[R] Using ts and timeSeries

[UnitRoot]

Hello. I have been working on a project which involves random number
generation and unit root test. After I generate the numbers (I am generating
stock returns using rnorm(1000,0,0.25),  I want to check if the series has a
unit root or not. But before I should modify the data to time series. My
question is which of the functions ts or timeSeries more appropriate to test
unit root using Augmented Dickey Fuller test?
Thank you.

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Re: [R] Pointers in R

[Jamie Olson]

One thing to keep in mind is the no side effects rule in R.  Almost all
variables are copied on assignment/modification.  Of course, there's also
lazy instantiation, so the copy isn't actually constructed unless it needs
to be, but this can impact the expected performance of more complicated data
structures built in pure R.

Jamie Olson
School of Computer Science
Carnegie Mellon University
5000 Forbes Ave.
Pittsburgh, PA 15213
jfol...@cs.cmu.edu


On Sun, Jun 12, 2011 at 7:36 PM, Jeff Newmiller jdnew...@dcn.davis.ca.uswrote:

 Lists are recursive and heterogenous in R. Just assign the values to
 elements in lists and assign those lists to elements in other lists to build
 your tree.
 ---
 Jeff Newmiller The . . Go Live...
 DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go...
 Live: OO#.. Dead: OO#.. Playing
 Research Engineer (Solar/Batteries O.O#. #.O#. with
 /Software/Embedded Controllers) .OO#. .OO#. rocks...1k
 ---
 Sent from my phone. Please excuse my brevity.

 Jaimin Dave davejaim...@gmail.com wrote:

 Hello Everyone,
 I am new to R and would like to create a quad tree in R. However the
 problem
 is that I don't think R has pointers. Is there any way to create a tree in
 R?

 Thanks

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Re: [R] How can I write methods for 'as()'?

[Jamie Olson]

Since nobody else has respond, I thought I'd take a stab.  Maybe if I'm
wrong enough somebody will correct me, but my understanding is that that
kind of situation, ie the pain of getting the correct method called when
there is a dependency on the type of more than one argument is part of the
motivation behind the S4 classes.  So maybe it really just is that kludgy
for s3 classes.

Jamie Olson
School of Computer Science
Carnegie Mellon University
5000 Forbes Ave.
Pittsburgh, PA 15213
jfol...@cs.cmu.edu


On Mon, Jun 6, 2011 at 12:19 PM, Janko Thyson 
janko.thyson.rst...@googlemail.com wrote:

 Okay, I found something that is working, but it looks and feels pretty
 awkward as the method def and method lookup takes place in one function ;-)

 setRefClass(A, fields=list(X=numeric))
 setRefClass(B, contains=A, fields=list(Y=character))

 mySetAs - function(
 from,
 to
 ){
 if(!existsMethod(f=coerce, signature=c(from=class(from), to=to))){
 setAs(from=class(from), to=to,
  def=function(from, to){
 out - getRefClass(to)$new(X=from)
 return(out)
 }
 )
 }
  mthd - selectMethod(coerce, signature=c(from=class(from),
 to=to), useInherited= c(from=TRUE, to=TRUE))
 out - mthd(from=from, to=to)
 return(out)
 }

 a - mySetAs(from=1:5, to=A)
 a$X
 b - mySetAs(from=1:5, to=B)
 b$X

 I'm sure there are much better ways. I'd appreciate any comments
 whatsoever.

 Best regards,
 Janko

 On 06.06.2011 17:46, Janko Thyson wrote:
  Somehow I don't see my own postings in the list, so sorry for replying
  to my own message and not the one that went out to the list.
 
  I got a little further and I think I found exactly the thing that is
  bothering me: how to get extended method dispatch going in 'setAs()':
 
  setRefClass(A, fields=list(X=numeric))
  setRefClass(B, contains=A, fields=list(Y=character))
 
  setAs(from=numeric, to=A,
  def=function(from,to){
  out - getRefClass(to)$new(X=from)
  return(out)
  }
  )
  a - as(1:5, A)
  a$X
 
  b - as(1:5, B)
 
  My problem is the last statement (b - as(1:5, B) which fails. I
  want to get around having to write new 'setAs' methods for all classes
  extending class 'A'. If 'B' inherits from 'A', shouldn't it then be
  possible to tell 'setAs' to look for the next suitable method, i.e.
  the method defined for 'A'? I tried 'NextMethod()' inside the body of
  'setAs' but that didn't work out.
 
  Thanks a lot,
  Janko
 
  On 06.06.2011 17:15, Janko Thyson wrote:
  Dear list,
 
  I wonder how to write methods for the function 'as' in the sense that
  I can call 'as(object, Class, strict=TRUE, ext)' and let method
  dispatch figure out the correct method.
 
  AFAIU, there is a difference between, e.g. 'as.data.frame' and the
  methods of 'as()' as stated above since the former depends on arg 'x'
  instead of 'object', 'Class' etc.?
 
   methods(as)
   as.data.frame
 
  I have to admit that I'm not really familiar with the S3 style of
  defining methods as I have been coding in S4 a lot, but my first
  attempt was to write something like this:
 
  as.myClass - function(x, ...){
  if(is(x, data.frame){
  x - as.list(x)
  }
  if(is(x, character){
  x - as.list(x)
  }
  ...
  out - getRefClass(myClass)$new(X=x)
  return(out)
  }
 
  But that way I'd have to explicitly call 'as.myClass(x)' whereas I'd
  simply like to type 'as(x, myClass)'.
  Also, how is it possible to have method dispatch recognize two
  signature arguments in S3? I.e., how can I define something like
  'as.data.frame.character' in order to have explicit sub methods for
  all the data types of 'x' so I wouldn't have to process them all in
  the definition of 'as.myClass' as I did above?
 
  Thanks for your help,
  Janko

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Re: [R] Traversing KD-tree (or equivalent) for radius-based search

[Jamie Olson]

There aren't a whole lot of more complex data structures available as R
packages.  My impression is that pure R implementation offers
dissatisfactory performance and native (or e.g. java)  implementations end
up inconsistent with R's no side effects principles.  I'd suggest building
an R interface to whatever spatial data structures you want, e.g. k-d or
r(+/*)-trees.

Jamie Olson
School of Computer Science
Carnegie Mellon University
5000 Forbes Ave.
Pittsburgh, PA 15213
jfol...@cs.cmu.edu


On Thu, Jun 2, 2011 at 8:43 PM, Andrea Taverna a.t...@libero.it wrote:

 Hi,

 I'm trying to implement the DBSCAN algorithm to get O(N*LogN) complexity
 and I'd need a spatial tree of some sort (kd,r,bd..), or a function that
 computes radius-based search on spatial data, i.e. given a radius eps finds
 ALL the points which fall in the corresponding hypersphere centered on the
 current examined point.  Is there a package with this features?

 So far I found RANN and other packages whose name I can't remember (I don't
 have them with me a.t.m.), and they all seemed to offer a nearest-neighbour
 search, which asks for an upper limit to the number of points to be found,
 but no direct access to the spatial tree they used . The algorithms they
 provide are built around that limit and setting it to large values makes
 their execution impractical.
 OTOH, as far as I have understood, DBSCAN needs to know all the points in
 the eps-neighbourhood, or will create too many clusters, especially if there
 are really high-density region, as it happened when I used RANN's nn2
 function in my implementation.

 thanks in advance,

 Andrea Taverna

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Re: [R] Standard deviation and Mean

[Uwe Ligges]

On 14.06.2011 22:29, Daniel Malter wrote:

Hi,

pick up any introductory manual of which there are many online. It so
happens that the functions for mean and sd are called mean() and sd(). If
you want to know how to use them type ?mean or ?sd in the R-prompt and hit
enter.



... and some rants back. Can you please

1. also reply to the original sender who may not be subscribed to the 
list and hence never receives the answer
2. quote the messages you are referring to, mailing list readers of this 
R-help *mailing list* won't see it otherwise


Uwe Ligges






Daniel

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Re: [R] How to generate bivariate exponential distribution?

[Petr Savicky]

On Tue, Jun 14, 2011 at 08:40:00AM -0700, xuyongdeng wrote:
  Any one know is there any package or function to generate bivariate
 exponential distribution? I gusee there should be three parameters, two rate
 parameters and one correlation parameter. I just did not find any function
 available on R.  Any suggestion is appreciated. 

Do you have a specific bivariate exponential distribution in mind?
If not, then try the following

  n - 1000
  lambda1 - 2
  lambda2 - 3
  common - 1
  x1 - rexp(n, rate=lambda1-common)
  x2 - rexp(n, rate=lambda2-common)
  z - rexp(n, rate=common)
  y1 - pmin(x1, z)
  y2 - pmin(x2, z)

The variables y1, y2 have exponential distribution with rates
lambda1, lambda2 and they are positively correlated, if 

  0  common  min(lambda1, lambda2)

The correlation increases with increasing common.

Petr Savicky.

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Re: [R] Side by side scatter plots with specified regression lines

[ONKELINX, Thierry]

Dear Sigrid,

This is very easy with the ggplot2 package

install.packages(ggplot2)
library(ggplot2)
ggplot(data = Your.Data.Frame, aes(x = YEAR, y = YIELD, colour = TREATMENT)) + 
geom_point() + geom_smooth(method = lm) + facet_wrap(~Country)

Best regards,

Thierry

 -Oorspronkelijk bericht-
 Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
 Namens Sigrid
 Verzonden: zondag 12 juni 2011 21:47
 Aan: r-help@r-project.org
 Onderwerp: [R] Side by side scatter plots with specified regression lines
 
 I am new and self taught in R, so please bear with me.
 
 I want to create two scatter plots side by side. The data set includes
 measurements from two different countries with 7 treatments over a timeline
 (x-axis).
 
 Problem 1
 I want to have each plot to include the data from one of the countries with
 7 regression lines of the treatments, but I do no know how to divide the data
 between them. This is how I created one plot with all the data.
 
  plot(YEAR,YIELD,col=red,xlab=Year,ylab=Yield,xlim=c(1,4),ylim=c(
  1,150))
 
 Problem 2
 The models I've found to describe the regression lines of the treatments seems
 to be different than the default ablines that R creates. I have the values of 
 the
 exact values of intercepts and slopes, but does not know how to add them to
 the graph. This is what I got so far.
 
  abline(lm(YIELD[TREATMENT==A]~YEAR[TREATMENT==A]),lty=2,col=1)
 
 I hope this is enough to give me some pointers, otherwise I will try to 
 elaborate.
 
 Thank you for your help.
 
 --
 View this message in context: http://r.789695.n4.nabble.com/Side-by-side-
 scatter-plots-with-specified-regression-lines-tp3592473p3592473.html
 Sent from the R help mailing list archive at Nabble.com.
 
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[R] Odp: problems with plots in loop (corrected Email)

[Petr PIKAL]

Hi

r-help-boun...@r-project.org napsal dne 14.06.2011 18:28:42:

 Andreas Betz ab...@portola.com 
 Odeslal: r-help-boun...@r-project.org
 
 14.06.2011 18:28
 
 Komu
 
 r-help@r-project.org
 
 Kopie
 
 Předmět
 
 [R] problems with plots in loop (corrected Email)
 
 Dear helpers,
 
 
 
 In an attempt to use a loop to generate graphs in a for loop in run into
 a problem. The plan is to fill each page with eight graphs (mfrow =
 c(4,2)) in to two columns. Only the buttom graphs ( meaning every fourth
 graph) have  tick labels on the x  axis to preserve space. I used an if
  Else statement to achieve that. 
 
 The problem is that the first eight graphs are skipped when I run the
 loop, the other graphs are fine.  However, all graphs can be generated
 individually. This indicates that the data sets are fine.
 
 
 
 
 
 Pulling the results from a nonlinear regression to more than 90 data
 sets from a list (resultslist[[i]])
 
 Generating the first derivative and multiply it by 100 to adjust for the
 scale. 
 
 
 
 Here is the code. 
 
 
 
 pdf('F:/diffnormal_16001a.pdf')
 par(mfcol = c(4,2))
 for(i in 1: 92){
 fit  -  NULL
 der -  NULL
 Zeit  - NULL
 Zeit - seq(0,300)
 try(guess - predict(resultslist_1600[[i]]) )
 if(class(guess) == try-error) {next}
 fit - smooth.spline(Zeit, guess)
 der - 100*(predict(fit, Zeit, deriv = 1))$y
   if((i/4)%%1 ==0){par(mar =c(4,4,0, 0) + 0.1)}
   else{par(mar =c(0,4,0, 0) + 0.1)}

Maybe the else just moved due to mailer but it shall be on the same line 
as if statement to be executed. But in that case you would obtain error 
message.

However when I get rid of all I do not have from your code
pdf(test.pdf)
par(mfcol = c(4,2))
for(i in 1: 8){
 fit  -  NULL
der -  NULL
 Zeit  - NULL
 Zeit - seq(0,300)
   if((i/4)%%1 ==0){par(mar =c(4,4,0, 0) + 0.1)}   else{par(mar =c(0,4,0, 
0) + 0.1)}
   leg = paste(Data_, i,sep = )
   plot(Zeit, Zeit, pch = 19, cex=3, ylab = Signal, axes = F)
   if((i/4)%%1 ==0){axis(1, at = seq(0, 360, length = 6), label = c(), 
font = 2)}
   mtext(side = 3, leg, line = -2)}

dev.off()

It seems to me that plotting works as expected.

Regards
Petr

   leg = paste(Data_, i,sep = )
   plot(resultslist_1600[[i]], type = all, pch = ., ylab = Signal,
 log = , axes = F)
   lines(Zeit, der)
   if((i/4)%%1 ==0){axis(1, at = seq(0, 360, length = 6), label = c(),
 font = 2)}
   axis(2, at =
 pretty(na.omit(eval(parse(text=paste(bleeder1600[,,i,],sep =),
 label = c())
   mtext(side = 3, leg, line = -2)}
 
 
 
 I understand, that I am supposed to submit working code. However, I deal
 with a fairly comprehensive data set and I have to generate a large list
 using another R package. Thus I explain what it does:
 
 
 
 Generate time points(Zeit - seq(0,300)
 
 Pulling the results from a nonlinear regression to more than 90 data
 sets from a list (resultslist_1600[[i]])
 
 Test if prediction can be done
 
 try(guess - predict(resultslist_1600[[i]]) )
 if(class(guess) == try-error) {next}
 (it turned out it can be done for all data set, as I am able to generate
 the graphs for each data set from the command line individually.
 
 Generating the first derivative and multiply it by 100 to adjust for the
 scale. 
 
 fit - smooth.spline(Zeit, guess)
 
 der - 100*(predict(fit, Zeit, deriv = 1))$y
 
 
 
 The problem appears to be hidden in the mfrow statement.
 
 I spent quite a bit of time on this problem, thus any help or new ideas
 would be very much appreciated.
 
 
 
 Thank you
 
 
 
 Andreas Betz
 
 Scientist
 
 andreasbetz@ Dear helpers,
 
 
 
 In an attempt to use a loop to generate graphs in a for loop in run into
 a problem. The plan is to fill each page with eight graphs (mfrow =
 c(4,2)) in to two columns. Only the buttom graphs ( meaning every fourth
 graph) have  tick labels on the x  axis to preserve space. I used an if
  Else statement to achieve that. 
 
 The problem is that the first eight graphs are skipped when I run the
 loop, the other graphs are fine.  However, all graphs can be generated
 individually. This indicates that the data sets are fine.
 
 
 
 
 
 Pulling the results from a nonlinear regression to more than 90 data
 sets from a list (resultslist[[i]])
 
 Generating the first derivative and multiply it by 100 to adjust for the
 scale. 
 
 
 
 Here is the code. 
 
 
 
 pdf('F:/diffnormal_16001a.pdf')
 par(mfcol = c(4,2))
 for(i in 1: 92){
 fit  -  NULL
 der -  NULL
 Zeit  - NULL
 Zeit - seq(0,300)
 try(guess - predict(resultslist_1600[[i]]) )
 if(class(guess) == try-error) {next}
 fit - smooth.spline(Zeit, guess)
 der - 100*(predict(fit, Zeit, deriv = 1))$y
   if((i/4)%%1 ==0){par(mar =c(4,4,0, 0) + 0.1)}
   else{par(mar =c(0,4,0, 0) + 0.1)}
   leg = paste(Data_, i,sep = )
   plot(resultslist_1600[[i]], type = all, pch = ., ylab = Signal,
 log = , axes = F)
   lines(Zeit, der)
   if((i/4)%%1 ==0){axis(1, at = seq(0, 360, length = 6), label = c(),
 font = 2)}
  

Re: [R] plotting on an image

[Johann Kim]

Thanks Greg,

Using rasterImage and the steps you described  works fine!
I agree with the distraction! But my purpose is to superimpose a heat map of 
eye tracking data on the original picture...

Thanks!




- original message 

Subject: RE: [R] plotting on an image
Sent: Wed, 15 Jun 2011
From: Greg Snowgreg.s...@imail.org

 If you are willing to prepend a step then you could:
 
 1. Create an empty plot using your data and type='n' (or just plot the data,
 the points will be overwritten), you may want to set the asp argument, or
 explicitly do the xlim and ylim arguments.
 2. Add the graphic using the rasterImage function
 3. Use functions such as points or lines (or others that add to existing
 plots) to plot you data on top of the image.
 
 If you need certain points within the image to correspond to certain
 coordinates then the locator and updateusr (TeachingDemos package) may be of
 help.
 
 But in all of this, make sure that you really want to do this, often (but
 not always) putting an image in the background is chartjunk that distracts
 more than helps.
 
 -- 
 Gregory (Greg) L. Snow Ph.D.
 Statistical Data Center
 Intermountain Healthcare
 greg.s...@imail.org
 801.408.8111
 
 
  -Original Message-
  From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
  project.org] On Behalf Of Johann Kim
  Sent: Monday, June 13, 2011 12:33 PM
  To: r-help@r-project.org
  Subject: [R] plotting on an image
  
  Hello all,
  
  has someone please a few hints about how to
  1.st: draw an image (preferrably a jpg) and then
  2nd: plot() on that image
  
  I am using a mac - and after searching and trying  different ways (I
  have installed EBImage) I now would like to ask for help...
  
  Thanks!
  Johann
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  and provide commented, minimal, self-contained, reproducible code.
 

--- original message end 

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[R] Reshaping data with xtabs reorders my rows

[filip.biele...@rega.kuleuven.be]

Dear,

I have a data frame melted from a list of matrices (melt from reshape
package), then I impute some missing values and then want to tabulate
the data again to the array (or list, doesn't matter) of matrices form.
However when using xtabs function it orders my rows alphabetically and
apparently doesn't take reorder = FALSE option or anything like it.
Is there anything I can do to stop it from doing so?

Relevant parts of code:

matrices.m - melt(combined_list)

matrices.m_i[is.na(matrices.m_i$value),]$value - predictions

matrices  - xtabs(value ~ location + variable + week, data =
matrices.m_i)

-- 
while(!succeed) { try(); }

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Re: [R] heatmap with values

[Jim Lemon]

On 06/14/2011 08:53 PM, Agustin Lobo wrote:

Hi!
I'm displaying a contingency table with heatmap():

svm.predPix.tabla


svm.predPix CC DD LL NN NN2
 CC  22  0  3  8   3
 DD   0 27  0  1   0
 LL   1  1 90  3   7
 NN   2  0  1 11   4
 NN2  0  0  5  1  20



heatmap(svm.predPix.tabla[5:1,], Rowv=NA, Colv=NA,col =

rev(heat.colors(32)), scale=column,  margins=c(5,10))

and I'm happy with the plot except that I would like to have the actual
values displayed within each cell.  Any help on how
to achieve this?

Data in
https://sites.google.com/site/openfiles2/home/svm.predPix.tabla.rda


Hi Agustin,
I was going to send the code from color2D.matplot to display the values, 
but it required so much modification that I would suggest:


library(plotrix)
color2D.matplot(svm.predPix.tabla,show.values=TRUE,axes=FALSE,
 xlab=,ylab=)
axis(1,at=0.5:4.5,labels=rownames(svm.predPix.tabla))
axis(2,at=4.5:0.5,labels=colnames(svm.predPix.tabla))

Jim

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[R] Fitting a choice model (Bradley-Terry generalization)

[David Scott]

I have some data I would like to model which involves choice of food by 
dung beetles.


There are a number of experiments  where in each case, there are five 
choices. Overall there are more than 5 different foods being compared 
(including a placebo) and different experiments use different comparisons.


The problem is a generalization of Bradley-Terry but it differs from 
some generalizations in that the comparisons are not pairwise, and they 
don't produce a full ordering, just that one is preferred to the other 
four possibilities.


I have had a look at the BradleyTerry2, eba, pmr and MLCM packages, none 
of which appear to provide the required functionality. I have also 
looked at a number of papers (Hunter, 2004; Firth, 2005; Huang Weng and 
Lin, 2006; and Fujimoto, Hino and Murata 2011). I think fitting using 
maximum likelihood should be possible, but would welcome any pointers to 
useful code,  relevant ideas, or similar analyses.


David Scott

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Re: [R] Fitting a choice model (Bradley-Terry generalization)

[Achim Zeileis]

On Wed, 15 Jun 2011, David Scott wrote:

I have some data I would like to model which involves choice of food by 
dung beetles.


There are a number of experiments where in each case, there are five 
choices. Overall there are more than 5 different foods being compared 
(including a placebo) and different experiments use different 
comparisons.


The problem is a generalization of Bradley-Terry but it differs from 
some generalizations in that the comparisons are not pairwise, and they 
don't produce a full ordering, just that one is preferred to the other 
four possibilities.


In some cases such comparisons are coded with undecided for those 
comparisons that are not fully ranked. Alternatively, sometimes they are 
also coded with NAs.


For example, if A is preferred over B and C, this may be coded as:
A  B, A  C, B = C or A  B, A  C, NA.

I have had a look at the BradleyTerry2, eba, pmr and MLCM packages, none 
of which appear to provide the required functionality.


That depends what you think the required functionality is. If it is 
dealing with NAs, then some certainly have the functionality. Similarly, 
dealing with undecided is provided in several implementations.


Additionally, to the packages above, the prefmod package provides a 
Bradley-Terry models as well as pattern models which might be interesting 
for you. The VGAM package can estimate Bradley-Terry models, see
http://www.jstatsoft.org/v32/i10/. Finally, the psychotree package 
provides a class paircomp for representing paired comparison data, to
estimate Bradley-Terry models, and in particular to assess the influence 
of covariates on such a model by recursive partitioning (see 
example(bttree, package = psychotree)).


hth,
Z

I have also looked at a 
number of papers (Hunter, 2004; Firth, 2005; Huang Weng and Lin, 2006; and 
Fujimoto, Hino and Murata 2011). I think fitting using maximum likelihood 
should be possible, but would welcome any pointers to useful code,  relevant 
ideas, or similar analyses.


David Scott

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[R] Query regarding auto arima

[siddharth arun]

I am using AUTO ARIMA for forecasting. But it is not detecting 'seasonality
term' of its own for any data.
Is there any other method by which we can detect seasonality and its
frequency for any data?

Is there any method through which seasonality and its frequency can be
automatically detected from ACF plot?


-- 
Siddharth Arun,
4th Year Undergraduate student
Industrial Engineering and Management,
IIT Kharagpur

[[alternative HTML version deleted]]

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[R] Legend in lattice

[Julio Rojas]

Dear all, I have been working in a plot based on figure 5.6 of the Lattice book 
(http://lmdvr.r-forge.r-project.org/figures/figures.html).

I
 have already modified it to include the size of the circles as another 
variable, but I would like to modify the legend to show it (like they do
 it in http://www.jstatsoft.org/v15/i05/paper). I have divided my variable in 
intervals:

DATA$s_Shape_2 - cut(x=DATA$Shape_inde,breaks=c(0.999,1.2,1.4,1.7,2.1,5))


I
 will use the index of the categories as the radius of the circles. Now,
 following the aforementioned figure 5.6, I have the following code:

NDVI.breaks - do.breaks(range(DATA.ord$NDVI), 50)

xyplot(Y_Center_P~X_Center_P|Level,data=DATA.ord,col = black,aspect = iso,
    fill.color = DATA.ord$c_RNDVI, cex = 
DATA$s_Shape_2,
    panel = function(x, y, fill.color, cex,..., 
subscripts) {
    fill - fill.color[subscripts]
    cex - cex[subscripts]
    panel.grid(h = -1, v = -1)
    panel.xyplot(x, y, pch = 21, fill = 
fill, cex = cex,...)
    },
    legend =
    list(right = list(fun = draw.colorkey, args 
= list(key = list(col = rainbow,
 at = NDVI.breaks), draw = FALSE)))
)


How can I add the categories with circles of their respective sizes to the 
legend?

I hope this is enough information to have some help. If not, let me know.

Regards.

Julio.


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[R] Count occurances in integers (or strings)

[Jay]

Hi,

I have a dataframe column from which I want to calculate the number of
1's in each entry. Some column values could, for example, be
0001001000 and 111.

To get the number of occurrences from a string I use this:
sum(unlist(strsplit(mydata[,my_column], )) == 1)

However, as my data is not in string form.. How do I convert it? l
tried:
lapply(mydata[,my_column],toString)

but I do not seem to get it right (or at least I do not understand the
output format).

Also, are there other options? Can I easily calculate the occurrences
directly from the integers?

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[R] Find values from one dataframe between certain values in another dataframe

[mwege]

Hi all,

I have a 2 files, one with a series of beginning and end times of animal
dives in (lets call it dives). The other file is readings from a time-depth
recorder, there is a datetime reading for every second, a temperature and
light level reading (lets call it tdr). 

Now I want to say from the file dives, using the start and end time of the
dives, go to the tdr file and find me the temperature values that falls
inbetween the times of the start and end times.

So this is what I have written:

dives-read.csv(file)#table with begin and end times of dives
tdr-read.csv(file2)#table with every second readings of temp and
light level


dat-{} #just create empty thing
for (i in 1:nrow(dives)){
st-x[i,1] # start datetime value for dive[i]. Column one is the start of
dive datetime value
ed-x[i,4] # end datetime value for dive[i] Column 4 is the end of the
dive datetime value
f-which(tdr[,2]=st  tdr[,2]=ed) # find row numbers where tdr datetimes
is between start  
#end times of dive[i].
Column 2 is the datetime value
z1-tdr[f,5] # extract temperature values
maxtemp-max(z1) #out of those values find the max value
dat-rbind(dat,maxtemp) #add that row onto a dat
}

The problem is when I just want to check wat f is it keeps telling me
interger(0). It says that there are no values in tdr that falls between the
start and end of the dives. But there is, I have checked. 

I reckoned that it has to do with the format of the datetime values. I
couldnt find how to convert it to a numeric value. At the moment my datetime
values are in a POSIXct format defined as follows

 as.POSIXct(strptime((dive$begindive),'%Y/%m/%d %H:%M:%S'),tz=GMT)

I have also tried to sort the tdr data first in ascending order
tdr - tdr[order(tdr$datetime),] #sort z according to dtime

I have even tried to convert to numeric format in excell (all three datetime
values) and then use those numerics in R but it still doesnt want to work.

The problem lies in the f-which(tdr[,2]=st  tdr[,2]=ed)  line.
It doenst find any values from tdr that are between st and ed. But there
definately is, I have done a manual check.

Any help is appreciated, thank you.
Mia 

--
View this message in context: 
http://r.789695.n4.nabble.com/Find-values-from-one-dataframe-between-certain-values-in-another-dataframe-tp3599033p3599033.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Query regarding auto arima

[Achim Zeileis]

On Wed, 15 Jun 2011, siddharth arun wrote:


I am using AUTO ARIMA for forecasting.


I assume you mean function auto.arima() from package forecast.


But it is not detecting 'seasonality term' of its own for any data.


Yes, it does so, if you supply a time series object with a frequency  1.

Is there any other method by which we can detect seasonality and its 
frequency for any data?


Is there any method through which seasonality and its frequency can be
automatically detected from ACF plot?


Usually you _know_ the frequency (i.e., 12 for monthly and 4 for quarterly 
data etc.).


For example for the famous monthly AirPassengers series:

library(forecast)
auto.arima(AirPassengers)

which returns a seasonal ARIMA model. See ?auto.arima for tweaking its 
arguments as well as the accompanying paper


  http://www.jstatsoft.org/v27/i03/

for more details.
Z



--
Siddharth Arun,
4th Year Undergraduate student
Industrial Engineering and Management,
IIT Kharagpur

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[R] Rmpi installation

[Unger, Kristian, Dr.]

Hi there

I am trying to install Rmpi (version 0.5-4) on our 8-core SUSE Linux
Enterprise Server 11 SP1. I read all I could find about Rmpi installation
but still cannot get it working.

Here the last command that I used:

R CMD INSTALL
--configure.args=--with-Rmpi-include=/usr/lib64/mpi/gcc/openmpi/include \
--with-Rmpi-libpath=/usr/lib64/mpi/gcc/openmpi/lib64 \
--with-Rmpi-type=OPENMPI Rmpi_0.5-4.tar.gz

Resulting in the following:

zytosrv01dmi:/home/unger/R_projects/OS # R CMD INSTALL
--configure-args=--with-Rmpi-include=/usr/lib64/mpi/gcc/openmpi/include
\--with-Rmpi-libpath=/usr/lib64/mpi/gcc/openmpi/lib64 \
--with-Rmpi-type=OPENMPI Rmpi_0.5-4.tar.gz
* installing to library Œ/usr/local/lib64/R/library¹
* installing *source* package ŒRmpi¹ ...
checking for gcc... gcc
checking for C compiler default output file name... a.out
checking whether the C compiler works... yes
checking whether we are cross compiling... no
checking for suffix of executables...
checking for suffix of object files... o
checking whether we are using the GNU C compiler... yes
checking whether gcc accepts -g... yes
checking for gcc option to accept ISO C89... none needed
checking how to run the C preprocessor... gcc -E
checking for grep that handles long lines and -e... /usr/bin/grep
checking for egrep... /usr/bin/grep -E
checking for ANSI C header files... yes
checking for sys/types.h... yes
checking for sys/stat.h... yes
checking for stdlib.h... yes
checking for string.h... yes
checking for memory.h... yes
checking for strings.h... yes
checking for inttypes.h... yes
checking for stdint.h... yes
checking for unistd.h... yes
checking mpi.h usability... no
checking mpi.h presence... no
checking for mpi.h... no
Try to find libmpi or libmpich ...
checking for main in -lmpi... no
libmpi not found. exiting...
ERROR: configuration failed for package ŒRmpi¹
* removing Œ/usr/local/lib64/R/library/Rmpi¹

So obviously libmpi is not found. Doing a locate search for libmpi
(straight after updatedb) shows:

zytosrv01dmi:/home/unger/R_projects/OS # locate libmpi
/opt/mpich/ch-p4/lib64/libmpich.a
/opt/mpich/ch-p4/lib64/libmpichf90.a
/opt/mpich/ch-p4/lib64/libmpichf90nc.a
/opt/mpich/ch-p4/lib64/libmpichfarg.a
/opt/mpich/ch-p4/lib64/libmpichfsup.a
/opt/mpich/ch-p4mpd/lib64/libmpich.a
/opt/mpich/ch-p4mpd/lib64/libmpichf90.a
/opt/mpich/ch-p4mpd/lib64/libmpichf90nc.a
/opt/mpich/ch-p4mpd/lib64/libmpichfarg.a
/opt/mpich/ch-p4mpd/lib64/libmpichfsup.a
/usr/lib64/mpi/gcc/openmpi/lib64/libmpi_cxx.la
/usr/lib64/mpi/gcc/openmpi/lib64/libmpi_cxx.so
/usr/lib64/mpi/gcc/openmpi/lib64/libmpi_cxx.so.0
/usr/lib64/mpi/gcc/openmpi/lib64/libmpi_cxx.so.0.0.0
/usr/lib64/mpi/gcc/openmpi/lib64/libmpi_f77.la
/usr/lib64/mpi/gcc/openmpi/lib64/libmpi_f77.so
/usr/lib64/mpi/gcc/openmpi/lib64/libmpi_f77.so.0
/usr/lib64/mpi/gcc/openmpi/lib64/libmpi_f77.so.0.0.0
/usr/lib64/mpi/gcc/openmpi/lib64/libmpi_f90.la
/usr/lib64/mpi/gcc/openmpi/lib64/libmpi_f90.so
/usr/lib64/mpi/gcc/openmpi/lib64/libmpi_f90.so.0
/usr/lib64/mpi/gcc/openmpi/lib64/libmpi_f90.so.0.0.0
/usr/lib64/mpi/gcc/openmpi/lib64/libmpi.la
/usr/lib64/mpi/gcc/openmpi/lib64/libmpi.so
/usr/lib64/mpi/gcc/openmpi/lib64/libmpi.so.0
/usr/lib64/mpi/gcc/openmpi/lib64/libmpi.so.0.0.0

What lets me assume that the path to libmpi is
/usr/lib64/mpi/gcc/openmpi/lib64 which I already included in the options...


How can I get this working?

I would highly appreciate any help on this!

Best wishes

Kristian




Dr. Kristian Unger


Arbeitsgruppenleiter Integrative Biologie / Head of Integrative Biology
Group
Abteilung für Strahlenzytogenetik / Research Unit of Radiation Cytogenetics


Helmholtz Zentrum München
Deutsches Forschungszentrum für Gesundheit und Umwelt (GmbH)
Ingolstädter Landstr. 1
85764 Neuherberg
www.helmholtz-muenchen.de
Aufsichtsratsvorsitzende: MinDir´in Bärbel Brumme-Bothe
Geschäftsführer: Prof. Dr. Günther Wess und Dr. Nikolaus Blum
Registergericht: Amtsgericht München HRB 6466
USt-IdNr: DE 129521671

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

Re: [R] Still have problems with tcltk in R 64 bit

[Arnaud Mosnier]

I agree that this is a really outdated source but I did not find the
way to tell R using correctly the tcl version included (at least for
the 64 bit version).
If I remove the environment variables, things work for R 32 bit (it
uses the tcl version included), but it does not work in R 64 bit.

Where are the configuration files used to define the path to each tcl version ?

Arnaud

2011/6/14 Uwe Ligges lig...@statistik.tu-dortmund.de:


 On 14.06.2011 22:01, Arnaud Mosnier wrote:

 I achieve to make tcltk work on R 64 installing Active tcltk8.5 64bit
 version then setting windows environment variables as in
 http://www.sciviews.org/_rgui/tcltk/InstallRTclTk.html.

 Don't read outdated sources but the manuals.

 The R binary distribution comes with tcltk under Windows (in ${R_HOME}/tcl)
 for both 32-bit and 64-bit and will user a different tcl if you set
 environment variables.

 Hence the easiest thing is just to tell R not to use your otehrwise set
 environment variabes and use its own tcl version.

 Uwe Ligges



 But now, it uses only this 64 bit version and thus do not work anymore
 in R 32 bit !

 In my case, it solves my problem as I will probably use only R 64bit
 but I do not like to end with an half solution.

 Arnaud

 2011/6/14 Peter Langfelderpeter.langfel...@gmail.com:

 On Tue, Jun 14, 2011 at 12:47 PM, Adrienne Woottenamwoo...@ncsu.edu
  wrote:

 Taking a quick look for it, it seems that they have replaced it with
 tcltk2.  I just did the installation with the same version in windows
 and it auto loaded the tcltk package and I never installed that
 package to begin with.  I would try it with tcltk2 and see if you get
 the package to install appropriately.  I'm not sure why tcltk isn't on
 CRAN anymore, it makes no sense not to have both tcltk and tcltk2, but
 here's hoping this helps you out.

 A

 FWIF, tcltk still exists but is now part of the standard R
 distribution (core packages is the term I think?) and as such is
 installed automatically when you install R. Therefore it is not
 available from CRAN.

 Peter


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Re: [R] Rmpi installation

[Hugo Mildenberger]

Hmm, 

looks like there was a trailing blank after the backslash and before end of 
line, 
resulting in --with-Rmpi-libpath possibly not recognised: 

   \--with-Rmpi-libpath=/usr/lib64/mpi/gcc/openmpi/lib64 \

I also doubt there is real need to escape newlines within a string. But another 
possible problem source is that according to R CMD INSTALL --help, the 
parameter 
is called  --configure-args, not configure.args. 


$ R CMD INSTALL  
--configure-args=--with-Rmpi-include=/usr/lib64/mpi/gcc/openmpi/include  
--with-Rmpi-libpath=/usr/lib64/mpi/gcc/openmpi/lib64 --
with-Rmpi-type=OPENMPI Rmpi_0.5-4.tar.gz

Best



On Wednesday 15 June 2011 14:25:46 Unger, Kristian, Dr. wrote:
 Hi there
 
 I am trying to install Rmpi (version 0.5-4) on our 8-core SUSE Linux
 Enterprise Server 11 SP1. I read all I could find about Rmpi installation
 but still cannot get it working.
 
 Here the last command that I used:
 
 R CMD INSTALL
 --configure.args=--with-Rmpi-include=/usr/lib64/mpi/gcc/openmpi/include \
 --with-Rmpi-libpath=/usr/lib64/mpi/gcc/openmpi/lib64 \
 --with-Rmpi-type=OPENMPI Rmpi_0.5-4.tar.gz
 
 Resulting in the following:
 
 zytosrv01dmi:/home/unger/R_projects/OS # R CMD INSTALL
 --configure-args=--with-Rmpi-include=/usr/lib64/mpi/gcc/openmpi/include
 \--with-Rmpi-libpath=/usr/lib64/mpi/gcc/openmpi/lib64 \
 --with-Rmpi-type=OPENMPI Rmpi_0.5-4.tar.gz
 * installing to library Œ/usr/local/lib64/R/library¹
 * installing *source* package ŒRmpi¹ ...
 checking for gcc... gcc
 checking for C compiler default output file name... a.out
 checking whether the C compiler works... yes
 checking whether we are cross compiling... no
 checking for suffix of executables...
 checking for suffix of object files... o
 checking whether we are using the GNU C compiler... yes
 checking whether gcc accepts -g... yes
 checking for gcc option to accept ISO C89... none needed
 checking how to run the C preprocessor... gcc -E
 checking for grep that handles long lines and -e... /usr/bin/grep
 checking for egrep... /usr/bin/grep -E
 checking for ANSI C header files... yes
 checking for sys/types.h... yes
 checking for sys/stat.h... yes
 checking for stdlib.h... yes
 checking for string.h... yes
 checking for memory.h... yes
 checking for strings.h... yes
 checking for inttypes.h... yes
 checking for stdint.h... yes
 checking for unistd.h... yes
 checking mpi.h usability... no
 checking mpi.h presence... no
 checking for mpi.h... no
 Try to find libmpi or libmpich ...
 checking for main in -lmpi... no
 libmpi not found. exiting...
 ERROR: configuration failed for package ŒRmpi¹
 * removing Œ/usr/local/lib64/R/library/Rmpi¹
 
 So obviously libmpi is not found. Doing a locate search for libmpi
 (straight after updatedb) shows:
 
 zytosrv01dmi:/home/unger/R_projects/OS # locate libmpi
 /opt/mpich/ch-p4/lib64/libmpich.a
 /opt/mpich/ch-p4/lib64/libmpichf90.a
 /opt/mpich/ch-p4/lib64/libmpichf90nc.a
 /opt/mpich/ch-p4/lib64/libmpichfarg.a
 /opt/mpich/ch-p4/lib64/libmpichfsup.a
 /opt/mpich/ch-p4mpd/lib64/libmpich.a
 /opt/mpich/ch-p4mpd/lib64/libmpichf90.a
 /opt/mpich/ch-p4mpd/lib64/libmpichf90nc.a
 /opt/mpich/ch-p4mpd/lib64/libmpichfarg.a
 /opt/mpich/ch-p4mpd/lib64/libmpichfsup.a
 /usr/lib64/mpi/gcc/openmpi/lib64/libmpi_cxx.la
 /usr/lib64/mpi/gcc/openmpi/lib64/libmpi_cxx.so
 /usr/lib64/mpi/gcc/openmpi/lib64/libmpi_cxx.so.0
 /usr/lib64/mpi/gcc/openmpi/lib64/libmpi_cxx.so.0.0.0
 /usr/lib64/mpi/gcc/openmpi/lib64/libmpi_f77.la
 /usr/lib64/mpi/gcc/openmpi/lib64/libmpi_f77.so
 /usr/lib64/mpi/gcc/openmpi/lib64/libmpi_f77.so.0
 /usr/lib64/mpi/gcc/openmpi/lib64/libmpi_f77.so.0.0.0
 /usr/lib64/mpi/gcc/openmpi/lib64/libmpi_f90.la
 /usr/lib64/mpi/gcc/openmpi/lib64/libmpi_f90.so
 /usr/lib64/mpi/gcc/openmpi/lib64/libmpi_f90.so.0
 /usr/lib64/mpi/gcc/openmpi/lib64/libmpi_f90.so.0.0.0
 /usr/lib64/mpi/gcc/openmpi/lib64/libmpi.la
 /usr/lib64/mpi/gcc/openmpi/lib64/libmpi.so
 /usr/lib64/mpi/gcc/openmpi/lib64/libmpi.so.0
 /usr/lib64/mpi/gcc/openmpi/lib64/libmpi.so.0.0.0
 
 What lets me assume that the path to libmpi is
 /usr/lib64/mpi/gcc/openmpi/lib64 which I already included in the options...
 
 
 How can I get this working?
 
 I would highly appreciate any help on this!
 
 Best wishes
 
 Kristian
 
 
 
 
 Dr. Kristian Unger
 
 
 Arbeitsgruppenleiter Integrative Biologie / Head of Integrative Biology
 Group
 Abteilung für Strahlenzytogenetik / Research Unit of Radiation Cytogenetics
 
 
 Helmholtz Zentrum München
 Deutsches Forschungszentrum für Gesundheit und Umwelt (GmbH)
 Ingolstädter Landstr. 1
 85764 Neuherberg
 www.helmholtz-muenchen.de
 Aufsichtsratsvorsitzende: MinDir´in Bärbel Brumme-Bothe
 Geschäftsführer: Prof. Dr. Günther Wess und Dr. Nikolaus Blum
 Registergericht: Amtsgericht München HRB 6466
 USt-IdNr: DE 129521671
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide 

Re: [R] Heatmap in R and/or ggplot2

[JiHO]

On Tue, Jun 14, 2011 at 19:56, idris idris.r...@gmail.com wrote:
 Follow up question: My data contains x, y, height, and day.

 I want to create the heatmap for each day, keeping the color coding
 consistent.

 I've created an example with 2 days, and you can see the charts below.

 Notice that the legend changes from day 1 to day 2. How can I make the
 legend consistent?

 Also, my real data contains hundreds of days. Is there a way to create a
 'movie' or a sequence of the heatmaps in chronological order of days?

 The way my code works now is obviously very naive as it just repeats the
 same code for day 1 and day 2. Is there a better way to do this?

legend: use the limit argument of scale_fill_gradientn and set it to
the max and min of height across all days

movie: there is no way to do that in R alone (that I know of). you can:
1- use the jpeg or png device with a name including a special code
which increases every time a new plot is produced (this is the
default)
2- plot the successive plots in a for loop
3- turn the sequence of images into a movie using  specialized software.

For step 3 you could use guicktime on Mac OS X or mencoder on Linux/OS
X. I don't know about Windows.

Since mencoder works on the command line, you can call it from R and I
have code to ease that:
https://gitorious.org/r/r-utils/blobs/master/lib_movie.R
but you should get familiar with mencoder a little bit before trying
to read/understand it.

JiHO
---
http://maururu.net

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Re: [R] Rmpi installation

[Unger, Kristian, Dr.]

Thank you very much Hugo. Using the command as suggested results exactly
the same error:

# R CMD INSTALL
--configure-args=--with-Rmpi-include=/usr/lib64/mpi/gcc/openmpi/include
--with-Rmpi-libpath=/usr/lib64/mpi/gcc/openmpi/lib64
--with-Rmpi-type=OPENMPI Rmpi_0.5-4.tar.gz
* installing to library ‘/usr/local/lib64/R/library’
* installing *source* package ‘Rmpi’ ...
checking for gcc... gcc
checking for C compiler default output file name... a.out
checking whether the C compiler works... yes
checking whether we are cross compiling... no
checking for suffix of executables...
checking for suffix of object files... o
checking whether we are using the GNU C compiler... yes
checking whether gcc accepts -g... yes
checking for gcc option to accept ISO C89... none needed
checking how to run the C preprocessor... gcc -E
checking for grep that handles long lines and -e... /usr/bin/grep
checking for egrep... /usr/bin/grep -E
checking for ANSI C header files... yes
checking for sys/types.h... yes
checking for sys/stat.h... yes
checking for stdlib.h... yes
checking for string.h... yes
checking for memory.h... yes
checking for strings.h... yes
checking for inttypes.h... yes
checking for stdint.h... yes
checking for unistd.h... yes
checking mpi.h usability... no
checking mpi.h presence... no
checking for mpi.h... no
Try to find libmpi or libmpich ...
checking for main in -lmpi... no
libmpi not found. exiting...
ERROR: configuration failed for package ‘Rmpi’
* removing ‘/usr/local/lib64/R/library/Rmpi’


Best wishes

Kristian


Dr. Kristian Unger


Arbeitsgruppenleiter Integrative Biologie / Head of Integrative Biology
Group
Abteilung für Strahlenzytogenetik / Research Unit of Radiation
Cytogenetics

Tel.: +49-89-3187-3515

Mob.: +49-160-90641879





Am 15.06.11 15:12 schrieb Hugo Mildenberger unter
hugo.mildenber...@web.de:

Hmm,

looks like there was a trailing blank after the backslash and before end
of line,
resulting in --with-Rmpi-libpath possibly not recognised:

   \--with-Rmpi-libpath=/usr/lib64/mpi/gcc/openmpi/lib64 \

I also doubt there is real need to escape newlines within a string. But
another
possible problem source is that according to R CMD INSTALL --help, the
parameter
is called  --configure-args, not configure.args.


$ R CMD INSTALL
--configure-args=--with-Rmpi-include=/usr/lib64/mpi/gcc/openmpi/include
--with-Rmpi-libpath=/usr/lib64/mpi/gcc/openmpi/lib64 --
with-Rmpi-type=OPENMPI Rmpi_0.5-4.tar.gz

Best



On Wednesday 15 June 2011 14:25:46 Unger, Kristian, Dr. wrote:
 Hi there

 I am trying to install Rmpi (version 0.5-4) on our 8-core SUSE Linux
 Enterprise Server 11 SP1. I read all I could find about Rmpi
installation
 but still cannot get it working.

 Here the last command that I used:

 R CMD INSTALL

--configure.args=--with-Rmpi-include=/usr/lib64/mpi/gcc/openmpi/include
\
 --with-Rmpi-libpath=/usr/lib64/mpi/gcc/openmpi/lib64 \
 --with-Rmpi-type=OPENMPI Rmpi_0.5-4.tar.gz

 Resulting in the following:

 zytosrv01dmi:/home/unger/R_projects/OS # R CMD INSTALL
 --configure-args=--with-Rmpi-include=/usr/lib64/mpi/gcc/openmpi/include
 \--with-Rmpi-libpath=/usr/lib64/mpi/gcc/openmpi/lib64 \
 --with-Rmpi-type=OPENMPI Rmpi_0.5-4.tar.gz
 * installing to library Œ/usr/local/lib64/R/library¹
 * installing *source* package ŒRmpi¹ ...
 checking for gcc... gcc
 checking for C compiler default output file name... a.out
 checking whether the C compiler works... yes
 checking whether we are cross compiling... no
 checking for suffix of executables...
 checking for suffix of object files... o
 checking whether we are using the GNU C compiler... yes
 checking whether gcc accepts -g... yes
 checking for gcc option to accept ISO C89... none needed
 checking how to run the C preprocessor... gcc -E
 checking for grep that handles long lines and -e... /usr/bin/grep
 checking for egrep... /usr/bin/grep -E
 checking for ANSI C header files... yes
 checking for sys/types.h... yes
 checking for sys/stat.h... yes
 checking for stdlib.h... yes
 checking for string.h... yes
 checking for memory.h... yes
 checking for strings.h... yes
 checking for inttypes.h... yes
 checking for stdint.h... yes
 checking for unistd.h... yes
 checking mpi.h usability... no
 checking mpi.h presence... no
 checking for mpi.h... no
 Try to find libmpi or libmpich ...
 checking for main in -lmpi... no
 libmpi not found. exiting...
 ERROR: configuration failed for package ŒRmpi¹
 * removing Œ/usr/local/lib64/R/library/Rmpi¹

 So obviously libmpi is not found. Doing a locate search for libmpi
 (straight after updatedb) shows:

 zytosrv01dmi:/home/unger/R_projects/OS # locate libmpi
 /opt/mpich/ch-p4/lib64/libmpich.a
 /opt/mpich/ch-p4/lib64/libmpichf90.a
 /opt/mpich/ch-p4/lib64/libmpichf90nc.a
 /opt/mpich/ch-p4/lib64/libmpichfarg.a
 /opt/mpich/ch-p4/lib64/libmpichfsup.a
 /opt/mpich/ch-p4mpd/lib64/libmpich.a
 /opt/mpich/ch-p4mpd/lib64/libmpichf90.a
 /opt/mpich/ch-p4mpd/lib64/libmpichf90nc.a
 

Re: [R] Still have problems with tcltk in R 64 bit

[Uwe Ligges]

Well, the R code in package tcltk for startup under Windows is, as you 
could have found out yourself easily:


.onLoad - function(lib, pkg)
{
packageStartupMessage(Loading Tcl/Tk interface ...,
  domain = R-tcltk, appendLF = FALSE)
if(!nzchar(tclbin - Sys.getenv(MY_TCLTK))) {
tclbin - file.path(R.home(), Tcl,
if(.Machine$sizeof.pointer == 8) bin64 
else bin)

if(!file.exists(tclbin))
stop(Tcl/Tk support files were not installed, call.=FALSE)
if(.Machine$sizeof.pointer == 8) {
lib64 - gsub(\\, /, file.path(R.home(), Tcl, lib64),
  fixed=TRUE)
Sys.setenv(TCLLIBPATH = lib64)
}
}
library.dynam(tcltk, pkg, lib, DLLpath = tclbin)
.C(tcltk_start, PACKAGE=tcltk)
addTclPath(system.file(exec, package = tcltk))
packageStartupMessage( , done, domain = R-tcltk)
invisible()
}


This tells us that if you do not have MY_TCLTK defined on startup of 
R, you probably forgot to select the tcltk files for 64 bit from the 
installer when installing your version of R.


Uwe Ligges




On 15.06.2011 14:28, Arnaud Mosnier wrote:

I agree that this is a really outdated source but I did not find the
way to tell R using correctly the tcl version included (at least for
the 64 bit version).
If I remove the environment variables, things work for R 32 bit (it
uses the tcl version included), but it does not work in R 64 bit.

Where are the configuration files used to define the path to each tcl version ?

Arnaud

2011/6/14 Uwe Liggeslig...@statistik.tu-dortmund.de:



On 14.06.2011 22:01, Arnaud Mosnier wrote:


I achieve to make tcltk work on R 64 installing Active tcltk8.5 64bit
version then setting windows environment variables as in
http://www.sciviews.org/_rgui/tcltk/InstallRTclTk.html.


Don't read outdated sources but the manuals.

The R binary distribution comes with tcltk under Windows (in ${R_HOME}/tcl)
for both 32-bit and 64-bit and will user a different tcl if you set
environment variables.

Hence the easiest thing is just to tell R not to use your otehrwise set
environment variabes and use its own tcl version.

Uwe Ligges




But now, it uses only this 64 bit version and thus do not work anymore
in R 32 bit !

In my case, it solves my problem as I will probably use only R 64bit
but I do not like to end with an half solution.

Arnaud

2011/6/14 Peter Langfelderpeter.langfel...@gmail.com:


On Tue, Jun 14, 2011 at 12:47 PM, Adrienne Woottenamwoo...@ncsu.edu
  wrote:


Taking a quick look for it, it seems that they have replaced it with
tcltk2.  I just did the installation with the same version in windows
and it auto loaded the tcltk package and I never installed that
package to begin with.  I would try it with tcltk2 and see if you get
the package to install appropriately.  I'm not sure why tcltk isn't on
CRAN anymore, it makes no sense not to have both tcltk and tcltk2, but
here's hoping this helps you out.

A


FWIF, tcltk still exists but is now part of the standard R
distribution (core packages is the term I think?) and as such is
installed automatically when you install R. Therefore it is not
available from CRAN.

Peter



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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.




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


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

Re: [R] Standard deviation and Mean

[Daniel Malter]

I post my replies through nabble. The second one, I can do. However, I would
assume that subscribers would not only see my reply, but also the original
reply, since the forum and email programs/platforms provide threaded msg-ing
these days, or not? 

The first seems to be an either/or option in nabble, either private msg or
for everyone to see, but not both.

Da.


Uwe Ligges-3 wrote:
 
 On 14.06.2011 22:29, Daniel Malter wrote:
 Hi,

 pick up any introductory manual of which there are many online. It so
 happens that the functions for mean and sd are called mean() and sd(). If
 you want to know how to use them type ?mean or ?sd in the R-prompt and
 hit
 enter.
 
 
 ... and some rants back. Can you please
 
 1. also reply to the original sender who may not be subscribed to the 
 list and hence never receives the answer
 2. quote the messages you are referring to, mailing list readers of this 
 R-help *mailing list* won't see it otherwise
 
 Uwe Ligges
 
 
 
 
 
 Daniel

 --
 View this message in context:
 http://r.789695.n4.nabble.com/Standard-deviation-and-Mean-tp3597521p3597734.html
 Sent from the R help mailing list archive at Nabble.com.

 __
 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.
 
 __
 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.
 

--
View this message in context: 
http://r.789695.n4.nabble.com/Standard-deviation-and-Mean-tp3597521p3599404.html
Sent from the R help mailing list archive at Nabble.com.

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

Re: [R] Legend in lattice

[Walmes Zeviani]

Júlio,

Your code is not reproducible, you doesn't provide any data. So I did a
minimal code that illustrates a possible procedure is the following

n - 30
da - data.frame(x=runif(n), y=runif(n), z=runif(n))
da$z - cut(da$z, seq(0,1,0.25))

require(lattice)
xyplot(y~x, da, cex=as.numeric(da$z), col=1,
   key=list(points=list(cex=1:4, pch=1), text=list(levels(da$z)),
columns=4))

Bests,
Walmes.

==
Walmes Marques Zeviani
LEG (Laboratório de Estatística e Geoinformação, 25.450418 S, 49.231759 W)
Departamento de Estatística - Universidade Federal do Paraná
fone: (+55) 41 3361 3573
VoIP: (3361 3600) 1053 1173
e-mail: wal...@ufpr.br
twitter: @walmeszeviani
homepage: http://www.leg.ufpr.br/~walmes
linux user number: 531218
==

[[alternative HTML version deleted]]

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

Re: [R] Using MLE Method to Estimate Regression Coefficients

[Prof. John C Nash]

The error msg puts it quite clearly -- the initial parameters 1,1,1,1 are 
inadmissible for
your function. You need to have valid initial parameters for the variable 
metric method
(option BFGS).

This is one of the main problems users have with any optimization method. It is 
ALWAYS a
good idea to actually evaluate your function outside of the optimizer i.e., 
simply put in
the initial parameters and find out what function value you get.

It should also be noted (as the package optimx does) that the VM and CG based 
methods
really don't do very well without analytic gradients.

JN


On 06/15/2011 06:00 AM, r-help-requ...@r-project.org wrote:
 Message: 46
 Date: Tue, 14 Jun 2011 13:04:55 -0400 (GMT-04:00)
 From: boyla...@earthlink.net
 To: r-help@r-project.org
 Subject: [R] Using MLE Method to Estimate Regression Coefficients
 Message-ID:
   
 11895593.1308071096125.javamail.r...@mswamui-andean.atl.sa.earthlink.net
   
 Content-Type: text/plain; charset=utf-8
 
 Good Afternoon,
 
   I am relatively new to R and have been trying to figure out how to estimate 
 regression coefficients using the MLE method.  Some background: I am trying 
 to examine scenarios in which certain estimators might be preferred to 
 others, starting with MLE.  I understand that MLE will (should) produce the 
 same results as Ordinary Least Squares if the assumption of normality holds. 
 That said, in the following code (my apologies up front for any lack of 
 elegance) I use the data from the printing press study (commonly used in the 
 QE and stats literature) to develop first and second order models using OLS.  
 Then, using some code I found online, I tried to use MLE to do the same 
 thing. However, I cannot get it to work, as I get an error in my attempt to 
 use the optim function.  I have been studying the optim function in R; I have 
 also explored the use of MLE in the R documentation via the stats4, MASS, and 
 a few other packages but to little avail.  My questions are as follows:
 
 1) Is there a particular error in the MLE code below that I am just not 
 seeing?
 2) Is there a simpler, more direct, or otherwise better way to approach it?
 3) Which package should I use for MLE regression?
 
 Sorry for the length and thanks in advance for any assistance someone might 
 have; I know your time is valuable.  I have pasted my code below but have 
 also attached as a .txt file.
 
 v/r,
 Greg
 Doctoral Student, 
 Dept. of Industrial Eng, Clemson University

[snip]
 
 # Now let's use the above function to estimate the model. 
 model - optim(c(1,1,1,1), llik.regress, method=BFGS, 
 control=list(fnscale=-1),
  hessian=TRUE)
 Error in optim(c(1, 1, 1, 1), llik.regress, method = BFGS, control = 
 list(fnscale = -1),  : 
   initial value in 'vmmin' is not finite
  
 
 --

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Re: [R] Rmpi installation

[Hugo Mildenberger]

Kristian,

I just tried that particular command here on a Gentoo system with 
openmpi-1.5.3 installed, because I wondered why the Rmpi configure 
script tests for main function in a shared library ... 

But I got a completly different output from configure. While the 
linker succeeds here, the package load test does not. However, on your 
system, may be the openmpi installation really is a kinda private one of 
gcc? I heard gcc makes use of openmpi internally. So is openmpi really 
installed? 

I just recognize that you are trying to use Rmpi_0.5-4.tar.gz while current 
version on  CRAN is Rmpi_0.5-9.tar.gz.

Best

Hugo

R CMD INSTALL --configure-args=--with-Rmpi-include=/usr/include  
--with-Rmpi-libpath=/usr/lib64/openmpi --with-Rmpi-type=OPENMPI 
Rmpi_0.5-9.tar.gz
* installing to library ‘/home/hm/R/x86_64-pc-linux-gnu-library/2.13’
* installing *source* package ‘Rmpi’ ...
checking for openpty in -lutil... no
checking for main in -lpthread... no
configure: creating ./config.status
config.status: creating src/Makevars
** libs
** libs
x86_64-pc-linux-gnu-gcc -std=gnu99 -I/usr/lib64/R/include -DPACKAGE_NAME=\\ 
-DPACKAGE_TARNAME=\\ -DPACKAGE_VERSION=\\ -
DPACKAGE_STRING=\\ -DPACKAGE_BUGREPORT=\\ -I/usr/include  -DMPI2 -DOPENMPI 
-I/usr/local/include-fpic  -O3 -pipe -march=core2 -
mtune=core2 -ggdb -c RegQuery.c -o RegQuery.o
x86_64-pc-linux-gnu-gcc -std=gnu99 -I/usr/lib64/R/include -DPACKAGE_NAME=\\ 
-DPACKAGE_TARNAME=\\ -DPACKAGE_VERSION=\\ -
DPACKAGE_STRING=\\ -DPACKAGE_BUGREPORT=\\ -I/usr/include  -DMPI2 -DOPENMPI 
-I/usr/local/include-fpic  -O3 -pipe -march=core2 -
mtune=core2 -ggdb -c Rmpi.c -o Rmpi.o
x86_64-pc-linux-gnu-gcc -std=gnu99 -I/usr/lib64/R/include -DPACKAGE_NAME=\\ 
-DPACKAGE_TARNAME=\\ -DPACKAGE_VERSION=\\ -
DPACKAGE_STRING=\\ -DPACKAGE_BUGREPORT=\\ -I/usr/include  -DMPI2 -DOPENMPI 
-I/usr/local/include-fpic  -O3 -pipe -march=core2 -
mtune=core2 -ggdb -c conversion.c -o conversion.o
x86_64-pc-linux-gnu-gcc -std=gnu99 -I/usr/lib64/R/include -DPACKAGE_NAME=\\ 
-DPACKAGE_TARNAME=\\ -DPACKAGE_VERSION=\\ -
DPACKAGE_STRING=\\ -DPACKAGE_BUGREPORT=\\ -I/usr/include  -DMPI2 -DOPENMPI 
-I/usr/local/include-fpic  -O3 -pipe -march=core2 -
mtune=core2 -ggdb -c internal.c -o internal.o
x86_64-pc-linux-gnu-gcc -std=gnu99 -shared -Wl,-O1 -Wl,--as-needed -o Rmpi.so 
RegQuery.o Rmpi.o conversion.o internal.o -L/usr/lib64/openmpi -lmpi 
-L/usr/lib64/R/lib -lR
installiert nach /home/hm/R/x86_64-pc-linux-gnu-library/2.13/Rmpi/libs
[...]
  ompi_mpi_init: orte_init failed
  -- Returned Not found (-13) instead of Success (0)


On Wednesday 15 June 2011 15:19:22 Unger, Kristian, Dr. wrote:
 Thank you very much Hugo. Using the command as suggested results exactly
 the same error:
 
 # R CMD INSTALL
 --configure-args=--with-Rmpi-include=/usr/lib64/mpi/gcc/openmpi/include
 --with-Rmpi-libpath=/usr/lib64/mpi/gcc/openmpi/lib64
 --with-Rmpi-type=OPENMPI Rmpi_0.5-4.tar.gz
 * installing to library ‘/usr/local/lib64/R/library’
 * installing *source* package ‘Rmpi’ ...
 checking for gcc... gcc
 checking for C compiler default output file name... a.out
 checking whether the C compiler works... yes
 checking whether we are cross compiling... no
 checking for suffix of executables...
 checking for suffix of object files... o
 checking whether we are using the GNU C compiler... yes
 checking whether gcc accepts -g... yes
 checking for gcc option to accept ISO C89... none needed
 checking how to run the C preprocessor... gcc -E
 checking for grep that handles long lines and -e... /usr/bin/grep
 checking for egrep... /usr/bin/grep -E
 checking for ANSI C header files... yes
 checking for sys/types.h... yes
 checking for sys/stat.h... yes
 checking for stdlib.h... yes
 checking for string.h... yes
 checking for memory.h... yes
 checking for strings.h... yes
 checking for inttypes.h... yes
 checking for stdint.h... yes
 checking for unistd.h... yes
 checking mpi.h usability... no
 checking mpi.h presence... no
 checking for mpi.h... no
 Try to find libmpi or libmpich ...
 checking for main in -lmpi... no
 libmpi not found. exiting...
 ERROR: configuration failed for package ‘Rmpi’
 * removing ‘/usr/local/lib64/R/library/Rmpi’
 
 
 Best wishes
 
 Kristian
 
 
 Dr. Kristian Unger
 
 
 Arbeitsgruppenleiter Integrative Biologie / Head of Integrative Biology
 Group
 Abteilung für Strahlenzytogenetik / Research Unit of Radiation
 Cytogenetics
 
 Tel.: +49-89-3187-3515
 
 Mob.: +49-160-90641879
 
 
 
 
 
 Am 15.06.11 15:12 schrieb Hugo Mildenberger unter
 hugo.mildenber...@web.de:
 
 Hmm,
 
 looks like there was a trailing blank after the backslash and before end
 of line,
 resulting in --with-Rmpi-libpath possibly not recognised:
 
\--with-Rmpi-libpath=/usr/lib64/mpi/gcc/openmpi/lib64 \
 
 I also doubt there is real need to escape newlines within a string. But
 another
 possible problem source is that according to R CMD INSTALL --help, the
 parameter
 is called  

Re: [R] Rmpi installation

[Unger, Kristian, Dr.]

Thanks Hugo.

I am pretty sure openmpi is installed:

# zypper se openmpi
Loading repository data...
Reading installed packages...

S | Name  | Summary | Type
--+---+-+---
i | openmpi   | A powerful implementaion of MPI | package
  | openmpi   | A powerful implementaion of MPI | srcpackage
i | openmpi-devel | A powerful implementaion of MPI | package



I got the same error message with the latest version available. The reason
why I took the somewhat older version is that I wanted to make sure that
it is not related to any libraries used by the newest version.

Best wishes

Kristian


Dr. Kristian Unger


Arbeitsgruppenleiter Integrative Biologie / Head of Integrative Biology
Group
Abteilung für Strahlenzytogenetik / Research Unit of Radiation
Cytogenetics

Tel.: +49-89-3187-3515

Mob.: +49-160-90641879





Am 15.06.11 16:16 schrieb Hugo Mildenberger unter
hugo.mildenber...@web.de:

Kristian,

I just tried that particular command here on a Gentoo system with
openmpi-1.5.3 installed, because I wondered why the Rmpi configure
script tests for main function in a shared library ...

But I got a completly different output from configure. While the
linker succeeds here, the package load test does not. However, on your
system, may be the openmpi installation really is a kinda private one of
gcc? I heard gcc makes use of openmpi internally. So is openmpi really
installed?

I just recognize that you are trying to use Rmpi_0.5-4.tar.gz while
current
version on  CRAN is Rmpi_0.5-9.tar.gz.

Best

Hugo

R CMD INSTALL --configure-args=--with-Rmpi-include=/usr/include
--with-Rmpi-libpath=/usr/lib64/openmpi --with-Rmpi-type=OPENMPI
Rmpi_0.5-9.tar.gz
* installing to library ‘/home/hm/R/x86_64-pc-linux-gnu-library/2.13’
* installing *source* package ‘Rmpi’ ...
checking for openpty in -lutil... no
checking for main in -lpthread... no
configure: creating ./config.status
config.status: creating src/Makevars
** libs
** libs
x86_64-pc-linux-gnu-gcc -std=gnu99 -I/usr/lib64/R/include
-DPACKAGE_NAME=\\ -DPACKAGE_TARNAME=\\ -DPACKAGE_VERSION=\\ -
DPACKAGE_STRING=\\ -DPACKAGE_BUGREPORT=\\ -I/usr/include  -DMPI2
-DOPENMPI -I/usr/local/include-fpic  -O3 -pipe -march=core2 -
mtune=core2 -ggdb -c RegQuery.c -o RegQuery.o
x86_64-pc-linux-gnu-gcc -std=gnu99 -I/usr/lib64/R/include
-DPACKAGE_NAME=\\ -DPACKAGE_TARNAME=\\ -DPACKAGE_VERSION=\\ -
DPACKAGE_STRING=\\ -DPACKAGE_BUGREPORT=\\ -I/usr/include  -DMPI2
-DOPENMPI -I/usr/local/include-fpic  -O3 -pipe -march=core2 -
mtune=core2 -ggdb -c Rmpi.c -o Rmpi.o
x86_64-pc-linux-gnu-gcc -std=gnu99 -I/usr/lib64/R/include
-DPACKAGE_NAME=\\ -DPACKAGE_TARNAME=\\ -DPACKAGE_VERSION=\\ -
DPACKAGE_STRING=\\ -DPACKAGE_BUGREPORT=\\ -I/usr/include  -DMPI2
-DOPENMPI -I/usr/local/include-fpic  -O3 -pipe -march=core2 -
mtune=core2 -ggdb -c conversion.c -o conversion.o
x86_64-pc-linux-gnu-gcc -std=gnu99 -I/usr/lib64/R/include
-DPACKAGE_NAME=\\ -DPACKAGE_TARNAME=\\ -DPACKAGE_VERSION=\\ -
DPACKAGE_STRING=\\ -DPACKAGE_BUGREPORT=\\ -I/usr/include  -DMPI2
-DOPENMPI -I/usr/local/include-fpic  -O3 -pipe -march=core2 -
mtune=core2 -ggdb -c internal.c -o internal.o
x86_64-pc-linux-gnu-gcc -std=gnu99 -shared -Wl,-O1 -Wl,--as-needed -o
Rmpi.so RegQuery.o Rmpi.o conversion.o internal.o -L/usr/lib64/openmpi
-lmpi
-L/usr/lib64/R/lib -lR
installiert nach /home/hm/R/x86_64-pc-linux-gnu-library/2.13/Rmpi/libs
[...]
  ompi_mpi_init: orte_init failed
  -- Returned Not found (-13) instead of Success (0)


On Wednesday 15 June 2011 15:19:22 Unger, Kristian, Dr. wrote:
 Thank you very much Hugo. Using the command as suggested results exactly
 the same error:

 # R CMD INSTALL
 --configure-args=--with-Rmpi-include=/usr/lib64/mpi/gcc/openmpi/include
 --with-Rmpi-libpath=/usr/lib64/mpi/gcc/openmpi/lib64
 --with-Rmpi-type=OPENMPI Rmpi_0.5-4.tar.gz
 * installing to library ‘/usr/local/lib64/R/library’
 * installing *source* package ‘Rmpi’ ...
 checking for gcc... gcc
 checking for C compiler default output file name... a.out
 checking whether the C compiler works... yes
 checking whether we are cross compiling... no
 checking for suffix of executables...
 checking for suffix of object files... o
 checking whether we are using the GNU C compiler... yes
 checking whether gcc accepts -g... yes
 checking for gcc option to accept ISO C89... none needed
 checking how to run the C preprocessor... gcc -E
 checking for grep that handles long lines and -e... /usr/bin/grep
 checking for egrep... /usr/bin/grep -E
 checking for ANSI C header files... yes
 checking for sys/types.h... yes
 checking for sys/stat.h... yes
 checking for stdlib.h... yes
 checking for string.h... yes
 checking for memory.h... yes
 checking for strings.h... yes
 checking for inttypes.h... yes
 checking for stdint.h... yes
 checking for unistd.h... yes
 checking mpi.h usability... no
 checking mpi.h presence... no
 checking for 

Re: [R] Still have problems with tcltk in R 64 bit

[Arnaud Mosnier]

I was pretty sure to have installed tcltk files for 64 bit from the
installer, but to be sure ...
- I removed previously created Environment variables (MY_TCLTK)
- I reinstalled R one more time ...  this time with the full
installation option

and ... it does not work (still works with R 32bit) !

- I also tried to define directly the path to the tcl version included
with R ... as the code you provided does ... thus MY_TCLTK =
C:\Program Files\R\R-2.13.0\Tcl\bin64

... still the same error in R 64 bit, and if I want to load tcltk in R
32, it gives me the following error (normal ... it tried to load a 64
bit application in a 32 bit environment).

Loading Tcl/Tk interface ...Error : .onLoad failed in loadNamespace()
for 'tcltk', details:
  call: inDL(x, as.logical(local), as.logical(now), ...)
  error: unable to load shared object 'C:/Program
Files/R/R-2.13.0/library/tcltk/libs/i386/tcltk.dll':
  LoadLibrary failure:  %1 is not a valid Win32 application.

Error: package/namespace load failed for 'tcltk'

I believe that the problem may come from the tcltk.dll file but I do
not understand how it can be broken each time when I installed other
versions ...

Arnaud


2011/6/15 Uwe Ligges lig...@statistik.tu-dortmund.de:
 Well, the R code in package tcltk for startup under Windows is, as you could
 have found out yourself easily:

 .onLoad - function(lib, pkg)
 {
    packageStartupMessage(Loading Tcl/Tk interface ...,
                          domain = R-tcltk, appendLF = FALSE)
    if(!nzchar(tclbin - Sys.getenv(MY_TCLTK))) {
        tclbin - file.path(R.home(), Tcl,
                            if(.Machine$sizeof.pointer == 8) bin64 else
 bin)
        if(!file.exists(tclbin))
            stop(Tcl/Tk support files were not installed, call.=FALSE)
        if(.Machine$sizeof.pointer == 8) {
            lib64 - gsub(\\, /, file.path(R.home(), Tcl, lib64),
                          fixed=TRUE)
            Sys.setenv(TCLLIBPATH = lib64)
        }
    }
    library.dynam(tcltk, pkg, lib, DLLpath = tclbin)
    .C(tcltk_start, PACKAGE=tcltk)
    addTclPath(system.file(exec, package = tcltk))
    packageStartupMessage( , done, domain = R-tcltk)
    invisible()
 }


 This tells us that if you do not have MY_TCLTK defined on startup of R,
 you probably forgot to select the tcltk files for 64 bit from the installer
 when installing your version of R.

 Uwe Ligges




 On 15.06.2011 14:28, Arnaud Mosnier wrote:

 I agree that this is a really outdated source but I did not find the
 way to tell R using correctly the tcl version included (at least for
 the 64 bit version).
 If I remove the environment variables, things work for R 32 bit (it
 uses the tcl version included), but it does not work in R 64 bit.

 Where are the configuration files used to define the path to each tcl
 version ?

 Arnaud

 2011/6/14 Uwe Liggeslig...@statistik.tu-dortmund.de:


 On 14.06.2011 22:01, Arnaud Mosnier wrote:

 I achieve to make tcltk work on R 64 installing Active tcltk8.5 64bit
 version then setting windows environment variables as in
 http://www.sciviews.org/_rgui/tcltk/InstallRTclTk.html.

 Don't read outdated sources but the manuals.

 The R binary distribution comes with tcltk under Windows (in
 ${R_HOME}/tcl)
 for both 32-bit and 64-bit and will user a different tcl if you set
 environment variables.

 Hence the easiest thing is just to tell R not to use your otehrwise set
 environment variabes and use its own tcl version.

 Uwe Ligges



 But now, it uses only this 64 bit version and thus do not work anymore
 in R 32 bit !

 In my case, it solves my problem as I will probably use only R 64bit
 but I do not like to end with an half solution.

 Arnaud

 2011/6/14 Peter Langfelderpeter.langfel...@gmail.com:

 On Tue, Jun 14, 2011 at 12:47 PM, Adrienne Woottenamwoo...@ncsu.edu
  wrote:

 Taking a quick look for it, it seems that they have replaced it with
 tcltk2.  I just did the installation with the same version in windows
 and it auto loaded the tcltk package and I never installed that
 package to begin with.  I would try it with tcltk2 and see if you get
 the package to install appropriately.  I'm not sure why tcltk isn't on
 CRAN anymore, it makes no sense not to have both tcltk and tcltk2, but
 here's hoping this helps you out.

 A

 FWIF, tcltk still exists but is now part of the standard R
 distribution (core packages is the term I think?) and as such is
 installed automatically when you install R. Therefore it is not
 available from CRAN.

 Peter


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 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide
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 PLEASE do read the posting guide
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Re: [R] Obtaining OLAP cubes using R

[Eric Lecoutre]

Ho Saravana,

I did have nearly the same issue some months ago -- you will find below a
good starting point.
I am quite sure there is a better way to achieve it or with better code,
but this should work!

Kind regards,

Eric


-
generateCube - function(x,simplify=TRUE,onlyComb=FALSE,...){
stopifnot(require(gtools))
xdf=as.data.frame(x)
out=vector(length=ncol(xdf),mode=list)
for (i in 1:ncol(xdf)){ # dimensions of the cube
 icomb - combinations(length(colnames(xdf)),i,colnames(xdf))
 out[[i]] - vector(length=nrow(icomb),mode=list)
 for (j in 1:nrow(icomb)){
  ijargs - icomb[j,]
  names(out[[i]])[j] - paste(ijargs,collapse=*)
  tmp -
   with(
   xdf,
   eval(parse(text=
paste(table(,paste(ijargs,collapse=,),),sep=)
))
  )
  if (simplify  i1) tmp-as.data.frame(ftable(tmp))
  out[[i]][[j]] - tmp
  }
 }
return(out)
}
M=diag(3)
M[1,2]=2
M - rbind(M,M[1,])
colnames(M) - paste(V,1:ncol(M),sep=)
out=generateCube(M)
out2=generateCube(M,simplify=FALSE)

out[[2]][[V1*V3]]
out2[[2]][[V1*V3]][0,1]
-




On 14 June 2011 14:59, Saravanan saravanan.thirumuruganat...@gmail.comwrote:

 Hello All,

 I have a dataset and I wish to obtain all possible data cuboids from it
 using R . For eg if my data frame is :

 ABC
 111
 121
 221

 The output intended is :
 A=1
 A=2
 B=1
 B=2
 C=1
 A=1,B=1
 A=1,B=2
 A=2,B=2
 A=1,C=1
 A=2,C=1
 B=1,C=1
 B=2,C=1
 A=1,B=1,C=1
 A=1,B=2,C=1
 A=2,B=2,C=1

 Are there any function(s) to do this in R ? I tried a combination of
 expand.grid and combn but the resulting code was very ugly and needed lot of
 hacks to make it work. I also tried to check the code for arules (which
 constructs similar itemsets) but unfortunately its code is in C and I am
 not very familiar in writing R extensions. Any pointers to functions will be
 much appreciated.

 Regards,
 Saravanan

 __
 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.htmlhttp://www.r-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.




-- 
Eric Lecoutre
Consultant - Business  Decision
Business Intelligence  Customer Intelligence

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

[R] Column of numbers added to dataframe when saving with read.csv

[Paolo Rossi]

I have a dataframe object having the following structure

 FinalOutput[1:3,]
 GasDays 2011-03-31 2010-09-30 2010-10-31 2010-11-30 2010-12-31
2011-01-31 2011-02-28
1 2006-10-01  217303553  221205033  222824639  217016511  216093460
216477468  216834021
2 2006-10-02  231158527  234565250  236004109  231467851  230100639
230079907  230734064
3 2006-10-03  282062314  285427832  286372163  282532055  280930498
281155966  281124614
After using
write.table(FinalOutput, paste(ModelComparison.csv, sep = ''), sep = ',')

the out put I get on the csv is



 GasDays 31/03/2011 30/09/2010 31/10/2010  30/11/2010
 31/12/201031/01/2011   28/02/2011 31/03/2011
1  01/10/2006 217303553.3 221205032.6
222824638.7   217016510.8
216093460   216477467.9   216834021217303553.3
2  02/10/2006 231158527.1 234565249.7
 236004108.7   231467850.7   230100639.1   230079907.4
230734064.4   231158527.1
3  03/10/2006 282062314.5 285427831.6
286372163   282532055.2   280930497.7
281155966 281124613.8   282062314.5


so essentially one column  full  of 1,2,3, ... is added to the file when
saving it.

Can someone pelase help me to get rid of it?

Thanks

Paolo




On 14 June 2011 13:59, Saravanan saravanan.thirumuruganat...@gmail.comwrote:

 Hello All,

 I have a dataset and I wish to obtain all possible data cuboids from it
 using R . For eg if my data frame is :

 ABC
 111
 121
 221

 The output intended is :
 A=1
 A=2
 B=1
 B=2
 C=1
 A=1,B=1
 A=1,B=2
 A=2,B=2
 A=1,C=1
 A=2,C=1
 B=1,C=1
 B=2,C=1
 A=1,B=1,C=1
 A=1,B=2,C=1
 A=2,B=2,C=1

 Are there any function(s) to do this in R ? I tried a combination of
 expand.grid and combn but the resulting code was very ugly and needed lot of
 hacks to make it work. I also tried to check the code for arules (which
 constructs similar itemsets) but unfortunately its code is in C and I am
 not very familiar in writing R extensions. Any pointers to functions will be
 much appreciated.

 Regards,
 Saravanan

 __
 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.htmlhttp://www.r-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.


[[alternative HTML version deleted]]

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

Re: [R] Column of numbers added to dataframe when saving with read.csv

[ONKELINX, Thierry]

Reading the helpfile (as the posting guide asks you to do) of write.table will 
solve your problem.

 -Oorspronkelijk bericht-
 Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
 Namens Paolo Rossi
 Verzonden: woensdag 15 juni 2011 16:52
 Aan: r-help@r-project.org
 Onderwerp: [R] Column of numbers added to dataframe when saving with
 read.csv
 
 I have a dataframe object having the following structure
 
  FinalOutput[1:3,]
  GasDays 2011-03-31 2010-09-30 2010-10-31 2010-11-30 2010-12-31
 2011-01-31 2011-02-28
 1 2006-10-01  217303553  221205033  222824639  217016511  216093460
 216477468  216834021
 2 2006-10-02  231158527  234565250  236004109  231467851  230100639
 230079907  230734064
 3 2006-10-03  282062314  285427832  286372163  282532055  280930498
 281155966  281124614
 After using
 write.table(FinalOutput, paste(ModelComparison.csv, sep = ''), sep = ',')
 
 the out put I get on the csv is
 
 
 
  GasDays 31/03/2011 30/09/2010 31/10/2010  30/11/2010
  31/12/201031/01/2011   28/02/2011 31/03/2011
 1  01/10/2006 217303553.3 221205032.6
 222824638.7   217016510.8
 216093460   216477467.9   216834021217303553.3
 2  02/10/2006 231158527.1 234565249.7
  236004108.7   231467850.7   230100639.1   230079907.4
 230734064.4   231158527.1
 3  03/10/2006 282062314.5 285427831.6
 286372163   282532055.2   280930497.7
 281155966 281124613.8   282062314.5
 
 
 so essentially one column  full  of 1,2,3, ... is added to the file when 
 saving it.
 
 Can someone pelase help me to get rid of it?
 
 Thanks
 
 Paolo
 
 
 
 
 On 14 June 2011 13:59, Saravanan
 saravanan.thirumuruganat...@gmail.comwrote:
 
  Hello All,
 
  I have a dataset and I wish to obtain all possible data cuboids from
  it using R . For eg if my data frame is :
 
  ABC
  111
  121
  221
 
  The output intended is :
  A=1
  A=2
  B=1
  B=2
  C=1
  A=1,B=1
  A=1,B=2
  A=2,B=2
  A=1,C=1
  A=2,C=1
  B=1,C=1
  B=2,C=1
  A=1,B=1,C=1
  A=1,B=2,C=1
  A=2,B=2,C=1
 
  Are there any function(s) to do this in R ? I tried a combination of
  expand.grid and combn but the resulting code was very ugly and needed
  lot of hacks to make it work. I also tried to check the code for
  arules (which constructs similar itemsets) but unfortunately its
  code is in C and I am not very familiar in writing R extensions. Any
  pointers to functions will be much appreciated.
 
  Regards,
  Saravanan
 
  __
  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.htmlhttp://www.r-project.org/p
  osting-guide.html and provide commented, minimal, self-contained,
  reproducible code.
 
 
   [[alternative HTML version deleted]]
 
 __
 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.

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

Re: [R] Column of numbers added to dataframe when saving with read.csv

[Sarah Goslee]

You need to add row.names=FALSE to your write.table() statement.

This and other useful options are documented in the help. And you'll
notice that that first column of row names appears in your R output as
well.

Sarah

On Wed, Jun 15, 2011 at 10:51 AM, Paolo Rossi
statmailingli...@googlemail.com wrote:
 I have a dataframe object having the following structure

  FinalOutput[1:3,]
     GasDays 2011-03-31 2010-09-30 2010-10-31 2010-11-30 2010-12-31
 2011-01-31 2011-02-28
 1 2006-10-01  217303553  221205033  222824639  217016511  216093460
 216477468  216834021
 2 2006-10-02  231158527  234565250  236004109  231467851  230100639
 230079907  230734064
 3 2006-10-03  282062314  285427832  286372163  282532055  280930498
 281155966  281124614
 After using
 write.table(FinalOutput, paste(ModelComparison.csv, sep = ''), sep = ',')

 the out put I get on the csv is



  GasDays 31/03/2011 30/09/2010     31/10/2010      30/11/2010
  31/12/2010            31/01/2011       28/02/2011         31/03/2011
 1              01/10/2006 217303553.3 221205032.6
 222824638.7       217016510.8
 216093460       216477467.9       216834021            217303553.3
 2              02/10/2006 231158527.1 234565249.7
  236004108.7       231467850.7       230100639.1       230079907.4
 230734064.4       231158527.1
 3              03/10/2006 282062314.5 285427831.6
 286372163       282532055.2       280930497.7
 281155966         281124613.8       282062314.5


 so essentially one column  full  of 1,2,3, ... is added to the file when
 saving it.

 Can someone pelase help me to get rid of it?

 Thanks

 Paolo

-- 
Sarah Goslee
http://www.functionaldiversity.org

__
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.

Re: [R] Column of numbers added to dataframe when saving with read.csv

[Ivan Calandra]

Hi Paolo,

Not sure to understand you well, but try with row.names=FALSE in your 
call to write.table()


HTH,
Ivan

Le 6/15/2011 16:51, Paolo Rossi a écrit :

I have a dataframe object having the following structure

  FinalOutput[1:3,]
  GasDays 2011-03-31 2010-09-30 2010-10-31 2010-11-30 2010-12-31
2011-01-31 2011-02-28
1 2006-10-01  217303553  221205033  222824639  217016511  216093460
216477468  216834021
2 2006-10-02  231158527  234565250  236004109  231467851  230100639
230079907  230734064
3 2006-10-03  282062314  285427832  286372163  282532055  280930498
281155966  281124614
After using
write.table(FinalOutput, paste(ModelComparison.csv, sep = ''), sep = ',')

the out put I get on the csv is



  GasDays 31/03/2011 30/09/2010 31/10/2010  30/11/2010
  31/12/201031/01/2011   28/02/2011 31/03/2011
1  01/10/2006 217303553.3 221205032.6
222824638.7   217016510.8
216093460   216477467.9   216834021217303553.3
2  02/10/2006 231158527.1 234565249.7
  236004108.7   231467850.7   230100639.1   230079907.4
230734064.4   231158527.1
3  03/10/2006 282062314.5 285427831.6
286372163   282532055.2   280930497.7
281155966 281124613.8   282062314.5


so essentially one column  full  of 1,2,3, ... is added to the file when
saving it.

Can someone pelase help me to get rid of it?

Thanks

Paolo




On 14 June 2011 13:59, Saravanansaravanan.thirumuruganat...@gmail.comwrote:


Hello All,

I have a dataset and I wish to obtain all possible data cuboids from it
using R . For eg if my data frame is :

ABC
111
121
221

The output intended is :
A=1
A=2
B=1
B=2
C=1
A=1,B=1
A=1,B=2
A=2,B=2
A=1,C=1
A=2,C=1
B=1,C=1
B=2,C=1
A=1,B=1,C=1
A=1,B=2,C=1
A=2,B=2,C=1

Are there any function(s) to do this in R ? I tried a combination of
expand.grid and combn but the resulting code was very ugly and needed lot of
hacks to make it work. I also tried to check the code for arules (which
constructs similar itemsets) but unfortunately its code is in C and I am
not very familiar in writing R extensions. Any pointers to functions will be
much appreciated.

Regards,
Saravanan

__
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.htmlhttp://www.r-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.


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



--
Ivan CALANDRA
PhD Student
University of Hamburg
Biozentrum Grindel und Zoologisches Museum
Dept. Mammalogy
Martin-Luther-King-Platz 3
D-20146 Hamburg, GERMANY
+49(0)40 42838 6231
ivan.calan...@uni-hamburg.de

**
http://www.for771.uni-bonn.de
http://webapp5.rrz.uni-hamburg.de/mammals/eng/1525_8_1.php

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

Re: [R] Column of numbers added to dataframe when saving with read.csv

[Paolo Rossi]

Sarah,

That is correct - thanks a lot for this to everyone who replied

Paolo

On 15 June 2011 16:03, Sarah Goslee sarah.gos...@gmail.com wrote:

 You need to add row.names=FALSE to your write.table() statement.

 This and other useful options are documented in the help. And you'll
 notice that that first column of row names appears in your R output as
 well.

 Sarah

 On Wed, Jun 15, 2011 at 10:51 AM, Paolo Rossi
 statmailingli...@googlemail.com wrote:
  I have a dataframe object having the following structure
 
   FinalOutput[1:3,]
  GasDays 2011-03-31 2010-09-30 2010-10-31 2010-11-30 2010-12-31
  2011-01-31 2011-02-28
  1 2006-10-01  217303553  221205033  222824639  217016511  216093460
  216477468  216834021
  2 2006-10-02  231158527  234565250  236004109  231467851  230100639
  230079907  230734064
  3 2006-10-03  282062314  285427832  286372163  282532055  280930498
  281155966  281124614
  After using
  write.table(FinalOutput, paste(ModelComparison.csv, sep = ''), sep =
 ',')
 
  the out put I get on the csv is
 
 
 
   GasDays 31/03/2011 30/09/2010 31/10/2010  30/11/2010
   31/12/201031/01/2011   28/02/2011 31/03/2011
  1  01/10/2006 217303553.3 221205032.6
  222824638.7   217016510.8
  216093460   216477467.9   216834021217303553.3
  2  02/10/2006 231158527.1 234565249.7
   236004108.7   231467850.7   230100639.1   230079907.4
  230734064.4   231158527.1
  3  03/10/2006 282062314.5 285427831.6
  286372163   282532055.2   280930497.7
  281155966 281124613.8   282062314.5
 
 
  so essentially one column  full  of 1,2,3, ... is added to the file when
  saving it.
 
  Can someone pelase help me to get rid of it?
 
  Thanks
 
  Paolo
 
 --
 Sarah Goslee
 http://www.functionaldiversity.org


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Re: [R] Rmpi installation

[Hugo Mildenberger]

Kristian,

these are the usual problems with binary distributions.  

Regarding 

   --with-Rmpi-include=/usr/lib64/mpi/gcc/openmpi/include

and configure output

  checking for mpi.h... no

... so does /usr/lib64/mpi/gcc/openmpi/include really exist? At least,
that appears to be a very unusual place to look for mpi.h (normally
to be found in /usr/include ).
 
And if you try to compile  link the attached demo program: does
the link phase succeed? Compile  link using

$ mpicc mtest.c -o mtest

Presumably you have already tried to run install.packages(Rmpi).


Kind regards

Hugo



On Wednesday 15 June 2011 16:22:07 Unger, Kristian, Dr. wrote:
 Thanks Hugo.
 
 I am pretty sure openmpi is installed:
 
 # zypper se openmpi
 Loading repository data...
 Reading installed packages...
 
 S | Name  | Summary | Type
 --+---+-+---
 i | openmpi   | A powerful implementaion of MPI | package
   | openmpi   | A powerful implementaion of MPI | srcpackage
 i | openmpi-devel | A powerful implementaion of MPI | package
 
 
 
 I got the same error message with the latest version available. The reason
 why I took the somewhat older version is that I wanted to make sure that
 it is not related to any libraries used by the newest version.
 
 Best wishes
 
 Kristian
 
 
 Dr. Kristian Unger
 
 
 Arbeitsgruppenleiter Integrative Biologie / Head of Integrative Biology
 Group
 Abteilung für Strahlenzytogenetik / Research Unit of Radiation
 Cytogenetics
 
 Tel.: +49-89-3187-3515
 
 Mob.: +49-160-90641879
 
 
 
 
 
 Am 15.06.11 16:16 schrieb Hugo Mildenberger unter
 hugo.mildenber...@web.de:
 
 Kristian,
 
 I just tried that particular command here on a Gentoo system with
 openmpi-1.5.3 installed, because I wondered why the Rmpi configure
 script tests for main function in a shared library ...
 
 But I got a completly different output from configure. While the
 linker succeeds here, the package load test does not. However, on your
 system, may be the openmpi installation really is a kinda private one of
 gcc? I heard gcc makes use of openmpi internally. So is openmpi really
 installed?
 
 I just recognize that you are trying to use Rmpi_0.5-4.tar.gz while
 current
 version on  CRAN is Rmpi_0.5-9.tar.gz.
 
 Best
 
 Hugo
 
 R CMD INSTALL --configure-args=--with-Rmpi-include=/usr/include
 --with-Rmpi-libpath=/usr/lib64/openmpi --with-Rmpi-type=OPENMPI
 Rmpi_0.5-9.tar.gz
 * installing to library ‘/home/hm/R/x86_64-pc-linux-gnu-library/2.13’
 * installing *source* package ‘Rmpi’ ...
 checking for openpty in -lutil... no
 checking for main in -lpthread... no
 configure: creating ./config.status
 config.status: creating src/Makevars
 ** libs
 ** libs
 x86_64-pc-linux-gnu-gcc -std=gnu99 -I/usr/lib64/R/include
 -DPACKAGE_NAME=\\ -DPACKAGE_TARNAME=\\ -DPACKAGE_VERSION=\\ -
 DPACKAGE_STRING=\\ -DPACKAGE_BUGREPORT=\\ -I/usr/include  -DMPI2
 -DOPENMPI -I/usr/local/include-fpic  -O3 -pipe -march=core2 -
 mtune=core2 -ggdb -c RegQuery.c -o RegQuery.o
 x86_64-pc-linux-gnu-gcc -std=gnu99 -I/usr/lib64/R/include
 -DPACKAGE_NAME=\\ -DPACKAGE_TARNAME=\\ -DPACKAGE_VERSION=\\ -
 DPACKAGE_STRING=\\ -DPACKAGE_BUGREPORT=\\ -I/usr/include  -DMPI2
 -DOPENMPI -I/usr/local/include-fpic  -O3 -pipe -march=core2 -
 mtune=core2 -ggdb -c Rmpi.c -o Rmpi.o
 x86_64-pc-linux-gnu-gcc -std=gnu99 -I/usr/lib64/R/include
 -DPACKAGE_NAME=\\ -DPACKAGE_TARNAME=\\ -DPACKAGE_VERSION=\\ -
 DPACKAGE_STRING=\\ -DPACKAGE_BUGREPORT=\\ -I/usr/include  -DMPI2
 -DOPENMPI -I/usr/local/include-fpic  -O3 -pipe -march=core2 -
 mtune=core2 -ggdb -c conversion.c -o conversion.o
 x86_64-pc-linux-gnu-gcc -std=gnu99 -I/usr/lib64/R/include
 -DPACKAGE_NAME=\\ -DPACKAGE_TARNAME=\\ -DPACKAGE_VERSION=\\ -
 DPACKAGE_STRING=\\ -DPACKAGE_BUGREPORT=\\ -I/usr/include  -DMPI2
 -DOPENMPI -I/usr/local/include-fpic  -O3 -pipe -march=core2 -
 mtune=core2 -ggdb -c internal.c -o internal.o
 x86_64-pc-linux-gnu-gcc -std=gnu99 -shared -Wl,-O1 -Wl,--as-needed -o
 Rmpi.so RegQuery.o Rmpi.o conversion.o internal.o -L/usr/lib64/openmpi
 -lmpi
 -L/usr/lib64/R/lib -lR
 installiert nach /home/hm/R/x86_64-pc-linux-gnu-library/2.13/Rmpi/libs
 [...]
   ompi_mpi_init: orte_init failed
   -- Returned Not found (-13) instead of Success (0)
 
 
 On Wednesday 15 June 2011 15:19:22 Unger, Kristian, Dr. wrote:
  Thank you very much Hugo. Using the command as suggested results exactly
  the same error:
 
  # R CMD INSTALL
  --configure-args=--with-Rmpi-include=/usr/lib64/mpi/gcc/openmpi/include
  --with-Rmpi-libpath=/usr/lib64/mpi/gcc/openmpi/lib64
  --with-Rmpi-type=OPENMPI Rmpi_0.5-4.tar.gz
  * installing to library ‘/usr/local/lib64/R/library’
  * installing *source* package ‘Rmpi’ ...
  checking for gcc... gcc
  checking for C compiler default output file name... a.out
  checking whether the C compiler works... yes
  checking whether we are cross compiling... no
  

[R] Problem auto.arima() in R

[siddharth arun]

I am using auto.arima() for forecasting.When I am using any in built data
such as AirPassangers it is capturing seasonality. But, If I am entering
data in any other format(in vector form or from an excel sheet) it is not
detecting seasonality.

Is there any specific format in which it detects seasonality or I am doing
some thing wrong?

Does data have to be entered in a specific format?

-- 
Siddharth Arun,
4th Year Undergraduate student
Industrial Engineering and Management,
IIT Kharagpur

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

Re: [R] Rmpi installation

[Unger, Kristian, Dr.]

Dear Hugo

I ran the command with the verbose switch and get the following output:

 mpicc mtest.c -ov mtest
mtest: In function `_start':
/usr/src/packages/BUILD/glibc-2.11.1/csu/../sysdeps/x86_64/elf/start.S:65:
multiple definition of `_start'
/usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../lib64/crt1.o:/usr/src/pack
ages/BUILD/glibc-2.11.1/csu/../sysdeps/x86_64/elf/start.S:65: first
defined here
mtest: In function `_fini':
(.fini+0x0): multiple definition of `_fini'
/usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../lib64/crti.o:initfini.c:(.
fini+0x0): first defined here
mtest:(.rodata+0x0): multiple definition of `_IO_stdin_used'
/usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../lib64/crt1.o:(.rodata.cst4
+0x0): first defined here
mtest: In function `__data_start':
(.data+0x0): multiple definition of `__data_start'
/usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../lib64/crt1.o:(.data+0x0):
first defined here
mtest: In function `main':
(.text+0xec): multiple definition of `main'
/tmp/ccUyk8e9.o:mtest.c:(.text+0x0): first defined here
mtest: In function `_init':
(.init+0x0): multiple definition of `_init'
/usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../lib64/crti.o:initfini.c:(.
init+0x0): first defined here
/usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../x86_64-suse-linux/bin/ld:
error in mtest(.eh_frame); no .eh_frame_hdr table will be created.
collect2: ld returned 1 exit status

Not sure if this makes any sense?


Best wishes

Kristian


Dr. Kristian Unger


Arbeitsgruppenleiter Integrative Biologie / Head of Integrative Biology
Group
Abteilung für Strahlenzytogenetik / Research Unit of Radiation
Cytogenetics

Tel.: +49-89-3187-3515

Mob.: +49-160-90641879





Am 15.06.11 17:25 schrieb Hugo Mildenberger unter
hugo.mildenber...@web.de:

Kristian,

these are the usual problems with binary distributions.

Regarding

   --with-Rmpi-include=/usr/lib64/mpi/gcc/openmpi/include

and configure output

  checking for mpi.h... no

... so does /usr/lib64/mpi/gcc/openmpi/include really exist? At least,
that appears to be a very unusual place to look for mpi.h (normally
to be found in /usr/include ).

And if you try to compile  link the attached demo program: does
the link phase succeed? Compile  link using

$ mpicc mtest.c -o mtest

Presumably you have already tried to run install.packages(Rmpi).


Kind regards

Hugo



On Wednesday 15 June 2011 16:22:07 Unger, Kristian, Dr. wrote:
 Thanks Hugo.

 I am pretty sure openmpi is installed:

 # zypper se openmpi
 Loading repository data...
 Reading installed packages...

 S | Name  | Summary | Type
 --+---+-+---
 i | openmpi   | A powerful implementaion of MPI | package
   | openmpi   | A powerful implementaion of MPI | srcpackage
 i | openmpi-devel | A powerful implementaion of MPI | package



 I got the same error message with the latest version available. The
reason
 why I took the somewhat older version is that I wanted to make sure that
 it is not related to any libraries used by the newest version.

 Best wishes

 Kristian

 
 Dr. Kristian Unger


 Arbeitsgruppenleiter Integrative Biologie / Head of Integrative Biology
 Group
 Abteilung für Strahlenzytogenetik / Research Unit of Radiation
 Cytogenetics

 Tel.: +49-89-3187-3515

 Mob.: +49-160-90641879





 Am 15.06.11 16:16 schrieb Hugo Mildenberger unter
 hugo.mildenber...@web.de:

 Kristian,
 
 I just tried that particular command here on a Gentoo system with
 openmpi-1.5.3 installed, because I wondered why the Rmpi configure
 script tests for main function in a shared library ...
 
 But I got a completly different output from configure. While the
 linker succeeds here, the package load test does not. However, on your
 system, may be the openmpi installation really is a kinda private one
of
 gcc? I heard gcc makes use of openmpi internally. So is openmpi really
 installed?
 
 I just recognize that you are trying to use Rmpi_0.5-4.tar.gz while
 current
 version on  CRAN is Rmpi_0.5-9.tar.gz.
 
 Best
 
 Hugo
 
 R CMD INSTALL --configure-args=--with-Rmpi-include=/usr/include
 --with-Rmpi-libpath=/usr/lib64/openmpi --with-Rmpi-type=OPENMPI
 Rmpi_0.5-9.tar.gz
 * installing to library ‘/home/hm/R/x86_64-pc-linux-gnu-library/2.13’
 * installing *source* package ‘Rmpi’ ...
 checking for openpty in -lutil... no
 checking for main in -lpthread... no
 configure: creating ./config.status
 config.status: creating src/Makevars
 ** libs
 ** libs
 x86_64-pc-linux-gnu-gcc -std=gnu99 -I/usr/lib64/R/include
 -DPACKAGE_NAME=\\ -DPACKAGE_TARNAME=\\ -DPACKAGE_VERSION=\\ -
 DPACKAGE_STRING=\\ -DPACKAGE_BUGREPORT=\\ -I/usr/include  -DMPI2
 -DOPENMPI -I/usr/local/include-fpic  -O3 -pipe -march=core2 -
 mtune=core2 -ggdb -c RegQuery.c -o RegQuery.o
 x86_64-pc-linux-gnu-gcc -std=gnu99 -I/usr/lib64/R/include
 -DPACKAGE_NAME=\\ 

Re: [R] Obtaining OLAP cubes using R

[Saravanan]

Hi Eric,

Your solution looks very close to my own. I also use expand.grid to get 
the cross product though. If I find an alternate way, I will send out my 
solution.

Thanks a lot for your suggestion !

Regards,
Saravanan

On 06/15/2011 09:34 AM, Eric Lecoutre wrote:
 Ho Saravana,
 I did have nearly the same issue some months ago -- you will find 
 below a good starting point.
 I am quite sure there is a better way to achieve it or with better 
 code,  but this should work!
 Kind regards,

 Eric
 -
 generateCube - function(x,simplify=TRUE,onlyComb=FALSE,...){
 stopifnot(require(gtools))
 xdf=as.data.frame(x)
 out=vector(length=ncol(xdf),mode=list)
 for (i in 1:ncol(xdf)){ # dimensions of the cube
  icomb - combinations(length(colnames(xdf)),i,colnames(xdf))
  out[[i]] - vector(length=nrow(icomb),mode=list)
  for (j in 1:nrow(icomb)){
   ijargs - icomb[j,]
   names(out[[i]])[j] - paste(ijargs,collapse=*)
   tmp -
with(
xdf,
eval(parse(text=
 paste(table(,paste(ijargs,collapse=,),),sep=)
 ))
   )
   if (simplify  i1) tmp-as.data.frame(ftable(tmp))
   out[[i]][[j]] - tmp
   }
  }
 return(out)
 }
 M=diag(3)
 M[1,2]=2
 M - rbind(M,M[1,])
 colnames(M) - paste(V,1:ncol(M),sep=)
 out=generateCube(M)
 out2=generateCube(M,simplify=FALSE)

 out[[2]][[V1*V3]]
 out2[[2]][[V1*V3]][0,1]
 -


 On 14 June 2011 14:59, Saravanan 
 saravanan.thirumuruganat...@gmail.com 
 mailto:saravanan.thirumuruganat...@gmail.com wrote:

 Hello All,

 I have a dataset and I wish to obtain all possible data cuboids
 from it using R . For eg if my data frame is :

 ABC
 111
 121
 221

 The output intended is :
 A=1
 A=2
 B=1
 B=2
 C=1
 A=1,B=1
 A=1,B=2
 A=2,B=2
 A=1,C=1
 A=2,C=1
 B=1,C=1
 B=2,C=1
 A=1,B=1,C=1
 A=1,B=2,C=1
 A=2,B=2,C=1

 Are there any function(s) to do this in R ? I tried a combination
 of expand.grid and combn but the resulting code was very ugly and
 needed lot of hacks to make it work. I also tried to check the
 code for arules (which constructs similar itemsets) but
 unfortunately its code is in C and I am not very familiar in
 writing R extensions. Any pointers to functions will be much
 appreciated.

 Regards,
 Saravanan

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




 -- 
 Eric Lecoutre
 Consultant - Business  Decision
 Business Intelligence  Customer Intelligence

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Re: [R] non-numeric argument to binary operator

[Peter Langfelder]

Write yourself an alternative function to table, for example like this:

tableOfGivenLevels = function(x, levels)
{
  n = length(levels)
  counts = rep(0, n);
  names(counts) = levels
  tab = table(x);
  counts[match(names(tab), levels)] = tab;
  counts;
}

x = sample(c(1:4), 20, replace = TRUE)
tableOfGivenLevels(x, c(0:6))

Then run your old code, the one with

 habs=t(apply(habs, 1, table))/b

but replace table by tableOfGivenLevels.

Peter


On Tue, Jun 14, 2011 at 10:43 PM, the_big_kowalski bkowal...@csumb.edu wrote:
 Most likely reason is that the number of unique values in the rows of
 habs is not the same.

 Yes, I think that is the problem, thank you.
 How would I write the code to include 0s in the matrix,
 ie, I want to have a record of when 1, 2, or 3 does not get sampled,
 to come up with a frequency of each value for each nn (which in this case is
 5)

 I tried 'factor' but what ends up happening is that the matrix gets reduced
 to zero
 after a couple of iterations.

 habs=t(apply(habs, 1, function(x) table(factor(x, levels = 1:3

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/non-numeric-argument-to-binary-operator-tp3598160p3598563.html
 Sent from the R help mailing list archive at Nabble.com.

 __
 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.




-- 
Sent from my Linux computer. Way better than iPad :)

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

Re: [R] Rmpi installation

[Hugo Mildenberger]

Dear Kristian,

   please run exactly

mpicc mtest.c -o mtest 
 
  If you really need it, add -v separately. mpicc is nothing but a compiler 
wrapper. 
  The -o switch specifies the outfile name, which has to follow immediately 
after -o, 
  with or without a blank character in between. I'm not sure about what happens 
  with -ov mtest. Probably an output file named v is produced on disk, in 
addition to 
  the file mtest stemming from a previous with -o mtest Depending on 
  mpicc's, gcc's and ld's version, gcc may instruct the linker to link the 
previously 
  produced executable mtest with the object newly compiled from mtest.c plus 
  several implicit libraries and startup objects, and output the result to v. 
  ld would then be faced with duplicate defined symbols and output a list of 
these 
  doublets.
  
  My proposal was meant as a test for an else correct openmpi installation, 
perhaps
  I should have said this. On successful completion, no message should be 
printed, but
  an executable named mtest should have been produced, which could then
  be run typing ./mtest. If the link process failed, then the openmpi 
installation was 
  inconsistent. 

Kind regards

Hugo
 
On Wednesday 15 June 2011 17:45:41 Unger, Kristian, Dr. wrote:
 Dear Hugo
 
 I ran the command with the verbose switch and get the following output:
 
  mpicc mtest.c -ov mtest
 mtest: In function `_start':
 /usr/src/packages/BUILD/glibc-2.11.1/csu/../sysdeps/x86_64/elf/start.S:65:
 multiple definition of `_start'
 /usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../lib64/crt1.o:/usr/src/pack
 ages/BUILD/glibc-2.11.1/csu/../sysdeps/x86_64/elf/start.S:65: first
 defined here
 mtest: In function `_fini':
 (.fini+0x0): multiple definition of `_fini'
 /usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../lib64/crti.o:initfini.c:(.
 fini+0x0): first defined here
 mtest:(.rodata+0x0): multiple definition of `_IO_stdin_used'
 /usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../lib64/crt1.o:(.rodata.cst4
 +0x0): first defined here
 mtest: In function `__data_start':
 (.data+0x0): multiple definition of `__data_start'
 /usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../lib64/crt1.o:(.data+0x0):
 first defined here
 mtest: In function `main':
 (.text+0xec): multiple definition of `main'
 /tmp/ccUyk8e9.o:mtest.c:(.text+0x0): first defined here
 mtest: In function `_init':
 (.init+0x0): multiple definition of `_init'
 /usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../lib64/crti.o:initfini.c:(.
 init+0x0): first defined here
 /usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../x86_64-suse-linux/bin/ld:
 error in mtest(.eh_frame); no .eh_frame_hdr table will be created.
 collect2: ld returned 1 exit status
 
 Not sure if this makes any sense?
 
 
 Best wishes
 
 Kristian
 
 
 Dr. Kristian Unger
 
 
 Arbeitsgruppenleiter Integrative Biologie / Head of Integrative Biology
 Group
 Abteilung für Strahlenzytogenetik / Research Unit of Radiation
 Cytogenetics
 
 Tel.: +49-89-3187-3515
 
 Mob.: +49-160-90641879
 
 
 
 
 
 Am 15.06.11 17:25 schrieb Hugo Mildenberger unter
 hugo.mildenber...@web.de:
 
 Kristian,
 
 these are the usual problems with binary distributions.
 
 Regarding
 
--with-Rmpi-include=/usr/lib64/mpi/gcc/openmpi/include
 
 and configure output
 
   checking for mpi.h... no
 
 ... so does /usr/lib64/mpi/gcc/openmpi/include really exist? At least,
 that appears to be a very unusual place to look for mpi.h (normally
 to be found in /usr/include ).
 
 And if you try to compile  link the attached demo program: does
 the link phase succeed? Compile  link using
 
 $ mpicc mtest.c -o mtest
 
 Presumably you have already tried to run install.packages(Rmpi).
 
 
 Kind regards
 
 Hugo
 
 
 
 On Wednesday 15 June 2011 16:22:07 Unger, Kristian, Dr. wrote:
  Thanks Hugo.
 
  I am pretty sure openmpi is installed:
 
  # zypper se openmpi
  Loading repository data...
  Reading installed packages...
 
  S | Name  | Summary | Type
  --+---+-+---
  i | openmpi   | A powerful implementaion of MPI | package
| openmpi   | A powerful implementaion of MPI | srcpackage
  i | openmpi-devel | A powerful implementaion of MPI | package
 
 
 
  I got the same error message with the latest version available. The
 reason
  why I took the somewhat older version is that I wanted to make sure that
  it is not related to any libraries used by the newest version.
 
  Best wishes
 
  Kristian
 
  
  Dr. Kristian Unger
 
 
  Arbeitsgruppenleiter Integrative Biologie / Head of Integrative Biology
  Group
  Abteilung für Strahlenzytogenetik / Research Unit of Radiation
  Cytogenetics
 
  Tel.: +49-89-3187-3515
 
  Mob.: +49-160-90641879
 
 
 
 
 
  Am 15.06.11 16:16 schrieb Hugo Mildenberger unter
  hugo.mildenber...@web.de:
 
  Kristian,
  
  I just tried that particular command here on 

Re: [R] BIZARRE results from wilcox.test()

[Daniel Malter]

I did not intend to bully you but rather tried to speak narrowly to the core
of the issue. In a sense the point was that the example you used to
illustrate the problem created part of the problem and that in a sensical
dataset you would not obtain nonsensical results. Secondly, my reply talked
to the solution of your problem by pointing out that the test uses the
normal approximation by standard, which is exactly that: an approximation,
implying that avoiding the approximation may solve the issue. This is what
exact=T does, as you figured out.

Daniel




genecleaner wrote:
 
 Dear Daniel and Sarah, 
 
 Thanks you for your rude replies .
 The script that I provided was only an example and to illustrate the
 problem. It makes perfectly sense to use the Wilcoxon test on my datasets.
 However, you replies were nonsensical, since you could not solve the
 problem but rather just bullied me.
 
 Anyway, this is the solution to the problem: the exact=TRUE statement
 should be added
 
 w - wilcox.test(c(1:50),(c(1:50)+100))
 w$p.value
 [1] 7.066072e-18
 w - wilcox.test(c(1:50),(c(1:50)+100), exact=TRUE)
 w$p.value
 [1] 1.982331e-29
 
 Best regards,
 genecleaner
 

--
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Re: [R] Rmpi installation

[Unger, Kristian, Dr.]

Hi Hugo

One step closer! I added the path to libmpi to ld.so.conf and ran ldconfig
(see
http://www.linuxforums.org/forum/programming-scripting/80405-linking-shared
-libraries.html).

Compiling the code you sent me works but running the file gets stuck with
the following message:

./mtest
librdmacm: couldn't read ABI version.
librdmacm: assuming: 4
libibverbs: Fatal: couldn't read uverbs ABI version.
CMA: unable to open /dev/infiniband/rdma_cm
--
WARNING: Failed to open OpenIB-cma [DAT_INTERNAL_ERROR:].
This may be a real error or it may be an invalid entry in the uDAPL
Registry which is contained in the dat.conf file. Contact your local
System Administrator to confirm the availability of the interfaces in
the dat.conf file.
--

Any suggestions?

Best wishes

Kristian




Dr. Kristian Unger


Arbeitsgruppenleiter Integrative Biologie / Head of Integrative Biology
Group
Abteilung für Strahlenzytogenetik / Research Unit of Radiation
Cytogenetics

Tel.: +49-89-3187-3515

Mob.: +49-160-90641879





Am 15.06.11 19:25 schrieb Hugo Mildenberger unter
hugo.mildenber...@web.de:

Dear Kristian,

   please run exactly

mpicc mtest.c -o mtest

  If you really need it, add -v separately. mpicc is nothing but a
compiler wrapper.
  The -o switch specifies the outfile name, which has to follow
immediately after -o,
  with or without a blank character in between. I'm not sure about what
happens
  with -ov mtest. Probably an output file named v is produced on disk,
in addition to
  the file mtest stemming from a previous with -o mtest Depending on
  mpicc's, gcc's and ld's version, gcc may instruct the linker to link
the previously
  produced executable mtest with the object newly compiled from mtest.c
plus
  several implicit libraries and startup objects, and output the result
to v.
  ld would then be faced with duplicate defined symbols and output a list
of these
  doublets.

  My proposal was meant as a test for an else correct openmpi
installation, perhaps
  I should have said this. On successful completion, no message should be
printed, but
  an executable named mtest should have been produced, which could then
  be run typing ./mtest. If the link process failed, then the openmpi
installation was
  inconsistent.

Kind regards

Hugo

On Wednesday 15 June 2011 17:45:41 Unger, Kristian, Dr. wrote:
 Dear Hugo

 I ran the command with the verbose switch and get the following output:

  mpicc mtest.c -ov mtest
 mtest: In function `_start':

/usr/src/packages/BUILD/glibc-2.11.1/csu/../sysdeps/x86_64/elf/start.S:65
:
 multiple definition of `_start'

/usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../lib64/crt1.o:/usr/src/pa
ck
 ages/BUILD/glibc-2.11.1/csu/../sysdeps/x86_64/elf/start.S:65: first
 defined here
 mtest: In function `_fini':
 (.fini+0x0): multiple definition of `_fini'

/usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../lib64/crti.o:initfini.c:
(.
 fini+0x0): first defined here
 mtest:(.rodata+0x0): multiple definition of `_IO_stdin_used'

/usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../lib64/crt1.o:(.rodata.cs
t4
 +0x0): first defined here
 mtest: In function `__data_start':
 (.data+0x0): multiple definition of `__data_start'

/usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../lib64/crt1.o:(.data+0x0)
:
 first defined here
 mtest: In function `main':
 (.text+0xec): multiple definition of `main'
 /tmp/ccUyk8e9.o:mtest.c:(.text+0x0): first defined here
 mtest: In function `_init':
 (.init+0x0): multiple definition of `_init'

/usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../lib64/crti.o:initfini.c:
(.
 init+0x0): first defined here

/usr/lib64/gcc/x86_64-suse-linux/4.3/../../../../x86_64-suse-linux/bin/ld
:
 error in mtest(.eh_frame); no .eh_frame_hdr table will be created.
 collect2: ld returned 1 exit status

 Not sure if this makes any sense?


 Best wishes

 Kristian

 
 Dr. Kristian Unger


 Arbeitsgruppenleiter Integrative Biologie / Head of Integrative Biology
 Group
 Abteilung für Strahlenzytogenetik / Research Unit of Radiation
 Cytogenetics

 Tel.: +49-89-3187-3515

 Mob.: +49-160-90641879





 Am 15.06.11 17:25 schrieb Hugo Mildenberger unter
 hugo.mildenber...@web.de:

 Kristian,
 
 these are the usual problems with binary distributions.
 
 Regarding
 
--with-Rmpi-include=/usr/lib64/mpi/gcc/openmpi/include
 
 and configure output
 
   checking for mpi.h... no
 
 ... so does /usr/lib64/mpi/gcc/openmpi/include really exist? At
least,
 that appears to be a very unusual place to look for mpi.h (normally
 to be found in /usr/include ).
 
 And if you try to compile  link the attached demo program: does
 the link phase succeed? Compile  link using
 
 $ mpicc mtest.c -o mtest
 
 Presumably you have already tried to run install.packages(Rmpi).
 
 
 Kind regards
 

Re: [R] Problem auto.arima() in R

[Ben Bolker]

siddharth arun sid.arun91 at gmail.com writes:

 
 I am using auto.arima() for forecasting.When I am using any in built data
 such as AirPassangers it is capturing seasonality. But, If I am entering
 data in any other format(in vector form or from an excel sheet) it is not
 detecting seasonality.
 
 Is there any specific format in which it detects seasonality or I am doing
 some thing wrong?
 
 Does data have to be entered in a specific format?
 
  
Unfortunately, this is far too vague a question for us to answer.
Please read the posting guide and give us some more details about
what you are trying to do, preferably as a small reproducible example.
Also, be aware that auto.arima() is a function in the 
contributed 'forecast' package: this would be useful information to include.

  Ben Bolker

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[R] Year cost optimisation

[Bart Joosen]

Hi,

I have a data file with all our purchases from last year,
it contains the unit price, count, and total dollars spend.

Now I'm looking for some way to classify all our purchases to find out
which purchases are the best ones to find cheaper alternatives.
for example: if we only buy 1 item of a product which costs 5 dollar, we
maybe can find a cheaper product of 4.5 dollar, but this won't make it on
our year budget. On the other hand, there may be some products of 2.5
dollar, but we purchased 1000ths of them, so 2.3 dollar/product can be a
significant saving.

Instead of searching for each of our products a cheaper alternative (takes a
awfull lot of time), I would like to concentrate on a top 20 for example.
Maybe calculate the pct contribution to the total budget, and take the top
20?
Any other ideas for an approach?

Here an example dataframe:

dat  - data.frame(item=1:160,
unit_price=round(c(runif(80,0,100), runif(80,250,1000)),2),
count=round(c(runif(40, 1,10), runif(40,20,1000),runif(40,
1,10), runif(40,20,1000)),0),
total=0)
dat$total = dat$count*dat$unit_price


Bart

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[R] Error bars

[Anna Harris]

Hi, 

Can anyone help with plotting vertical error bars on a bar graph. 

I have tried following examples online and in the big R book and writing my own 
function but I have been unsuccessful and don’t really have an understanding 
of what it is I am doing. I have calculated my standard errors so basically 
just need to draw the bars on the graph but just don’t have a clue!!!

I don’t even know what information people will need to help me..

The code for my graph is:

barplot(tapply(Sporangia,list(Host,Isolate),mean),xlab=Isolate,ylab=Mean 
Sporangia per Needle,col=c(grey39,grey64,grey89),beside=T)
col=c(grey39,grey64,grey89)
legend(topright,inset=.05,title=Host,c(European Larch,Hybrid 
Larch,Japanese Larch),fill=col,horiz=FALSE)

Any help would be greatly appreciated, 

Anna


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Re: [R] Reshaping data with xtabs reorders my rows

[filip.biele...@rega.kuleuven.be]

Dnia 2011-06-15, o godz. 12:05:01
Jim Lemon j...@bitwrit.com.au napisał(a):

 On 06/15/2011 06:46 PM, filip.biele...@rega.kuleuven.be wrote:
  Dear,
 
  I have a data frame melted from a list of matrices (melt from
  reshape package), then I impute some missing values and then want
  to tabulate the data again to the array (or list, doesn't matter)
  of matrices form. However when using xtabs function it orders my
  rows alphabetically and apparently doesn't take reorder = FALSE
  option or anything like it. Is there anything I can do to stop it
  from doing so?
 
  Relevant parts of code:
 
  matrices.m- melt(combined_list)
 
  matrices.m_i[is.na(matrices.m_i$value),]$value- predictions
 
  matrices- xtabs(value ~ location + variable + week, data =
  matrices.m_i)
 
 Hi filip,
 Have you tried specifically classing your variables as factors and
 using the ordered argument?
 
 Jim
 
Dear,

That's exacly what the problem was - the factor levels are by default
in alphabetical order, so I had to fix that by:

levels(matrices.m_i$location) -
unique(as.character(matrices.m_i$location))

levels(matrices.m_i$variable) -
unique(as.character(matrices.m_i$variable))

Then xtabs works as advertised.

Cheers, 
filip

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[R] plot with two y axes BUT unaligned x axis

[bjmjarrett]

Hi all,

I have scoured the archives of this forum but nothing quite seems to fit the
bill...

I would like to plot a graph displaying two variables (y axes) that share
date as the x axis. However, the date values for each variable are not the
same - for example, some parasitoids were not released on days that
collections from the trap took place, whilst sometimes releases did occur on
the same day. I would like to align them. My code is:


CollectionDate-as.Date(CollectionDate,%m/%d/%Y) # days that had
collections
release.date.Total-as.Date(release.date.Total,%m/%d/%Y) # days that had
releases

par(mar=c(5,4,4,4)+0.1)

plot(CollectionDate[data$Trap==DPAU1],TotalP[data$Trap==DPAU1],type=n,ylim=c(0,80),xlab=Time,ylab=Number
of egg masses) # I am using one trap of many

lines(CollectionDate[data$Trap==DPAU1],TotalP[data$Trap==DPAU1],lwd=2)

par(new=T)

plot(release.date.Total[data$Trap==DPAU1],parasitoid.total[data$Trap==DPAU1],axes=F,type=h,lty=1,col=red,xlab=,ylab=,lwd=1.2)

axis(4,las=1)

mtext(side=4,line=3,Total parasitoids released)
###

As in the above code, I suppressed the axes for the second plot, but if I
don't the x axis (date) has two tick marks for each year.

Is there a way to align the date objects to begin with without compromising
the graphs themselves?

Thanks in advance,

Ben

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Re: [R] xyplot Legend Title and Position

[Justin McBride]

Dennis,

Thanks for your suggestion, but that is not exactly what I was after.
I was trying to get the legend in the margin on the top right of the
page and not in the plot frame.  Is there a way to do this?

Thanks,
Justin

On Tue, Jun 14, 2011 at 6:03 PM, Dennis Murphy djmu...@gmail.com wrote:
 Hi:

 Part of the problem is that you have a point in the upper right corner
 of your plot, so one way around it is to expand the y-range. Try this:

 xyplot(Yield ~ Date,
      groups=Machine, ylim = c(7, 22),
      auto.key=list(title=Machine, corner = c(0.95, 1), cex=1.0),
       par.settings = list(superpose.symbol=list(pch = 0:18, cex=1)),
      )

 HTH,
 Dennis

 On Tue, Jun 14, 2011 at 4:36 PM, Justin McBride crazyhaw...@gmail.com wrote:
 Dear R Community,

 I'm using xyplot in Lattice with a legend and a title on the legend.
 The title on legend is being cut off, as can be seen by running the
 code below.   The legend is on the right, but I would like to get to
 the top right of the graphics window.  Is there a way to get the
 legend title to display correctly and move the whole legend up the the
 top right?

 Thanks,
 Justin

 ### R code

 library(lattice)
 Yield=c(16, 17, 11, 8, 16, 18)
 Date = c(1, 1, 2, 3, 4, 5)
 Machine = c(1, 3, 2, 2, 3, 1)


 xyplot(Yield ~ Date,
       groups=Machine,
       auto.key=list(title=Machine, space = right, cex=1.0),
        par.settings = list(superpose.symbol=list(pch = 0:18, cex=1)),
       )

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[R] Escape sequence in eval ()

[Franc Lucas]

   Hello,
   I am wondering how to get the quotation marks into a variable expression. I
   can't escape it with the backslash \ ...
   Example:
   I can access my data frame via
   TABLE$2011-01-02$columnD
   Now I want to do this automatically.. (with a for loop)..
   a - TABLE
   b -  \2011-01-02\ 
   c - columnD
   acessmytable - paste(a,b,c, sep=$)
   parse(text=accessmytable)
   eval(accessmytable)
   It will return
   [1] TABLE$\2011-01-02\$close
   but I need
   TABLE$2011-01-02$close
   I highly appreciate your help!
   Sincerely,
   Franc
   BS Student, University of Mannheim

   Schon gehört? WEB.DE hat einen genialen Phishing-Filter in die
   Toolbar eingebaut! [1]http://produkte.web.de/go/toolbar

References

   1. http://produkte.web.de/go/toolbar
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[R] specifying interactions in a gam model with by

[Jean V Adams]

I?m confused by the difference in the fit of a gam model (in package mgcv) 
when I specify an interaction in different ways.  I would appreciate it if 
someone could explain the cause of these differences.

For example:

x - c(105, 124, 124, 124, 144, 144, 150, 176, 178, 178, 
206, 206, 212, 215, 215, 227, 229, 229, 229, 234, 
234, 254, 254, 290, 290, 303, 334, 334, 334, 344, 
345, 345)
y - c(0.31, 1.41, 2.87, 1.92, 0.31, 0.31, 0.31, 0.31, 0.31, 0.31, 
0.31, 0.31, 0.31, 1.92, 0.31, 0.31, 0.31, 0.31, 0.31, 0.31, 
0.31, 2.08, 2.28, 1.59, 2.13, 2.77, 3.97, 4.54, 4.35, 3.6, 
5.2, 4.6)
loc - c(S, N, S, S, N, N, S, S, N, N, 
N, N, S, N, S, S, N, S, S, N, 
N, N, N, N, N, S, N, S, S, N, 
S, S)
locf - as.factor(loc)
locN - as.numeric(loc==N)
locS - as.numeric(loc==S)

# fit the model with a separate smooth for loc = N and loc = S
fit1 - gam(y ~ locf + s(x, by=locN) + s(x, by=locS))

# fit the model with a separate smooth for each level of the factor loc (N 
and S)
fit2 - gam(y ~ locf + s(x, by=locf))

# The shape of the relations are similar, but the vertical locations are 
different,
# and the size of the standard errors of the smooth are different
windows()
par(mfrow=c(2, 2), mar=c(4, 4, 2, 1))
plot(fit1)
plot(fit2)

# The R-sq., deviance explained, GCV score, and scale est. are the same,
# but the estimates and degrees of freedom are different
summary(fit1)
summary(fit2)

I'm using R version 2.13.0 (2011-04-13) and mgcv version 1.7-6 on Windows 
XP.

Thanks for your help.

Jean


`·.,,  (((º   `·.,,  (((º   `·.,,  (((º

Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East Steinfest Road
Antigo, WI 54409  USA
http://www.glsc.usgs.gov  (GLSC web site)
jvad...@usgs.gov  (E-mail)
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[R] Correlations by subgroups

[jfdawson]

I'm hoping there is a simple answer to this - it seems that there should be,
but I can't figure it out.

I have a matrix/data frame with three variables of interest - V1, V2, V3.
One, V1, is a factor with x levels (x may be a large number); I want to
calculate  the correlation between the other two (i.e. cor(V2,V3)) for each
level, and store it as a vector of length x.

I should think this should be possible using a function like tapply, but I
cannot work out what the syntax would be - everything I have tried produces
either an error, or just repeats the correlations between V2  V3 on the
whole matrix.

Could anyone suggest what I should be doing?

Many thanks,
Jeremy

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Re: [R] Heatmap in R and/or ggplot2

[idris]

Thanks for the tip on the limit attribute on scale_fill_gradientn.

I'll check out mencoder and let you know if I use your code for the movie.

Cheers


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[R] Problem in reading Missing and Measure values, using read.spss().

[Smart Guy]

Hello All,
I am using read.spss() to read a SPSS dataset into R data.frame.
However I am not able to read user defined MISSING values when it
defined as range in SPSS variable view. I am also not able to read the
value from the MEASURE column in the SPSS variable view to determine
whether a SPSS column is nominal or ordinal. Can anyone point me to
the right direction to read these values from SPSS dataset by using
read.spss() or any another function like read.spss().
Thanks in advance for your help.

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[R] Count occurances in integers (or strings)

[Jean V Adams]

If I understand you correctly, your data column is not a character.
class(mydata[,my_column])

If it's numeric or a factor, this should work

# convert it to character to split apart the individual digits
# then convert the digits to numeric, and calculate the sum
sapply(strsplit(as.character(mydata[, my_column]), ), function(x) 
sum(as.numeric(x)))

Jean


`·.,,  (((º   `·.,,  (((º   `·.,,  (((º

Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East Steinfest Road
Antigo, WI 54409  USA
http://www.glsc.usgs.gov  (GLSC web site)


-

Jay josip.2000 at gmail.com 
Wed Jun 15 11:09:25 CEST 2011

Hi,

I have a dataframe column from which I want to calculate the number of
1's in each entry. Some column values could, for example, be
0001001000 and 111.

To get the number of occurrences from a string I use this:
sum(unlist(strsplit(mydata[,my_column], )) == 1)

However, as my data is not in string form.. How do I convert it? l
tried:
lapply(mydata[,my_column],toString)

but I do not seem to get it right (or at least I do not understand the
output format).

Also, are there other options? Can I easily calculate the occurrences
directly from the integers?

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[R] Problems with nls

[Christopher Hulme-Lowe]

I'm trying to fit the Bass Diffusion Model using the nls function in R but
I'm running into a strange problem. The model has either two or three
parameters, depending on how it's parameterized, p (coefficient of
innovation), q (coefficient of immitation), and sometimes m (maximum market
share). Regardless of how I parameterize the model I get an error saying
that the step factor has decreased below it's minimum. I have tried
re-setting the minimum in nls.controls but that doesn't seem to fix the
problem. Likewise, I have run through a variety of start values in the past
few days, all to no avail. Looking at the trace output it appears that R
believes I always have one more parameter than I actually have (i.e. when
the model is parameterized with p and q R seems to be seeing three
parameters, when m is also included R seems to be seeing four). My
experience with nls is limited, can someone explain to me why it's doing
this? I've included the data set I'm working with (published in Michalakelis
et al. 2008) and some example code.

## Assign relevant variables
adoption -
c(167000,273000,531000,938000,2056452,3894103,5932090,7963742,9314687,10469060,11393302,11976340)
time - seq(from = 1,to = 12, by = 1)
## Models
Bass.Model - adoption ~ ((p + q)^2/p) * (exp(-(p + q) * time)/((q / p) *
exp(-(p + q) * time) + 1)^2)
## Starting Parameters
Bass.Params - list(p = 0.1, q = 0.1)
## Model fitting
Bass.Fit - nls(formula = Bass.Model, start = Bass.Params, algorithm =
plinear, trace = TRUE)

Chris Hulme-Lowe
University of Minnesota
Department of Psychology
Quant. Methods and Psychometrics

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[R] varimp_in_party_package

[Jinrui Xu]

Hello everyone,

I use the following command lines to get important variable from  
training dataset.



data.controls - cforest_unbiased(ntree=500, mtry=3)
data.cforest - cforest(V1~.,data=rawinput,controls=data.controls)
data.cforest.varimp - varimp(data.cforest, conditional = TRUE)

I got error: Error in model.matrix.default(as.formula(f),data =  
blocks): term 1 would require 4e+17 columns


I changed data dimension to 150. The problem still exists. So, I guess  
there are other problems. Please give me some help or hints. Thanks!


jinrui,

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[R] Cross-correlogram error message?

[William M Forney]

   Hello,
   I am using the spline.correlog function of the ncf library, and am getting
   the   error   cannot  allocate  vector  of  size  832.2  Mb.   Using
   memory.limit(), I already increased the size to 4Gb.  I thought it might
   be related to the storage space, but I have over 14Gb on my harddrive.  I
   have attached the line of code and error message below any suggestions?
   Thanks in advance for any assistance with this matter, Will
   _
crosscor7 - spline.correlog(x=X, y=Y, z=SOLAR7_3M, w=TIR_3M, resamp = 5)
   Error: cannot allocate vector of size 832.2 Mb
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[R] PC biplot display

[Alyssa Cirtwill]

Hello,

I am creating biplots based on the prcomp principal components  
analysis command. I would like to display only the principal component  
vectors and the variables, not the data. Can anyone suggest how to do  
this?


Thanks,

Alyssa

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[R] function optimization

[navishkumarb]

Hello 
I would like to optimize a function which is as follows. 


nc.adj - function(nc, G) {


  x = a + G + (b/(G^2 + (c - G)^2)) - nc
  return(x)
}

Can I just know how to get the optimized values of a,b,c for given G and nc
using optim/optimize function. 


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Re: [R] How to include multiple random effects in 'lme' function

[karena]

Can anyone help me with this?

thank you in advance.

Karena

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[R] RES: Problem in reading Missing and Measure values, using read.spss().

[Filipe Leme Botelho]

---BeginMessage---
Hi,

I never tried to read SPSS data, but to read a csv extracted from excel
for instance, you can do something like

read.csv(example, na.strings='#N/A')

if that is the case, then your NAs will be identified and properly read
while loading data.

HTH. Cheers,
Filipe Botelho

-Mensagem original-
De: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
Em nome de Smart Guy
Enviada em: quarta-feira, 15 de junho de 2011 11:37
Para: r-help@r-project.org
Assunto: [R] Problem in reading Missing and Measure values,using
read.spss().

Hello All,
I am using read.spss() to read a SPSS dataset into R data.frame.
However I am not able to read user defined MISSING values when it
defined as range in SPSS variable view. I am also not able to read the
value from the MEASURE column in the SPSS variable view to determine
whether a SPSS column is nominal or ordinal. Can anyone point me to
the right direction to read these values from SPSS dataset by using
read.spss() or any another function like read.spss().
Thanks in advance for your help.

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[R] prediction intervals using R

[Dave Evens]

Dear members,

I'm fitting linear model using lm which has numerous auto-regressive terms as 
well as other explanatory variables. In order to calculate prediction 
intervals, i've used a for-loop as the auto-regressive parameters need to be 
updated each time so that a new forecast and corresponding prediction interval 
can be calculated.

I'm fitting a number of these models which have different values for the 
response variable and possibly different explanatory variables. The response is 
temperature in fahrenheit (F), and the different models are for cities. So each 
city has its own fitted linear model for temperature. I'm assuming that they're 
independent models for the time being, I want to combine the results across all 
cities and have overall prediction intervals. Because I assuming that they're 
independent can I just add together the degrees of freedom from each model 
(i.e. total degrees of freedom=df1+df2+...) and the variance-covariance 
matrices (i.e. V=V1+V2+...) in order to calcalate the overall prediction 
intervals?

Any help would be most appreciated.

Regards,
Dave

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Re: [R] Error bars

[Savitri N Appana]

You might try 'bargraph.CI' in R pkg 'sciplot'.

?bargraph.CI

HTH,
Savi

 Anna Harris annaharrisish...@hotmail.com 6/15/2011 1:00 PM 
Hi, 

Can anyone help with plotting vertical error bars on a bar graph. 

I have tried following examples online and in the big R book and writing my own 
function but I have been unsuccessful and don’t really have an understanding of 
what it is I am doing. I have calculated my standard errors so basically just 
need to draw the bars on the graph but just don’t have a clue!!!

I don’t even know what information people will need to help me..

The code for my graph is:

barplot(tapply(Sporangia,list(Host,Isolate),mean),xlab=Isolate,ylab=Mean 
Sporangia per Needle,col=c(grey39,grey64,grey89),beside=T)
col=c(grey39,grey64,grey89)
legend(topright,inset=.05,title=Host,c(European Larch,Hybrid 
Larch,Japanese Larch),fill=col,horiz=FALSE)

Any help would be greatly appreciated, 

Anna


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Re: [R] Correlations by subgroups

[Ista Zahn]

I have to confess that plyr has made me lazy about remembering tapply,
by, aggregate et al., so I'm no help there. But if you want to use
plyr it's just

ddply(dat, .(V1), summarize, cor.v2.v3 = cor(V2, V3))

Best,
Ista

On Wed, Jun 15, 2011 at 10:31 AM, jfdawson j.f.daw...@aston.ac.uk wrote:
 I'm hoping there is a simple answer to this - it seems that there should be,
 but I can't figure it out.

 I have a matrix/data frame with three variables of interest - V1, V2, V3.
 One, V1, is a factor with x levels (x may be a large number); I want to
 calculate  the correlation between the other two (i.e. cor(V2,V3)) for each
 level, and store it as a vector of length x.

 I should think this should be possible using a function like tapply, but I
 cannot work out what the syntax would be - everything I have tried produces
 either an error, or just repeats the correlations between V2  V3 on the
 whole matrix.

 Could anyone suggest what I should be doing?

 Many thanks,
 Jeremy

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-- 
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Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org

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Re: [R] Correlations by subgroups

[Daniel Malter]

x-c(1,1,1,1,1,2,2,2,2,2)
y-rnorm(10)
z-y+rnorm(10)
by(data.frame(y,z),factor(x),cor)

hth,
Daniel


jfdawson wrote:
 
 I'm hoping there is a simple answer to this - it seems that there should
 be, but I can't figure it out.
 
 I have a matrix/data frame with three variables of interest - V1, V2, V3.
 One, V1, is a factor with x levels (x may be a large number); I want to
 calculate  the correlation between the other two (i.e. cor(V2,V3)) for
 each level, and store it as a vector of length x.
 
 I should think this should be possible using a function like tapply, but I
 cannot work out what the syntax would be - everything I have tried
 produces either an error, or just repeats the correlations between V2  V3
 on the whole matrix.
 
 Could anyone suggest what I should be doing?
 
 Many thanks,
 Jeremy
 

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Re: [R] R string functions

[Daniel Malter]

x-'GTTACTGGTACC'
table(strsplit(x,''))

hth,
Daniel


karena wrote:
 
 Hi, 
 
 I have a string GGCCCAATCGCAATTCCAATT
 
 What I want to do is to count the percentage of each letter in the string,
 what string functions can I use to count the number of each letter
 appearing in the string?
 
 For example, the letter A appeared 6 times, letter T appeared 5 times,
 how can I use a string function to get the these number?
 
 thanks,
 
 karena
 

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Re: [R] Count occurances in integers (or strings)

[Steven Kennedy]

This should work:

mydata$my_column-as.character(mydata$my_column)
sum(unlist(strsplit(mydata[,my_column], )) == 1)


On Wed, Jun 15, 2011 at 7:09 PM, Jay josip.2...@gmail.com wrote:
 Hi,

 I have a dataframe column from which I want to calculate the number of
 1's in each entry. Some column values could, for example, be
 0001001000 and 111.

 To get the number of occurrences from a string I use this:
 sum(unlist(strsplit(mydata[,my_column], )) == 1)

 However, as my data is not in string form.. How do I convert it? l
 tried:
 lapply(mydata[,my_column],toString)

 but I do not seem to get it right (or at least I do not understand the
 output format).

 Also, are there other options? Can I easily calculate the occurrences
 directly from the integers?

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Re: [R] Problems with nls

[Gabor Grothendieck]

On Wed, Jun 15, 2011 at 11:06 AM, Christopher Hulme-Lowe
hulme...@umn.edu wrote:
 I'm trying to fit the Bass Diffusion Model using the nls function in R but
 I'm running into a strange problem. The model has either two or three
 parameters, depending on how it's parameterized, p (coefficient of
 innovation), q (coefficient of immitation), and sometimes m (maximum market
 share). Regardless of how I parameterize the model I get an error saying
 that the step factor has decreased below it's minimum. I have tried
 re-setting the minimum in nls.controls but that doesn't seem to fix the
 problem. Likewise, I have run through a variety of start values in the past
 few days, all to no avail. Looking at the trace output it appears that R
 believes I always have one more parameter than I actually have (i.e. when
 the model is parameterized with p and q R seems to be seeing three
 parameters, when m is also included R seems to be seeing four). My
 experience with nls is limited, can someone explain to me why it's doing
 this? I've included the data set I'm working with (published in Michalakelis
 et al. 2008) and some example code.

 ## Assign relevant variables
 adoption -
 c(167000,273000,531000,938000,2056452,3894103,5932090,7963742,9314687,10469060,11393302,11976340)
 time - seq(from = 1,to = 12, by = 1)
 ## Models
 Bass.Model - adoption ~ ((p + q)^2/p) * (exp(-(p + q) * time)/((q / p) *
 exp(-(p + q) * time) + 1)^2)
 ## Starting Parameters
 Bass.Params - list(p = 0.1, q = 0.1)
 ## Model fitting
 Bass.Fit - nls(formula = Bass.Model, start = Bass.Params, algorithm =
 plinear, trace = TRUE)

Using the default nls algorithm (which means we must specify m in the
formula and in the starting values) rather than plinear and using
commonly found p and q for starting values:

 Bass.Model - adoption ~ m * ((p + q)^2/p) * (exp(-(p + q) * time)/((q / p) *
+ exp(-(p + q) * time) + 1)^2)
 nls(formula = Bass.Model, start = c(p = 0.03, q = 0.4, m = max(adoption)))
Nonlinear regression model
  model:  adoption ~ m * ((p + q)^2/p) * (exp(-(p + q) * time)/((q/p)
*  exp(-(p + q) * time) + 1)^2)
   data:  parent.frame()
p q m
2.70842174019e-03 4.56307730094e-01 1.02730314877e+08
 residual sum-of-squares: 2922323788247

Number of iterations to convergence: 14
Achieved convergence tolerance: 3.05692430520e-06


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[R] Print the summary of a model to file

[Diviya Smith]

Hi there,

I am having a strange problem. I am running nls on some data.

#data

x - -(1:100)/10

y - 100 + 10 * (exp(-x / 2)


Using nls I fit an exponential model to this data and get a great fit


summary(fit)

Formula: wcorr ~ (Y0 + a * exp(m1 * -dist/100))


Parameters:

 Estimate Std. Error t value Pr(|t|)

Y0 -0.0001821  0.0002886  -0.6310.528

a   0.1669675  0.0015223 109.678   2e-16 ***

m1 10.8045840  0.1575957  68.559   2e-16 ***

---

Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1


Residual standard error: 0.007244 on 997 degrees of freedom


Number of iterations to convergence: 7

Achieved convergence tolerance: 4.214e-06



I would like to print this summary to a file but when I use cat it adds a
lot of the intermediate values and does not give me the Pvalues of the
t-test. Can you suggest a good way to print this to a file? I have used-


cat(paste(a), file=testing.txt, append = FALSE)


output in file:

wcorr ~ (Y0 + a * exp(m1 * -dist/100))

 c(0.168587967690081, 0.0410369885074373, 0.015604934198253,
0.00651400975192817, -0.00198358204504023, -0.00297164058538482,
0.00200403258235102, -0.00585536622301963, -0.00677164279240891,
-0.00663060500378226, -0.00429206279972985, -0.00752682816527953,
-0.00772771510594769, -0.0140235396260263, -0.00905111970710343,
-0.00527727528681396, -0.00637982823781946, -0.0101696023470134,
-0.0101074232949501, -0.00715411863549439, -0.00662351777569056,
-0.00284945195584713, -0.00401775422983583, -0.00833925944560227,

 0.00724377249911614

 c(3, 997)

 c(0.001587401493019, 3.32626693693668e-05, 0.412938194338572,
3.32626693693668e-05, 0.0441665508678859, 2.86654588531843,
0.412938194338572, 2.86654588531843, 473.324480704723)

 nls(formula = wcorr ~ (Y0 + a * exp(m1 * -dist/100)), start = pa1, control
= list(maxiter = 1000, tol = 1e-05, minFactor = 0.0009765625, printEval =
FALSE, warnOnly = TRUE), algorithm = default, trace = FALSE)

 list(isConv = TRUE, finIter = 7, finTol = 4.21396379219722e-06, stopCode =
0, stopMessage = converged)

 list(maxiter = 1000, warnOnly = TRUE)

 NULL

 c(-0.00018212632256334, 0.166967461792613, 10.8045839935834,
0.000288607886496774, 0.00152233960007251, 0.157595671762849,
-0.631051094181987, 109.678196497457, 68.5588878978997, 0.528151752824769,
0, 0)

 c(-0.00018212632256334, 0.166967461792613, 10.8045839935834,
0.000288607886496774, 0.00152233960007251, 0.157595671762849,
-0.631051094181987, 109.678196497457, 68.5588878978997, 0.528151752824769,
0, 0)

Thanks,

Diviya

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[R] R string functions

[karena]

Hi, 

I have a string GGCCCAATCGCAATTCCAATT

What I want to do is to count the percentage of each letter in the string,
what string functions can I use to count the number of each letter appearing
in the string?

For example, the letter A appeared 6 times, letter T appeared 5 times,
how can I use a string function to get the these number?

thanks,

karena

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Re: [R] Problems with nls

[Gabor Grothendieck]

On Wed, Jun 15, 2011 at 5:35 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
 On Wed, Jun 15, 2011 at 11:06 AM, Christopher Hulme-Lowe
 hulme...@umn.edu wrote:
 I'm trying to fit the Bass Diffusion Model using the nls function in R but
 I'm running into a strange problem. The model has either two or three
 parameters, depending on how it's parameterized, p (coefficient of
 innovation), q (coefficient of immitation), and sometimes m (maximum market
 share). Regardless of how I parameterize the model I get an error saying
 that the step factor has decreased below it's minimum. I have tried
 re-setting the minimum in nls.controls but that doesn't seem to fix the
 problem. Likewise, I have run through a variety of start values in the past
 few days, all to no avail. Looking at the trace output it appears that R
 believes I always have one more parameter than I actually have (i.e. when
 the model is parameterized with p and q R seems to be seeing three
 parameters, when m is also included R seems to be seeing four). My
 experience with nls is limited, can someone explain to me why it's doing
 this? I've included the data set I'm working with (published in Michalakelis
 et al. 2008) and some example code.

 ## Assign relevant variables
 adoption -
 c(167000,273000,531000,938000,2056452,3894103,5932090,7963742,9314687,10469060,11393302,11976340)
 time - seq(from = 1,to = 12, by = 1)
 ## Models
 Bass.Model - adoption ~ ((p + q)^2/p) * (exp(-(p + q) * time)/((q / p) *
 exp(-(p + q) * time) + 1)^2)
 ## Starting Parameters
 Bass.Params - list(p = 0.1, q = 0.1)
 ## Model fitting
 Bass.Fit - nls(formula = Bass.Model, start = Bass.Params, algorithm =
 plinear, trace = TRUE)

 Using the default nls algorithm (which means we must specify m in the
 formula and in the starting values) rather than plinear and using
 commonly found p and q for starting values:

 Bass.Model - adoption ~ m * ((p + q)^2/p) * (exp(-(p + q) * time)/((q / p) *
 + exp(-(p + q) * time) + 1)^2)
 nls(formula = Bass.Model, start = c(p = 0.03, q = 0.4, m = max(adoption)))
 Nonlinear regression model
  model:  adoption ~ m * ((p + q)^2/p) * (exp(-(p + q) * time)/((q/p)
 *      exp(-(p + q) * time) + 1)^2)
   data:  parent.frame()
                p                 q                 m
 2.70842174019e-03 4.56307730094e-01 1.02730314877e+08
  residual sum-of-squares: 2922323788247

 Number of iterations to convergence: 14
 Achieved convergence tolerance: 3.05692430520e-06

and if its important to you to use plinear then try absorbing
(p+q)^2/p into m which means you will have to back that out from .lin
to calculate m:

 Bass.Model - adoption ~ (exp(-(p + q) * time)/((q / p) *
+ exp(-(p + q) * time) + 1)^2)
 nls(formula = Bass.Model, start = c(p = 0.03, q = 0.4), alg = plinear)
Nonlinear regression model
  model:  adoption ~ (exp(-(p + q) * time)/((q/p) * exp(-(p + q) *
time) +  1)^2)
   data:  parent.frame()
p q  .lin
2.70843216557e-03 4.56307130968e-01 7.99164076337e+09
 residual sum-of-squares: 2922323788276

Number of iterations to convergence: 12
Achieved convergence tolerance: 2.56069064476e-06

-- 
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GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com

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Re: [R] R string functions

[Peter Alspach]

Tena koe Karena

Try:

table(strsplit(GGCCCAATCGCAATTCCAATT, ''))

HTH ...

Peter Alspach

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of karena
 Sent: Thursday, 16 June 2011 8:37 a.m.
 To: r-help@r-project.org
 Subject: [R] R string functions
 
 Hi,
 
 I have a string GGCCCAATCGCAATTCCAATT
 
 What I want to do is to count the percentage of each letter in the
 string,
 what string functions can I use to count the number of each letter
 appearing
 in the string?
 
 For example, the letter A appeared 6 times, letter T appeared 5
 times,
 how can I use a string function to get the these number?
 
 thanks,
 
 karena
 
 --
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 functions-tp3600484p3600484.html
 Sent from the R help mailing list archive at Nabble.com.
 
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Re: [R] Print the summary of a model to file

[Kyle .]

You could try the sink function! I use that from time to time:
?sink


Kyle H. Ambert
Fellow, National Library of Medicine
Department of Medical Informatics  Clinical Epidemiology
Oregon Health  Science University 
ambe...@gmail.com
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Re: [R] Problems with nls

[Daniel Malter]

There may be two issues here. The first might be that, if I understand the
Bass model correctly, the formula you are trying to estimate is the adoption
in a given time period. What you supply as data, however, is the cumulative
adoption by that time period.

The second issue might be that the linear algorithm may fail and that it may
be preferable to use Newton-Raphson (the standard) as this may provide
better values in the iterations.

If you do both, i.e., you do NLS on period adoption and use Newton-Raphson,
you get an estimate. Though, I am of course not sure whether that is
correct in the sense that it is what you would expect to find.


adoption - 
c(167000,273000,531000,938000,2056452,3894103,5932090,7963742,9314687,10469060,11393302,11976340)
 
time - seq(from = 1,to = 12, by = 1)

adoption2-c(0,adoption[1:(length(adoption)-1)])
S-(adoption-adoption2)/max(adoption)

## Models 
Bass.Model - S ~ M*((p + q)^2/p) * (exp(-(p + q) * time)/((q / p) * 
exp(-(p + q) * time) + 1)^2) 
## Starting Parameters 
Bass.Params - list(p = 0.1, q = 0.1, M=1) 
## Model fitting 
Bass.Fit - nls(formula = Bass.Model, start = Bass.Params)
summary(Bass.Fit)


c.hulmelowe wrote:
 
 I'm trying to fit the Bass Diffusion Model using the nls function in R but
 I'm running into a strange problem. The model has either two or three
 parameters, depending on how it's parameterized, p (coefficient of
 innovation), q (coefficient of immitation), and sometimes m (maximum
 market
 share). Regardless of how I parameterize the model I get an error saying
 that the step factor has decreased below it's minimum. I have tried
 re-setting the minimum in nls.controls but that doesn't seem to fix the
 problem. Likewise, I have run through a variety of start values in the
 past
 few days, all to no avail. Looking at the trace output it appears that R
 believes I always have one more parameter than I actually have (i.e. when
 the model is parameterized with p and q R seems to be seeing three
 parameters, when m is also included R seems to be seeing four). My
 experience with nls is limited, can someone explain to me why it's doing
 this? I've included the data set I'm working with (published in
 Michalakelis
 et al. 2008) and some example code.
 
 ## Assign relevant variables
 adoption -
 c(167000,273000,531000,938000,2056452,3894103,5932090,7963742,9314687,10469060,11393302,11976340)
 time - seq(from = 1,to = 12, by = 1)
 ## Models
 Bass.Model - adoption ~ ((p + q)^2/p) * (exp(-(p + q) * time)/((q / p) *
 exp(-(p + q) * time) + 1)^2)
 ## Starting Parameters
 Bass.Params - list(p = 0.1, q = 0.1)
 ## Model fitting
 Bass.Fit - nls(formula = Bass.Model, start = Bass.Params, algorithm =
 plinear, trace = TRUE)
 
 Chris Hulme-Lowe
 University of Minnesota
 Department of Psychology
 Quant. Methods and Psychometrics
 
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Re: [R] R string functions

[Steve Lianoglou]

Hi,

On Wed, Jun 15, 2011 at 4:37 PM, karena dr.jz...@gmail.com wrote:
 Hi,

 I have a string GGCCCAATCGCAATTCCAATT

 What I want to do is to count the percentage of each letter in the string,
 what string functions can I use to count the number of each letter appearing
 in the string?

 For example, the letter A appeared 6 times, letter T appeared 5 times,
 how can I use a string function to get the these number?

The replies you've already received are already helpful ... in
addition to them, though, I'd suggest you check out the Biostrings
package from bioconductor since it looks like you are working with
DNA:

http://www.bioconductor.org/packages/release/bioc/html/Biostrings.html

There are many (many^2) things already implemented in that package
that you will likely want to do with genomic sequences, and done so in
a memory-and-performance efficient manner.

For this particular example:

R library(Biostrings)
R x - DNAString(GGCCCAATCGCAATTCCAATT)
R oligonucleotideFrequency(x, 1)
A C G T
6 7 7 5

## And just for fun:
R oligonucleotideFrequency(x, 2)
AA AC AG AT CA CC CG CT GA GC GG GT TA TC TG TT
 3  0  0  3  3  3  1  0  0  2  5  0  0  2  0  2

Depending on how much genomic/sequence stuff you are planning to do,
it could be worth your while to invest some time looking into various
functionality the Biostrings (and IRanges) package(s) provides for
you.

Hope that helps,
-steve

-- 
Steve Lianoglou
Graduate Student: Computational Systems Biology
 | Memorial Sloan-Kettering Cancer Center
 | Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact

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Re: [R] Problems with nls

[Gabor Grothendieck]

On Wed, Jun 15, 2011 at 6:05 PM, Daniel Malter dan...@umd.edu wrote:
 There may be two issues here. The first might be that, if I understand the
 Bass model correctly, the formula you are trying to estimate is the adoption
 in a given time period. What you supply as data, however, is the cumulative
 adoption by that time period.

 The second issue might be that the linear algorithm may fail and that it may
 be preferable to use Newton-Raphson (the standard) as this may provide
 better values in the iterations.

 If you do both, i.e., you do NLS on period adoption and use Newton-Raphson,
 you get an estimate. Though, I am of course not sure whether that is
 correct in the sense that it is what you would expect to find.


 adoption -
 c(167000,273000,531000,938000,2056452,3894103,5932090,7963742,9314687,10469060,11393302,11976340)
 time - seq(from = 1,to = 12, by = 1)

 adoption2-c(0,adoption[1:(length(adoption)-1)])
 S-(adoption-adoption2)/max(adoption)

 ## Models
 Bass.Model - S ~ M*((p + q)^2/p) * (exp(-(p + q) * time)/((q / p) *
 exp(-(p + q) * time) + 1)^2)
 ## Starting Parameters
 Bass.Params - list(p = 0.1, q = 0.1, M=1)
 ## Model fitting
 Bass.Fit - nls(formula = Bass.Model, start = Bass.Params)
 summary(Bass.Fit)


If your hypothesis regarding the cumulative vs. adoptions is correct
then it may be that poster wants this:

 S - diff(adoption)
 time - seq_along(S)
 Bass2 - S ~ m * ((p + q)^2/p) * (exp(-(p + q) * time)/((q / p) *
+ exp(-(p + q) * time) + 1)^2)
 nls(formula = Bass2, start = c(p = 0.03, q = 0.4, m = max(S)))
Nonlinear regression model
  model:  S ~ m * ((p + q)^2/p) * (exp(-(p + q) * time)/((q/p) *
exp(-(p +  q) * time) + 1)^2)
   data:  parent.frame()
p q m
8.65635536465e-03 6.52817192695e-01 1.23485254536e+07
 residual sum-of-squares: 321990186229

Number of iterations to convergence: 16
Achieved convergence tolerance: 8.10600476229e-06

 # or equivalently in terms of plinear where S, time and Bass2 are
 # as written just above
 m - 1 # set m to 1 since we are using .lin instead
 nls(formula = Bass2, start = c(p = 0.03, q = 0.4), alg = plinear)
Nonlinear regression model
  model:  S ~ m * ((p + q)^2/p) * (exp(-(p + q) * time)/((q/p) *
exp(-(p +  q) * time) + 1)^2)
   data:  parent.frame()
p q  .lin
8.65637919209e-03 6.52816636341e-01 1.23485299874e+07
 residual sum-of-squares: 321990186247

Number of iterations to convergence: 9
Achieved convergence tolerance: 5.6090474901e-06




-- 
Statistics  Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com

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Re: [R] R program writing standard/practices

[Santosh]

Dear Experts,
Thanks to all those who responded!  More requests for suggestions/thoughts:

Along with the above conventions/styles of writing code (as provided in your
links), there is always a tug of war on agreement on the scope/depth of a
program..

Also, there is a carry-over of Java- / C|C++ - style of programming
techniques to catch/trap errors. R programming software inherently has very
nice error trap mechanisms etc, which obviate explicit error trap
programming in many cases.

So, while one does agree that programs with error trap mechanisms are more
robust, the key question remains about drawing the line between simplicity
of a program versus complex robust program.

Simplicity helps when there is a resource crunch and verifying/validating
robust programs would require users/teams to have deeper knowledge of R,
which may not always be available!

Would highly appreciate your ideas on ways to improving code quality, easier
code verification under resource crunch situations.

Regards,
Santosh

On Fri, Jun 10, 2011 at 10:04 AM, vioravis viora...@gmail.com wrote:

 Check this out:

 http://www1.maths.lth.se/help/R/RCC/

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[R] When models and anova(model) disagree...

[Rob James]

I have a situation where the parameter estimates from lrm identify a 
binary predictor variable (X) as clearly non-significant (p0.3), but 
the ANOVA of that same model gives X  a chi^2-df rank of  200, and 
adjudicates X  and one interaction of X and a continuous measure as 
highly significant.  The N is massive and X has two categories, each 
with  100,000 observations.  I would expect X to have a significant 
impact on the outcome.


The full model includes a large number of continuous (coded with rcs 
with 3 knots)  and categorical variables, as well as a plethora of 
interactions between the categorical and continuous variables. Only one 
of the interactions between the binary variable and the other 
categorical or continuous variables is statistically significant.


Can anyone offer a suggestion on what might explain this discordance?

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[R] Trouble with compound functions---differential equations

[Aimee Jones]

Hi all,
My apologies if this message is incredibly inept but I am very new to both
computer programming and to R.

I am working with the odesolve add-on and have the following function
defined

RVF_Single - function(t, x, p)
within the script I also have the following functions defined:

T1-function(t) {T1-27.5-12.5*cos(2*pi*t/365)}
and

B1-function(T1,t) {B1-dnorm(T1(t),mean=22.5,sd=3.3)}



When the script is run it doesn't return an error message but the graphs
returned are wrong. When I input plot(T1,0,3650) it returns the plot of
T1 as expected---a series of waves between 15 and 40, BUT when I input
plot(B1,0,3650) I get an error message of Error in 2 * pi * t : 't' is
missing.




Can anyone advise as to why t registers for function T1 but disappears for
function B1?


Thank you,


Aimee

ps: If it's relevant I'm using R64 (R 2.11.1) on a Mac

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Re: [R] Correlations by subgroups

[jfdawson]

Many thanks - this is exactly what I was hoping for. Very helpful!

Best wishes,
Jeremy

On 15 Jun 2011, at 21:22, Ista Zahn-2 [via R] wrote:

I have to confess that plyr has made me lazy about remembering tapply,
by, aggregate et al., so I'm no help there. But if you want to use
plyr it's just

ddply(dat, .(V1), summarize, cor.v2.v3 = cor(V2, V3))

Best,
Ista

On Wed, Jun 15, 2011 at 10:31 AM, jfdawson [hidden 
email]x-msg://177/user/SendEmail.jtp?type=nodenode=3600456i=0 wrote:

 I'm hoping there is a simple answer to this - it seems that there should be,
 but I can't figure it out.

 I have a matrix/data frame with three variables of interest - V1, V2, V3.
 One, V1, is a factor with x levels (x may be a large number); I want to
 calculate  the correlation between the other two (i.e. cor(V2,V3)) for each
 level, and store it as a vector of length x.

 I should think this should be possible using a function like tapply, but I
 cannot work out what the syntax would be - everything I have tried produces
 either an error, or just repeats the correlations between V2  V3 on the
 whole matrix.

 Could anyone suggest what I should be doing?

 Many thanks,
 Jeremy

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--
Ista Zahn
Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.orghttp://yourpsyche.org/

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Re: [R] R string functions

[karena]

Thank all you guys for the great help~. I appreciate

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Re: [R] Correlations by subgroups

[jfdawson]

Many thanks - this is very helpful indeed!

Best wishes,
Jeremy

On 15 Jun 2011, at 21:58, Daniel Malter [via R] wrote:

x-c(1,1,1,1,1,2,2,2,2,2)
y-rnorm(10)
z-y+rnorm(10)
by(data.frame(y,z),factor(x),cor)

hth,
Daniel

jfdawson wrote:
I'm hoping there is a simple answer to this - it seems that there should be, 
but I can't figure it out.

I have a matrix/data frame with three variables of interest - V1, V2, V3. One, 
V1, is a factor with x levels (x may be a large number); I want to calculate  
the correlation between the other two (i.e. cor(V2,V3)) for each level, and 
store it as a vector of length x.

I should think this should be possible using a function like tapply, but I 
cannot work out what the syntax would be - everything I have tried produces 
either an error, or just repeats the correlations between V2  V3 on the whole 
matrix.

Could anyone suggest what I should be doing?

Many thanks,
Jeremy



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[R] Bayesian Credible Intervals for a Proportion

[tinazilla]

I am trying to calculate Bayesian Credible Intervals for a proportion
(disease prevalence values to be more specific) and am having trouble using
R to do this. I am working with ncredint() function but have not had success
with it. Please help!

Example: 
Positive samples = 3
Total sampled = 10
Prevalence = 0.3

pvec - seq(1,10,by=1)
npost = dbinom(pvec,10,prob=0.3, log=FALSE)
ncredint(pvec, npost, tol=0.01, verbose=FALSE)

But I don't get the right Bayesian CI. What am I doing wrong?

Thanks!
Tina

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Re: [R] Trouble with compound functions---differential equations

[Rolf Turner]

On 16/06/11 11:07, Aimee Jones wrote:

Hi all,
My apologies if this message is incredibly inept but I am very new to both
computer programming and to R.

I am working with the odesolve add-on and have the following function
defined

RVF_Single- function(t, x, p)
within the script I also have the following functions defined:

T1-function(t) {T1-27.5-12.5*cos(2*pi*t/365)}
and

B1-function(T1,t) {B1-dnorm(T1(t),mean=22.5,sd=3.3)}


Actually your code should read:

T1-function(t) {27.5-12.5*cos(2*pi*t/365)}

and

B1-function(T1,t) {dnorm(T1(t),mean=22.5,sd=3.3)}

i.e. don't assign the value that you calculate in the code; this
is the value ***returned*** by the function.  What you is in effect
harmless here, but it is confusing and could cause problems in
other contexts.

When the script is run it doesn't return an error message but the graphs
returned are wrong. When I input plot(T1,0,3650) it returns the plot of
T1 as expected---a series of waves between 15 and 40, BUT when I input
plot(B1,0,3650) I get an error message of Error in 2 * pi * t : 't' is
missing.

Can anyone advise as to why t registers for function T1 but disappears for
function B1?


Well, T1() a function of ***t*** only (where t is the variable against which
you expect the values of T1() to be plotted.  Whereas, B1 is a function of
two variables T1 and t, which confuses things.

Note that by calling plot() in this way you are in fact calling 
plot.function()

which is in fact a wrapper for curve().  As has been discussed recently on
this list, the syntax for curve() is a bit delicate.

A workaround for your problem is:

plot(function(t){B1(T1,t)},0,3650)

HTH

cheers,

Rolf Turner

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Re: [R] Trouble with compound functions---differential equations

[Berend Hasselman]

Aimee Jones wrote:
 
 Hi all,
 My apologies if this message is incredibly inept but I am very new to both
 computer programming and to R.
 
 I am working with the odesolve add-on and have the following function
 defined
 
 RVF_Single - function(t, x, p)
 within the script I also have the following functions defined:
 
 T1-function(t) {T1-27.5-12.5*cos(2*pi*t/365)}
 and
 
 B1-function(T1,t) {B1-dnorm(T1(t),mean=22.5,sd=3.3)}
 
 When the script is run it doesn't return an error message but the graphs
 returned are wrong. When I input plot(T1,0,3650) it returns the plot
 of
 T1 as expected---a series of waves between 15 and 40, BUT when I input
 plot(B1,0,3650) I get an error message of Error in 2 * pi * t : 't' is
 missing.
 
 Can anyone advise as to why t registers for function T1 but disappears for
 function B1?
 

Because B1 is a function with 2 arguments.
plot calls B1 with 1 argument, which will be argument T1. So t is missing
since it hasn't received a value.

Redefine B1 as 

B1-function(t) {B1-dnorm(T1(t),mean=22.5,sd=3.3)}

and you will get your plot.

Berend


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Re: [R] When models and anova(model) disagree...

[Greg Snow]

The individual tests on coefficients in logistic regression are generally based 
on a Wald test statistic.  Unfortunately there is a bit of a paradox possible 
in this case where the coefficient is highly significant, but due to a 
flattening of the likelihood the standard error is overestimated and the 
p-value ends up non-significant.  The anova function uses the likelihood ratio 
test which is not affected by this and is the more trustworthy statistic to use.

I assume that you are using the lrm function from the rms package, the book 
that that package goes along with gives more detail (including the name of the 
paradox which I don't remember at the moment (and my copy of the book is 
currently 40 miles away)).

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of Rob James
Sent: Wednesday, June 15, 2011 4:18 PM
To: r-help@r-project.org
Subject: [R] When models and anova(model) disagree...


I have a situation where the parameter estimates from lrm identify a 
binary predictor variable (X) as clearly non-significant (p0.3), but 
the ANOVA of that same model gives X  a chi^2-df rank of  200, and 
adjudicates X  and one interaction of X and a continuous measure as 
highly significant.  The N is massive and X has two categories, each 
with  100,000 observations.  I would expect X to have a significant 
impact on the outcome.

The full model includes a large number of continuous (coded with rcs 
with 3 knots)  and categorical variables, as well as a plethora of 
interactions between the categorical and continuous variables. Only one 
of the interactions between the binary variable and the other 
categorical or continuous variables is statistically significant.

Can anyone offer a suggestion on what might explain this discordance?

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Re: [R] Bayesian Credible Intervals for a Proportion

[Greg Snow]

It would help us to help you if you told us which package ncredint is in (it is 
not in any of the packages that is installed on the computer I am presently 
using, but could be in multiple others).

Does this interval match what you are expecting?

 library(TeachingDemos)
 hpd(qbeta, shape1=4, shape2=8)
[1] 0.09337233 0.58795256



-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of tinazilla
Sent: Wednesday, June 15, 2011 6:42 PM
To: r-help@r-project.org
Subject: [R] Bayesian Credible Intervals for a Proportion

I am trying to calculate Bayesian Credible Intervals for a proportion
(disease prevalence values to be more specific) and am having trouble using
R to do this. I am working with ncredint() function but have not had success
with it. Please help!

Example: 
Positive samples = 3
Total sampled = 10
Prevalence = 0.3

pvec - seq(1,10,by=1)
npost = dbinom(pvec,10,prob=0.3, log=FALSE)
ncredint(pvec, npost, tol=0.01, verbose=FALSE)

But I don't get the right Bayesian CI. What am I doing wrong?

Thanks!
Tina

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