date:20120807

On Mon, Aug 6, 2012 at 7:35 PM, Marc Schwartz marc_schwa...@me.com wrote:
 is.letter - function(x) grepl([[:alpha:]], x)
 is.number - function(x) grepl([[:digit:]], x)

Quick follow-up question.

I'm always reluctant to create functions that would resemble the
method of a function (here, is() ), but would in fact not be a genuine
method. So would there be any incompatibility between is() and
is.letter(), given that the latter is not a method of the former?
Is it good (or acceptable) practice to define is.letter() as above?
Would is_letter() be better?

Regards
Liviu

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Re: [R] Passing arguments to a function within a function ...

2012-08-07 Thread Rui Barradas


Hello,

Your problem seems simple, if I understand it correctly. Just add an 
extra argument to the caller function, drep().



drep - function(pop.conc = x, sicklist = adult.CP.mortality,
par = cental estimate, ...) { # This extra 
argument

dr.out = by(x, x$Name, function(x) {
z = as.data.frame((x))

dose.response(pop = z$pop * sum(z[, adult.CP.mortality$Xpop]),
conc = z$conc,
cases = z$cases,
relationship = sicklist$relationship,
beta = sicklist$beta,
Xpop = sicklist$Xpop,
par = par,  # Note the difference (!)
verbose = FALSE)
})
names(dr.out) - paste(sicklist$end.point, : , names(dr.out), sep 
= )

dr.out
}


In this case, the new argument has a default value.When the callee 
dose.response() is called, the first 'par' is the name one of it's 
arguments, the second 'par' is a value, the passed to value. The first 
is a name, it does not have a value and can not, for instance, be 
printed, only the second can. They are completely different in nature.


And the same for other arguments, like 'base.conc'.

Hope this helps,

Rui Barradas

Em 07-08-2012 09:52, christiaan pauw escreveu:

Hallo Everybody

How do you specify arguments for a function used within another function?

Here is my problem:

I am reconstructing a calculator for the burden of disease due to air
pollution from publications and tools published by the WHO. The
calculations make use of published dose-response relationships for
particular health end-points. This is then applied to populations with
known or estimated levels of exposure and incidence rates to calcute
the number of cases of each end-point attributable to each pollutant.
I have functions that work on their own but when is have to use the
one within the other, I don't know how to specify its arguments

Here are example data and the functions:


## Example data frame with population, concentration and cases ##

x = data.frame(Name = LETTERS[1:10],
pop=sample(x=1000:1,size=10),
Xbabies = 0.106,
Xkids = 0.232,
Xteens = 0.375,
Xadults = 0.235,
Xaged = 0.52,
cases = sample(x=100:500,size=10),
conc = sample(x=20:125,size=10)
)

## Two of the published dose-response relationships

adult.CP.mortality = list(end.point = Cardiopulmanory mortality in
adults over 30,
   pollutant = PM10,
   relationship = log-linear,
   beta = c(0.0562,0.1551,0.2541),
   Xpop =
c(Xbabies,Xkids,Xteens,Xadults,Xaged)[4:5])

adult.LC.mortality = list(end.point = Lung Cancer mortality in adults over 30,
   pollutant = PM10,
   relationship = log-linear,
   beta = c(0.0856, 0.2322,0.3787),
   Xpop =
c(Xbabies,Xkids,Xteens,Xadults,Xaged)[4:5])

## Generic function to calculculate the Attributable cases for a
pollutant in a population
dose.response - function(pop,
   Xpop = 1,
   conc,
   base.conc=7.5,
   relationship = c(linear,log-linear)[1],
   beta=c(0.0006,0.0008,0.0010),
   cases=NULL,
   incidence.rate=NULL,
   par = c(low estimate,cental
estimate,high estimate),
   verbose = FALSE
   ){
   # Turn case rate into case number
   d - cases
   if(verbose==TRUE) message(d = , d)
   if(verbose==TRUE) message(pop = , pop)
   if(verbose==TRUE) message(beta = , beta)
   if(verbose==TRUE) message(conc = , conc)
   if(verbose==TRUE) message(base.coc = , base.conc)
   if(verbose==TRUE) message(relationship = , relationship)
   if(is.null(cases)==TRUE) {d - incidence.rate*pop} # use incidence
rate if cases are not available
   if(relationship == linear) {RR = exp(beta*(conc-base.conc))}
#RR=exp[beta(X-Xo)]
   if(relationship == log-linear) {RR = ((conc+1)/(base.conc+1))^beta}
   if(verbose==TRUE) message(RR = , RR)
   AF =  (RR-1)/RR  #AF=(RR-1)/RR
   if(verbose==TRUE) message(AF = , AF)
   AM =  AF * d #AM = AF * cases
   if(verbose==TRUE) message(AM = , AM)
   out = data.frame(beta=beta,cases=d,
RelativeRisk=RR,AttributableFraction=AF,
AttributableInsidence=AM)
   rownames(out) = c(low estimate,cental estimate,high estimate)
   if(verbose==TRUE) message(dimentions of out  = , dim(out))
   out.idx = na.omit(match(par,rownames(out)))
   out[out.idx,]
}

## Function using the  published dose-response relationships with the
generic function
drep - function(pop.conc=x,sicklist=adult.CP.mortality,...){

Re: [R] How to convert data to 'normal' if they are in the form of standard scientific notations?

2012-08-07 Thread Tal Galili

Try:

options(scipen = 9)


Contact
Details:---
Contact me: tal.gal...@gmail.com |  972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--




On Tue, Aug 7, 2012 at 11:11 AM, Petr PIKAL petr.pi...@precheza.cz wrote:

 Hi

 
  Dear Jean
 
  Thanks a lot for your help.
 
  The reason I did not provide producible code is that my work started
 with
  reading in some large csv files, e.g. the data is not created by myself.
   But the data is from the same data provider so I would expect to
 receive
  data in exactly same data format.
 
 
  I use read.csv to read the data in. My major curious is that by using
  exactly same code as I provided in my email, e.g. 'as.factor' why one of
  them work (e.g. convert the numerical data to factor) but the other  one
  remains numerical with scientific notation?  So, in R, how do I check if
  the data format are different for these two files in their original csv
  files, which  might cause the different results..?
 
  Also I tried your code and created some reproducible examples, but still
  can not make it work as in your example

 a-c(2.0e+9,2.1e+9)
 is.numeric(a)
 [1] TRUE
 a.f-factor(a)
 is.numeric(a.f)
 [1] FALSE
 is.factor(a.f)
 [1] TRUE

 so factor comes correctly

 print(a,digits=12)
 [1] 2.0e+09 2.1e+09

 print(a, digits=21)
 [1] 20.000 21.000

 b-c(3000,3100)
 print(b,digits=5)
 [1] 3.0e+07 3.1e+07

 So the printed result depends probably on your local setting.
 See also ?options help page. And maybe also ?format and ?sprintf

 Regards
 Petr


 
 
   a-c(2.0e+9,2.1e+9) print(a,digits=4)[1] 20 21  # I
  expected to see 2.0e+9 here...? print(a,digits=7)[1] 20
  21  # Think here I should expect same 2.0e+9?
 getOption(digits)
  # Checking my default number of digits now..[1] 7 b-c(3000,
  3100) print(b)[1] 3000 3100   # This is what I expected

  to see print(b,digits=5)[1] 3000 3100   # I'm so confused why
 it
  is not working, e.g. printing 3.0e+9! getOption(digits)   # checking
  again, but now I would expect it has being changed to 5[1] 7
 
 
  Any thoughts please...?
 
  Thanks
  HJ
 
 
  On Mon, Aug 6, 2012 at 7:04 PM, Jean V Adams jvad...@usgs.gov wrote:
 
   HJ,
  
   You don't provide any reproducible code, so I had to make up my own.
  
   dat - data.frame(a=letters[1:5], x=c(20110911001084, 20110911001084,
   20110911001084, 20110911001084, 20110911001084),
   y=c(2.10004e+12, 2.10004e+12, 2.10004e+12, 2.10004e+12,
   2.10004e+12))
  
   In my example, the long numbers print out without scientific notation.
  
   dat
 a  x y
   1 a 20110911001084 210004000
   2 b 20110911001084 210004000
   3 c 20110911001084 210004000
   4 d 20110911001084 210004000
   5 e 20110911001084 210004000
  
   I can make it print with scientific notation using the digits argument
 to
   the print() function.
  
   print(dat, digits=3)
 ax   y
   1 a 2.01e+13 2.1e+12
   2 b 2.01e+13 2.1e+12
   3 c 2.01e+13 2.1e+12
   4 d 2.01e+13 2.1e+12
   5 e 2.01e+13 2.1e+12
  
   What is your default number of digits?
   getOption(digits)
  
   Jean
  
  
   HJ YAN yhj...@googlemail.com wrote on 08/06/2012 11:14:17 AM:
  
   
Dear R users
   
I read two csv data files into R and  called them Tem1 and Tem5.
   
For the first column, data in Tem1 has 13 digits where in Tem5 there
 are
   14
digits for each observation.
   
Originally there are 'numerical' as can be seen in my code below.
 But
   how
can I display/convert them using other form rather than scientific
notations which seems a standard/default?
   
 I want them to be in the form like '20110911001084', but I'm very
   confused
why when I used 'as.factor' call it works for my 'Tem1' but not for
'Tem5'...??
   
   
Many thanks!
   
HJ
   
 Tem1[1:5,1][1] 2.10004e+12 2.10004e+12 2.10004e+12 2.10004e+12 2.
10004e+12 Tem5[1:5,1][1] 2.011091e+13 2.011091e+13 2.011091e+13 2.
  
011091e+13 2.011091e+13 class(Tem1[1:5,1])[1] numeric class(Tem5
[1:5,1])[1] numeric as.factor(Tem1[1:5,1])[1] 2.10004e+12 2.
10004e+12 2.10004e+12 2.10004e+12 2.10004e+12
Levels: 2.10004e+12 as.factor(Tem5[1:5,1])[1] 20110911001084
20110911001084 20110911001084 20110911001084 20110911001084
Levels: 20110911001084
  
 
 [[alternative HTML version deleted]]
 
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Re: [R] Passing arguments to a function within a function ...

2012-08-07 Thread christiaan pauw

Thanks Rui

It works.

On 7 August 2012 11:34, Rui Barradas ruipbarra...@sapo.pt wrote:
 Hello,

 Your problem seems simple, if I understand it correctly. Just add an extra
 argument to the caller function, drep().

Christiaan Pauw
Nova Institute
www.nova.org.za

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Re: [R] Force evaluation of a symbol when a function is created

2012-08-07 Thread Duncan Murdoch

Here's one more way.  It seems to me this is the most R-like way to do 
what you want:


  multiply_by_Y - function(Y) {
force(Y)
function(x) x*Y
  }

  F - multiply_by_Y(3)

The force call forces Y to be evaluated at that point, so its value is 
fixed from that point forward.


Duncan Murdoch

On 12-08-06 5:07 PM, Schoenfeld, David Alan,Ph.D.,Biostatistics wrote:


I am porting a program in matlab to R,
The problem is that Matlab has a feature where symbols that aren't arguments 
are evaluated immediately.
That is:
Y=3
F=@(x) x*Y

Will yield a function such that F(2)=6.
If later say. Y=4 then F(2) will still equal 6.

R on the other hand has lazy evaluation.
F-function(x){x*Y}
Will do the following
Y=3
F(2)=6
Y=4
F(2)=8.
Does anyone know of away to defeat lazy evaluation in R so that I can easily 
simulate the Matlab behavior.  I know that I can live without this in ordinary 
programming but it would make my port much easier.

Thanks.




The information in this e-mail is intended only for the ...{{dropped:14}}

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Re: [R] Overlay Histogram

2012-08-07 Thread li li

Thanks a lot.
Hannah

2012/8/6 R. Michael Weylandt michael.weyla...@gmail.com

 See

 example(layout)

 for one idea. I think you might also want to look into rug plots.

 Best,
 Michael

 On Mon, Aug 6, 2012 at 2:40 PM, li li hannah@gmail.com wrote:
  Dear all,
For two sets of random variables, say, x -  rnorm(1000, 10, 10) and  y
  - rnorm(1000. 3, 20).
  Is there any way to overlay the histograms (and density curves) of x and
 y
  on the plot of y vs. x?
  The histogram of x is on the x axis and that of y is on the y axis.
The density curve here is to approximate the shape of the distribution
  and does not have to have area 1.
 Thank you in advance.
Hannah
 
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Re: [R] add leading zeros

Hello


On Fri, Jul 27, 2012 at 6:54 AM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
 Much easier than you think:

 x - c(1L, 9000L)

 sprintf(%05i,x)

For anyone interested, I came up with a small wrapper for the above:
add.lead - function(x, width=max(nchar(x))){
sprintf(paste('%0', width, 'i', sep=''), x)
}

 x - c(1L, 15L, 234L, 9000L)
 (xa - add.lead(x))
[1] 0001 0015 0234 9000
 nchar(xa)
[1] 4 4 4 4
 (xb - add.lead(x, 5))
[1] 1 00015 00234 09000
 nchar(xb)
[1] 5 5 5 5
 (xc - add.lead(x, 15))
[1] 001 015 234 0009000
 nchar(xc)
[1] 15 15 15 15


Regards
Liviu

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[R] Error with convUL (PBSmapping)

2012-08-07 Thread niandra

Hi all,

I'm trying to use the function convUL from the Package PBSmapping but i get i 
weird error with my data but also with the example in the convUL's help. 

Here is the example script with the error message: 
data(nepacLL, package=PBSmapping)
#--- set the zone attribute
#--- use a zone that is most central to the mapped region
attr(nepacLL, zone) - 6
#--- convert and plot the result
nepacUTM - convUL(nepacLL)
Error in .C(convUL, inXY = as.double(inXY), inVerts = as.integer(inVerts),  : 
  C symbol name convUL not in DLL for package PBSmapping

someone can help me?
Thanks
[[alternative HTML version deleted]]

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Re: [R] Parallel runs of an external executable with snow in local

2012-08-07 Thread Uwe Ligges




On 06.08.2012 17:07, Xavier Portell/UPC wrote:

Thanks Uwe but, actually, I did so.

Since #8220;filetorun.exe#8221; looks in the current folder for 
#8220;input.txt#8221;, I tried moving all needed files to a newly created temporary folder 
#8220;tmp.id#8221; (say, tmp.1) and running the executable. This works fine by doing it 
directly from the windows command line but not by doing it from R, since using:

#
System(#8220;C:/Users/#8230;/currentworkdirectory/temp.1/filetorun.exe#8221;)
#

Causes #8220;filetorun.exe#8221; to look for #8220;input.txt#8221; in 
#8220;C:/Users/#8230;/currentworkdirectory#8221;. So there#8217;s no point on moving files 
to a folder, it seems that input file must be situated in the current R work directory. Does anybody know how 
to avoid this behaviour?


Errr, you mean you cannot specify an input for filetorun.exe?

If so, you can ask R to switch to use another working directory for each 
instance.


Best,
Uwe Ligges









I hope I#8217;ve explained that clearly,

Xavier Portell Canal, PhD candidate.
Department of Agri-food Engineering,
Universitat Politècnica de Catalunya

-Uwe Ligges lig...@statistik.tu-dortmund.de ha escrit: -

Per a: Xavier Portell/UPC xavier.port...@upc.edu
De: Uwe Ligges lig...@statistik.tu-dortmund.de
Data: 05/08/2012 07:46PM
a/c: r-help@r-project.org
Assumpte: Re: [R] Parallel runs of an external executable with snow in local



On 03.08.2012 19:21, Xavier Portell/UPC wrote:

Hi everyone,

I'm aiming to run an external executable (say filetorun.EXE) in parallel. The external executable collect needed data from a 
file, say input.txt and, in turn,generates several output files, say output.txt. I need to generate 
input.txt, run the executable and keep input.txt and output.txt. I'm using Windows 7, R 
version 2.15.1 (2012-06-22) on RStudio and platform: i386.pc.mingw32/i386 (32-bit).

My first attempt was a R code which, by using
System(filetorun.EXE, intern = F, ignore.stdout = F,
ignore.stderr = F, wait = T, input = NULL,
show.output.on.console = T, minimized = F, invisible = T))
, ran the executable and kept required files to a conveniently named folder. 
After that I changed my previous R script so I could use the function 
lapply().This script apparently worked fine.

Finally, I tried to parallelize the problem by using snow and parLapply(). The 
resulting script looks like this:

## Not run
#
library(snow)cl - makeCluster(3, type = SOCK)
clusterExport(cl,list('param.esp','copy.files','for12.template','program.executor'))
parLapply(cl,a.list,a.function))stopCluster(cl)
#
##End not run

Although it runs, the parallelized version is messing up the input parameters to pass to the 
executable (see table below, where parameters P1 and P2 are considered. .s comes from 
the serial code and .p from the parallelized one):
s r P1.s P2.s P1.p P2.p
1 1 1  1.0 3.00  2.0 3.00
2 2 1  1.5 3.00  2.0 3.75
3 3 1  2.0 3.00  2.0 3.00
4 4 1  1.0 3.75  1.5 3.00
5 5 1  1.5 3.75  1.5 3.00
6 6 1  2.0 3.75  2.0 3.75

My first thought to avoid the described behaviour was creating a temporary file, say tmp.id with id being an identification run number, 
and copying filetorun.EXE and Input.txt to tmp.id. However, while doing so, I realised that although running the 
correct filetorun.EXE copy (i.e., the one in tmp.id) R looks for input.txt in the work directory.



Not sure about the real setup, but you can actually specify the path,
not only filenames.

Uwe Ligges





I've been looking thoroughly for a solution but I got nothing.

Thanks for any help in advance,


Xavier Portell Canal

PhD candidate
Department of Agri-food engineering,
Universitat Politècnica de Catalunya

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Re: [R] process evaluation packages (slightly off topic)

Search on R package Gantt, where you will find, among others,

http://cran.r-project.org/web/packages/plan/plan.pdf

-- Bert

On Tue, Aug 7, 2012 at 1:28 AM, Petr PIKAL petr.pi...@precheza.cz wrote:
 Dear all

 I need to perform some process evaluation. Sorry for not posting data and
 code - I do not have any, I ask only for pointing me to correct direction.

 Suppose I have several connected processes P1, P2, ..., Pn. Each process
 takes some time and have some capacity (let say like preparing a dinner
 for several persons - only one stove, limited capacity of utensils,
 heating and cooling takes some time) and some processes can by cyclic (fry
 onion in pan, put it aside, in the same pan fry meat, put an onion and
 some water and simmer for a while...).

 I can prepare some oriented graph (paper/pencil) or Word or drawing
 programme, I can also evaluate whole process by shading spreadsheet cells
 but those two tasks are not connected.

 Is there any R package/other software suitable for simplifying or helping
 in such tasks? E.g. When I prepare oriented graph with capacity and time
 for each node is there any automatic way to transfer this graph to
 timeline to see how long whole process will take, where are bottlenecks or
 so?

 Thank you

 Petr

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-- 

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

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Re: [R] Force evaluation of a symbol when a function is created

Duncan, et.al:

Many thanks: let the closure do the work automatically rather than
manually manipulating it.

However, in the spirit of the OP's original request, I believe the
call would be:

Y - 3  ## That is, Y gets a value at some prior point, perhaps 
programmatically.
   F - multiply_by_Y(Y) # ... F picks up this value implicitly -- no need 
 for explicit assignment.

But then there is no need for force(), is there?

 fy - function(Y)function(x) x*Y
 Y - 2
 F - fy(Y)
 F(5)
[1] 10
 Y - 3
 F(5)
[1] 10
 G - fy(Y)
 G(5)
[1] 15

That is, one simply relies on lexical scoping/closures to retain the
value of  Y used as a free variable in  function(x)x*Y when it is
defined. No need to explicitly force() it. If wrong, I would be
grateful for correction. This appears to me to duplicate the Matlab
behavior rather closely.


-- Bert

On Tue, Aug 7, 2012 at 3:48 AM, Duncan Murdoch murdoch.dun...@gmail.com wrote:
 Here's one more way.  It seems to me this is the most R-like way to do what
 you want:

   multiply_by_Y - function(Y) {
 force(Y)
 function(x) x*Y
   }

   F - multiply_by_Y(3)

 The force call forces Y to be evaluated at that point, so its value is
 fixed from that point forward.

 Duncan Murdoch

 On 12-08-06 5:07 PM, Schoenfeld, David Alan,Ph.D.,Biostatistics wrote:


 I am porting a program in matlab to R,
 The problem is that Matlab has a feature where symbols that aren't
 arguments are evaluated immediately.
 That is:
 Y=3
 F=@(x) x*Y

 Will yield a function such that F(2)=6.
 If later say. Y=4 then F(2) will still equal 6.

 R on the other hand has lazy evaluation.
 F-function(x){x*Y}
 Will do the following
 Y=3
 F(2)=6
 Y=4
 F(2)=8.
 Does anyone know of away to defeat lazy evaluation in R so that I can
 easily simulate the Matlab behavior.  I know that I can live without this in
 ordinary programming but it would make my port much easier.

 Thanks.




 The information in this e-mail is intended only for the ...{{dropped:14}}

 __
 R-help@r-project.org 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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Internal Contact Info:
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Re: [R] Overlay Histogram

I cannot see any obvious way to do this.  Ben Gunter's suggestion re layout 
makes sense.  Here is a version using grid and ggplot2.  Note I shamelessly 
stole code to due it.


library(ggplot2)
library(grid)

dd - data.frame(x =  rnorm(1000, 10, 10), 
   y =  rnorm(1000, 3, 20))
#  From https://stat.ethz.ch/pipermail/r-help/2011-June/280588.html

p1  -  ggplot(dd, aes(x=x)) +
geom_histogram(aes(y=..density..), fill=red, colour=black)+
geom_density(colour=black, adjust=4) +
opts(title=Normal Random Sample)

p2  -  ggplot(dd, aes(x=y)) +
geom_histogram(aes(y=..density..), fill=blue, colour=black)+
geom_density(colour=black, adjust=4) +
opts(title=Normal Random Sample) +
coord_flip()

# From StackOverflow 
http://stackoverflow.com/questions/9490482/combined-plot-of-ggplot2-not-in-a-single-plot-using-par-or-layout-functio

# this simplifies the vp statement 
# otherwise we would have to use something like
# print(plot5 , vp = viewport(layout.pos.row = 2, layout.pos.col = 1:2))  
# and so on for the rest of the plots.
vplayout - function(x, y) viewport(layout.pos.row = x, layout.pos.col = y)

grid.newpage()
pushViewport(viewport(layout = grid.layout(2, 1)))
print(p1, vp = vplayout(1, 1))   
print(p2, vp = vplayout(2, 1))




John Kane
Kingston ON Canada


 -Original Message-
 From: hannah@gmail.com
 Sent: Mon, 6 Aug 2012 15:40:55 -0400
 To: r-help@r-project.org
 Subject: [R] Overlay Histogram
 
 Dear all,
   For two sets of random variables, say, x -  rnorm(1000, 10, 10) and  y
 - rnorm(1000. 3, 20).
 Is there any way to overlay the histograms (and density curves) of x and
 y
 on the plot of y vs. x?
 The histogram of x is on the x axis and that of y is on the y axis.
   The density curve here is to approximate the shape of the distribution
 and does not have to have area 1.
Thank you in advance.
   Hannah
 
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Re: [R] Plotting Where People Live on a U.S. Map

2012-08-07 Thread Lopez, Dan

Hi Ricardo,

I wonder if this is like the other site I found recently on the web: 
http://www.gpsvisualizer.com/geocoder/

Thanks for sending me this resource. I will check it out.

Dan

From: Ricardo Pietrobon [mailto:rpie...@duke.edu]
Sent: Monday, August 06, 2012 10:44 PM
To: Lopez, Dan
Cc: Sarah Goslee; R help (r-help@r-project.org)
Subject: Re: [R] Plotting Where People Live on a U.S. Map

Dan, google refine http://goo.gl/AeKml can actually transform zip codes into 
longitude/latitude - http://goo.gl/1HDWb will show you how to do this from 
street adresses, but it should also work from city names -- i think it will 
allocate a default long/lat for a city, but not sure of the exact mechanism
On Fri, Aug 3, 2012 at 1:10 PM, Lopez, Dan 
lopez...@llnl.govmailto:lopez...@llnl.gov wrote:
Thank you!
Dan

From: Sarah Goslee 
[mailto:sarah.gos...@gmail.commailto:sarah.gos...@gmail.com]
Sent: Thursday, August 02, 2012 5:51 PM
To: Lopez, Dan
Cc: R help (r-help@r-project.orgmailto:r-help@r-project.org)
Subject: Re: [R] Plotting Where People Live on a U.S. Map

Hi Dan,

For question 1, yes you'll need geographic coordinates. I thinknit's possible 
to get a shapefile of zip codes, but maybe someone else will know the details.

For #2, you  probably want maps instead of map, and you need to load a package 
before you can use it:
install.packages(maps)
library(maps)
and then your code.

Sarah

On Thursday, August 2, 2012, Lopez, Dan wrote:
Hi,

QUESTION TOPIC #1
I have some data I want to plot on a map. But what I have are home addresses: 
street, City, State, complete postal code--i.e 95377-1234. Is there a way to 
plot this data or do I need latitudinal and longitude coordinates? If so how do 
I convert them? Is there a package that will do the conversion in R?

QUESTION TOPIC #2
I was trying to experiment with this code that I found at the site below but 
got a message that indicated that the map function is not found. So I tried 
installing the maps package but got the below message. Is there an alternative 
way of doing this (please refer to URL below)?

# The message I got:
 install.packages(map)
Warning message:
package 'map' is not available (for R version 2.15.0)

# The code I tried to run:
states - data.frame(map(state, plot=FALSE)[c(x,y)])
colnames(states) - c(Lon,Lat)
ggplot(states, aes(x=Lon, y=Lat)) + geom_path()
+ geom_point(alpha=0.6,size=0.3,data=subway)

# Where I got the code from and also an image of what I am attempting to do 
(please enter this in your URL)
http://www.google.com/imgres?um=1hl=enbiw=1790bih=845tbm=ischtbnid=4rMjXYA_w1qDiM:imgrefurl=http://www.informaniac.net/docid=SJqcsPghztrj0Mimgurl=http://lh5.ggpht.com/_yBbodrC25kU/Ta6Ifqr0ZLI/AAABRCg/98rIF-kMMns/map%25255B7%25255D.pngw=512h=319ei=mgsbUIzqJuKbiAL5v4DQDgzoom=1iact=hcvpx=176vpy=477dur=5741hovh=177hovw=285tx=110ty=113sig=117496213270544868088page=2tbnh=125tbnw=200start=32ndsp=40ved=1t:429,r:0,s:32,i:175

Dan

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http://www.stringpage.com
http://www.sarahgoslee.com
http://www.functionaldiversity.org

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[R] Styling gridExtra's title and left labels

2012-08-07 Thread Alastair

Hi,

I'm using the gridExtra package to combine some graphs like in the
arrangeGrob example. Each of the graphs has a title but they appear much
larger than the overall combined plot title and left axis label. Does anyone
know how I can control the style / size of the gridExtra labels? 

library(gridExtra)
library(ggplot2)
dsamp - diamonds[sample(nrow(diamonds), 1000), ] 

p1 - qplot(carat, price, data=dsamp, colour=clarity)
p2 - qplot(carat, price, data=dsamp, colour=clarity, geom=path)


tmp - ggplot_gtable(ggplot_build(p1))
leg - which(sapply(tmp$grobs, function(x) x$name) == guide-box)
legend - tmp$grobs[[leg]]

grid.arrange(arrangeGrob(p1 + theme(legend.position=none) +
ggtitle(Scatter Plot),
p2 + theme(legend.position=none) +
ggtitle(Line Plot),
main =this is a title,
left = This is my global Y-axis
title), legend, 
 widths=unit.c(unit(1, npc) - legend$width,
legend$width), nrow=1)





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[R] Overlapping a Plot with Dataframe

2012-08-07 Thread mhimanshu

Hello Everyone,

I am trying to overlap a plot with a data set in the form of a data frame. 

Its very easy to overlap the data using points function. But the only
problem I am facing is Standard deviation bar on the plot.

data - data.frame(
  x = c(3.00,2.00,3.80,2.40,2.00),
  error = c(0.0,0.4,1.1,0.7,0.5)
)

I tried plotrix, segments but they are making a new plot with data points n
error bar.
I want to Overlap this data sets on a plot with its Error bar.

Can anyone help me with it.
Many Thanks,
Himanshu



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[R] time plot

2012-08-07 Thread Karan Anand

hi,
i am new to using R if someone can please tell me how to change the
angel of the graph to a 45 degree angel.

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Re: [R] ERROR : cannot allocate vector of size (in MB GB)

2012-08-07 Thread Akkara, Antony (GE Energy, Non-GE)

How is possible to split a .csv file in terms of size (in KiloByte) ? 

-Original Message-
From: jim holtman [mailto:jholt...@gmail.com] 
Sent: Tuesday, July 24, 2012 11:30 PM
To: Akkara, Antony (GE Energy, Non-GE)
Cc: r-help@r-project.org
Subject: Re: [R] ERROR : cannot allocate vector of size (in MB  GB)

try this:

input - file(yourLargeCSV, r)
fileNo - 1
repeat{
myLines - readLines(input, n=10) # 100K lines / file
if (length(myLines) == 0) break
writeLines(myLines, sprintf(output%03d.csv, fileNo))
fileNo - fileNo + 1
}
close(input)

On Tue, Jul 24, 2012 at 9:45 AM, Rantony antony.akk...@ge.com wrote:
 Hi,

 Here in R, I need to load a huge file(.csv) , its size is 200MB. [may 
 come more than 1GB sometimes].
 When i tried to load into a variable it taking too much of time and 
 after that when i do cbind by groups, getting an error like this

  Error: cannot allocate vector of size 82.4 Mb 

 My requirement is, spilt data from Huge-size-file(.csv) to no. of 
 small csv files.
 Here i will give no of lines to be 'split by' as input.

 Below i give my code
 ---
 SplitLargeCSVToMany -
function(DataMatrix,Destination,NoOfLineToGroup)
 {
 test - data.frame(read.csv(DataMatrix))

 # create groups No.of rows
 group - rep(1:NROW(test),
each=NoOfLineToGroup)
 new.test - cbind(test, group=group)
 new.test2 - new.test
 new.test2[,ncol(new.test2)] - NULL

 # now get indices to write out
 indices - split(seq(nrow(test)), new.test[, 
 'group'])

 # now write out the files
 for (i in names(indices))
 {
 write.csv(new.test2[indices[[i]],], 
 file=paste(Destination,data., i, .csv, sep=),row.names=FALSE)
 }
 }

 -
 My system Configuration is,
 Intel Core2 Duo
 speed : 3GHz
 2 GB RAM
 OS: Windows-XP [ServicePack-3]
 ---

 Any hope to solve this issue ?

 Thanks in advance,
 Antony.

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 Nabble.com.

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--
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Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

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[R] Problem with global variable building a package

2012-08-07 Thread Eva Prieto Castro



Hi,

Â 

My name is
Eva and this is my first message here. My English is not very good, but I hope
you can understand my question, in the context of an academic project.

Â 

I have
developed several functions in R and the
idea is that the user can access functions in order to:

Â 

1)Â Â Â Â Â 
Alter
parameters in relation with data and type of analysis.

2)Â Â Â Â Â 
Run
statistical analysis (Text and pdf files with results will be generated).

3)Â Â Â Â Â 
View
the value of the most important parameters.

Â 

All the
parameters I need are stored in a list object, and this list is used in all the
functions along the cycle performed by the user, but I would like the user does
not need to pass the name of the list as argument when he/she runs the different
functions, so I think I need to treat the list as a global variable.

Â 

Firstly, I
used the global assignment operator (â-â); secondly, I used âgetâ and 
âassignâ
functions and even I used a new.env() in order to use a new environment
exclusively for my list. However, when I try to build a package with all my
functions I donât reach this end, because of an error in parse process.

Â 

My question
for you is the following: taking into account that my end is to build a
package, what can I do with this âglobalâ list?. How can I treat it?.

Â 

Thanks in
advance.

Â 

Regards,

Eva

Â 


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Re: [R] label_wrap_gen question

2012-08-07 Thread Brian Diggs


On 8/6/2012 9:07 PM, vd3000 wrote:

Hi, all

I am trying to use the label_wrap_gen function in this website.
https://github.com/hadley/ggplot2/wiki/labeller
I tried to make a long name like this

Light and heavy good vehicles (diesel) -\nGVX

f2 = facet_grid(vehicle ~ ., labeller=label_wrap_gen(width=15))

eventually, I got something like this in my label...
*Light and heavy
good vehicles
(diesel) - GVX*
I suppose the -n could break GVX to the next row but it failed...
Is it a bug? or it has been overpowered by width=15?? so -n could not
function well?

Eventually I tried f2 = facet_grid(vehicle ~.)

The -n did work and I got
*Light and heavy  good vehicles (diesel) -
GVX*

But it also failed because I could not show all the label properly...

Anyone has idea about this?
It is freaking me out~ I am sorry I am stupid on R
Thanks in advance.

VD


label_wrap_gen uses strwrap to do the heavy lifting of figuring out 
how to actually split the text into lines. From the help page of 
strwrap, Whitespace (space, tab or newline characters) in the input is 
destroyed., so this is documented behavior. I don't see an option to 
strwrap to suppress this behavior.



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Senior Research Associate, Department of Surgery
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[R] R enquire

2012-08-07 Thread Manuel Montesino San Martin

Dear Madam or Mister,

My name is Manuel Montesino and I am a PhD student at the University of 
Copenhagen. I am not a programmer myself, but my thesis requires the combined 
use of R and a model. I would be very grateful if you help me with the 
following issue;

I am trying to combine a small piece of code written in Fortran 77 with R. The 
F77 code aims to select data from a library and write it into a new file (1) 
The F77 code has being compiled using gFortran and the statements declared at 
(2). Once everything is done, the dynamic library is loaded and run in R using 
the piece of code showed at (3).Then, a runtime error appears when R is asked 
to run the subroutine (weathersel), R closes down and the selected weather 
data file is created. Exploring the error a bit further shows that the loop 
stops at the line 1917. The subroutine works for lower number of lines and the 
subroutine works if only the last lines of the weather library are explored. 
This seems to me that there is a limit in the number of iterations that R can 
handle depending on the memory available at the computer. However, I am not 
sure. Why appears the runtime error? Is it true that there is a limitation in 
the loops that R can handle? Is it because a piece of F77!
  code?

Thank you very much in advance for your help and I hope to hear from you soon.

Best wishes,

Manuel.


(1)F77 code
subroutine weathersel(GRD,YR,SWJD,HRVJD)
C==
C --
C --
C
  INTEGER :: GRID,YEAR,JDAY,A,B,C,D
  REAL :: LAT,LONG,MINT,MAXT,TEMP
  REAL :: PREC,WIND,RAD,RH,DELTA,ET
C---
C ** Identifying the variables
C
  A=GRD
  B=YR
  C=SWJD
  D=HRVJD
C
C-
C  ** Opening the files
C
  open(unit=16,file='Weather library\weather.dat',status='old') 
Open the file with the weather data.
  open(unit=18,file='weathersel.dat')   
   Create a new file.
C
C-
C  ** Crating the library
C
DO I=1,12922
 From the first 
to the last row in the weather data...
  READ(16,*)GRID,LAT,LONG,YEAR,JDAY,MINT,MAXT,  
  ...read the weather variables in the row...
  TEMP,PREC,WIND,RAD,RH,DELTA,ET
  IF(GRID.EQ.A)THEN 
 ...if the read 
location equals to the solicited location...
IF(YEAR.EQ.B.AND.JDAY.GT.C.AND.JDAY.LT.D)THEN   
 (and other options)
 WRITE(18,*)JDAY,MAXT,MINT,TEMP,RH,PREC,WIND,RAD
   ...then write the data in the new file in that order...
   END IF
 END IF
ENDDO
C
   END  
  
...end the program


(2)gFortran compilation:


gfortran -c -fdefault-real-8 -Wall weathrsel.for
gfortran -shared -o WEATHRSEL.DLL weathrsel.o
del *.o
pause



(3)R code calling the dynamic library


#Load the library
dyn.load(WEATHRSEL.DLL)

# Define the parameters
GRD-23 # Grid
YR-1992   # Year
SWJD-210  # Showing julian day
HRVJD-260 # Harvesting julian day

# Run the code
.Fortran('weathersel',GRD,YR,SWJD,HRVJD)

# Stop using the library
dyn.unload(WEATHRSEL.DLL)


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[R] Unable to download R package

2012-08-07 Thread ODowd, Darragh

Hi,

I'm new to R. I'm using the R Gui and have tried to install additional packeges.
When I choose Packages/Install package(s) I get the following error...

utils:::menuInstallPkgs() Warning: unable to access index for repository 
http://cran.ma.imperial.ac.uk/bin/windows/contrib/2.15
Warning: unable to access index for repository 
http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.15
Error in install.packages(NULL, .libPaths()[1L], dependencies = NA, type = 
type) :
no packages were specified

Can someone


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Re: [R] how to write out a tree file with bootstrap from phangorn package

2012-08-07 Thread Klaus Schliep

Dear Jian-Feng,

you can use the function plotBS. plotBS plots a tree and adds the
support values (in %). This function also silently returns this tree:
tree -  plotBS(fit$tree, bs)
# You can export than this tree using write.tree or write.nexus, e.g.
write.tree(tree)

Regards,
Klaus


On 8/7/12, Mao Jianfeng jianfeng@gmail.com wrote:
 Dear R-helpers and Klaus,

 I would like to know how to write out a tree file with bootstrap from
 phangorn package. That tree file could be in newick format or others.

 I am new for phylogenetic operation in R. Could you please give me any
 directions on that? Thanks in advance.

 Best wishes,
 Jian-Feng,

 
 # as a example
 # I accomplished 1000 bootstrap simulation on a fit object (a maximum
 likelihood tree object)
 # how could I output a tree in newick format for tuning outside R?

 bs - bootstrap.pml(fit, bs=1000, optNni=T, optInv=T, multicore=T)


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[R] Hist function

2012-08-07 Thread Dominic Roye

Hello everyone,


First i explain my aim. I would like to calculate the frecuency of
temperature (datos.mx1, columns 3-6)  for each month with the thresholds of
table lf.med and lc.med.

numero1 - hist(subset(datos.mx1[,3],
datos.mx1$Mes==1),plot=FALSE,breaks=c(min(datos.mx1[,3]),lf.med[1,2],lc.med[1,2],max(datos.mx1[,3])))
# First month and only the first weather station (column 3)

I now have to apply the above command for all 12 months at all the weather
stations. I've been trying to get with a for-loop, but without success.

How can i do it the best way?

After this command, i'm only interested in the part of $counts. I hope
someone can help me. Any help is greatly appreciated!!!  Thank you,
everybody`!!

 datos.mx1
  Fecha Mes Serra.da.Foladoira Santiago Sergude Rio.Do.Sol
101.01.2006   14.1 14.4 9.35.7
202.01.2006   15.5  9.811.07.6
303.01.2006   19.9 13.012.79.9
404.01.2006   19.7 13.312.79.7
505.01.2006   1   10.2 12.213.18.0
606.01.2006   14.2 14.210.25.4
707.01.2006   11.0  6.9 3.23.2
.
.
...etc.

lf.med
   Mes Serra.da.Foladoira  Santiago   Sergude Rio.Do.Sol
11   7.833702  8.811945  8.122162   7.899513
22   7.766978  8.592693  8.041483   7.887327
33   7.864609  8.861457  8.385405   8.080646
44   8.204098  8.986149  8.803744   8.438141
55   8.716766  9.317366  9.385637   8.977087
66   9.377651  9.874497 10.194340   9.686213
77   9.838470 10.055430 10.447219   9.991618
88  10.046204 10.099868 10.520872  10.098330
99   9.780911 10.038629 10.156048   9.851299
10  10   9.425308  9.779886  9.646881   9.496192
11  11   8.417168  9.346191  8.791033   8.549046
12  12   7.968706  8.795800  8.038208   8.017603

lc.med
   Mes Serra.da.Foladoira Santiago  Sergude Rio.Do.Sol
11   21.29353 26.62562 23.71784   22.04786
22   21.97787 27.07213 25.68340   23.39020
33   22.69749 27.42612 26.97889   24.33741
44   24.49683 29.21054 29.76726   26.59280
55   25.24233 29.83056 30.75132   27.28277
66   26.46742 30.70485 33.02328   28.77892
77   27.72731 30.24450 34.28162   29.97767
88   27.64356 30.84850 34.52161   30.00666
99   27.21998 31.18390 33.19776   29.54618
10  10   25.35830 28.83195 29.63192   26.42831
11  11   22.37895 26.38226 25.69239   23.47575
12  12   21.44064 26.09142 23.99676   22.26208

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[R] What is the difference between probe level data and summarized data

2012-08-07 Thread ali_protocol

Hi all,

What is the difference between probe level data and summarized data.
Thanks in advance.



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[R] (no subject)

2012-08-07 Thread bob jackson

hi, 
I'm working on a research project where we're looking at the changing 
resonances in femur bones during hip replacement operations. basically, I've 
got a spreadsheet with one column listing frequencies in 5Hz bands and 3 
columns showing the amplitude of that frequency when the bone is loose, medium 
to tight (referring to the tension of the reamer/chisel creating a cavity for 
the hip replacement). 
I'm coming at this more from a sound perspective so I would really appreciate 
some advice from anyone with a handle on stats, could correlograms be used in 
presenting this data? When working with a large number of bone samples, can 
anyone recommend a good method of recognizing and mapping these changes?
All the best, 
Bob__
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[R] predicting test dataset response from training dataset with randomForest

2012-08-07 Thread Mojca ZELNIKAR




Hi

I am new to R so I apologize if this is trivial.

I am trying to predict the resistance or susceptibility of my  
sequences to a certain drug with a randomForest function from a file  
with amino acids on each of the positions in the protein. I ran the  
following:



library(randomForest)

path - C:\\...
path2 - ...
name - ...

actualFileName - paste(path, path2, name, .txt, sep=)

# reading in the training dataset
dat1 - read.table(actualFileName, header=TRUE, sep=\t,  
colClasses=character)


head(dat1)

  X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15 X16 X17 ... SR
1  M  K  V  K  L  L  V  L  L   C   T   F   T   A   T   Y   A ... suscep
2  M  K  V  K  L  L  V  L  L   C   T   F   A   A   T   Y   A ... suscep
3  M  K  V  K  L  L  V  L  L   C   T   F   T   A   T   Y   A ... resist
...



# some of the important sites identified by fisher test
dat1$X13 - as.factor(dat1$X13)
dat1$X52 - as.factor(dat1$X52)
dat1$X53 - as.factor(dat1$X53)
dat1$X64 - as.factor(dat1$X64)
dat1$X85 - as.factor(dat1$X85)
dat1$X99 - as.factor(dat1$X99)
dat1$X111 - as.factor(dat1$X111)
dat1$X142 - as.factor(dat1$X142)
dat1$X157 - as.factor(dat1$X157)
dat1$X158 - as.factor(dat1$X158)
dat1$X162 - as.factor(dat1$X162)
dat1$X169 - as.factor(dat1$X169)
dat1$X200 - as.factor(dat1$X200)
dat1$X202 - as.factor(dat1$X202)
dat1$X203 - as.factor(dat1$X203)
dat1$X205 - as.factor(dat1$X205)
dat1$X206 - as.factor(dat1$X206)
dat1$X209 - as.factor(dat1$X209)
dat1$X210 - as.factor(dat1$X210)
dat1$X225 - as.factor(dat1$X225)
dat1$X269 - as.factor(dat1$X269)
dat1$X283 - as.factor(dat1$X283)
dat1$X290 - as.factor(dat1$X290)
dat1$X432 - as.factor(dat1$X432)
dat1$X434 - as.factor(dat1$X434)
dat1$X455 - as.factor(dat1$X455)
dat1$X467 - as.factor(dat1$X467)
dat1$X512 - as.factor(dat1$X512)
dat1$SR - as.factor(dat1$SR)



dat1.rf -randomForest(SR ~ X13+ X52+ X53+ X64+ X85+ X99+ X111+  
X142+ X157+ X158+ X162+ X169+ X200+
+ X202+ X203+ X205+ X206+ X209+ X210+ X225+ X269+ X283+ X290+ X432+  
X434+ X455+ X467+ X512, data=dat1, importance=TRUE,

+ proximity=TRUE, varUsed=TRUE, ntree=5000, varImpPlot=TRUE)



print(dat1.rf)



varImpPlot(dat1.rf)

varUsed(dat1.rf, by.tree=FALSE, count=TRUE)




MDSplot(dat1.rf, dat1$SR, palette=rep(1, 2),

+   pch=as.numeric(dat1$SR))



path3 - C:\\Users...
path4 - ...
name2 - ...

# reading in the test dataset
actualFileName2 - paste(path3, path4, name2, .txt, sep=)

dat2 - read.table(actualFileName2, header=TRUE, sep=\t,  
colClasses=character)






dat2$X13 - as.factor(dat2$X13)
dat2$X52 - as.factor(dat2$X52)
dat2$X53 - as.factor(dat2$X53)
dat2$X64 - as.factor(dat2$X64)
dat2$X85 - as.factor(dat2$X85)
dat2$X99 - as.factor(dat2$X99)
dat2$X111 - as.factor(dat2$X111)
dat2$X142 - as.factor(dat2$X142)
dat2$X157 - as.factor(dat2$X157)
dat2$X158 - as.factor(dat2$X158)
dat2$X162 - as.factor(dat2$X162)
dat2$X169 - as.factor(dat2$X169)
dat2$X200 - as.factor(dat2$X200)
dat2$X202 - as.factor(dat2$X202)
dat2$X203 - as.factor(dat2$X203)
dat2$X205 - as.factor(dat2$X205)
dat2$X206 - as.factor(dat2$X206)
dat2$X209 - as.factor(dat2$X209)
dat2$X210 - as.factor(dat2$X210)
dat2$X225 - as.factor(dat2$X225)
dat2$X269 - as.factor(dat2$X269)
dat2$X283 - as.factor(dat2$X283)
dat2$X290 - as.factor(dat2$X290)
dat2$X432 - as.factor(dat2$X432)
dat2$X434 - as.factor(dat2$X434)
dat2$X455 - as.factor(dat2$X455)
dat2$X467 - as.factor(dat2$X467)
dat2$X512 - as.factor(dat2$X512)
dat2$SR - as.factor(dat2$SR)



dat2.pred-predict(dat1.rf, dat2, type=response, norm.votes=TRUE,  
predict.all=FALSE, proximity=FALSE, nodes=FALSE)


Error in predict.randomForest(dat1.rf, dat2, type = response,  
norm.votes = TRUE,  :

  New factor levels not present in the training data


The thing is that each of the amino acid positions in the training  
dataset is present also in the training dataset. So I don't know how  
to deal with the error.


Thank you very much.

Kind regards,

Mojca Zelnikar

--
The University of Edinburgh is a charitable body, registered in
Scotland, with registration number SC005336.

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Re: [R] How to convert data to 'normal' if they are in the form of standard scientific notations?

Hi,

This might also helps.
a-c(2.0e+9,2.1e+9)
is.numeric(a)
#[1] TRUE
 format(a,sci=FALSE)
#[1] 20 21
A.K.





- Original Message -
From: Petr PIKAL petr.pi...@precheza.cz
To: HJ YAN yhj...@googlemail.com
Cc: r-help@r-project.org
Sent: Tuesday, August 7, 2012 4:11 AM
Subject: Re: [R] How to convert data to 'normal' if they are in the form of 
standard scientific notations?

Hi

 
 Dear Jean
 
 Thanks a lot for your help.
 
 The reason I did not provide producible code is that my work started 
with
 reading in some large csv files, e.g. the data is not created by myself.
  But the data is from the same data provider so I would expect to 
receive
 data in exactly same data format.
 
 
 I use read.csv to read the data in. My major curious is that by using
 exactly same code as I provided in my email, e.g. 'as.factor' why one of
 them work (e.g. convert the numerical data to factor) but the other  one
 remains numerical with scientific notation?  So, in R, how do I check if
 the data format are different for these two files in their original csv
 files, which  might cause the different results..?
 
 Also I tried your code and created some reproducible examples, but still
 can not make it work as in your example

a-c(2.0e+9,2.1e+9)
is.numeric(a)
[1] TRUE
a.f-factor(a)
is.numeric(a.f)
[1] FALSE
is.factor(a.f)
[1] TRUE

so factor comes correctly

print(a,digits=12)
[1] 2.0e+09 2.1e+09

print(a, digits=21)
[1] 20.000 21.000

b-c(3000,3100)
print(b,digits=5)
[1] 3.0e+07 3.1e+07

So the printed result depends probably on your local setting.
See also ?options help page. And maybe also ?format and ?sprintf

Regards
Petr


 
 
  a-c(2.0e+9,2.1e+9) print(a,digits=4)[1] 20 21  # I 
 expected to see 2.0e+9 here...? print(a,digits=7)[1] 20 
 21  # Think here I should expect same 2.0e+9? 
getOption(digits)
 # Checking my default number of digits now..[1] 7 b-c(3000,
 3100) print(b)[1] 3000 3100       # This is what I expected 

 to see print(b,digits=5)[1] 3000 3100   # I'm so confused why 
it 
 is not working, e.g. printing 3.0e+9! getOption(digits)   # checking 
 again, but now I would expect it has being changed to 5[1] 7
 
 
 Any thoughts please...?
 
 Thanks
 HJ
 
 
 On Mon, Aug 6, 2012 at 7:04 PM, Jean V Adams jvad...@usgs.gov wrote:
 
  HJ,
 
  You don't provide any reproducible code, so I had to make up my own.
 
  dat - data.frame(a=letters[1:5], x=c(20110911001084, 20110911001084,
          20110911001084, 20110911001084, 20110911001084),
          y=c(2.10004e+12, 2.10004e+12, 2.10004e+12, 2.10004e+12,
  2.10004e+12))
 
  In my example, the long numbers print out without scientific notation.
 
  dat
    a              x             y
  1 a 20110911001084 210004000
  2 b 20110911001084 210004000
  3 c 20110911001084 210004000
  4 d 20110911001084 210004000
  5 e 20110911001084 210004000
 
  I can make it print with scientific notation using the digits argument 
to
  the print() function.
 
  print(dat, digits=3)
    a        x       y
  1 a 2.01e+13 2.1e+12
  2 b 2.01e+13 2.1e+12
  3 c 2.01e+13 2.1e+12
  4 d 2.01e+13 2.1e+12
  5 e 2.01e+13 2.1e+12
 
  What is your default number of digits?
  getOption(digits)
 
  Jean
 
 
  HJ YAN yhj...@googlemail.com wrote on 08/06/2012 11:14:17 AM:
 
  
   Dear R users
  
   I read two csv data files into R and  called them Tem1 and Tem5.
  
   For the first column, data in Tem1 has 13 digits where in Tem5 there 
are
  14
   digits for each observation.
  
   Originally there are 'numerical' as can be seen in my code below. 
But
  how
   can I display/convert them using other form rather than scientific
   notations which seems a standard/default?
  
    I want them to be in the form like '20110911001084', but I'm very
  confused
   why when I used 'as.factor' call it works for my 'Tem1' but not for
   'Tem5'...??
  
  
   Many thanks!
  
   HJ
  
Tem1[1:5,1][1] 2.10004e+12 2.10004e+12 2.10004e+12 2.10004e+12 2.
   10004e+12 Tem5[1:5,1][1] 2.011091e+13 2.011091e+13 2.011091e+13 2.
 
   011091e+13 2.011091e+13 class(Tem1[1:5,1])[1] numeric class(Tem5
   [1:5,1])[1] numeric as.factor(Tem1[1:5,1])[1] 2.10004e+12 2.
   10004e+12 2.10004e+12 2.10004e+12 2.10004e+12
   Levels: 2.10004e+12 as.factor(Tem5[1:5,1])[1] 20110911001084
   20110911001084 20110911001084 20110911001084 20110911001084
   Levels: 20110911001084
 
 
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[R] Save reload list objects

2012-08-07 Thread robgriffin247

Hi,
I think this is probably ( hopefully) very easy but I the ideas I have
tried don't work.

Simply, I have 2 very large lists in my workspace which take an awfully long
time to make, each containing the information from 12500+ lmer models.
I would like to now save these lists as files (txt, csv... whatever suits
best) on to my computer/usb.
Then I would like to be able to reload them in to the workspace on another
computer (the second computer will not be able to run the script which makes
the list m1-
lapply(split(narrow,narrow$gene),function(x)lmer(value~sex+(1|line:sex)+(1|line),data=x))
) in exactly the same layout,/state/condition as they were when they were
made (exact replicas basically) so that it does not affect subsequent code
in my script .
Then I can just silence the original script line which makes the list and
just have a couple of lines which save and load each one to the same object
name as the original lists.

So the commands I need help for should able to *save the lists *rg.lmer and
rg.lmer2 separately
And then *load the lists as lists, /not a df/,*  in to another workspace

Possibly looking towards sink() but not sure and couldn't figure out how
this works...

Thanks in advance
Rob



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[R] Sum of vector elements

2012-08-07 Thread number10

Hi, Is it possible to avoid using do and while loops to calculate the sum of
the elements of a vector until the appearance of the first positive element.



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[R] Interpreting predictions of svm

2012-08-07 Thread Camomille

Hi, I have some difficulties in interpreting the prediction of a svm model
using the package e1071.

y1 is the variable I want to predict. It is of type factor and has got two
levels:   50% and  50%.
z is the dataset.

 model - svm(y1 ~ ., data = z,type=C-classification, cross=10)
 model

Call:
svm(formula = y1 ~ ., data = z, type = C-classification, cross = 10)


Parameters:
   SVM-Type:  C-classification
 SVM-Kernel:  radial
   cost:  1
  gamma:  0.07142857

Number of Support Vectors:  68

 pred - predict(model,newdata=z,probability=TRUE,decision.values = TRUE)
 table(pred)
pred
 50%  50%
  414 0

The results of pred is not what I intended to get as, I expected this
type of result:

50%  50%

  50%8925
 50% 38262

What should I do?




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Re: [R] trouble with looping for effect of sampling interval increase

2012-08-07 Thread White, William Patrick

My apologies, here is a sample dataset generator:

#Running sum Test Data
Coin - c(-1,1)
flips=sample(Coin, 1000, replace=T)
Runningsum -cumsum (flips)
#A deactivated plot
#plot (Runningsum)
Test - cbind (Runningsum)
datasetORIGINAL  - cbind (Runningsum)






From: Jean V Adams [jvad...@usgs.gov]
Sent: Monday, August 06, 2012 1:33 PM
To: White, William Patrick
Cc: r-help@r-project.org
Subject: Re: [R] trouble with looping for effect of sampling interval increase

You would make it much easier for R-help readers to solve your problem if you 
provided a small example data set with your code, so that we could reproduce 
your results and troubleshoot the issues.

Jean


Naidraug white@wright.edu wrote on 08/05/2012 09:08:25 AM:

 I've looked everywhere and tinkered for three days now, so I figure asking
 might be good.
 So here's a general rundown of what I am trying to get my code to do I am
 giving you the whole rundown because I need a solution that retain certain
 ways of doing things because they give me the information i need.
 I want to examine the effect of increasing my sampling interval on my data.
 Example: what if instead of sampling every hour I sampled every two, oh
 yeah, how about every three?.. etc ad nausea.  How I want to do this is to
 take the data I have now, add an index  to it, that contains counters. Those
 counters will look something like 1,2,1,2,.. for the first one,
 1,2,3,1,2,3.. for the next one. I have a lot of them, like say a thousand...
 Then for each column in the index my loops should start in the first column,
 run only the ones, store that, then run the twos, and store that in the same
 column of output in a different row. Then move to the next column run the
 ones, store in the next column of output, run the twos, store in the next
 row of that column, run the threes, etc on out until there is no more. I
 want to use this index for a number of reasons. The first is that after this
 I will be going back through and using a different method for sub-sampling
 but keeping all else the same. So all I have to do there is change the way I
 generate the index. The second is that it allows me to run  many subsamples
 and see their range.  So the code I have made, generates my index, and does
 the heavy lifting all correctly, as well as my averages, and quartiles, but
 a look at the head () of my key output (IntervalBetas)  shows that something
 has gone a miss. You have to look close to catch it.  The values generated
 for each row of output are identical, this should not be the case, as row
 one of the first output column should be generated from all values indexed
 by a one in the first column, whereas in column two there are different
 values indexed by the number one. I've checked about everything I can think
 of, done print() on my loop sequence things (those little i and j) and
 wiggled about everything. I am flummoxed. I think the bit that is messing up
 is in here :
 #Here is the loop for betas from sampling interval increase
  c - WHOLESIZE[2]-1
  for (i in 1:c)
  {
  x - length(unique(index[,i]))

  for (j in 1:x)
  {

  data - WHOLE [WHOLE[,x]==j,1]

 But also here is the whole code in case I am wrong that that is the problem
 area:

 #loop for making index


  #clean dataset of empty cells
  dataset - na.omit (datasetORIGINAL)
  #how messed up was the data?
  holeyDATA - datasetORIGINAL - dataset

  D - dim(dataset)

 #what is the smallest sample?
 tinysample - 100




 #how long is the dataset?
  datalength - length (dataset)


  #MD - how many divisions

 MD - datalength/tinysample

  #clear things up for the index loop
  WHOLE - NULL
 index - NULL
  #do the index loop

  for (a in 1:MD)
  {
  index - cbind (index, rep (1:a, length = D[1]))
  }
 index - subset(index, select = -c(1) )

  #merge dataset and index loop
  WHOLE - cbind (dataset, index)

  WHOLESIZE - dim (WHOLE)

 #Housekeeping before loops
 IntervalBetas - NULL


 IntervalBetas - c(NA,NA)
 IntervalBetas - as.data.frame (IntervalBetas)
 IntervalLowerQ - NULL
 IntervalUpperQ - NULL
 IntervalMean - NULL
 IntervalMedian - NULL

 #Here is the loop for betas from sampling interval increase
  c - WHOLESIZE[2]-1
  for (i in 1:c)
  {
  x - length(unique(index[,i]))

  for (j in 1:x)
  {

  data - WHOLE [WHOLE[,x]==j,1]




  #get power spectral density

  PSDPLOT - spectrum (data, detrend = TRUE, plot = FALSE)
  frequency - PSDPLOT$freq
  PSD - PSDPLOT$spec
  #log transform the power spectral density
  Logfrequency - log(frequency)
  LogPSD- log(PSD)
  #fit my line to the data
  Line - lm (LogPSD ~ Logfrequency)
  #store the slope of the line
  Betas - rbind (Betas, -coef(Line)[2])

 #Get values on the curve shape
 BSkew - skew (Betas)
 BMean - mean (Betas)
 BMedian - median (Betas)
 Q - quantile (Betas)


 #store curve shape values
 IntervalLowerQ - rbind (IntervalLowerQ , Q[2])
 IntervalUpperQ - rbind (IntervalUpperQ , Q[4])
 IntervalSkew - rbind (IntervalSkew ,

Re: [R] time plot

What graph?  

Please read the posting guidelines and provide some example data and some code 
showing what you are doing.

John Kane
Kingston ON Canada


 -Original Message-
 From: anand.kara...@gmail.com
 Sent: Tue, 7 Aug 2012 12:24:50 +0530
 To: r-help@r-project.org
 Subject: [R] time plot
 
 hi,
 i am new to using R if someone can please tell me how to change the
 angel of the graph to a 45 degree angel.
 
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Re: [R] What is the difference between probe level data and summarized data

No idea.  What are you doing?  What packages are you  using?
What function is (presumably) giving you these results?


John Kane
Kingston ON Canada


 -Original Message-
 From: mohammadianalimohammad...@gmail.com
 Sent: Tue, 7 Aug 2012 01:12:08 -0700 (PDT)
 To: r-help@r-project.org
 Subject: [R] What is the difference between probe level data and
 summarized data
 
 Hi all,
 
 What is the difference between probe level data and summarized data.
 Thanks in advance.
 
 
 
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 View this message in context:
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Re: [R] Save reload list objects

2012-08-07 Thread Ista Zahn

Hi Rob,

Unless I've missed something you are just looking for ?save and ?load.

Best,
Ista

On Tue, Aug 7, 2012 at 9:20 AM, robgriffin247 rg.rfo...@hotmail.co.uk wrote:
Hi,
I think this is probably ( hopefully) very easy but I the ideas I have
tried don't work.

Simply, I have 2 very large lists in my workspace which take an awfully long
time to make, each containing the information from 12500+ lmer models.
I would like to now save these lists as files (txt, csv... whatever suits
best) on to my computer/usb.
Then I would like to be able to reload them in to the workspace on another
computer (the second computer will not be able to run the script which makes
the list m1-
lapply(split(narrow,narrow$gene),function(x)lmer(value~sex+(1|line:sex)+(1|line),data=x))
) in exactly the same layout,/state/condition as they were when they were
made (exact replicas basically) so that it does not affect subsequent code
in my script .
Then I can just silence the original script line which makes the list and
just have a couple of lines which save and load each one to the same object
name as the original lists.

So the commands I need help for should able to *save the lists *rg.lmer and
rg.lmer2 separately
And then *load the lists as lists, /not a df/,* in to another workspace

Possibly looking towards sink() but not sure and couldn't figure out how
this works...

Thanks in advance
Rob

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Re: [R] Overlapping a Plot with Dataframe

What are you actually plotting?   

John Kane
Kingston ON Canada

 -Original Message-
 From: bioinfo.himan...@gmail.com
 Sent: Tue, 7 Aug 2012 05:57:06 -0700 (PDT)
 To: r-help@r-project.org
 Subject: [R] Overlapping a Plot with Dataframe

 Hello Everyone,

 I am trying to overlap a plot with a data set in the form of a data
 frame.

 Its very easy to overlap the data using points function. But the only
 problem I am facing is Standard deviation bar on the plot.

 data - data.frame(
   x = c(3.00,2.00,3.80,2.40,2.00),
   error = c(0.0,0.4,1.1,0.7,0.5)
 )

 I tried plotrix, segments but they are making a new plot with data points
 n
 error bar.
 I want to Overlap this data sets on a plot with its Error bar.

 Can anyone help me with it.
 Many Thanks,
 Himanshu

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Re: [R] Unable to download R package

2012-08-07 Thread Jeff Newmiller

Try a different repository?
---
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--- 
Sent from my phone. Please excuse my brevity.

ODowd, Darragh darragh.od...@ilim.com wrote:

Hi,

I'm new to R. I'm using the R Gui and have tried to install additional
packeges.
When I choose Packages/Install package(s) I get the following error...

utils:::menuInstallPkgs() Warning: unable to access index for
repository http://cran.ma.imperial.ac.uk/bin/windows/contrib/2.15
Warning: unable to access index for repository
http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.15
Error in install.packages(NULL, .libPaths()[1L], dependencies = NA,
type = type) :
no packages were specified

Can someone


--
**
Irish Life Investment Managers Limited is regulated by the Central Bank
of Ireland.  Irish Life Investment Managers Limited Registered Office:
Beresford Court, Beresford Place, Dublin 1.  Registered in Ireland
Number 116000

While Irish Life Investment Managers uses reasonable efforts to ensure
that the information contained in this email is current, accurate and
complete at the date of publication, no representations or warranties
are made (express or implied) as to the reliability, accuracy or
completeness of such information. Irish Life Investment Managers
therefore cannot be held liable for any loss arising directly or
indirectly from the use of, or any action taken in reliance on, any
information contained in this email. 

This material is for information only and does not constitute an offer
or recommendation to buy or sell any investment, or subscribe to any
investment management or advisory service. It is intended for the use
of institutional and other professional investors. Past performance is
not indicative of future results. The value of funds we manage may fall
as well as rise.

This email and any files transmitted with it are
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Re: [R] Sum of vector elements

I'd do something like this:

x - sample(seq(-10, 10))

sum(x[seq_len(which.max(x  0)])

Though others might have more direct solutions.

which.max() gets you the index of the first time x  0 -- seq_len
gives you numbers 1 to that index -- then just subset and sum like
normal.

Best,
Michael

On Tue, Aug 7, 2012 at 7:37 AM, number10 havana_dr...@ymail.com wrote:
 Hi, Is it possible to avoid using do and while loops to calculate the sum of
 the elements of a vector until the appearance of the first positive element.



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Re: [R] Unable to download R package

I can access the Mac tree of the ICL repos and it seems unlikely that
both ICL and Prof Ripley's site would be down at the same time.

OP: are you having firewall issues?

Michael

On Tue, Aug 7, 2012 at 10:58 AM, Jeff Newmiller
jdnew...@dcn.davis.ca.us wrote:
 Try a different repository?
 ---
 Jeff NewmillerThe .   .  Go Live...
 DCN:jdnew...@dcn.davis.ca.usBasics: ##.#.   ##.#.  Live Go...
   Live:   OO#.. Dead: OO#..  Playing
 Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
 /Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
 ---
 Sent from my phone. Please excuse my brevity.

 ODowd, Darragh darragh.od...@ilim.com wrote:

Hi,

I'm new to R. I'm using the R Gui and have tried to install additional
packeges.
When I choose Packages/Install package(s) I get the following error...

utils:::menuInstallPkgs() Warning: unable to access index for
repository http://cran.ma.imperial.ac.uk/bin/windows/contrib/2.15
Warning: unable to access index for repository
http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.15
Error in install.packages(NULL, .libPaths()[1L], dependencies = NA,
type = type) :
no packages were specified

Can someone


--
**
Irish Life Investment Managers Limited is regulated by the Central Bank
of Ireland.  Irish Life Investment Managers Limited Registered Office:
Beresford Court, Beresford Place, Dublin 1.  Registered in Ireland
Number 116000

While Irish Life Investment Managers uses reasonable efforts to ensure
that the information contained in this email is current, accurate and
complete at the date of publication, no representations or warranties
are made (express or implied) as to the reliability, accuracy or
completeness of such information. Irish Life Investment Managers
therefore cannot be held liable for any loss arising directly or
indirectly from the use of, or any action taken in reliance on, any
information contained in this email.

This material is for information only and does not constitute an offer
or recommendation to buy or sell any investment, or subscribe to any
investment management or advisory service. It is intended for the use
of institutional and other professional investors. Past performance is
not indicative of future results. The value of funds we manage may fall
as well as rise.

This email and any files transmitted with it are
confide...{{dropped:16}}

__
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Re: [R] (no subject)

1. This is completely off topic. This is an R help list, not a
statistical advice list like stats.stackexchange.com.

2. I would strongly recommend that you abandon internet statistical
advice lists and seek local statistical help. You simply do not appear
to have sufficient statistical background to formulate an appropriate
analysis for what appears to be complex data. That you think you can
do so appears, itself, to be a measure of your statistical ignorance.*
Of course, that's just my perception...

Cheers,
Bert

* Note: This is not meant in any sense as an insult, merely my
perception of the state of your statistical knowledge based on your
post. I am, for example, ignorant of audio physics, dentistry,
plumbing, JAVA, ...

On Tue, Aug 7, 2012 at 4:15 AM, bob jackson jackbob...@yahoo.com wrote:
hi,
I'm working on a research project where we're looking at the changing
resonances in femur bones during hip replacement operations. basically, I've
got a spreadsheet with one column listing frequencies in 5Hz bands and 3
columns showing the amplitude of that frequency when the bone is loose,
medium to tight (referring to the tension of the reamer/chisel creating a
cavity for the hip replacement).
I'm coming at this more from a sound perspective so I would really appreciate
some advice from anyone with a handle on stats, could correlograms be used in
presenting this data? When working with a large number of bone samples, can
anyone recommend a good method of recognizing and mapping these changes?
All the best,
Bob
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Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

Re: [R] Unable to download R package

Olympics--they are all off partying?

John Kane
Kingston ON Canada

 -Original Message-
 From: michael.weyla...@gmail.com
 Sent: Tue, 7 Aug 2012 11:03:19 -0500
 To: jdnew...@dcn.davis.ca.us
 Subject: Re: [R] Unable to download R package

 I can access the Mac tree of the ICL repos and it seems unlikely that
 both ICL and Prof Ripley's site would be down at the same time.

 OP: are you having firewall issues?

 Michael

 On Tue, Aug 7, 2012 at 10:58 AM, Jeff Newmiller
 jdnew...@dcn.davis.ca.us wrote:
 Try a different repository?
 ---
 Jeff NewmillerThe .   .  Go
 Live...
 DCN:jdnew...@dcn.davis.ca.usBasics: ##.#.   ##.#.  Live
 Go...
   Live:   OO#.. Dead: OO#..  Playing
 Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
 /Software/Embedded Controllers)   .OO#.   .OO#.
 rocks...1k
 ---
 Sent from my phone. Please excuse my brevity.

 ODowd, Darragh darragh.od...@ilim.com wrote:

 Hi,

 I'm new to R. I'm using the R Gui and have tried to install additional
 packeges.
 When I choose Packages/Install package(s) I get the following error...

 utils:::menuInstallPkgs() Warning: unable to access index for
 repository http://cran.ma.imperial.ac.uk/bin/windows/contrib/2.15
 Warning: unable to access index for repository
 http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.15
 Error in install.packages(NULL, .libPaths()[1L], dependencies = NA,
 type = type) :
 no packages were specified

 Can someone

 --
 **
 Irish Life Investment Managers Limited is regulated by the Central Bank
 of Ireland.  Irish Life Investment Managers Limited Registered Office:
 Beresford Court, Beresford Place, Dublin 1.  Registered in Ireland
 Number 116000

 While Irish Life Investment Managers uses reasonable efforts to ensure
 that the information contained in this email is current, accurate and
 complete at the date of publication, no representations or warranties
 are made (express or implied) as to the reliability, accuracy or
 completeness of such information. Irish Life Investment Managers
 therefore cannot be held liable for any loss arising directly or
 indirectly from the use of, or any action taken in reliance on, any
 information contained in this email.

 This material is for information only and does not constitute an offer
 or recommendation to buy or sell any investment, or subscribe to any
 investment management or advisory service. It is intended for the use
 of institutional and other professional investors. Past performance is
 not indicative of future results. The value of funds we manage may fall
 as well as rise.

 This email and any files transmitted with it are
 confide...{{dropped:16}}

 __
 R-help@r-project.org mailing list
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 PLEASE do read the posting guide
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Re: [R] What is the difference between probe level data and summarized data

2012-08-07 Thread Martin Morgan


On 08/07/2012 01:12 AM, ali_protocol wrote:

Hi all,

What is the difference between probe level data and summarized data.
Thanks in advance.


sounds like microarrays, for which http://biostars.org *or* the 
Bioconductor mailing list http://bioconductor.org/help/mailing-list 
might be appropriate.


Martin




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[R] NADA Package: Referencing Data Frame Columns


  The sample data sets that come with the NADA package are limited to one or
two variables and a censored measurement indicator column. I try to mimic
examples using my data but keep missing the target.

  My water chemistry data is available in two formats: long (as seen in a
database table) and wide (as seen in a spreadsheet). The two structures are:

str(chem)
'data.frame':   65349 obs. of  8 variables:
 $ site: Factor w/ 64 levels D-1,D-2,D-3,..: 1 1 1 1 1 1 1 ...
 $ sampdate: Date, format: 2007-12-12 2007-12-12 ...
 $ era : Factor w/ 2 levels Post,Pre: 1 1 1 1 1 1 1 1 1 1 ...
 $ param   : Factor w/ 64 levels AgDis,AgTot,..: 2 4 5 7 11 15 25 ...
 $ quant   : num  1.30e-04 1.06e-01 2.31e+02 1.13e-02 5.00e-03 ...
 $ ceneq1  : logi  TRUE FALSE FALSE FALSE TRUE FALSE ...
 $ floor   : num  0 0.106 231 0.0113 0 100 0 1.43 0 0.0239 ...
 $ ceiling : num  1.30e-04 1.06e-01 2.31e+02 1.13e-02 5.00e-03 2.39e-02 ...

and

str(chem.cast)
'data.frame':   56938 obs. of  70 variables:
 $ site : Factor w/ 64 levels D-1,D-2,D-3,..: 1 1 1 1 1 ...
 $ sampdate : Date, format: 2007-12-12 2007-12-12 ...
 $ era  : Factor w/ 2 levels Post,Pre: 1 1 1 1 1 1 1 1 1 1 ...
 $ ceneq1   : logi  TRUE FALSE FALSE FALSE TRUE FALSE ...
 $ floor: num  0 0.106 231 0.0113 0 100 0 1.43 0 0.0239 ...
 $ ceiling  : num  1.30e-04 1.06e-01 2.31e+02 1.13e-02 5.00e-03 ...
 $ AgDis: num  NA NA NA NA NA NA NA NA NA NA ...
 $ AgTot: num  0.00013 NA NA NA NA NA NA NA NA NA ...
 $ AlDis: num  NA NA NA NA NA NA NA NA NA NA ...
 $ AlTot: num  NA 0.106 NA NA NA NA NA NA NA NA ...
 $ Alk  : num  NA NA 231 NA NA NA NA NA NA NA ...
 $ AsDis: num  NA NA NA NA NA NA NA NA NA NA ...
  and so on.

  I do not know if the latter is appropriate; that is, that the ceneq1,
floor, and ceiling values are available for each site, sampdate, and
chemical.

  Is the appropriate way to use the NADA methods for analyses and plotting
to subset each chemical separately from the 'chem' data frame? Or, is there
a syntax other than, for example,

cenboxplot(chemVdis, chem$ceneq1, chem$era)
Error in cenros(obs[group == i], cen[group == i]) :
  error in evaluating the argument 'obs' in selecting a method for function
'ros': Error: object 'Vdis' not found

  I get the same error when trying to use the 'chem.cast' data frame.

Rich

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Re: [R] Multiple Comparisons-Kruskal-Wallis-Test: kruskal{agricolae} and kruskalmc{pgirmess} don't yield the same results although they should do (?)

2012-08-07 Thread greatest.possible.newbie

I see.. So apparently the different functions are doing the same thing! :-)
Besides I didn't know the groups should have about the same size.
Thank you four your time Mr. Dalgaard.



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[R] Rcolorbrewer Package

2012-08-07 Thread li li

Hi all,
   I am trying to download the Rcolorbrewer package from Cran
http://cran.r-project.org/web/packages/RColorBrewer//index.html

It seems the files have been removed. Does anyone know where can
I download the package?
   Thanks.
 Hannah

[[alternative HTML version deleted]]

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Re: [R] Rcolorbrewer Package

The link works just fine for me and I can download the package if I
take out one of the repeated slashes.

Best,
Michael

On Tue, Aug 7, 2012 at 12:08 PM, li li hannah@gmail.com wrote:
 Hi all,
I am trying to download the Rcolorbrewer package from Cran
 http://cran.r-project.org/web/packages/RColorBrewer//index.html

 It seems the files have been removed. Does anyone know where can
 I download the package?
Thanks.
  Hannah

 [[alternative HTML version deleted]]

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Re: [R] Sum of vector elements

2012-08-07 Thread Jeff Newmiller

Another approach... not exactly more direct, but perhaps more robust and more 
general:

tmp - aggregate(x,list(lvl=cumsum(abs(diff(c(FALSE,x0), FUN=sum)
ans - tmp[0==tmp$lvl,x]

abs(diff()) finds transitions, FALSE forces level zero to represent negative 
numbers
cumsum marks levels (groups of positive and not-positive numbers)
aggregate does the summation
ans may be empty if x started with positive numbers.
---
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Sent from my phone. Please excuse my brevity.

R. Michael Weylandt michael.weyla...@gmail.com wrote:

I'd do something like this:

x - sample(seq(-10, 10))

sum(x[seq_len(which.max(x  0)])

Though others might have more direct solutions.

which.max() gets you the index of the first time x  0 -- seq_len
gives you numbers 1 to that index -- then just subset and sum like
normal.

Best,
Michael

On Tue, Aug 7, 2012 at 7:37 AM, number10 havana_dr...@ymail.com
wrote:
 Hi, Is it possible to avoid using do and while loops to calculate the
sum of
 the elements of a vector until the appearance of the first positive
element.



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Re: [R] Rcolorbrewer Package

install.packages(RColorBrewer)  seems to work for me. 

John Kane
Kingston ON Canada

 -Original Message-
 From: hannah@gmail.com
 Sent: Tue, 7 Aug 2012 13:08:11 -0400
 To: r-help@r-project.org
 Subject: [R] Rcolorbrewer Package

 Hi all,
I am trying to download the Rcolorbrewer package from Cran
 http://cran.r-project.org/web/packages/RColorBrewer//index.html

 It seems the files have been removed. Does anyone know where can
 I download the package?
Thanks.
  Hannah

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Re: [R] Sum of vector elements

2012-08-07 Thread William Dunlap

I'll add my suggestion, which first extracts from x the elements
up to but not including the first positive one and then sums them:
   f3 - function(x) { sum(x[cumsum(x0) == 0]) }
Michaels suggestion was
   sum(x[seq_len(which.max(x  0)])
but that includes the first positive element so I think it should be
at least
   f1 - function(x) sum(x[seq_len(which.max(x0)-1)])
(but that still fails in some cases).
Jeff's was
  f2 - function (x) {
  tmp - aggregate(x, list(lvl = cumsum(abs(diff(c(FALSE, x  0), FUN = 
sum)
  tmp[0 == tmp$lvl, x]
  }   

To test them, make a list of datasets
 xs - list(allNeg=c(-1, -3, -9), allPos=c(1,3,9), firstPos=c(1,-3,-9), 
 firstNeg=c(-1,-3,+9), misc=c(-1,-3,9,-27,-81))
 sapply(xs, function(x)try(f1(x)))
  allNeg   allPos firstPos firstNeg misc 
   000   -4   -4 
 sapply(xs, function(x)try(f2(x)))
$allNeg
[1] -13

$allPos
numeric(0)

$firstPos
numeric(0)

$firstNeg
[1] -4

$misc
[1] -4

 sapply(xs, function(x)try(f3(x)))
  allNeg   allPos firstPos firstNeg misc 
 -1300   -4   -4


Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf
 Of Jeff Newmiller
 Sent: Tuesday, August 07, 2012 10:25 AM
 To: R. Michael Weylandt; number10
 Cc: r-help@r-project.org
 Subject: Re: [R] Sum of vector elements
 
 Another approach... not exactly more direct, but perhaps more robust and more 
 general:
 
 tmp - aggregate(x,list(lvl=cumsum(abs(diff(c(FALSE,x0), FUN=sum)
 ans - tmp[0==tmp$lvl,x]
 
 abs(diff()) finds transitions, FALSE forces level zero to represent negative 
 numbers
 cumsum marks levels (groups of positive and not-positive numbers)
 aggregate does the summation
 ans may be empty if x started with positive numbers.
 ---
 Jeff NewmillerThe .   .  Go Live...
 DCN:jdnew...@dcn.davis.ca.usBasics: ##.#.   ##.#.  Live Go...
   Live:   OO#.. Dead: OO#..  Playing
 Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
 /Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
 ---
 Sent from my phone. Please excuse my brevity.
 
 R. Michael Weylandt michael.weyla...@gmail.com wrote:
 
 I'd do something like this:
 
 x - sample(seq(-10, 10))
 
 sum(x[seq_len(which.max(x  0)])
 
 Though others might have more direct solutions.
 
 which.max() gets you the index of the first time x  0 -- seq_len
 gives you numbers 1 to that index -- then just subset and sum like
 normal.
 
 Best,
 Michael
 
 On Tue, Aug 7, 2012 at 7:37 AM, number10 havana_dr...@ymail.com
 wrote:
  Hi, Is it possible to avoid using do and while loops to calculate the
 sum of
  the elements of a vector until the appearance of the first positive
 element.
 
 
 
  --
  View this message in context:
 http://r.789695.n4.nabble.com/Sum-of-vector-elements-tp4639395.html
  Sent from the R help mailing list archive at Nabble.com.
 
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Re: [R] NADA Package: Referencing Data Frame Columns

On Tue, Aug 7, 2012 at 11:26 AM, Rich Shepard rshep...@appl-ecosys.com wrote:
   The sample data sets that come with the NADA package are limited to one or
 two variables and a censored measurement indicator column. I try to mimic
 examples using my data but keep missing the target.

   My water chemistry data is available in two formats: long (as seen in a
 database table) and wide (as seen in a spreadsheet). The two structures are:

 str(chem)
 'data.frame':   65349 obs. of  8 variables:
  $ site: Factor w/ 64 levels D-1,D-2,D-3,..: 1 1 1 1 1 1 1 ...
  $ sampdate: Date, format: 2007-12-12 2007-12-12 ...
  $ era : Factor w/ 2 levels Post,Pre: 1 1 1 1 1 1 1 1 1 1 ...
  $ param   : Factor w/ 64 levels AgDis,AgTot,..: 2 4 5 7 11 15 25 ...
  $ quant   : num  1.30e-04 1.06e-01 2.31e+02 1.13e-02 5.00e-03 ...
  $ ceneq1  : logi  TRUE FALSE FALSE FALSE TRUE FALSE ...
  $ floor   : num  0 0.106 231 0.0113 0 100 0 1.43 0 0.0239 ...
  $ ceiling : num  1.30e-04 1.06e-01 2.31e+02 1.13e-02 5.00e-03 2.39e-02 ...

 and

 str(chem.cast)
 'data.frame':   56938 obs. of  70 variables:
  $ site : Factor w/ 64 levels D-1,D-2,D-3,..: 1 1 1 1 1 ...
  $ sampdate : Date, format: 2007-12-12 2007-12-12 ...
  $ era  : Factor w/ 2 levels Post,Pre: 1 1 1 1 1 1 1 1 1 1 ...
  $ ceneq1   : logi  TRUE FALSE FALSE FALSE TRUE FALSE ...
  $ floor: num  0 0.106 231 0.0113 0 100 0 1.43 0 0.0239 ...
  $ ceiling  : num  1.30e-04 1.06e-01 2.31e+02 1.13e-02 5.00e-03 ...
  $ AgDis: num  NA NA NA NA NA NA NA NA NA NA ...
  $ AgTot: num  0.00013 NA NA NA NA NA NA NA NA NA ...
  $ AlDis: num  NA NA NA NA NA NA NA NA NA NA ...
  $ AlTot: num  NA 0.106 NA NA NA NA NA NA NA NA ...
  $ Alk  : num  NA NA 231 NA NA NA NA NA NA NA ...
  $ AsDis: num  NA NA NA NA NA NA NA NA NA NA ...
   and so on.

   I do not know if the latter is appropriate; that is, that the ceneq1,
 floor, and ceiling values are available for each site, sampdate, and
 chemical.

   Is the appropriate way to use the NADA methods for analyses and plotting
 to subset each chemical separately from the 'chem' data frame? Or, is there
 a syntax other than, for example,

 cenboxplot(chemVdis, chem$ceneq1, chem$era)
 Error in cenros(obs[group == i], cen[group == i]) :
   error in evaluating the argument 'obs' in selecting a method for function
 'ros': Error: object 'Vdis' not found

   I get the same error when trying to use the 'chem.cast' data frame.



Take a look at with()

Michael

 Rich

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Re: [R] Problem with global variable building a package

On Tue, Aug 7, 2012 at 2:09 AM, Eva Prieto Castro evapcas...@yahoo.es wrote:


 Hi,



 My name is
 Eva and this is my first message here. My English is not very good, but I hope
 you can understand my question, in the context of an academic project.



 I have
 developed several functions in R and the
 idea is that the user can access functions in order to:



 1)
 Alter
 parameters in relation with data and type of analysis.

 2)
 Run
 statistical analysis (Text and pdf files with results will be generated).

 3)
 View
 the value of the most important parameters.



 All the
 parameters I need are stored in a list object, and this list is used in all 
 the
 functions along the cycle performed by the user, but I would like the user 
 does
 not need to pass the name of the list as argument when he/she runs the 
 different
 functions, so I think I need to treat the list as a global variable.



 Firstly, I
 used the global assignment operator (“-“); secondly, I used “get” and 
 “assign”
 functions and even I used a new.env() in order to use a new environment
 exclusively for my list. However, when I try to build a package with all my
 functions I don’t reach this end, because of an error in parse process.



 My question
 for you is the following: taking into account that my end is to build a
 package, what can I do with this “global” list?. How can I treat it?.

Hi Eva,

I was actually doing something similar just the other day -- for a
quick and dirty solution, you can use assignInNamespace() or
assignInMyNamespace() but the help pages suggest CRAN won't look
kindly thereupon. To avoid that, I simply used a hidden [starting with
a dot] variable in the user's global environment (i.e., I defined it
on package load with the .onAttach() function and then just referenced
it when needed) but the best practices solution is probably to use
options() or something similar. See, e.g., setCompilerOptions() of the
compiler package or setRmetricsOptions() of the Rmetrics bundle.

Best,
Michael





 Thanks in
 advance.



 Regards,

 Eva




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Re: [R] Error with convUL (PBSmapping)

This looks like a fairly low level problem that you might need the
maintainer to look at. To get contact information, type

maintainer(PBSmapping)

at the R prompt.

Best,
Michael

On Tue, Aug 7, 2012 at 7:21 AM, niandra rmaill...@yahoo.it wrote:
 Hi all,

 I'm trying to use the function convUL from the Package PBSmapping but i get 
 i weird error with my data but also with the example in the convUL's help.

 Here is the example script with the error message:
 data(nepacLL, package=PBSmapping)
 #--- set the zone attribute
 #--- use a zone that is most central to the mapped region
 attr(nepacLL, zone) - 6
 #--- convert and plot the result
 nepacUTM - convUL(nepacLL)
 Error in .C(convUL, inXY = as.double(inXY), inVerts = as.integer(inVerts),  
 :
   C symbol name convUL not in DLL for package PBSmapping

 someone can help me?
 Thanks
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Re: [R] Hist function

On Tue, Aug 7, 2012 at 8:05 AM, Dominic Roye dominic.r...@gmail.com wrote:
 Hello everyone,


 First i explain my aim. I would like to calculate the frecuency of
 temperature (datos.mx1, columns 3-6)  for each month with the thresholds of
 table lf.med and lc.med.

 numero1 - hist(subset(datos.mx1[,3],
 datos.mx1$Mes==1),plot=FALSE,breaks=c(min(datos.mx1[,3]),lf.med[1,2],lc.med[1,2],max(datos.mx1[,3])))
 # First month and only the first weather station (column 3)

 I now have to apply the above command for all 12 months at all the weather
 stations. I've been trying to get with a for-loop, but without success.

 How can i do it the best way?

I'd suggest the tapply() function. If you could use dput() to make a
small reproducible example of your data, I'll help you set up the
exact syntax, but I think it's something like this:

tapply(datos.mx1, datos.mx1$Mes, function(x) hist(x[,3], plot = FALSE,
breaks = c(min(x[,3], lf.med[1,2], lc.med[1,2], max(x[,3]

Best,
Michael


 After this command, i'm only interested in the part of $counts. I hope
 someone can help me. Any help is greatly appreciated!!!  Thank you,
 everybody`!!

 datos.mx1
   Fecha Mes Serra.da.Foladoira Santiago Sergude Rio.Do.Sol
 101.01.2006   14.1 14.4 9.35.7
 202.01.2006   15.5  9.811.07.6
 303.01.2006   19.9 13.012.79.9
 404.01.2006   19.7 13.312.79.7
 505.01.2006   1   10.2 12.213.18.0
 606.01.2006   14.2 14.210.25.4
 707.01.2006   11.0  6.9 3.23.2
 .
 .
 ...etc.

lf.med
Mes Serra.da.Foladoira  Santiago   Sergude Rio.Do.Sol
 11   7.833702  8.811945  8.122162   7.899513
 22   7.766978  8.592693  8.041483   7.887327
 33   7.864609  8.861457  8.385405   8.080646
 44   8.204098  8.986149  8.803744   8.438141
 55   8.716766  9.317366  9.385637   8.977087
 66   9.377651  9.874497 10.194340   9.686213
 77   9.838470 10.055430 10.447219   9.991618
 88  10.046204 10.099868 10.520872  10.098330
 99   9.780911 10.038629 10.156048   9.851299
 10  10   9.425308  9.779886  9.646881   9.496192
 11  11   8.417168  9.346191  8.791033   8.549046
 12  12   7.968706  8.795800  8.038208   8.017603

lc.med
Mes Serra.da.Foladoira Santiago  Sergude Rio.Do.Sol
 11   21.29353 26.62562 23.71784   22.04786
 22   21.97787 27.07213 25.68340   23.39020
 33   22.69749 27.42612 26.97889   24.33741
 44   24.49683 29.21054 29.76726   26.59280
 55   25.24233 29.83056 30.75132   27.28277
 66   26.46742 30.70485 33.02328   28.77892
 77   27.72731 30.24450 34.28162   29.97767
 88   27.64356 30.84850 34.52161   30.00666
 99   27.21998 31.18390 33.19776   29.54618
 10  10   25.35830 28.83195 29.63192   26.42831
 11  11   22.37895 26.38226 25.69239   23.47575
 12  12   21.44064 26.09142 23.99676   22.26208

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Re: [R] How to convert data to 'normal' if they are in the form of standard scientific notations?



Hi, 

Try these also:

 a-c(2.0e+9,2.1e+9) 
 is.numeric(a)
[1] TRUE
formatC(a,format=fg)
#[1] 20 21
 formatC(a,format=f)
#[1] 20. 21.

 formatC(a,format=f,digits=0)
#[1] 20 21
A.K.


- Original Message -
From: HJ YAN yhj...@googlemail.com
To: Jean V Adams jvad...@usgs.gov
Cc: r-help@r-project.org
Sent: Monday, August 6, 2012 7:05 PM
Subject: Re: [R] How to convert data to 'normal' if they are in the form of 
standard scientific notations?

Dear Jean

Thanks a lot for your help.

The reason I did not provide producible code is that my work started with
reading in some large csv files, e.g. the data is not created by myself.
But the data is from the same data provider so I would expect to receive
data in exactly same data format.


I use read.csv to read the data in. My major curious is that by using
exactly same code as I provided in my email, e.g. 'as.factor' why one of
them work (e.g. convert the numerical data to factor) but the other  one
remains numerical with scientific notation?  So, in R, how do I check if
the data format are different for these two files in their original csv
files, which  might cause the different results..?

Also I tried your code and created some reproducible examples, but still
can not make it work as in your example


 a-c(2.0e+9,2.1e+9) print(a,digits=4)[1] 20 21  # I expected 
 to see 2.0e+9 here...? print(a,digits=7)[1] 20 21  # Think 
 here I should expect same 2.0e+9? getOption(digits)      # Checking my 
 default number of digits now..[1] 7 b-c(3000,3100) print(b)[1] 
 3000 3100       # This is what I expected to see 
 print(b,digits=5)[1] 3000 3100   # I'm so confused why it is not 
 working, e.g. printing 3.0e+9! getOption(digits)   # checking again, but 
 now I would expect it has being changed to 5[1] 7


Any thoughts please...?

Thanks
HJ


On Mon, Aug 6, 2012 at 7:04 PM, Jean V Adams jvad...@usgs.gov wrote:

 HJ,

 You don't provide any reproducible code, so I had to make up my own.

 dat - data.frame(a=letters[1:5], x=c(20110911001084, 20110911001084,
         20110911001084, 20110911001084, 20110911001084),
         y=c(2.10004e+12, 2.10004e+12, 2.10004e+12, 2.10004e+12,
 2.10004e+12))

 In my example, the long numbers print out without scientific notation.

 dat
   a              x             y
 1 a 20110911001084 210004000
 2 b 20110911001084 210004000
 3 c 20110911001084 210004000
 4 d 20110911001084 210004000
 5 e 20110911001084 210004000

 I can make it print with scientific notation using the digits argument to
 the print() function.

 print(dat, digits=3)
   a        x       y
 1 a 2.01e+13 2.1e+12
 2 b 2.01e+13 2.1e+12
 3 c 2.01e+13 2.1e+12
 4 d 2.01e+13 2.1e+12
 5 e 2.01e+13 2.1e+12

 What is your default number of digits?
 getOption(digits)

 Jean


 HJ YAN yhj...@googlemail.com wrote on 08/06/2012 11:14:17 AM:

 
  Dear R users
 
  I read two csv data files into R and  called them Tem1 and Tem5.
 
  For the first column, data in Tem1 has 13 digits where in Tem5 there are
 14
  digits for each observation.
 
  Originally there are 'numerical' as can be seen in my code below.  But
 how
  can I display/convert them using other form rather than scientific
  notations which seems a standard/default?
 
   I want them to be in the form like '20110911001084', but I'm very
 confused
  why when I used 'as.factor' call it works for my 'Tem1' but not for
  'Tem5'...??
 
 
  Many thanks!
 
  HJ
 
   Tem1[1:5,1][1] 2.10004e+12 2.10004e+12 2.10004e+12 2.10004e+12 2.
  10004e+12 Tem5[1:5,1][1] 2.011091e+13 2.011091e+13 2.011091e+13 2.

  011091e+13 2.011091e+13 class(Tem1[1:5,1])[1] numeric class(Tem5
  [1:5,1])[1] numeric as.factor(Tem1[1:5,1])[1] 2.10004e+12 2.
  10004e+12 2.10004e+12 2.10004e+12 2.10004e+12
  Levels: 2.10004e+12 as.factor(Tem5[1:5,1])[1] 20110911001084
  20110911001084 20110911001084 20110911001084 20110911001084
  Levels: 20110911001084


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Re: [R] multiple comparisons for GAMs

2012-08-07 Thread Bird_Girl

I have looked into using glht (‘multcomp’ package) to do multiple comparisons
for a model fit with GAM {mgcv} but after reading the description of the
‘multcomp’ package, I believe this method only applies to parametric models
and linear hypotheses.  When I ran the code glht(model,linfct….) I got an
error message (see below), but I am not sure if that was the result of my
data being input in the incorrect format or the result of the test not
working on the type of model I used.  Does anyone know if there is an
equivalent procedure for non-parametric models, or one that can be used to
test for significant differences in curves modeled using GAMs? 

I have included a highly simplified, mini dataset that approximates the
shape of the relationships exhibited by my own much larger data set, as well
as the code I used.  In a nutshell, I would like to test whether the shape
of the relationship between light and log_abundance at a height of 200m is
significantly different from the relationship between light and
log_abundance at a height of 400m.


 require (lattice)
 require (mgcv)
 require(multcomp)
 species=read.csv(Book1.csv, header=TRUE, sep=,, quote=\,fill=TRUE)
 attach(species)
 names(species)
 species
  height light log_abundance
1200m  0.15  0.20
2200m  0.23  0.28
3200m  0.38  0.30
4200m  0.41  0.47
5200m  0.52  0.48
6200m  0.63  0.42
7200m  0.71  0.37
8200m  0.85  0.35
9200m  0.96  0.40
10   200m  1.05  0.45
11   200m  1.16  0.37
12   200m  1.23  0.30
13   200m  1.39  0.26
14   200m  1.47  0.15
15   400m  0.12  0.10
16   400m  0.25  0.09
17   400m  0.36  0.12
18   400m  0.42  0.14
19   400m  0.60  0.24
20   400m  0.65  0.28
21   400m  0.74  0.37
22   400m  0.86  0.40
23   400m  0.93  0.35
24   400m  1.06  0.25
25   400m  1.15  0.15
26   400m  1.24  0.18
27   400m  1.37  0.40
28   400m  1.48  0.57

coplot(log_abundance~light|height)
 model1=gam(log_abundance~s(light)+height+te(light,by=height,k=6))
 plot(model1,trans=function(x)exp(x)/(1+exp(x)),shade=T,pages=1)
 summary (model1)
 glht(model1,linfct=mcp(height=Tukey))

Error in linfct[[nm]] %*% C : 
  requires numeric/complex matrix/vector arguments





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Re: [R] NADA Package: Referencing Data Frame Columns


On Tue, 7 Aug 2012, R. Michael Weylandt wrote:


Take a look at with()


Michael,

  Thank you. I will.

Rich

--
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Applied Ecosystem Services, Inc.   |Helping Ensure Our Clients' Futures
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Re: [R] Package to remove collinear variables

2012-08-07 Thread John C Frain

For background have a look at http://en.wikipedia.org/wiki/Multicollinearity.

I have also used

Regression Diagnostics: Identifying Influential Data and Sources of
Collinearity (Wiley Series in Probability and Statistics) by David A.
Belsley, Edwin Kuh and Roy E. Welsch

Sections 1.9 to 1.12 of
Hands-On Intermediate Econometrics Using R: Templates for Extending
Dozens of Practical Examples [With CDROM] by Hrishikesh D. Vinod
(2008)

Basically how you proceed depends a lot on what you are trying to achieve.

Best Regards John

On 5 August 2012 23:04, Roberto Moscetti rmosce...@unitus.it wrote:
Hi,
thank you for your help. I know, I need to learn enough statistics to
understand how to process my data. The reason because of I write on this
forum is to ask to people a way to learn.
I am a postharvest researcher and statistic is not my main field, so I try
to do my best.

Do you know a book (or literature) than can help me?

Thank you very much for your time and suggestions.

Best regards,
Roberto

Il 05/08/2012 12:55, Jeff Newmiller ha scritto:

There is no magic bullet (package) for your problem. You must either
learn enough statistics to understand how to analyze your data, or consult
with someone who does.

FWIW collinearity is not in general amenable to automatic removal.
However, you can identify which inputs are collinear with each other, and
omit the redundant ones next iteration of your analysis, using (for example)
the approach suggested by Uwe. Deciding WHICH of the redundant inputs is
most appropriate to keep is the part computers are not so good at... that is
where you must be smarter or more creative than the computer.

Also, it would help you get responses if you included the context (earlier
discussion) in your replies.. most people do not use Nabble here. Reading
and following the requests in the footer of every message will also help.

---
Jeff NewmillerThe . . Go
Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live
Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#.
rocks...1k

---
Sent from my phone. Please excuse my brevity.

Roberto rmosce...@unitus.it wrote:

I do not know, because I tried to use rfe function (Backwards Feature
Selection, Caret Package) to select wavelengths useful for a prediction
model. Otherwise, rfe function give me back a lot of warning messages
about
collinearity between variables.

So, I do not know if your script can be useful.
I tried to use VIF-Regression to select variables, but rfe function
advise
me with the same warning messages again.

What do you think about that?

Thank you very much for your help.

Best,
Roberto

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Ireland
www.tcd.ie/Economics/staff/frainj/home.html
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mailto:fra...@gmail.com

Re: [R] NADA Package: Referencing Data Frame Columns


On Tue, 7 Aug 2012, David Winsemius wrote:


cenboxplot(chemVdis, chem$ceneq1, chem$era)
___^
Incorrect use of '' when '$' was intended.


  Mea culpa, David! Thanks for catching my typo.

Rich

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Re: [R] issue with nzchar() ?

On Mon, Aug 6, 2012 at 5:27 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
 On Mon, Aug 6, 2012 at 9:53 AM, Liviu Andronic landronim...@gmail.com wrote:
 On Mon, Aug 6, 2012 at 4:48 PM, Liviu Andronic landronim...@gmail.com 
 wrote:
 string, something that I find strange. At best NA is the equivalent of
 an empty string.

 Certainly not to my mind, unless you think that zero and NA should be
 the same for integers and doubles as well. NA (in whatever form) is,
 to my mind, _unknown_ which is very different than knowing 0.

This is a tricky question and I don't have a strong opinion yet.


 I'm not sure why that's the case, but it's documented on the help page
 (under value):

  For ‘nchar’, an integer vector giving the sizes of each element,
  currently always ‘2’ for missing values (for ‘NA’).

I most certainly missed this bit in the help page.


 My guess is that it's this way for back-compatability from a time when
 there probably wasn't a proper NA_character_ (that's the parser
 literal for a character NA) and they really were just NA (the
 string) -- perhaps in some far distant R 3.0 we'll see
 nchar(NA_character_) = NA_integer_

As David has also suggested (and Bert alluded), it may be worth having
a nchar(..., returnNA=FALSE) argument, which if TRUE would return NA
when it encounters NA values in the original vector.

Thank you all for the comments. Regards
Liviu

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[R] Bayesian estimates for the 1st-order Spatial Autoregressive model

2012-08-07 Thread CARTIER, ADRIAN AW (AG/1000)

Greetings:







I am a relatively new user to R. I was wondering if anyone is familiar with 
MATLAB's far_g() function.  If yes, is there an R equivalent to this?  I would 
like to have the ability to input as my observation vector continuous values. I 
noticed that there was something close in R, sar_probit_mcmc(), but I can only 
use a binary vector as my observation vector.







If my question sounds a bit confusing I am pasting below the MATLAB description 
of the function so you can see what it is that I am looking for.







function results = far_g(y,W,ndraw,nomit,prior) % PURPOSE: Bayesian estimates 
for the 1st-order Spatial autoregressive model



%  y = rho*W*y + e,e = N(0,sige*V),



%  V = diag(v1,v2,...vn), r/vi = ID chi(r)/r, r = Gamma(m,k)



%  sige = gamma(nu,d0)



%  rho = Uniform(rmin,rmax), or rho = beta(a1,a2);











Thanks in advance,


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[R] Decimal number

HI

i have a little problem please help me to solve it
 
this is the code in R:

 beta0
[1] 64.90614
 beta1
[1] 17.7025
 beta
[1] 17 64

her beta- c(beta0, beta1)

thank you in advance
hafida



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Re: [R] Styling gridExtra's title and left labels

2012-08-07 Thread baptiste auguie

Hi,

You can use a grob instead of a text string, e.g

main = textGrob(Title goes here, gp=gpar(fontsize=24))

HTH,

b.

On 7 August 2012 03:49, Alastair alastair.and...@gmail.com wrote:
 Hi,

 I'm using the gridExtra package to combine some graphs like in the
 arrangeGrob example. Each of the graphs has a title but they appear much
 larger than the overall combined plot title and left axis label. Does anyone
 know how I can control the style / size of the gridExtra labels?

 library(gridExtra)
 library(ggplot2)
 dsamp - diamonds[sample(nrow(diamonds), 1000), ]

 p1 - qplot(carat, price, data=dsamp, colour=clarity)
 p2 - qplot(carat, price, data=dsamp, colour=clarity, geom=path)


 tmp - ggplot_gtable(ggplot_build(p1))
 leg - which(sapply(tmp$grobs, function(x) x$name) == guide-box)
 legend - tmp$grobs[[leg]]

 grid.arrange(arrangeGrob(p1 + theme(legend.position=none) +
 ggtitle(Scatter Plot),
 p2 + theme(legend.position=none) +
 ggtitle(Line Plot),
 main =this is a title,
 left = This is my global Y-axis
 title), legend,
  widths=unit.c(unit(1, npc) - legend$width,
 legend$width), nrow=1)





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Re: [R] NADA Package: Referencing Data Frame Columns


On Tue, 7 Aug 2012, R. Michael Weylandt wrote:


Take a look at with()


Michael,

  Works like a charm! Thanks again for the pointer.

Rich

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Re: [R] test if elements of a character vector contain letters

On Mon, Aug 6, 2012 at 7:35 PM, Marc Schwartz marc_schwa...@me.com wrote:
 is.letter - function(x) grepl([[:alpha:]], x)
 is.number - function(x) grepl([[:digit:]], x)


Another follow-up. To test for (non-)alphanumeric one would do the following:
 x - c(letters, 1:26, '+', '-', '%^')
 x[1:10] - paste(x[1:10], 1:10, sep='')
 x
 [1] a1  b2  c3  d4  e5  f6  g7  h8  i9  j10 k
l   m   n
[15] o   p   q   r   s   t   u   v   w   x   y
z   1   2
[29] 3   4   5   6   7   8   9   10  11  12  13
14  15  16
[43] 17  18  19  20  21  22  23  24  25  26  +
-   %^
 xb - grepl([[:alnum:]],x)  ##test for alphanumeric chars
 x[xb]
 [1] a1  b2  c3  d4  e5  f6  g7  h8  i9  j10 k
l   m   n
[15] o   p   q   r   s   t   u   v   w   x   y
z   1   2
[29] 3   4   5   6   7   8   9   10  11  12  13
14  15  16
[43] 17  18  19  20  21  22  23  24  25  26
 xb - grepl([[:punct:]],x)  ##test for non-alphanumeric chars
 x[xb]
[1] +   -   %^


More regex rules are available on the Wiki [1]. Regards
Liviu

[1] http://en.wikipedia.org/wiki/Regular_expression

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Re: [R] Decimal number

hello arun
 her beta can be considered simply as an object. 

 when i wrote beta
I want to get the number of beta and beta with the full part after the
comma


hafida



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[R] Trying to build up a user interface with the R tcltk package

2012-08-07 Thread Julio Sergio Santana

I'm trying to build up a user inteface using the R tcltk component. Since the 
documentation for this R library is scarce and poor, I have decided to use 
only the .Tcl function to pass commands to the tcl interpreter. So I wrote
my 
first very simple tcltk program, hoping to run it from inside R and get the 
nice Tk window, with the following R instruction:

   - .Tcl('source smalltst.tcl')
   - Tcl

Eventhough, neither R nor tcl complain about the instruction, I don't get 
the tk window. The tcltk program, stored in smalltst.tcl, which, by the 
way displays a simple tk window when run from tclsh, is as follows:

#! /usr/bin/tclsh
package require Tk
ttk::frame .c -padding 3 3 12 12
ttk::frame .c.f -borderwidth 5 -relief sunken -width 200 -height 100
ttk::label .c.namelbl -text Name
grid .c -column 0 -row 0 -sticky nsew
grid .c.f -column 0 -row 0 -columnspan 3 -rowspan 2 -sticky nsew
grid .c.namelbl -column 3 -row 0 -columnspan 2 -sticky nw -padx 5

Do you have any comments on this?

Thanks,

 --Sergio.



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Re: [R] add leading zeros

2012-08-07 Thread David Winsemius



On Aug 7, 2012, at 3:55 AM, Liviu Andronic wrote:


Hello


On Fri, Jul 27, 2012 at 6:54 AM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:

Much easier than you think:

x - c(1L, 9000L)

sprintf(%05i,x)


For anyone interested, I came up with a small wrapper for the above:
add.lead - function(x, width=max(nchar(x))){
   sprintf(paste('%0', width, 'i', sep=''), x)
}


Thanks, Liviu;

Your post prompted me to add a variant in my .Rprofile that adds  
leading zeros to numeric-date values in ddmm format which lost  
them because they were imported as integers (because I forgot to use  
colClasses.)


add.dt0 - function(x, width=8 ){
sprintf(paste('%0', width, switch(is.numeric(x)+1, 's', 'i'),  
sep=''), x)

 }

Can also be used for Excel (character) output that omits the leading 0  
in date fields that were output in csv-format as  m/dd/yy.


#character input
dts - (7/12/08,  11/11/11)
 as.Date(add.dt0(dts), format=%d/%m/%y, origin=1970-01-01)
#[1] 2008-12-07 2011-11-11

# numeric input
 as.Date(add.dt0(1122011), format=%d%m%Y, origin=1970-01-01)
#[1] 2011-12-01

 as.Date(1122011, format=%d%m%Y, origin=1970-01-01)
[1] NA
 as.Date(add.dt0(1122011), format=%d%m%Y, origin=1970-01-01)
[1] 2011-12-01
 as.Date(01122011, format=%d%m%Y, origin=1970-01-01)
[1] NA
 as.Date(add.dt0(01122011), format=%d%m%Y, origin=1970-01-01)
[1] 2011-12-01


--
David.



x - c(1L, 15L, 234L, 9000L)
(xa - add.lead(x))

[1] 0001 0015 0234 9000

nchar(xa)

[1] 4 4 4 4

(xb - add.lead(x, 5))

[1] 1 00015 00234 09000

nchar(xb)

[1] 5 5 5 5

(xc - add.lead(x, 15))
[1] 001 015 234  
0009000

nchar(xc)

[1] 15 15 15 15


Regards
Liviu



David Winsemius, MD
Alameda, CA, USA

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[R] looking for accessibility help (blind student)

2012-08-07 Thread Peter Petto

I will have a blind student in my AP Statistics class this year.

I'm thinking about using R as his calculator. It seems like it's core text-in 
text-out nature may match screen reading software well. And I'd like to explore 
directing visualizations to tactile graphics output.

I'm wondering if anyone could give me suggestions (people, websites, 
organizations) where I could discover anything that could help me help him.

Thanks so much.

  Peter++

===

Peter Petto ppe...@ppetto.com
Lakewood High School Math   cell: 440.249.4289
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[R] summing and combining rows

2012-08-07 Thread Christopher R. Dolanc


Hello,

I have a data set that needs to be combined so that rows are summed by a 
group based on a certain variable. I'm pretty sure rowsum() or rowsums() 
can do this but it's difficult for me to figure out how it will work for 
my data based on the examples I've read.


My data are structured like this:

PlotSizeClassStems
12   Class31
12   Class43
17   Class35
17   Class42

I simply want to sum the size classes by plot and create a new data 
frame with a size class called Class34 or with the SizeClass variable 
removed. I actually do have other size classes that I want to leave 
alone, but combine 3 and 4, so if I could figure out how to do this by 
creating a new class, that would be preferable.


I've also attached a more detailed sample of data.

Thanks,
Chris Dolanc

--
Christopher R. Dolanc
Post-doctoral Researcher
University of Montana and UC-Davis

  Data   Plot  Lat LatCat Elevation ElevCatType SizeClass 
Stems Area   Density
5171   VTM  39C16 39.76282   Lat6  1500   Elev1 ConiferClass3   
  0 751.5347   0.0
5172   VTM  39C16 39.76282   Lat6  1500   Elev1 ConiferClass4   
  1 751.5347  13.30611
5771   VTM  39F11 39.57721   Lat6   500   Elev1 ConiferClass3   
  0 751.5347   0.0
5772   VTM  39F11 39.57721   Lat6   500   Elev1 ConiferClass4   
  0 751.5347   0.0
5787   VTM  39F13 39.56214   Lat6  1500   Elev1 ConiferClass3   
  3 694.0784   0.0
5788   VTM  39F13 39.56214   Lat6  1500   Elev1 ConiferClass4   
  1 694.0784   0.0
5795   VTM  39F14 39.54522   Lat6   900   Elev1 ConiferClass3   
  1 763.9850  13.08926
5796   VTM  39F14 39.54522   Lat6   900   Elev1 ConiferClass4   
  2 763.9850   0.0
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[R] Error using ddply inside user-defined function

2012-08-07 Thread Jennifer Sabatier

Hi All,

I *think* it's ddply because the function recognizes vr1, etc, in other
parts of the function.

Here's some code:

# create dataset

PROV.PM.FBCTS - c(0.00 ,0.00, 33205.19, 25994.56, 23351.37, 26959.56
,27632.58, 26076.24, 0.00, 0.00 , 6741.42, 18665.09 ,18129.59 ,21468.39
,21294.60 ,22764.82, 26076.73)
FBCTS.INV.TOT - c(0 ,  0,  958612,  487990,  413344,  573347,  870307,
552681 ,  0, 0 , 163831 , 400161 , 353000,  358322 , 489969,  147379,
1022769)
FBCTS.REC.TOT - c(0   ,0, 1638818  ,297119 , 299436  ,414164 , 515735,
 529001   ,0, 0 , 427341 , 625893  ,437854 , 407091,  425119 ,   8663,
  0)
MECH.NAME - c(Mechanism A,Mechanism A,Mechanism A,Mechanism
A,Mechanism A,Mechanism A,Mechanism A,Mechanism A, Mechanism
B,Mechanism B,Mechanism B,Mechanism B,Mechanism B,Mechanism
B,Mechanism B,Mechanism B,Mechanism B,)

vn - data.frame(MECH.NAME, PROV.PM.FBCTS, FBCTS.INV.TOT, FBCTS.REC.TOT)


# create function

allocation - function(vr1, vr2, vr3)
{

d - ddply(vn, MECH.NAME, summarise, SUM = vr1 + vr2 + vr3)

vn - merge(vn, d, by.x=MECH.NAME, by.y=MECH.NAME, all=T)
new - (vr1+vr2+vr3) / vn$SUM
return(new)

}

# run function

vn$PROV.PM.FBCTS - allocation(PROV.PM.FBCTS, FBCTS.INV.TOT, FBCTS.REC.TOT)

Here's the error:

 vn$PROV.PM.FBCTS - allocation(PROV.PM.FBCTS, FBCTS.INV.TOT,
FBCTS.REC.TOT)
Error in eval(expr, envir, enclos) : object 'vr1' not found


# If I pull the calculations out of the function it works:


d - ddply(vn, MECH.NAME, summarise, SUM = sum(PROV.PM.FBCTS,
FBCTS.INV.TOT, FBCTS.REC.TOT))
vn - merge(vn, d, by.x=MECH.NAME, by.y=MECH.NAME, all=T)
vn$PROV.PM.FBCTS - (PROV.PM.FBCTS +  FBCTS.INV.TOT + FBCTS.REC.TOT) /
vn$SUM


 vn
 MECH.NAME PROV.PM.FBCTS FBCTS.INV.TOT FBCTS.REC.TOT SUM
1  Mechanism A0. 0 0 7713774
2  Mechanism A0. 0 0 7713774
3  Mechanism A0.34103091958612   1638818 7713774
4  Mechanism A0.10515004487990297119 7713774
5  Mechanism A0.09543077413344299436 7713774
6  Mechanism A0.13151418573347414164 7713774
7  Mechanism A0.18326628870307515735 7713774
8  Mechanism A0.14360783552681529001 7713774
9  Mechanism B0. 0 0 5402533
10 Mechanism B0. 0 0 5402533
11 Mechanism B0.11067280163831427341 5402533
12 Mechanism B0.19337580400161625893 5402533
13 Mechanism B0.14974155353000437854 5402533
14 Mechanism B0.14565046358322407091 5402533
15 Mechanism B0.17332290489969425119 5402533
16 Mechanism B0.03309685147379  8663 5402533
17 Mechanism B0.19413964   1022769 0 5402533


Can anyone help me figure out why I can't put this calculation in a
function?  Is this a systemic thing I should look for in the future?

I know there are other ways to do what I want, but I am not very curious
about this.

Best,

Jen

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[R] Execution of a function

Hi

i have  aproblem withe execution of my function
 
first, i wrote my function in the script of R 
nom_fonction - function(arg1[=expr1], arg2[=expr2], ...){ 
bloc d'instructions
}

 when i want to have the result i mean the laste instruction in the bloc of
 instruction ,  i try to
wrote the name of function

source(aj.fun)
Error in readLines(file, warn = FALSE) : 'con' is not a connection

return(aj.fun)
Error: no function to return from, jumping to top level

thanks for helping
hafida, univ of algeria 

 




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[R] help to program my function

HI

i have a problem please help me to solve it:
http://r.789695.n4.nabble.com/file/n4639434/aj.pdf aj.pdf 

i want to calculate the vecteur a[j] where j: 1...8

this is the code in R:

aj.fun - function(j, i, X, z, E, beta0, beta1){
+ n - length(X)
+ iX - order(X)
+ iz - order(z)
+ e1 - -(beta)*z[ iz[1:(i - 1)] ]
+ numer - E[j] - sum( X[ iX[1:(i - 1)] ] * exp(e1) )
+ e2 - -(beta)*z[ iz[i:n] ]
+ denom - sum( exp(e2) )
+ numer/denom
+ }

 iX-order(X)
 iX
 [1] 75 37 29 60 73 20 69 55 30 70 72 38 26 35 65 61 74 50 71 57 25 54 64 76
56
[26] 58 48 67 46 63 28 62 36 49 47 66  1 42 41 19 39 43 22 51 68 33 27  9 15
11
[51] 10 59 32 40 45 44 52 16 18 34  4 53 21 23 31  7  6 13 14 12 17 24  5  8 
2
[76]  3

 iZ-order(Z)
 iZ
 [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
25
[26] 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49
50
[51] 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74
75
[76] 76

e1 - -(beta)*Z[ iZ[1:(i - 1)] ]
Warning message:
In 1:(i - 1) : numerical expression has 76 elements: only the first used
 e1
[1]  -442 -1664

 numer - E[j] - sum( X[ iX[1:(i - 1)] ] * exp(e1))
Warning message:
In 1:(i - 1) : numerical expression has 76 elements: only the first used
 numer
[1] 9.5 9.5 9.5 9.5 9.5 9.5 9.5 9.5

 e2 - -(beta)*Z[ iZ[i:n] ]
Warning message:
In i:n : numerical expression has 76 elements: only the first used
 e2
 [1]  -442 -1664  -442 -1792  -476 -1792  -476 -1792  -510 -1920  -510 -1920
[13]  -510 -1920  -510 -1920  -510 -1920  -510 -2048  -544 -2048  -544 -2048
[25]  -544 -2048  -544 -2048  -544 -2048  -544 -2048  -544 -2048  -578 -2176
[37]  -578 -2176  -578 -2176  -578 -2176  -578 -2176  -578 -2176  -578 -2176
[49]  -578 -2176  -578 -2176  -578 -2304  -612 -2304  -612 -2304  -612 -2304
[61]  -612 -2304  -612 -2304  -612 -2304  -612 -2304  -646 -2432  -646 -2432
[73]  -646 -2432  -646 -2432
 denom - sum( exp(e2) )
 numer/denom
[1] 4.313746e+192 4.313746e+192 4.313746e+192 4.313746e+192 4.313746e+192
[6] 4.313746e+192 4.313746e+192 4.313746e+192

my problem that the vecteur a[j] could not have the same number!!!
 

thank you in advance
hafida



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Re: [R] summing and combining rows

On Tue, Aug 7, 2012 at 12:47 PM, Christopher R. Dolanc
crdol...@ucdavis.edu wrote:
 Hello,

 I have a data set that needs to be combined so that rows are summed by a
 group based on a certain variable. I'm pretty sure rowsum() or rowsums() can
 do this but it's difficult for me to figure out how it will work for my data
 based on the examples I've read.

 My data are structured like this:

 PlotSizeClassStems
 12   Class31
 12   Class43
 17   Class35
 17   Class42

 I simply want to sum the size classes by plot

Which variable do you want to sum? You'll want to use something like

aggregate( XXX ~ YYY, data = dat.name, FUN = sum)

to sum variable XXX by classes YYY.

See ? aggregate for more details and worked examples.

Cheers,
Michael

  and create a new data frame
 with a size class called Class34 or with the SizeClass variable removed. I
 actually do have other size classes that I want to leave alone, but combine
 3 and 4, so if I could figure out how to do this by creating a new class,
 that would be preferable.

 I've also attached a more detailed sample of data.

 Thanks,
 Chris Dolanc

 --
 Christopher R. Dolanc
 Post-doctoral Researcher
 University of Montana and UC-Davis


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Re: [R] test if elements of a character vector contain letters

On Tue, Aug 7, 2012 at 4:28 AM, Liviu Andronic landronim...@gmail.com wrote:
 On Mon, Aug 6, 2012 at 7:35 PM, Marc Schwartz marc_schwa...@me.com wrote:
 is.letter - function(x) grepl([[:alpha:]], x)
 is.number - function(x) grepl([[:digit:]], x)

 Quick follow-up question.

 I'm always reluctant to create functions that would resemble the
 method of a function (here, is() ), but would in fact not be a genuine
 method. So would there be any incompatibility between is() and
 is.letter(), given that the latter is not a method of the former?
 Is it good (or acceptable) practice to define is.letter() as above?
 Would is_letter() be better?

It certainly won't cause problems if you never define anything of
class letter or number.


 Regards
 Liviu


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Re: [R] Execution of a function

On Tue, Aug 7, 2012 at 11:26 AM, hafida hafida...@hotmail.fr wrote:
 Hi

i have  aproblem withe execution of my function

first, i wrote my function in the script of R
nom_fonction - function(arg1[=expr1], arg2[=expr2], ...){
 bloc d'instructions
 }

 when i want to have the result i mean the laste instruction in the bloc of
 instruction ,  i try to
wrote the name of function

source(aj.fun)
 Error in readLines(file, warn = FALSE) : 'con' is not a connection

source() takes a file name, not a function name.

Cheers,
Michael


return(aj.fun)
 Error: no function to return from, jumping to top level

Don't use return() outside of functions.



 thanks for helping
 hafida, univ of algeria






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[R] Re-grouping data in R

2012-08-07 Thread Abraham Mathew

I have a data frame with a column of values that I want to bucket (group)
into specific levels.

 str(dat)'data.frame': 3678 obs. of  39 variables:
 $ id  : int  23 76 129 156 166 180 200 214 296 344 ...
 $ final_purchase_amount   : Factor w/ 32 levels
\\N,1082,1109,..: 1 1 1 1 1 1 1 1 1 1 ...


So I ran the following to produce new levels, one for values from 100
to 400, 401 to 1000, and 1001+.


dat$final_purchase_amount- NA
dat$final_purchase_amount[dat$final_purchase_amount %in%
levels(dat$final_purchase_amount)[c(8,9,11,12,13,15,16,17,18,19,20,21)]]
- 100 to 400
dat$final_purchase_amount[dat$final_purchase_amount %in%
levels(dat$final_purchase_amount)[c(22,23,24,25,26,27,28,29,30,31,32)]]
- 401 to 1000
dat$final_purchase_amount[dat$final_purchase_amount %in%
levels(dat$final_purchase_amount)[c(2,3,4,5,6,7,10,14)]] - 1001 +
dat$final_purchase_amount - factor(dat$final_purchase_amount)
levels(dat$final_purchase_amount)
table(dat$final_purchase_amount)



However, this doesn't seem to produce any levels and returns the following.


 levels(dat$final_purchase_amount)character(0)



Can anyone point to what I'm doing wrong.



Thanks!


-- 
*Abraham Mathew
Statistical Analyst
www.amathew.com
720-648-0108
@abmathewks*

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[R] What is this called? lapply(datum,[[,ColumnName)

2012-08-07 Thread Kevin Chang

Hello R users



I recently learned how to use this command:


lapply(datum,[[,ColumnName)

 

Unfortunately, I don't know how exactly it works, what it's called (in
particular the [[ part], and what other things you can do with it
(retrieve multiple columns?).


Given datum is a list of dataframes with the same column, but different
number of rows,

 

lapply(datum,[[,ColumnName),

 

Returns you a list of a particular column form each of your dataframes from
datum.

 

I think this is a really useful command, and would like to learn more about
it.
Unfortunately I don't know what subject or topic it is under and cannot look
it up in a book.

 

Thank you,

 

Kevin

 

Master of Science Student

Department of Food, Resource and Agricultural Economics

University of Guelph
Office:519-824-4120 ext. 58528

Mobile:  226-979-2813

 http://fare.uoguelph.ca/users/kchang01
http://fare.uoguelph.ca/users/kchang01 



 


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Re: [R] Decimal number

2012-08-07 Thread Eloi Mercier


Hi hafida,

I fail to understand the question. Could you elaborate please ?

Is this what you want ?

 beta0 = 64.90614
 beta1 = 17.7025
 beta-c(beta0, beta1)
beta
[1] 64.90614 17.70250

As a side note, please keep in mind that R doesn't allow white space 
within variable name (name it her_beta instead).


Cheers,

Eloi


Salut Hafida,

je ne comprend pas la question. Peux tu détailler stp ?
Si c'est pour afficher les valeurs complètes de beta0 et beta1 alors 
l'exemple ci-dessus devrait faire l'affaire.


Autrement, R n'autorise pas la présence d'espace dans le nom des variables.

Cordialement,

Eloi

On 12-08-07 01:37 PM, hafida wrote:

hello arun
  her beta can be considered simply as an object.


when i wrote beta
I want to get the number of beta and beta with the full part after the

comma


hafida



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--
Eloi Mercier
Bioinformatics PhD Student, UBC
Paul Pavlidis Lab
2185 East Mall
University of British Columbia
Vancouver BC V6T1Z4

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Re: [R] NADA Package: Referencing Data Frame Columns

2012-08-07 Thread David Winsemius



On Aug 7, 2012, at 11:31 AM, R. Michael Weylandt wrote:

On Tue, Aug 7, 2012 at 11:26 AM, Rich Shepard rshep...@appl-ecosys.com 
 wrote:
 The sample data sets that come with the NADA package are limited  
to one or
two variables and a censored measurement indicator column. I try to  
mimic

examples using my data but keep missing the target.
snipped

 Is the appropriate way to use the NADA methods for analyses and  
plotting
to subset each chemical separately from the 'chem' data frame? Or,  
is there

a syntax other than, for example,



_

cenboxplot(chemVdis, chem$ceneq1, chem$era)
___^
Incorrect use of '' when '$' was intended.


Error in cenros(obs[group == i], cen[group == i]) :
 error in evaluating the argument 'obs' in selecting a method for  
function

'ros': Error: object 'Vdis' not found

 I get the same error when trying to use the 'chem.cast' data frame.




Take a look at with()

Michael




David Winsemius, MD
Alameda, CA, USA

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Re: [R] process evaluation packages (slightly off topic)

2012-08-07 Thread Rui Barradas


Hello,

There's also, apparently for simple needs, plotrix::gantt.chart
See an example in
http://addictedtor.free.fr/graphiques/sources/source_74.R

To transfer graphs to timelines, etc, (or anything graph related) 
package igraph.


Hope this helps,

Rui Barradas

Em 07-08-2012 14:46, Bert Gunter escreveu:

Search on R package Gantt, where you will find, among others,

http://cran.r-project.org/web/packages/plan/plan.pdf

-- Bert

On Tue, Aug 7, 2012 at 1:28 AM, Petr PIKAL petr.pi...@precheza.cz wrote:

Dear all

I need to perform some process evaluation. Sorry for not posting data and
code - I do not have any, I ask only for pointing me to correct direction.

Suppose I have several connected processes P1, P2, ..., Pn. Each process
takes some time and have some capacity (let say like preparing a dinner
for several persons - only one stove, limited capacity of utensils,
heating and cooling takes some time) and some processes can by cyclic (fry
onion in pan, put it aside, in the same pan fry meat, put an onion and
some water and simmer for a while...).

I can prepare some oriented graph (paper/pencil) or Word or drawing
programme, I can also evaluate whole process by shading spreadsheet cells
but those two tasks are not connected.

Is there any R package/other software suitable for simplifying or helping
in such tasks? E.g. When I prepare oriented graph with capacity and time
for each node is there any automatic way to transfer this graph to
timeline to see how long whole process will take, where are bottlenecks or
so?

Thank you

Petr

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Re: [R] What is this called? lapply(datum,[[,ColumnName)

I... cannot look it up in a book.
Baloney!

Read the R Language Definition manual section on indexing in the
Evaluation of Expressions Chapter. Then ?lapply and ?[[ -- and the
relevant sections in The Introduction to R manual.

However, I will admit that it's tricky. The point to grasp --
described at various levels of detail in the above manuals and in
numerous other resources that you can check out on the CRAN website --
is that y[[a]] is just syntactic sugar for [[(y,a) . R is a
functional language (mostly). So your lapply statement is just
lapplying the [[ function(extraction operator ) to the list of data
frames.

Cheers,
Bert

On Tue, Aug 7, 2012 at 12:06 PM, Kevin Chang kchan...@uoguelph.ca wrote:
Hello R users

I recently learned how to use this command:

lapply(datum,[[,ColumnName)

Unfortunately, I don't know how exactly it works, what it's called (in
particular the [[ part], and what other things you can do with it
(retrieve multiple columns?).

Given datum is a list of dataframes with the same column, but different
number of rows,

lapply(datum,[[,ColumnName),

Returns you a list of a particular column form each of your dataframes from
datum.

I think this is a really useful command, and would like to learn more about
it.
Unfortunately I don't know what subject or topic it is under and cannot look
it up in a book.

Thank you,

Kevin

Master of Science Student

Department of Food, Resource and Agricultural Economics

University of Guelph
Office:519-824-4120 ext. 58528

Mobile: 226-979-2813

http://fare.uoguelph.ca/users/kchang01
http://fare.uoguelph.ca/users/kchang01

[[alternative HTML version deleted]]

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

Re: [R] Decimal number

Hello Hafida,

So, I guess the decimal number problem is solved. 


I'll check on it.
A.K.






From: hafida goual hafida...@hotmail.fr
To: smartpink...@yahoo.com 
Sent: Tuesday, August 7, 2012 4:58 PM
Subject: RE: [R] Decimal number



thank you  a lot

please can you visite my post titled  executed function.

hafida



 Date: Tue, 7 Aug 2012 13:55:00 -0700
 From: smartpink...@yahoo.com
 Subject: Re: [R] Decimal number
 To: hafida...@hotmail.fr
 CC: r-help@r-project.org
 
 Hello,
 
   beta0-64.90614
  beta1-17.7025
  herbeta-c(beta0,beta1)
  herbeta
 [1] 64.90614 17.70250
 
 So, here you have herbeta (or is it simply beta) assigned to the vector 
 containing values of beta0 and beta1.  
 Your statement
 I want to get the number of beta and beta with the full part after the
 comma 
 is confusing.
 A.K.
 
 
 
 
 
 - Original Message -
 From: hafida hafida...@hotmail.fr
 To: r-help@r-project.org
 Cc: 
 Sent: Tuesday, August 7, 2012 4:37 PM
 Subject: Re: [R] Decimal number
 
 hello arun
 her beta can be considered simply as an object. 
 
  when i wrote beta
 I want to get the number of beta and beta with the full part after the
 comma
 
 
 hafida
 
 
 
 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Decimal-number-tp4639428p4639452.html
 Sent from the R help mailing list archive at Nabble.com.
 
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Re: [R] Execution of a function

HI,
Suppose, I have a function like this:

fun1-function(x){
 ifelse(x[1]0,
 sum(x[min(which(x0)):(which.max(x0)-1)]),
sum(x[min(which(x0)):max(which(x0))])
)
 }

#And I save it as aj.fun (-- filename)
#Then,
source(aj.fun)
 fun1
function(x){
 ifelse(x[1]0,
 sum(x[min(which(x0)):(which.max(x0)-1)]),
sum(x[min(which(x0)):max(which(x0))])
)
 }

a1-c(-5,-2,4,2,-4,-6,3)

 fun1(a1)
#[1] -7

Hope this helps.
A.K.


- Original Message -
From: hafida hafida...@hotmail.fr
To: r-help@r-project.org
Cc: 
Sent: Tuesday, August 7, 2012 12:26 PM
Subject: [R] Execution of a function

Hi

i have  aproblem withe execution of my function

first, i wrote my function in the script of R 
nom_fonction - function(arg1[=expr1], arg2[=expr2], ...){ 
bloc d'instructions
}

 when i want to have the result i mean the laste instruction in the bloc of
 instruction ,  i try to
wrote the name of function

source(aj.fun)
Error in readLines(file, warn = FALSE) : 'con' is not a connection

return(aj.fun)
Error: no function to return from, jumping to top level

thanks for helping
hafida, univ of algeria 






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[R] reshape2's dcast() Adds NAs to Data Frame


  I need to understand how and why dcast() adds NAs to a data frame that
contained no missing values.

  The database table of chemical concentrations has all missing values
removed because they cannot contribute to data analyses. The structure of
the R data frame of these data have no NA values, and neither does the data
frame resulting from applying the reshape2 melt() function to it. However,
the data frame produced by the dcast() function does contain NAs for all
chemicals. I assume this is because of the syntax I used:

chem.cast - dcast(chem.melt, site + sampdate + era + ceneq1 + floor +
ceiling ~ param)

  How should I reshape the data frame from long to wide without adding these
spurious NAs?

Rich

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[R] Setting Number of Displayed Digits


  Some chemicals have concentrations at or below the method detection limit
(MDL; 'less-thans') of 0.005 mg/L. When I look at the data frame that is how
the concentration is displayed. But, when I ask for a summary() of that data
frame column only 0 is displayed.

  How can I adjust the number of digits displayed by functions such as
summary()?

Rich

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Re: [R] help to program my function

2012-08-07 Thread Rui Barradas


Hello,

Your problem is something else, before returning a vector of only the 
same values there are warning messages.

Inline.

Em 07-08-2012 18:14, hafida escreveu:

HI


i have a problem please help me to solve it:

http://r.789695.n4.nabble.com/file/n4639434/aj.pdf aj.pdf


i want to calculate the vecteur a[j] where j: 1...8
this is the code in R:
aj.fun - function(j, i, X, z, E, beta0, beta1){

+ n - length(X)
+ iX - order(X)
+ iz - order(z)
+ e1 - -(beta)*z[ iz[1:(i - 1)] ]
+ numer - E[j] - sum( X[ iX[1:(i - 1)] ] * exp(e1) )
+ e2 - -(beta)*z[ iz[i:n] ]
+ denom - sum( exp(e2) )
+ numer/denom
+ }


iX-order(X)
iX

  [1] 75 37 29 60 73 20 69 55 30 70 72 38 26 35 65 61 74 50 71 57 25 54 64 76
56
[26] 58 48 67 46 63 28 62 36 49 47 66  1 42 41 19 39 43 22 51 68 33 27  9 15
11
[51] 10 59 32 40 45 44 52 16 18 34  4 53 21 23 31  7  6 13 14 12 17 24  5  8
2
[76]  3


iZ-order(Z)
iZ

  [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
25
[26] 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49
50
[51] 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74
75
[76] 76


e1 - -(beta)*Z[ iZ[1:(i - 1)] ]

Warning message:
In 1:(i - 1) : numerical expression has 76 elements: only the first used


This means that 'i' is a vector of length 76, which, in this context, is 
illegal. The variable 'i' must be a vector of only ONE value. If you 
want to compute the function for several (76) values do it one at a 
time, the function aj.fun is not vectorized.


Hope this helps,

Rui Barradas


e1

[1]  -442 -1664


numer - E[j] - sum( X[ iX[1:(i - 1)] ] * exp(e1))

Warning message:
In 1:(i - 1) : numerical expression has 76 elements: only the first used

numer

[1] 9.5 9.5 9.5 9.5 9.5 9.5 9.5 9.5


e2 - -(beta)*Z[ iZ[i:n] ]

Warning message:
In i:n : numerical expression has 76 elements: only the first used

e2

  [1]  -442 -1664  -442 -1792  -476 -1792  -476 -1792  -510 -1920  -510 -1920
[13]  -510 -1920  -510 -1920  -510 -1920  -510 -2048  -544 -2048  -544 -2048
[25]  -544 -2048  -544 -2048  -544 -2048  -544 -2048  -544 -2048  -578 -2176
[37]  -578 -2176  -578 -2176  -578 -2176  -578 -2176  -578 -2176  -578 -2176
[49]  -578 -2176  -578 -2176  -578 -2304  -612 -2304  -612 -2304  -612 -2304
[61]  -612 -2304  -612 -2304  -612 -2304  -612 -2304  -646 -2432  -646 -2432
[73]  -646 -2432  -646 -2432

denom - sum( exp(e2) )
numer/denom

[1] 4.313746e+192 4.313746e+192 4.313746e+192 4.313746e+192 4.313746e+192
[6] 4.313746e+192 4.313746e+192 4.313746e+192


my problem that the vecteur a[j] could not have the same number!!!
  


thank you in advance
hafida



--
View this message in context: 
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[R] Which R function for GLMM with binary response, nested random factors with temporal correlation?

2012-08-07 Thread Andrew Digby


Despite lots of investigation, I haven't found any R packages might be suitable 
for the following problem. I'd be very grateful for suggestions. 

I have three-way nested data, with a series of measures (obs) taken in quick 
succession (equal time spacing) from each subject on different days. The 
measures taken on the same day are temporally correlated, so I'd like to use an 
AR1 correlation structure for those, but treat subjects and days as nested 
random factors (random intercept) since there is little temporal correlation 
between days. The response is binary.

So I need a GLMM with a correlation structure. I've tried using GEE, but the R 
packages can't cope with multilevel nested data. The only R function I've found 
that can do this is glmmPQL. 

m - glmmPQL(y ~ f1 * f2 * f3 + (1|subj/day), correlation=corAR1(form 
=~obsno|subj/day))

f1 - f3 are fixed factors

However, PQL estimation is not recommended for binary response data. With no 
AIC and unreliable p values, model selection seems impossible! So my question 
is:

1) are there any other functions which are suitable for a GLMM with multilevel 
nested random effects and a AR1 correlation structure? Or is MCMC the only 
option?
2) to make things more complicated, I'd also like to include a varFunc variance 
structure to cope with heterogeneity. Is this possible in ML methods in R? I'd 
also like to extend to a multinomial response at a later stage. 

GEE seems the best bet, but I come unstuck with the three-way nested factors. 

Thanks for your help.

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[R] Setting Number of Displayed Digits


  Some chemicals have concentrations at or below the method detection limit
(MDL; 'less-thans') of 0.005 mg/L. When I look at the data frame that is how
the concentration is displayed. But, when I ask for a summary() of that data
frame column only 0 is displayed.

  How can I adjust the number of digits displayed by functions such as
summary()?

Rich

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Re: [R] Which R function for GLMM with binary response, nested random factors with temporal correlation?

Have you posted on R-sig-mixed-models? That would be more likely to
yield useful responses than here.

-- Bert

On Tue, Aug 7, 2012 at 3:02 PM, Andrew Digby andrewdi...@mac.com wrote:

Despite lots of investigation, I haven't found any R packages might be
suitable for the following problem. I'd be very grateful for suggestions.

I have three-way nested data, with a series of measures (obs) taken in quick
succession (equal time spacing) from each subject on different days. The
measures taken on the same day are temporally correlated, so I'd like to use
an AR1 correlation structure for those, but treat subjects and days as nested
random factors (random intercept) since there is little temporal correlation
between days. The response is binary.

So I need a GLMM with a correlation structure. I've tried using GEE, but the
R packages can't cope with multilevel nested data. The only R function I've
found that can do this is glmmPQL.

m - glmmPQL(y ~ f1 * f2 * f3 + (1|subj/day), correlation=corAR1(form
=~obsno|subj/day))

f1 - f3 are fixed factors

However, PQL estimation is not recommended for binary response data. With no
AIC and unreliable p values, model selection seems impossible! So my question
is:

1) are there any other functions which are suitable for a GLMM with
multilevel nested random effects and a AR1 correlation structure? Or is MCMC
the only option?
2) to make things more complicated, I'd also like to include a varFunc
variance structure to cope with heterogeneity. Is this possible in ML methods
in R? I'd also like to extend to a multinomial response at a later stage.

GEE seems the best bet, but I come unstuck with the three-way nested factors.

Thanks for your help.

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

[R] r-forge down?

2012-08-07 Thread Alexander Shenkin

Hi Folks,

I've looked around, haven't found anything, and I'm not sure where else
to check.  I haven't visited R-forge (http://r-forge.r-project.org) in a
long time.  Now that I'm trying, it seems to be down.  Anyone know if
this is a temporary condition, and if so, when it's expected to rise again?

Thanks,
Allie

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Re-grouping data in R

2012-08-07 Thread David L Carlson

Your first command erases all the data in that column:

dat$final_purchase_amount- NA

so when you refer to it later, it consists of only NAs.

--
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77843-4352


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of Abraham Mathew
 Sent: Tuesday, August 07, 2012 1:57 PM
 To: r-help@r-project.org
 Subject: [R] Re-grouping data in R
 
 I have a data frame with a column of values that I want to bucket
 (group)
 into specific levels.
 
  str(dat)'data.frame':   3678 obs. of  39 variables:
  $ id  : int  23 76 129 156 166 180 200 214 296
 344 ...
  $ final_purchase_amount   : Factor w/ 32 levels
 \\N,1082,1109,..: 1 1 1 1 1 1 1 1 1 1 ...
 
 
 So I ran the following to produce new levels, one for values from 100
 to 400, 401 to 1000, and 1001+.
 
 
 dat$final_purchase_amount- NA
 dat$final_purchase_amount[dat$final_purchase_amount %in%
 levels(dat$final_purchase_amount)[c(8,9,11,12,13,15,16,17,18,19,20,21)]
 ]
 - 100 to 400
 dat$final_purchase_amount[dat$final_purchase_amount %in%
 levels(dat$final_purchase_amount)[c(22,23,24,25,26,27,28,29,30,31,32)]]
 - 401 to 1000
 dat$final_purchase_amount[dat$final_purchase_amount %in%
 levels(dat$final_purchase_amount)[c(2,3,4,5,6,7,10,14)]] - 1001 +
 dat$final_purchase_amount - factor(dat$final_purchase_amount)
 levels(dat$final_purchase_amount)
 table(dat$final_purchase_amount)
 
 
 
 However, this doesn't seem to produce any levels and returns the
 following.
 
 
  levels(dat$final_purchase_amount)character(0)
 
 
 
 Can anyone point to what I'm doing wrong.
 
 
 
 Thanks!
 
 
 --
 *Abraham Mathew
 Statistical Analyst
 www.amathew.com
 720-648-0108
 @abmathewks*
 
   [[alternative HTML version deleted]]
 
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 guide.html
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Re: [R] r-forge down?

2012-08-07 Thread Roy Mendelssohn

http://www.rforge.net

-Roy

On Aug 7, 2012, at 3:25 PM, Alexander Shenkin wrote:

 Hi Folks,
 
 I've looked around, haven't found anything, and I'm not sure where else
 to check.  I haven't visited R-forge (http://r-forge.r-project.org) in a
 long time.  Now that I'm trying, it seems to be down.  Anyone know if
 this is a temporary condition, and if so, when it's expected to rise again?
 
 Thanks,
 Allie
 
 __
 R-help@r-project.org 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.

**
The contents of this message do not reflect any position of the U.S. 
Government or NOAA.
**
Roy Mendelssohn
Supervisory Operations Research Analyst
NOAA/NMFS
Environmental Research Division
Southwest Fisheries Science Center
1352 Lighthouse Avenue
Pacific Grove, CA 93950-2097

e-mail: roy.mendelss...@noaa.gov (Note new e-mail address)
voice: (831)-648-9029
fax: (831)-648-8440
www: http://www.pfeg.noaa.gov/

Old age and treachery will overcome youth and skill.
From those who have been given much, much will be expected 
the arc of the moral universe is long, but it bends toward justice -MLK Jr.

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] r-forge down?

2012-08-07 Thread Alexander Shenkin

Thanks Roy,

That's one R-forge site, but my impression is that the
r-forge.r-project.org site is a different one (and it's the one that
hosts the package I'm interested in).  Please correct me if those two
sites are somehow related.

thanks,
allie

On 8/7/2012 5:29 PM, Roy Mendelssohn wrote:
 http://www.rforge.net

 -Roy

 On Aug 7, 2012, at 3:25 PM, Alexander Shenkin wrote:

 Hi Folks,

 I've looked around, haven't found anything, and I'm not sure where else
 to check.  I haven't visited R-forge (http://r-forge.r-project.org) in a
 long time.  Now that I'm trying, it seems to be down.  Anyone know if
 this is a temporary condition, and if so, when it's expected to rise again?

 Thanks,
 Allie

 __
 R-help@r-project.org 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.
 **
 The contents of this message do not reflect any position of the U.S. 
 Government or NOAA.
 **
 Roy Mendelssohn
 Supervisory Operations Research Analyst
 NOAA/NMFS
 Environmental Research Division
 Southwest Fisheries Science Center
 1352 Lighthouse Avenue
 Pacific Grove, CA 93950-2097

 e-mail: roy.mendelss...@noaa.gov (Note new e-mail address)
 voice: (831)-648-9029
 fax: (831)-648-8440
 www: http://www.pfeg.noaa.gov/

 Old age and treachery will overcome youth and skill.
 From those who have been given much, much will be expected 
 the arc of the moral universe is long, but it bends toward justice -MLK Jr.


__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Setting Number of Displayed Digits