start, format=%d-%b-%y) =
 as.Date(22-Apr-04, format = %d-%b-%y), select=c(therms,temp,days))
  before2sw - subset(before2sw2, temp  65)

Well, the first thing that I would do is to convert 'start' to a Date:

  energy.data$start - as.Date(energy.data$start, format = %d-%b-%y)

Then create your cutoff dates:

  Start - as.Date(22-Apr-04, format = %d-%b-%y)
  End - as.Date(01-Sep-04, format = %d-%b-%y)

Then you can do this:

  New.DF - subset(energy.data, 
  (start = Start)  (start  End)  (days  65),
  select = c(therms, temp, days))

and then add:

  NewDF$HDD - with(NewDF, (therms / (65 - temp) * days))

You can also do the calculations and add a new column to the full data
frame and simply add 'HDD' to the 'select' argument in subset().

Note that if you don't want to modify the original data frame, which is
something that I tend to avoid for a variety of reasons, you can copy it
to another first and then run the above steps on the copy for subsequent
analysis.

You want to generally avoid using attach(), as it can have deleterious
and not immediately evident side-effects.  Review the Details section
of ?attach and note some of the effects seen in the Examples there.

HTH,

Marc Schwartz

 Is it also possible to combine in this step:
 
 attach(before2sw)
 before2sw.HDD - therms / (65 - temp) * days
 
 My data looks like this:
  head(energy.data)
   start therms   gas KWHs elect temp days
 1 10-Jun-98  9 16.84  613 63.80   75   40
 2 20-Jul-98  6 15.29  721 74.21   76   29
 3 18-Aug-98  7 15.73  597 62.22   76   29
 4 16-Sep-98 42 35.81  460 43.98   70   33
 5 19-Oct-98105 77.28  314 31.45   57   29
 6 17-Nov-98106 77.01  342 33.86   48   30
 
 
 Thanks for your suggestions and advice. I'm continuing to enjoy learning
 R.
 
 -Kevin

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[R] Adding shading to background of Time series plot

[jmbucci]

Hello All,

Thank you in advance.

I have looked up and down in the archives to see if I could find some help
on moving the polygons to the background of a ts.plot. I've tried
panel.first in the gpars(...) arguments and trellis.focus, but I get
errors. Am I missing something easy here? Here is my script:

ts.plot(tsdjia, tstenyield,  gpars = list(xlab =  , ylab =  ,
axes=FALSE, col=c(12,14)))
mtext(Time, side=1, line=3, cex = 0.75, font = 1)
mtext(Yield, side=4, line=1, cex = 0.75, font = 1)
mtext(Dow Jones Level, side=2, line=2, cex = 0.7, font = 1)

polygon(cord.x2,cord.y1,col = gray85)
polygon(cord.x3,cord.y1,col = gray85)
polygon(cord.x4,cord.y1,col = gray85)
polygon(cord.x5,cord.y1,col = gray85)
polygon(cord.x6,cord.y1,col = gray85)

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[R] A faster way to calculate Trace?

[YONGWAN CHUN]

I want to know how to get trace of product of matrices **faster** when the 
matrices are really big. Unfortunately the matrices are not symmetric. If 
anybody know how to get the trace of it, please help me. An example is as below.

n - 2500
a - matrix(rnorm(n*n),n,n)
b - matrix(rnorm(n*n),n,n)
tr1 - sum(diag(a %*% b))
tr2 - sum(diag(a %*% b %*% a %*% b))

Thanks,

Yongwan Chun

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Re: [R] Get the names of the columns in a tserie

[lvdtime]

Please, I need this information, it's important for my work

Thank you.


lvdtime wrote:
 
 Hello everybody,
 
 I'm a beginner in R, and I'm currently working on Tseries (analysis of a
 portfolio)
 
 I imported the data like this (library tseries) :
 X-read.ts(X.dat, start=c(1995,1), frequency=261, header=T, sep=;)
 
 There is a header which contains the names of each column (codes of
 shares)
 
 I'd like to know if it is possible to get the names of the columns (to
 display it in the plots : ylab=name of the col)
 To summarize, I wonder if a code like label(X[,i]) exists...
 
 
 Thank you for your help
 
 
 LVDTime
 
 

-- 
View this message in context: 
http://www.nabble.com/Get-the-names-of-the-columns-in-a-tserie-tf2511355.html#a7022436
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] A faster way to calculate Trace?

[Berwin A Turlach]

G'day Yongwan,

 YC == YONGWAN CHUN [EMAIL PROTECTED] writes:

YC I want to know how to get trace of product of matrices
YC **faster** when the matrices are really big. Unfortunately the
YC matrices are not symmetric. If anybody know how to get the
YC trace of it, please help me. An example is as below.  
The first one is quite simple to speed up:

 n - 2500
 a - matrix(rnorm(n*n),n,n)
 b - matrix(rnorm(n*n),n,n)
 sum(diag(a %*% b))
[1] 1890.638

 tb - t(b)
 sum(a*tb)
[1] 1890.638

For the second one, you may try:

  sum(diag(a %*% b %*% a %*% b))
[1] 10668786
 cmat - a %*% b
 sum(cmat*t(cmat))
[1] 10668786

It gives somewhat a speedup, since you only have to multiply two huge
matrices once instead of thrice, but I wonder whether further
improvements are possible.

Hope this helps.

Cheers,

Berwin

== Full address 
Berwin A Turlach  Tel.: +61 (8) 6488 3338 (secr)   
School of Mathematics and Statistics+61 (8) 6488 3383 (self)  
The University of Western Australia   FAX : +61 (8) 6488 1028
35 Stirling Highway   
Crawley WA 6009e-mail: [EMAIL PROTECTED]
Australiahttp://www.maths.uwa.edu.au/~berwin

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Re: [R] Tukey-Kramer test

[Spencer Graves]

  RSiteSearch(Tukey-Kramer) produced 7 hits for me just now.  Have 
you looked at these?  If yes, please provide a commented, minimal, 
self-contained, reproducible example of something you've tried, 
explaining why it did not meet your needs, as suggested in the posting 
guide www.R-project.org/posting-guide.html. 

  Hope this helps. 
  Spencer Graves

A.R. Criswell wrote:
 Hello All,

 I found the TukeyHSD() function. Is there a Tukey-Kramer test for
 unbalanced data?

 Andrew

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[R] barcharts with 2 different axes

[Geoff Russell]

Dear UseRs,

I'm having trouble drawing a barchart  with 2 quantities in different units
as follows with
one unit on the top axis and the other unit on the bottom. E.g.,

   x Tonnes | .
.  .

Row 1 Tonnes1  | xx
Row 1 Litres1 | yyy

Row 2 Tonnes1  | x
Row 2 Litres1 | 

   y Litres|.
.  .


Can anybody provide an example please?

Cheers,
Geoff Russell

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Re: [R] Adding shading to background of Time series plot

[Gabor Grothendieck]

There is an example here using plot

   http://www.mayin.org/ajayshah/KB/R/html/g5.html

and an example here using xyplot.zoo:


   library(zoo)
   ?xyplot.zoo

that would also work with xyplot.ts



On 10/26/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
 Hello All,

 Thank you in advance.

 I have looked up and down in the archives to see if I could find some help
 on moving the polygons to the background of a ts.plot. I've tried
 panel.first in the gpars(...) arguments and trellis.focus, but I get
 errors. Am I missing something easy here? Here is my script:

 ts.plot(tsdjia, tstenyield,  gpars = list(xlab =  , ylab =  ,
 axes=FALSE, col=c(12,14)))
 mtext(Time, side=1, line=3, cex = 0.75, font = 1)
 mtext(Yield, side=4, line=1, cex = 0.75, font = 1)
 mtext(Dow Jones Level, side=2, line=2, cex = 0.7, font = 1)

 polygon(cord.x2,cord.y1,col = gray85)
 polygon(cord.x3,cord.y1,col = gray85)
 polygon(cord.x4,cord.y1,col = gray85)
 polygon(cord.x5,cord.y1,col = gray85)
 polygon(cord.x6,cord.y1,col = gray85)

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[R] VGAM package released on CRAN

[Thomas Yee]

Dear useRs,

upon request, the VGAM package (currently version 0.7-1) has been
officially released on CRAN (the package has been at my website
http://www.stat.auckland.ac.nz/~yee/VGAM for a number of years now).

VGAM implements a general framework for several classes of
regression models using iteratively reweighted least squares
(IRLS). The key ideas are Fisher scoring, generalized linear
and additive models, IRLS and vector smoothing. It is a large R
package that fits many models and distributions, usually by maximum
likelihood estimation. See below for a partial listing, or better,
http://www.stat.auckland.ac.nz/~yee/VGAM/doc/VGAMrefcard.pdf. Some
broad categories are categorical data analysis, LMS quantile regression,
extreme value models, univariate and continuous distributions.

The package is under continual development and definitely not complete
yet. Until then it should probably be considered 'alpha software'. Nothing
is cast in iron yet, so please expect lots of changes if you use it. I
hope to add more functionality to certain parts of the package, especially
relating to inference and model selection, soon.

Re. support, this is minimal. I first have to apologize that it won't be
possible for me to answer every e-mail I receive. Sorry about this. If you
use VGAM please consult the online help plus the incomplete documentation
found at my website. Additionally, other technical details can be found
in the journal articles. On the positive side, a monographs is in the
making, however it is expected to be finished in quite a while :)

Suggestions, bug reports and other comments are very welcome.

Best regards,

Thomas Yee

Thomas W. Yee, Telephone: 64-9-3737599 extn 86857
Department of Statistics,  Fax:   64-9-3737000 or 3677149
University of Auckland,E-mail:[EMAIL PROTECTED]
Private Bag 92019,
Auckland 1001, New Zealand.



Information on package 'VGAM'

Description:

Package:   VGAM
Version:   0.7-1
Date:  2006-10-24
Title: Vector Generalized Linear and Additive Models
Author:Thomas W. Yee [EMAIL PROTECTED]
Maintainer:Thomas Yee [EMAIL PROTECTED]
Depends:   R (= 2.4.0), splines, methods, stats, stats4
Description:   Vector generalized linear and additive models, and
   associated models (Reduced-Rank VGLMs, Quadratic
   RR-VGLMs, Reduced-Rank VGAMs). This package fits many
   models and distribution by maximum likelihood estimation
   (MLE) or penalized MLE. Also fits constrained ordination
   models in ecology.
URL:   http://www.stat.auckland.ac.nz/~yee/VGAM

Index:

AA.Aa.aaThe AA-Aa-aa Blood Group System
AB.Ab.aB.ab The AB-Ab-aB-ab Blood Group System
AB.Ab.aB.ab2The AB-Ab-aB-ab2 Blood Group System
ABO The ABO Blood Group System
acatOrdinal Regression with Adjacent Categories
Probabilities
benini  Benini Distribution Family Function
betaII  Beta Distribution of the Second Kind
betabin.ab  Beta-binomial Distribution Family Function
betabinomialBeta-binomial Distribution Family Function
betaff  The 2-parameter Beta Distribution Family Function
betageometric   Beta-geometric Distribution Family Function
betaprime   The Beta-Prime Distribution
bilogistic4 Bivariate Logistic Distribution Family Function
binom2.or   Bivariate Logistic Regression
binom2.rho  Bivariate Probit Model
binomialff  Binomial Family Function
bisaBirnbaum-Saunders Distribution Family Function
bratBradley Terry Model
bratt   Bradley Terry Model With Ties
calibrate.qrrvglm   Calibration for CQO, UQO and CAO models
cao Fitting Constrained Additive Ordination (CAO)
cauchy1 Cauchy Distribution Family Function
cdf.lmscreg Cumulative Distribution Function for LMS
Quantile Regression
cexpon  Censored Exponential Distribution
cgumbel Censored Gumbel Distribution
chisq   Chi-squared Distribution
cnormal1Censored Normal Distribution
cqo Fitting Constrained Quadratic Ordination (CQO)
cratio  Ordinal Regression with Continuation Ratios
cumulative  Ordinal Regression with Cumulative Probabilities
dagum   Dagum Distribution Family Function
dcnormal1   Univariate Normal Distribution with Double
Censoring
deplot.lmscreg  Density Plot for LMS Quantile Regression
dgumbel The Gumbel Distribution
dirichlet   Fitting a Dirichlet Distribution
dirmultinomial  Fitting a Dirichlet-Multinomial Distribution
erlang  

Re: [R] avoiding a loop

[Ray Brownrigg]

On Wednesday 25 October 2006 07:36, Leeds, Mark (IED) wrote:
 I think I asked a similar question 3 years ago to the  Splus list and I
 think the answer was no or noone answered so noone should spend more
 than 5 minutes on this
 because it could definitely be a waste of time.

 My question is whether the function below can be rewritten without a for
 loop. apply is fine if it can be done that way but i doubt it. I call it
 a lot and would
 prefer to not loop.

 #---
 --

 constructLt-function(invector) {

 outvector-invector

  for ( i in 2:length(invector) ) {
  if ( invector[i]  1 ) {
   outvector[i]-invector[i]*outvector[i-1]
  }
 }

 return(outvector)

 }

Depending on the nature of your data, there is a faster way.  It still 
involves looping, but not over the entire vector.

Try the following:
constructLt - function(invector) {
  outvector - invector
  cs - cumsum(rle(invector  1)$lengths)
  if (invector[1]  1)
cs - c(1, cs)
  for (i in 0:(length(cs)%/%2 - 1)){
starti - cs[2*i + 1]
stopi - cs[starti + 1]
outvector[starti:stopi] - cumprod(invector[starti:stopi])
  }
  return(outvector)
}

It is in the order of 3 times as fast for random vectors of considerable 
length ( 1000).  For random vectors of length 50 it is about the same speed 
as the full looping algorithm.  However if the data is such that there are 
longer runs (than N(1, 1)), then you might expect a better speedup.

HTH
Ray Brownrigg

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[R] Dum: problem installing (Don't install at all)

[Daniel Sarrazin]

I'm on mandriva 2006

I did:
urpmi R-2.0.0-1mdk.i586.rpm

got the message:
Some asked package can't be intalled:
R-2.0.0-1mdk.i586 (due to unsatified)
Continue ? (Y/N)

Then after I try typping R on the shell and it does work. Of course there is 
no R reposatory

Thank

Daniel

##
Daniel Sarrazin
B.Sc Physics
[EMAIL PROTECTED]
##

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and provide commented, minimal, self-contained, reproducible code.

[R] How to call Fortran routin in R

[lan gao]

Hi, all,
 Can anyone give me steps how to call Fortan routine from Visual Fortran
in R?
 Many thanks

Gao

[[alternative HTML version deleted]]

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Header of dataframe

[Gregor Gorjanc]

Benjamin Dickgiesser dickgiesser at gmail.com writes:
...
 Is there an easy way to add a column name to the first column?

What about

summary.aggregate - function(y, ...)
{
  temp.mean - aggregate(y, FUN=mean, ...)
  temp.sd - aggregate(y, FUN=sd, ...)
  temp.length - aggregate(y, FUN=length, ...)
  temp - data.frame(cbind(col=names(y), mean=temp.mean$x, 
   stdev=temp.sd$x,n=temp.length$x))
}

or 

summary.aggregate - function(y, ...)
{
  temp.mean - aggregate(y, FUN=mean, ...)
  temp.sd - aggregate(y, FUN=sd, ...)
  temp.length - aggregate(y, FUN=length, ...)
  temp - data.frame(cbind(mean=temp.mean$x, 
   stdev=temp.sd$x,n=temp.length$x))
  rownames(temp) - names(y)
}

Gregor

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Re: [R] barcharts with 2 different axes

[Deepayan Sarkar]

On 10/26/06, Geoff Russell [EMAIL PROTECTED] wrote:
 Dear UseRs,

 I'm having trouble drawing a barchart  with 2 quantities in different units
 as follows with
 one unit on the top axis and the other unit on the bottom. E.g.,

x Tonnes | .
 .  .

 Row 1 Tonnes1  | xx
 Row 1 Litres1 | yyy

 Row 2 Tonnes1  | x
 Row 2 Litres1 | 

y Litres|.

So if I read this right, the relative heights of the xxx bars and the
yyy bars, which you have placed side by side, are in different units
and thus their comparison is meaningless. It's doable, but do you
really want this sort of graph? A better and much easier to implement
design seems to be something like

  +-+-+
Row 1 | xx  | yyy |
Row 2 | x   | |
  +-+-+
 Tonnes1   Litres1

 Can anybody provide an example please?

Please provide some example data (the form of which will matter).

-Deepayan

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