date:20100209



On Feb 9, 2010, at 4:27 AM, Roslina Zakaria wrote:


Hi R-users,


I have this code below and I understand the error message but do not  
know how to correct it.  My question is how do I get rid of “with  
absolute error  7.5e-06” attach to value of cdf so that I can carry  
out the calculation.


Read the help page. The Value section says that integrate returns a  
list. Access the list by name of desired component:


cdf$value  # should get you the desired number.

--
David.


integrand - function(z)
{ alp  - 2.0165
  rho  - 0.868

# simplified expressions
  a  - alp-0.5
  c1 - sqrt(pi)/(gamma(alp)*(1-rho)^alp)
  c2 - sqrt(rho)/(1-rho)
  t1 - exp(-z/(1-rho))
  t2 - (z/(2*c2))^a
  bes1   - besselI(z*c2,a)
  t1bes1 - t1*bes1
  c1*t1bes1*t2
}

z1  - 20
cdf - integrate(integrand, lower = 0, upper = z1,abs.tol = FALSE)
r   - runif(1)

## Newton iteration
z2  - z1 - (cdf - r)/integrand(z1)

Output


z1  - 20



cdf - integrate(integrand, lower = 0, upper = z1)



r   - runif(1)



z2  - z1 - (cdf - r)/integrand(z1)

Error in cdf - r : non-numeric argument to binary operator


source(.trPaths[5], echo=TRUE, max.deparse.length=1)



cdf

0.9996004 with absolute error  7.5e-06

Thank you so much for any help given.





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David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] Problem in Installing R Packages

2010-02-09 Thread Uwe Ligges

Please see the R for Windows FAQs if you have a problem for downloading 
packages from the internet via R.


Uwe Ligges


On 09.02.2010 10:12, mohsin ali wrote:


Dear Sir



I am facing the following problems in installing packages.

Would you please help in this regards?




install.packages(ctv)

Warning in install.packages(ctv) :
   'lib = C:/PROGRA~2/R/R-210~1.1/library' is not writable
--- Please select a CRAN mirror for use in this session ---
Warning: unable to access index for repository 
http://cran.ms.unimelb.edu.au/bin/windows/contrib/2.10
Warning: unable to access index for repository 
http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.10
Warning messages:
1: In open.connection(con, r) :
   unable to connect to 'cran.r-project.org' on port 80.
2: In getDependencies(pkgs, dependencies, available, lib) :
   package ‘ctv’ is not available




Best Regards

_
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Re: [R] convert R plots into annotated web-graphics

2010-02-09 Thread Duncan Temple Lang

Hi

While there is different level of support for SVG in the different browsers,
basic SVG (non-animation) does work on all of them (with a plugin for IE).

In addition to the 2 SVG packages on CRAN, there is SVGAnnotation at
www.omegahat.org/SVGAnnotation and that is quite a bit more powerful.
There is a link on that page to some examples that are similar to yours.

Imagemaps are a perfectly good way of achieving the interactivity you describe 
and Barry's
imagemap package should make this pretty straightforward.
If all you need is to have event handlers for regions, then imagemaps will be 
fine.
And some JavaScript code will allow you to connect the image map events to 
changing
characteristics of the table.

The rest of this mail is about richer approaches

However, there are other styles of interaction and animation that require 
working at the level of objects on the plot,
i.e. points, lines, text, etc.  When we have these objects at rendering time 
rather than pixels and regions,
we can, e.g., change the color of a point, changing its appearance (color or 
size), hide or move a point, etc.
You need this to do linked plots, for example, i.e. where we mouse over a point 
in one plot or the data table
and highlight the corresponding observations in other plots.

If you want this richer framework, you can generate the plot in R in such a way 
that it will be
displayed in your browser not as a PNG file, but with real objects being 
created within the
rendering.  The SVGAnnotation package does this reasonably comprehensively.

You can generate a plot in R that will be displayed on the JavaScript canvas.
Again, this will create objects and they can then be manipulated by JavaScript
event handlers that work on the plot elements and the table.
There is a prototype  of such an R-JavaScript canvas graphics devices
in the RGraphicsDevice package at www.omegahat.org/RGraphicsDevice.

Also, there is a beta-level Flash device that works at the object level and
allows an R programmer to annotate the resulting plot in either R or 
ActionScript.
(This is at www.omegahat.org/FlashMXML.)
There is another Flash graphics device for R at
https://r-forge.r-project.org/projects/swfdevice/
but this doesn't work at the object-level (at this point in time, at least).


Both the FlashMXML and JavaScript packages rely on the RGraphicsDevice
package and that could be fixed up minorily to handle font metric calculations
with more accuracy (e.g. using RFreetype).


Instead of using an HTML table and modifying it programmatically via CSS
properties, etc., you might use a widget.

   a DataTable widget from the Yahoo UI javascript library.
   a Flash DataGrid to display the data as an interactive table.


As I said, image maps are probably simplest if your needs are reasonably simple.
These other approaches allow for potentially richer Web-based graphics.



Barry Rowlingson wrote:
 On Sun, Feb 7, 2010 at 2:35 PM, Rainer Tischler rainer_...@yahoo.de wrote:
 Dear all,

 I would like to make a large scatter plot created with R available as an 
 interactive web graphic, in combination with additional text-annotations for 
 each data point in the plot. The idea is to present the text-annotations in 
 an HTML-table and inter-link the data points in the plot with their 
 corresponding entries in the table, i.e. when clicking on a data point in 
 the plot, the corresponding entry in the table should be highlighted or 
 centered and vice-versa, when clicking on a table-entry, the corresponding 
 point in the plot should be highlighted.

 I have seen that CRAN contains various R-packages for SVG-based output of 
 interactive graphics (with hyperlinks and tool-tip annotations for each data 
 point); however, SVG is not supported by all browsers. Is anybody aware of 
 another solution for this problem (maybe based on image-maps and javascript)?
 If you have alternative ideas for interlinking tabular annotations with 
 plotted data points, I would appreciate any recommendation/suggestion.
 (I work with R 2.8.1 on different 32-bit PCs with both Linux and Windows 
 operating systems).

 
  My 'imagemaps' package?
 
 https://r-forge.r-project.org/projects/imagemap/
 
 Barry
 
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Re: [R] Hypercube in R

2010-02-09 Thread Uwe Ligges




On 08.02.2010 21:36, Andrej Kastrin wrote:

Dear all,

Does anybody have an idea or suggestion how to construct (plot)
4-dimensional hypercube in R.




Maybe the ggobi interface package is of some help?

Uwe Ligges




Thanks in advance for any pointers.

Regards, Andrej

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[R] linear predictors and survreg function

2010-02-09 Thread carol white

Hi,
I calculated the linear predictors derived from weibull model using ovarian 
data sets. I calculated the linear predictors as the sum of covariates weighted 
by the weibull coefficients and compared to the linear predictors generated by 
survreg function. Why are they different? note that the first element of 
coefficients vector is intercept was excluded in my calculation.

Look forward to your reply,

Carol

--
data(ovarian)
library(survival)
survreg.obj = survreg(Surv(ovarian[,1],ovarian[,2])~age +resid.ds +rx 
+ecog.ps,ovarian, dist = weibull, scale = 1)
 survreg.obj$linear.predictors
 [1] 5.298074 5.108976 5.558852 7.584172 7.221841 7.202655 7.019320 6.764081
 [9] 6.011550 7.939097 7.174129 8.634805 6.783737 7.261585 8.955989 8.366687
[17] 7.970807 8.489844 8.302639 8.385361 7.553247 4.855690 7.851908 7.235689
[25] 6.616655 7.917497

***

lp = survreg.obj$coefficients[2:5]%*%t(ovarian[,3:6])
 lp
 1 2 3 4 5 6 7
[1,] -7.484549 -7.673647 -7.223771 -5.198451 -5.560782 -5.579968 -5.763303
 8 91011121314
[1,] -6.018542 -6.771073 -4.843526 -5.608495 -4.147818 -5.998886 -5.521038
15161718192021
[1,] -3.826634 -4.415936 -4.811816 -4.292779 -4.479984 -4.397262 -5.229376
2223242526
[1,] -7.926933 -4.930715 -5.546934 -6.165968 -4.865126

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[R] step and glmer

2010-02-09 Thread Alba


Is it possible to use the step() function with a glmer() as an object? I
obtain the following error message when I try to do it: Error in x$terms :
$ operator not defined for this S4 class.

I perform the glmer correctly but I can't do the step.

Thank you so much.
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Re: [R] step and glmer

2010-02-09 Thread ONKELINX, Thierry

glmer() nor lmer() work with the step function. You have to build your
models manually.





ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie  Kwaliteitszorg
Gaverstraat 4
9500 Geraardsbergen
Belgium

Research Institute for Nature and Forest
team Biometrics  Quality Assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium

tel. + 32 54/436 185
thierry.onkel...@inbo.be
www.inbo.be

To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher

The plural of anecdote is not data.
~ Roger Brinner

The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey
  

 -Oorspronkelijk bericht-
 Van: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] Namens Alba
 Verzonden: dinsdag 9 februari 2010 14:07
 Aan: r-help@r-project.org
 Onderwerp: [R] step and glmer
 
 
 Is it possible to use the step() function with a glmer() as 
 an object? I obtain the following error message when I try to 
 do it: Error in x$terms :
 $ operator not defined for this S4 class.
 
 I perform the glmer correctly but I can't do the step.
 
 Thank you so much.
 --
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Please do not print this message unnecessarily.

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Re: [R] contour persp

2010-02-09 Thread Peter Ehlers


Further to Uwe's answer: I suspect that you're not telling us
the whole story (that's why it's so useful to have reproducible
code).

Try this:

 x - y - 1:3
 z - outer(x, y)
 z[1,1] - 1/0
 persp(x,y,z)

which results in:
Error in persp.default(x, y, z) : invalid 'z' limits

So, are some your 'NaN's actually 'Inf's?

 -Peter Ehlers

Uwe Ligges wrote:



On 07.02.2010 22:46, Andrew Wang wrote:
I  have this data set that both x  y are ordered vectors of length 
600  700 respectively; z is a 600 by 700 matrix whose entry z[i,j] is 
either a missing value (indicated by 'NaN') or a real number between 0 
and 1.  The contour function


contour(x,y,z)

gives me a blank picture. I guess the reason is that most of z-entries 
are missing, only less than 1% are non missing.


Question (1)

Is there a way that I could manipulate the data or function to have 
the non-missing values plotted?


Also, trying function persp gives me this error message

persp(x,y,z)

Error in persp.default(x, y, z) : invalid 'z' limits

I look at the manual of persp. I guess, the error message comes from 
its internal call


zlim = range(z, na.rm = TRUE)

it appears to me that persp can't handle missing value yet its 
manual states clearly


z: a matrix containing the values to be plotted ('NA's are
   allowed).  Note that ‘x’ can be used instead of ‘z’
   for convenience.


Question (2)

Can persp handle missing values in z? if the answer is a  sounding 
yes, how should I do in my case?





Works for me:
persp(1:2, 1:2, matrix(c(1:3, NA), nrow=2))


Hence you really need to specify an example where it does not work - 
as the posting guide asks you to do anyway.


Uwe Ligges




Please help, Thanks!

Your frustrated

Andrew




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[R] performing the same commands on two different data sets

2010-02-09 Thread kayj


Hi All,


I would like to perform the same set of commands on the two different sets.
I would to avoid writing the same set of command twice so I was wondering if
there is an smart way to do this. I was thinking to use a loop as follows

Data-read.table(file=”data.txt”, header=T, sep=’\t’)
D1-Data[,1:30]
D2-Data[,30:60]

ID-c(“D1”,”D2”) 
for (i in 1:length(ID)){
data-ID[i]

}

I can not assign D1 or D2 as data-ID[i]? is there a way to do this?

Thanks


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[R] Problems installing stats (prcomp) package

2010-02-09 Thread syrvn


Hi,

since nearly 2 hours I try to install the stats package which includes the
prcomp package for
principal component analysis. I can not find the stats package via the R
paket-repository.
I found the following url (http://finzi.psych.upenn.edu/R/library/stats/)
which includes
2 files: stats.rdb and stats.rdx. I downloaded both files and tried to
install them via the R installer
choosing local packets (binary and source) but both variants failed.

Can anyone help?

best
mentor
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[R] Help on R functions

2010-02-09 Thread vikrant


Dear R Experts,
I have written a following function :-

myfunction- function(servername,dbname,dbtablename){
library(RODBC)
channel - odbcDriverConnect(driver=SQL Server;server=servername)
initdata- sqlQuery(channel,paste(select * from  dbname ..  dbtablename))
return(dim(initdata))
}
I have written this function which has input  parameters like servername
,dbname and dbtable to connect to Ms SQL server 2005 and get data from the
table.
Now when I run this function using the following command

myfunction(01wh155073,test_DB,test_vikrant)


The variable initdata should contain all the data. But it does not contain
any data and dim(initdata)) is NULL.

I dont know how to pass the strings as parameters  in the function. Do I
have done this correctly?

If I run the same commands directly from the command line I get the expected
data.

library(RODBC)
channel - odbcDriverConnect(driver=SQL Server;server=01wh155073)
initdata- sqlQuery(channel,paste(select * from  test_DB .. 
test_vikrant))
dim(initdata)

Then the initdata contains the data in the table test_vikrant. My question
is there a way to write above in a function which contains the above
parameters.
Please Help me...
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[R] Missing interaction effect in binomial GLMM with lmer

2010-02-09 Thread Weber, Sam

Dear all,

I was wondering if anyone could help solve a problem of a missing interaction 
effect!!

I carried out a 2 x 2 factorial experiment to see if eggs from 2 different 
locations (Origin =  1 or 2) had different hatching success under 2 different 
incubation schedules (Treat = 1 or 2). Six eggs were taken from 10 females 
(random = Female) at each location and split between the treatments, giving 30 
eggs from each location in each treatment.

Overall proportions hatching were as follows:

Treat
1 2
Origin
1   29/30 5/30
2   29/30   16/30


I made a binomial response in which hatching was a success and not-hatching was 
a failure, and analysed as a binomial GLMM. I'm particularly interested in the 
interaction between the two factors. An expression reproducing the raw data is 
attached at the end of the post in case it is helpful.

hatch.frame$success-cbind(hatch.frame$Hatched,hatch.frame$Nothatched)
model-lmer(success~Origin*Treat+(1|Female),family=binomial,method=ML,data=hatch.frame)
model2-update(model,~.-Origin:Treat)
anova(model,model2)

Data:
Models:
model2: success ~ Origin + Treat + (1 | Female)
model: success ~ Origin * Treat + (1 | Female)
Df AIC BIC   logLik  Chisq  Chi Df   Pr(Chisq)
model2  4  94.707 105.857 -43.353
model   5  95.350 109.287 -42.675 1.3572  1  0.244

model3-update(model2,~.-Origin)
anova(model2,model3)

Data:
Models:
model3: success ~ Treat + (1 | Female)
model2: success ~ Origin + Treat + (1 | Female)
   Df AIC BIC  logLik  Chisq Chi Df Pr(Chisq)
model3  3  98.863 107.225 -46.431
model2  4  94.707 105.857 -43.353 6.1558  10.01310 *
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

model4-update(model2,~.-Treat)
anova(model2,model4)

Data:
Models:
model4: success ~ Origin + (1 | Female)
model2: success ~ Origin + Treat + (1 | Female)
   Df AIC BIC  logLik  Chisq Chi Df Pr(Chisq)
model4  3 155.592 163.954 -74.796
model2  4  94.707 105.857 -43.353 62.885  1  2.191e-15 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
So the model implies that there is a very significant effect of treatment 
(reduced hatching at treatment 2) with a small effect of origin (improved 
hatching from origin 2). However the lack of interaction effect implies 
hatching was better for Origin 2 at both treatments, which looking at the raw 
values above does not seem to be the case. Identical numbers of eggs hatched 
from both Origins in Treatment 1, but much more from Origin 2 hatched at 
Treatment 2.

If you divide the analysis by treatments, Origin only has a significant effect 
on hatching under Treatment 2 and not with Treatment 1

Hot-data.frame(hatch.frame[hatch.frame$Treat==2,])
Cold-data.frame(hatch.frame[hatch.frame$Treat==1,])

#2
model-lmer(success~Origin+(1|Female),family=binomial,method=ML,data=Hot)
model2-update(model,~.-Origin)
anova(model,model2)
Data: Hot
Models:
model2: incubate ~ (1 | Code)
model: incubate ~ Origin + (1 | Code)
   Df AIC BIC  logLik  Chisq Chi Df Pr(Chisq)
model2  2  78.633  82.821 -37.316
model   3  73.697  79.980 -33.848 6.9357  1   0.008449 **



#1

model-lmer(success~Origin+(1|Female),family=binomial,method=ML,data=Cold)
model2-update(model,~.-Origin)
anova(model,model2)

Data: Cold
Models:
model2: incubate ~ (1 | Code)
model: incubate ~ Origin + (1 | Code)
   Df AIC BIC  logLik  Chisq Chi Df Pr(Chisq)
model2  2 21.5086 25.6973 -8.7543
model   3 23.3472 29.6302 -8.6736 0.1615  1 0.6878





So I can't understand where the interaction effect has gone in the full model?! 
I get the same result in a binomial GLM, without the random effect of Female 
i.e. a small effect of origin but no interaction with treatment. I'm sure I 
must be missing something here so I would be very grateful to anyone who can 
point out my mistakes. I've read previous R Help posts that suggest binomial 
GLM(M) can create problems when estimated probabilities are close to 0 or 1. In 
Treatment 1 hatching probability was 0.97 for both Origins, so could this be 
the source of the problem?



Thanks for your help



Sam Weber



University of Exeter
Centre for Ecology and Conservation
Tremough Campus
Penryn
Cornwall TR10 9EZ
UK




hatch.frame -
structure(list(Female = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L,
2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L,
4L, 5L, 5L, 5L, 5L, 5L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 7L, 7L, 7L,
7L, 7L, 7L, 8L, 8L, 8L, 8L, 8L, 8L, 9L, 9L, 9L, 9L, 9L, 9L, 10L,
10L, 10L, 10L, 10L, 10L, 11L, 11L, 11L, 11L, 11L, 11L, 12L, 12L,
12L, 12L, 12L, 12L, 13L, 13L, 13L, 13L, 13L, 13L, 14L, 14L, 14L,
14L, 14L, 14L, 15L, 15L, 15L, 15L, 15L, 15L, 16L, 16L, 16L, 16L,
16L, 16L, 17L, 17L, 17L, 17L, 17L, 17L, 18L, 18L, 18L, 18L, 18L,
18L, 19L, 19L, 19L,

[R] transparent concentric circles

2010-02-09 Thread Karin Lagesen

I have a data set which I would like to plot as a set of concentric
circles. The data represent a count of the number of characteristics
shared by various elements - an example would look like this:

1 100
2 75
3 50
4 25

I.e. all four sets share 25 characteristics, three of them share 50
characteristics, and so on.

I would like to plot these as concentric circles, with the circle size
preferentially being proportional to the size of the number of elements
(this is not a must, however). I would also like the colors of the circles
to become stronger/deeper as we progress to the innermost circle (which
would be the one containing the number of characteristics shared by all
four).

Can somebody point me to what I can use to do this?

Thanks!



Karin
-- 
Karin Lagesen, PhD

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Re: [R] R ggplot2 legend text left justify

Hi Paul,

That's a bug in the current version of ggplot.  I'm working on update
for later this week.

Hadley

On Mon, Feb 8, 2010 at 5:56 PM, Paul Sutcliffe psutc...@it.uts.edu.au wrote:


 In ggplot2 how do you justify the legend text ?
 In the example below the opts(legend.text = theme_text(size = 9,hjust=0))
  changes the size of the text OK but it remains right justified.


 mydata=data.frame(RowID=c(A,B,C),Name=c(long long long long long
 name,short name ,medium medium name),Speed=c(100,140,120))
 mydata
  RowID                          Name Speed
 1     A long long long long long name   100
 2     B                   short name    140
 3     C            medium medium name   120
 ggplot(mydata, aes(RowID,Speed))+ opts(legend.text = theme_text(size =
 9,hjust=0)) + geom_bar() + xlab(Name) + aes(colour = Name) +aes(fill=Name)


 Thank in advance
 Paul


 --
 Dr. Paul J Sutcliffe


 Casual Academic
 Faculty of Engineering and IT, University of Technology, Sydney (UTS) email:
 psutc...@it.uts.edu.au
 http://www-staff.it.uts.edu.au/~psutclif/
 CRICOS Provider Code 00099F

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Re: [R] performing the same commands on two different data sets

2010-02-09 Thread Kjetil Halvorsen

Did you try  data-ID[[i]]  instead of your data-ID[i]?

Kjetil

On Tue, Feb 9, 2010 at 11:19 AM, kayj kjaj...@yahoo.com wrote:

 Hi All,


 I would like to perform the same set of commands on the two different sets.
 I would to avoid writing the same set of command twice so I was wondering if
 there is an smart way to do this. I was thinking to use a loop as follows

 Data-read.table(file=”data.txt”, header=T, sep=’\t’)
 D1-Data[,1:30]
 D2-Data[,30:60]

 ID-c(“D1”,”D2”)
 for (i in 1:length(ID)){
 data-ID[i]

 }

 I can not assign D1 or D2 as data-ID[i]? is there a way to do this?

 Thanks


 --
 View this message in context: 
 http://n4.nabble.com/performing-the-same-commands-on-two-different-data-sets-tp1474452p1474452.html
 Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Help on R functions

2010-02-09 Thread Duncan Murdoch


On 09/02/2010 7:10 AM, vikrant wrote:

Dear R Experts,
I have written a following function :-

myfunction- function(servername,dbname,dbtablename){
library(RODBC)
channel - odbcDriverConnect(driver=SQL Server;server=servername)
initdata- sqlQuery(channel,paste(select * from  dbname ..  dbtablename))
return(dim(initdata))
}
I have written this function which has input  parameters like servername
,dbname and dbtable to connect to Ms SQL server 2005 and get data from the
table.
Now when I run this function using the following command

myfunction(01wh155073,test_DB,test_vikrant)


The variable initdata should contain all the data. But it does not contain
any data and dim(initdata)) is NULL.

I dont know how to pass the strings as parameters  in the function. Do I
have done this correctly?


No.  driver=SQL Server;server=servername is a literal string.  If you 
want to replace servername with the contents of the servername argument, use


paste(driver=SQL Server;server=, servername, sep=)

and similarly for your query.

Duncan Murdoch



If I run the same commands directly from the command line I get the expected
data.

library(RODBC)
channel - odbcDriverConnect(driver=SQL Server;server=01wh155073)
initdata- sqlQuery(channel,paste(select * from  test_DB .. 
test_vikrant))

dim(initdata)

Then the initdata contains the data in the table test_vikrant. My question
is there a way to write above in a function which contains the above
parameters.
Please Help me...


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Re: [R] Problems installing stats (prcomp) package

2010-02-09 Thread Sarah Goslee

The stats package should be installed automatically, and loaded by
default when you start R.

The search() command will show you which packages are loaded,
for example on my computer starting with R --vanilla:
 search()
[1] .GlobalEnvpackage:stats package:graphics
[4] package:grDevices package:utils package:datasets
[7] package:methods   Autoloads package:base

The stats package is already loaded.

If for some reason this is untrue for you, we need more information,
as requested in the posting guide, including OS and R version.

Sarah


On Tue, Feb 9, 2010 at 7:09 AM, syrvn ment...@gmx.net wrote:

 Hi,

 since nearly 2 hours I try to install the stats package which includes the
 prcomp package for
 principal component analysis. I can not find the stats package via the R
 paket-repository.
 I found the following url (http://finzi.psych.upenn.edu/R/library/stats/)
 which includes
 2 files: stats.rdb and stats.rdx. I downloaded both files and tried to
 install them via the R installer
 choosing local packets (binary and source) but both variants failed.

 Can anyone help?

 best
 mentor

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

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Re: [R] performing the same commands on two different data sets

2010-02-09 Thread Sarah Goslee

You skipped a step: the get() function allows you to use the name
of an object to load the object itself.

ID-c(“D1”,”D2”)

for (i in 1:length(ID)){
   data- get(ID[i])
   do stuff with data
}

Alternatively you could write a function that takes the object
as its argument.

myFunction - function(data) {
   do stuff with data
}

myFunction(D1)
myFunction(D2)

Naming your objects data and Data is a recipe for confusion, though.

Sarah

On Tue, Feb 9, 2010 at 9:19 AM, kayj kjaj...@yahoo.com wrote:

 Hi All,


 I would like to perform the same set of commands on the two different sets.
 I would to avoid writing the same set of command twice so I was wondering if
 there is an smart way to do this. I was thinking to use a loop as follows

 Data-read.table(file=”data.txt”, header=T, sep=’\t’)
 D1-Data[,1:30]
 D2-Data[,30:60]

 ID-c(“D1”,”D2”)
 for (i in 1:length(ID)){
 data-ID[i]

 }

 I can not assign D1 or D2 as data-ID[i]? is there a way to do this?

 Thanks


 --
 View this message in context: 
 http://n4.nabble.com/performing-the-same-commands-on-two-different-data-sets-tp1474452p1474452.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.




-- 
Sarah Goslee
http://www.stringpage.com
http://www.astronomicum.com
http://www.functionaldiversity.org

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Re: [R] Fast way to determine number of lines in a file

 I was looking for a fast line counter as well a while ago and ended up
 writing a small function in R:

  countLines() in the R.utils package

 At least at the time, it was faster than readLines() [for unknown
 reasons].  It is also more memory efficient.  It supports connections.
  I don't think it beats a system call to 'wc', though.  When there
 will be a faster solution available, it'll be calling that instead.
 It does not avoid reading the file twice.

Perfect - exactly what I was looking for.

Thanks!

Hadley

-- 
http://had.co.nz/

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Re: [R] ggplot2 stacked line plot

Hi Liam,

Your syntax is a little off.  You want:

p - ggplot2(~, aes(x = ~, y = ~, colour = Type)) +
  geom_area(aes(fill = Type), position = 'stack')

Position isn't an aesthetic.

Hadley

On Sun, Feb 7, 2010 at 10:40 PM, Liam Blanckenberg
liam.blanckenb...@gmail.com wrote:
 Hi all,

 I have been hunting around for hours trying to figure out how to
 generate a stacked line chart using ggplot2. This type of chart can be
 generated in excel 2007 by selecting: Chart type  Line  Stacked
 line. I can generate a stacked area chart using the following code:

   p - ggplot2(~, aes(x = ~, y = ~, colour = Type)) +
 geom_area(aes(position = 'stack', fill = Type))

 However, when I try and replicate this using the following code for geom_line:

   p - ggplot(~, aes(x = ~, y = ~, colour = Type)) +
 geom_line(aes(position = 'stack'))

 the resulting plot is not stacked - i.e. each 'Type' is plotted at its
 actual value rather than cumulatively to form a stacked chart... I
 have poured through Hadley's ggplot2 book (ggplot2: elegant graphics
 for data analysis), the R help list and also done general google
 searching but cannot find a way to generate this type of plot.

 R version: 2.9.2
 ggplot2 version: 0.8.5
 OS: windows 7 (64-bit).

 Any suggestions or assistance would be greatly appreciated.

 Regards,

 Liam

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Re: [R] transparent concentric circles

2010-02-09 Thread Barry Rowlingson

On Tue, Feb 9, 2010 at 2:20 PM, Karin Lagesen kar...@cbs.dtu.dk wrote:
 I have a data set which I would like to plot as a set of concentric
 circles. The data represent a count of the number of characteristics
 shared by various elements - an example would look like this:

 1 100
 2 75
 3 50
 4 25

 I.e. all four sets share 25 characteristics, three of them share 50
 characteristics, and so on.

 I would like to plot these as concentric circles, with the circle size
 preferentially being proportional to the size of the number of elements
 (this is not a must, however). I would also like the colors of the circles
 to become stronger/deeper as we progress to the innermost circle (which
 would be the one containing the number of characteristics shared by all
 four).

 Can somebody point me to what I can use to do this?

 help.search(circle)?

Have you tried any of those? Specifically:

plotrix::draw.circleDraw a circle.
shape::filledcircle adds colored circle to a plot
grid::grid.circle   Draw a Circle

 - assuming you have those packages loaded...

Barry

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[R] question about nlme...

2010-02-09 Thread Jeff Harring

I am looking for R code to be able to fit a linear-linear piecewise 
model with person-specific changepoint. I have searched the web, but 
have not been able to locate any code.


Below is my attempt at some code:

chgpt = function(a1,a2,a3,gam,wave){
yht=numeric(10)
y1=(wave = gam)*(a1+(a2*wave))
y2=(wave  gam)*((a1+(a2-a3)*gam)+a3*wave)
yhat=y1+y2
return(yht)
}

nl.dat - nlme(y ~ chgpt(a1,a2,a3,gam,wave)~wave|id,
   data = ndat,
   fixed = a1 + a2 + a3 + gam,
   random = a1 + a2 + a3,
   start = c(a1=10,a2=5,a3=1,gam=4.5))
summary(n1.dat)

I would appreciate any advice on this code or guidance as to where to 
search for example code to fit this nonlinear mixed effects model.


Many thanks,
Jeff Harring

--
**
Jeffrey R. Harring, Assistant Professor
Department of Measurement, Statistics  Evaluation (EDMS)
1230 Benjamin Building
University of Maryland
College Park, MD 20742-1115

Phone:  301.405.3630
Fax:301.314.9245
Email:  harr...@umd.edu
Web:http://www.education.umd.edu/EDMS/fac/Harring/webpage.html

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Re: [R] performing the same commands on two different data sets

2010-02-09 Thread kayj


thank you all for your help, the get() function worked perfectly.

-- 
View this message in context: 
http://n4.nabble.com/performing-the-same-commands-on-two-different-data-sets-tp1474452p1474515.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] step and glmer

2010-02-09 Thread Alba


Thank you for your rapid answer!
-- 
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http://n4.nabble.com/step-and-glmer-tp1474390p1474580.html
Sent from the R help mailing list archive at Nabble.com.

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[R] Interpretation of high order interaction terms.

I have difficulties in interpreting high order interaction terms in
high-way ANOVA.

According to Introductory Statistics with R by Peter Dalgaard (Section 12.5),

The exact definition of the interaction terms and the interpretation of their
associated regression coefficients can be elusive. Some peculiar things
happen if an interaction term is present but one or more of the main
effects are missing. The full details are probably best revealed through
experimentation.


However, I do want to understand the full details. I have checked a
few books, but I haven't found a book that describes high order
interactions in a satisfactory way. Besides learning interactions
through experimentation, could somebody let me know if there are any
detailed documentation on interactions?

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Re: [R] interpreting error estimate in SEM

2010-02-09 Thread Jarrett Byrnes

If it is not standardized, I have the following script that I use to  
calculate the r^2 for each variable of a model.  I do wonder, however,  
if there is a way to calculate a model-wide r^2 like index.

-Jarrett

##
#returns the r^2 for endogenous variables and error path coefficients
#
# Last updated 8/18/08
##

rsquare.sem-function (sem.obj){
output.info-list()

for (index in 1:length(sem.obj$S[,1])){
var.name-sem.obj$var.names[index]
obs.var-sem.obj$S[index,index]
est.var-sem.obj$P[index,index]
r.sq-1-(est.var/obs.var)
std.error.coef-sqrt(1-r.sq)

#if it's 0, or close to (due to rounding error), it's exogenous
if(r.sq  1e-10){
output.info-c(output.info, var.name, obs.var, est.var, r.sq,  
std.error.coef)
}

}

output.matrix-matrix(output.info, ncol=5, byrow=TRUE)
colnames(output.matrix)-c(Variable, Observed.Variance,  
Estimated.Variance, R^2, Standardized.Error.Coefficient)
output.matrix
}






Jarrett Byrnes
Postdoctoral Associate, Santa Barbara Coastal LTER
Marine Science Institute
University of California Santa Barbara
Santa Barbara, CA 93106-6150
http://www.lifesci.ucsb.edu/eemb/labs/cardinale/people/byrnes/index.html

On Feb 9, 2010, at 4:17 AM, John Fox wrote:

 Dear Kathryn,

 I assume that MWDError is the error variance associated with MWD. If  
 MWD is
 a standardized variable (i.e., with a variance of 1) then 59% of its
 variance is unaccounted for.

 I hope this helps,
 John

 
 John Fox
 Senator William McMaster
  Professor of Social Statistics
 Department of Sociology
 McMaster University
 Hamilton, Ontario, Canada
 web: socserv.mcmaster.ca/jfox

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org 
 ]
 On
 Behalf Of Kathryn Barto
 Sent: February-09-10 6:23 AM
 To: r-help@r-project.org
 Subject: [R] interpreting error estimate in SEM

 hi,

 I'm using the sem package, and I want to make sure I'm interpreting  
 the
 output correctly. Here is an excerpt of my output. When I report  
 the MWD
 error, should I say that 59% of the variation in MWD is not  
 explained by
 the
 model, or that 59% of the variation is explained, or something  
 entirely
 different.

 Parameter Estimates
   Estimate  Std Error z value  Pr(|z|)
 NplantRoots 0.462029   0.120803   3.82466 1.3095e-04
 plantRoots
 --
 - N
 ...
 CoError 1.11 0.151850   6.58553 4.5325e-11 Co -- Co
 carbonateError  1.09 0.156217   6.40142 1.5394e-10 carbonate
 --
 carbonate
 MWDError0.589788   0.092128   6.40185 1.5351e-10MWD  
 --
 MWD

 Thanks
 Kathryn

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[R] Resampling a grid to coarsen its resolution

2010-02-09 Thread Steve Murray


Dear all,

I have a grid (data frame) dataset at 0.5 x 0.5 degrees spatial resolution (720 
columns x 360 rows; regular spacing) and wish to coarsen this to a resolution 
of 2.5 x 2.5 degrees. A simple calculation which takes the mean of a block of 
points to form the regridded values would do the trick. Values which should be 
excluded from the calculation are - (unless all points within a block are 
-, in which case - should be returned as the 'new' cell).

How would I go about achieving this in R?

Any help or guidelines would be very much appreciated.

Many thanks,

Steve


  
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[R] Formula used to create new plot

2010-02-09 Thread Timothée Poisot

Dear list users,

I was wondering if there is any way to know, for an open graphical peripheric, 
the instruction that was used to call it. For example, if I create a plot using

plot(runif(10))

is there any way to have the call returned to me? That would be especially 
usefull to know it there are any log axes, for example

thanks
t



---
Timothée POISOT
-
Institut des Sciences de l'Evolution
Université Montpellier 2, CC 065
Place Eugène Bataillon
34095 Montpellier CEDEX 5
-
Phone   :   (+33)4 67 14 40 61
Fax :   (+33)4 67 14 40 61
E-mail  :   tpoi...@um2.fr
Web :   http://www.timotheepoisot.fr/
---


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[R] How can I rearange my dataframe

2010-02-09 Thread Alex Levitchi

Hello 
I am recently began to work with R, so I am not so experienced. 
But anyway I cannot find a clear way to process my dataframe which is a bigger 
one. 
It shows similar to this 

 name=c(A,B,C,B,C,C,C,B,C) 
 nicknames=c(A1,B1,C1,B2,C2,C3,C4,B3,C5) 
 value=c(4,5,9,2,7,6,3,6,7) 
 table=data.frame(cbind(name,nickname,value)) 
 table=data.frame(cbind(name,nicknames,value)) 
 table 
name nicknames value 
1 A A1 4 
2 B B1 5 
3 C C1 9 
4 B B2 2 
5 C C2 7 
6 C C3 6 
7 C C4 3 
8 B B3 6 
9 C C5 7 

So I have to rearrange it in the next way: 
- the first column should contain just unduplicated data, I did this, it is OK 
and it will look like 
1 A 
2 B 
3 C 

- the second column should contain different 'nicknames' which correspond to 
the single A, B or C 
name nickname value 
1 A A1 
2 B B1,B2,B3 
3 C C1,C2,C3,C4,C5 

-the third one should contain the mean value of the numbers which correspond to 
the same A, B or C 
1 A A1 mean(4) 
2 B B1,B2,B3 mean(5,2,6) 
3 C C1,C2,C3,C4,C5 mean(9,7,6,3,7) 

I did this using a loop 'for'. 
to be clear I created tree dataframes which correspond to each of columns, and 
finally will combine them 

 ulist=which(!duplicated(table$name)) # I extract the list of positions in 
 which I don't have duplications 
 name1=data.frame(table$name[ulist]) # I extract the list of unique names 
 nicknames1=data.frame(row.names(1:length(ulist))) # I create a dataframe of 
 dimension equal to unique list length 
 value1=data.frame(row.names(1:length(ulist))) # I create a dataframe of 
 dimension equal to unique list length 

 for(i in 1:length(ulist)) { 
position=which(as.character(name1[i,1])==table$name) 
nicknames1[i,1]=toString(table$nicknames[position]) 
value1[i,1]=mean(as.numeric(table$value[position])) 
} 
 fin=cbind(name1,nicknames1,value1) 
 colnames(fin)=c(NAME,NICKNAME,VALUE) 
 fin 
NAME NICKNAME VALUE 
1 A A1 3.00 
2 B B1, B2, B3 3.33 
3 C C1, C2, C3, C4, C5 5.20 

it works successfully. But in general I work with dataframes of high dimensions 
(tens thousands or more rows). 
So my loop works too slow (i.e., a dataframe of 2 rows and 3 columns is 
processed in about 10 minutes). 
I intend to integrate it into a function, so it is obvious that time will be 
even longer. 

If someone can advise me any possibility to modify which I have done or to the 
way I can do it, please give me a message. 

King regards to all guys who develop and maintain R sources for such dummies as 
me 
Alex Levitchi 



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Re: [R] Conditional sampling?

2010-02-09 Thread Olga Lyashevska


Hello all,

Here is my solution, in case someone else needs it.


I have a dataframe consisting of two columns.


col1-factor(c(a,a,b,b,c,c))
col2-factor(c(a,b,c,d,e,f))
somedf-data.frame(col1,col2)
somedf

col1   col2
1  a  d
2  a  e
3  b  f
4  b  g
5  c  h
6  c  i


sample(col1,2,replace=T)

[1] b c
Levels: a b c

Now I want to sample from col2, but I want to restrict a vector I am
going to sample from to only those elements that correspond to col1  
That
is, I want to take a sample from f, g, h, i.  Elements corresponding  
to

level a need to be dropped.

   col1col2

3  b  f
4  b  g
5  c  h
6  c  i




s - sample(col1,2,replace=T)
col2[col1 %in% s]

Cheers,
Olga

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[R] Nonparametric alternative to LDA

2010-02-09 Thread Robert Lonsinger

Can anybody recommend a nonparametric alternative to linear discriminant
function that may be available as a package in R?

Cheers,
Rob

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[R] Re-execute previous command

2010-02-09 Thread mnstn


Hello All,
In bash, to re-execute a command, say,  'gnuplot plot.gnu', one can type
!gnu. Is there a similar feature in the R console? For example:
plot(g1$mean,g2$mean)
.
.
.
Lines, many lines of commands
-Here I want to repeat the previous plot command.
!plot and plot(+TAB) do not work.

This question was asked in this forum back in 2006 but the answer seems to
have been lost somewhere in their discussions. This is the thread
http://n4.nabble.com/R-command-line-need-intelligent-command-history-recall-td793072.html#a793075
. Can some one comment if there is a way to do it?

Thank you,
MoonStone
-- 
View this message in context: 
http://n4.nabble.com/Re-execute-previous-command-tp1474629p1474629.html
Sent from the R help mailing list archive at Nabble.com.

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[R] Comparing means and trends in short time-series

2010-02-09 Thread Simone Vincenzi

Dear R-list, 
 I have a statistical problem with the comparison of short time-series,
representing densities of fish in different streams. For each stream (6 in
total, here below showed only part of the dataset) I have 8 years of density
data, as follows:

 

   year density stream

1  2000 0.51  stream1

2  2001 0.87  stream1

3  2002 0.68  stream1

4  2003 0.56  stream1

5  2004 0.50  stream1

6  2005 0.51  stream1

7  2006 1.07  stream1

8  2007 1.02  stream1

9  2000 0.23  stream2

10 2001 0.43  stream2

11 2002 0.56  stream2

12 2003 0.59  stream2

13 2004 0.52  stream2

14 2005 0.36  stream2

15 2006 0.28  stream2

16 2007 0.38  stream2

17 2000 0.07  stream3

18 2001 0.06  stream3

.

. 

There is a clear problem of non-indipendence of data, so this seem to
preclude a classical ANOVA and subsequent pairwise comparisons. What I would
like to test is: are mean densities different among streams from 2000 to
2007? Is it sufficient and appropriate a test of correlation (using rcorr of
the library Design, for example) to test if density trends in time are
similar among streams? 

Any help would be greatly appreciated.

 

Simone

 

Simone Vincenzi, PhD 
Department of Environmental Sciences
University of Parma



 


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Re: [R] How can I rearange my dataframe

2010-02-09 Thread jim holtman

try this:

 x - read.table(textConnection(name nicknames value
+ 1 A A1 4
+ 2 B B1 5
+ 3 C C1 9
+ 4 B B2 2
+ 5 C C2 7
+ 6 C C3 6
+ 7 C C4 3
+ 8 B B3 6
+ 9 C C5 7), header=TRUE)
 closeAllConnections()
 result - do.call(rbind, lapply(split(x, x$name), function(.name){
+ data.frame(name=.name$name[1], nicknames=paste(.name$nicknames,
collapse=','),
+ mean=mean(.name$value))
+ }))

 result
  name  nicknames mean
AA A1 4.00
BB   B1,B2,B3 4.33
CC C1,C2,C3,C4,C5 6.40



On Tue, Feb 9, 2010 at 11:24 AM, Alex Levitchi alex.levit...@cbm.fvg.it wrote:
 Hello
 I am recently began to work with R, so I am not so experienced.
 But anyway I cannot find a clear way to process my dataframe which is a 
 bigger one.
 It shows similar to this

 name=c(A,B,C,B,C,C,C,B,C)
 nicknames=c(A1,B1,C1,B2,C2,C3,C4,B3,C5)
 value=c(4,5,9,2,7,6,3,6,7)
 table=data.frame(cbind(name,nickname,value))
 table=data.frame(cbind(name,nicknames,value))
 table
 name nicknames value
 1 A A1 4
 2 B B1 5
 3 C C1 9
 4 B B2 2
 5 C C2 7
 6 C C3 6
 7 C C4 3
 8 B B3 6
 9 C C5 7

 So I have to rearrange it in the next way:
 - the first column should contain just unduplicated data, I did this, it is 
 OK and it will look like
 1 A
 2 B
 3 C

 - the second column should contain different 'nicknames' which correspond to 
 the single A, B or C
 name nickname value
 1 A A1
 2 B B1,B2,B3
 3 C C1,C2,C3,C4,C5

 -the third one should contain the mean value of the numbers which correspond 
 to the same A, B or C
 1 A A1 mean(4)
 2 B B1,B2,B3 mean(5,2,6)
 3 C C1,C2,C3,C4,C5 mean(9,7,6,3,7)

 I did this using a loop 'for'.
 to be clear I created tree dataframes which correspond to each of columns, 
 and finally will combine them

 ulist=which(!duplicated(table$name)) # I extract the list of positions in 
 which I don't have duplications
 name1=data.frame(table$name[ulist]) # I extract the list of unique names
 nicknames1=data.frame(row.names(1:length(ulist))) # I create a dataframe of 
 dimension equal to unique list length
 value1=data.frame(row.names(1:length(ulist))) # I create a dataframe of 
 dimension equal to unique list length

 for(i in 1:length(ulist)) {
 position=which(as.character(name1[i,1])==table$name)
 nicknames1[i,1]=toString(table$nicknames[position])
 value1[i,1]=mean(as.numeric(table$value[position]))
 }
 fin=cbind(name1,nicknames1,value1)
 colnames(fin)=c(NAME,NICKNAME,VALUE)
 fin
 NAME NICKNAME VALUE
 1 A A1 3.00
 2 B B1, B2, B3 3.33
 3 C C1, C2, C3, C4, C5 5.20

 it works successfully. But in general I work with dataframes of high 
 dimensions (tens thousands or more rows).
 So my loop works too slow (i.e., a dataframe of 2 rows and 3 columns is 
 processed in about 10 minutes).
 I intend to integrate it into a function, so it is obvious that time will be 
 even longer.

 If someone can advise me any possibility to modify which I have done or to 
 the way I can do it, please give me a message.

 King regards to all guys who develop and maintain R sources for such dummies 
 as me
 Alex Levitchi



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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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Re: [R] How can I rearange my dataframe



On Feb 9, 2010, at 11:24 AM, Alex Levitchi wrote:


Hello
I am recently began to work with R, so I am not so experienced.
But anyway I cannot find a clear way to process my dataframe which  
is a bigger one.

It shows similar to this


name=c(A,B,C,B,C,C,C,B,C)
nicknames=c(A1,B1,C1,B2,C2,C3,C4,B3,C5)
value=c(4,5,9,2,7,6,3,6,7)
table=data.frame(cbind(name,nickname,value))
table=data.frame(cbind(name,nicknames,value))
table

name nicknames value
1 A A1 4
2 B B1 5
3 C C1 9
4 B B2 2
5 C C2 7
6 C C3 6
7 C C4 3
8 B B3 6
9 C C5 7

So I have to rearrange it in the next way:
- the first column should contain just unduplicated data, I did  
this, it is OK and it will look like

1 A
2 B
3 C

- the second column should contain different 'nicknames' which  
correspond to the single A, B or C

name nickname value
1 A A1
2 B B1,B2,B3
3 C C1,C2,C3,C4,C5


Dataframes are not designed to hold irregular length items. Lists are  
the data structure best suited for this type of data. tapply() is one  
function useful for colecting elements of one structure based on the  
contents of another (name):


(I renamed your table object table1 to avoid confusion with the  
table function.)


 tapply(table1$nicknames, table1$name, list)
$A
[1] A1
Levels: A1 B1 B2 B3 C1 C2 C3 C4 C5

$B
[1] B1 B2 B3
Levels: A1 B1 B2 B3 C1 C2 C3 C4 C5

$C
[1] C1 C2 C3 C4 C5
Levels: A1 B1 B2 B3 C1 C2 C3 C4 C5

The process of tabulating has created factor variables which some  
would see as a good thing, but perhaps was not desired. Since you now  
have a lis,  you can sequentially apply the as.character function to  
recover only the character vectors:


lapply( tapply(table1$nicknames, table1$name, list), as.character)
$A
[1] A1

$B
[1] B1 B2 B3

$C
[1] C1 C2 C3 C4 C5

Then I saw the rest of your request, so forget the above and see if  
this two-liner looks a bit more simple.


 tcollapse - tapply(table1$nicknames, table1$name,  paste,   
collapse=, )

#gets you the strings separated by commas and spaces.

 cbind(names(tcollapse), tcollapse, lapply( tapply(table1$nicknames,  
table1$name, list), length)  )

  tcollapse
A A A1 1
B B B1, B2, B3 3
C C C1, C2, C3, C4, C5 5

You can obviously name them whatever you like.

--
David


-the third one should contain the mean value of the numbers which  
correspond to the same A, B or C

1 A A1 mean(4)
2 B B1,B2,B3 mean(5,2,6)
3 C C1,C2,C3,C4,C5 mean(9,7,6,3,7)

I did this using a loop 'for'.
to be clear I created tree dataframes which correspond to each of  
columns, and finally will combine them


ulist=which(!duplicated(table$name)) # I extract the list of  
positions in which I don't have duplications
name1=data.frame(table$name[ulist]) # I extract the list of unique  
names
nicknames1=data.frame(row.names(1:length(ulist))) # I create a  
dataframe of dimension equal to unique list length
value1=data.frame(row.names(1:length(ulist))) # I create a  
dataframe of dimension equal to unique list length



for(i in 1:length(ulist)) {

position=which(as.character(name1[i,1])==table$name)
nicknames1[i,1]=toString(table$nicknames[position])
value1[i,1]=mean(as.numeric(table$value[position]))
}

fin=cbind(name1,nicknames1,value1)
colnames(fin)=c(NAME,NICKNAME,VALUE)
fin

NAME NICKNAME VALUE
1 A A1 3.00
2 B B1, B2, B3 3.33
3 C C1, C2, C3, C4, C5 5.20

it works successfully. But in general I work with dataframes of high  
dimensions (tens thousands or more rows).
So my loop works too slow (i.e., a dataframe of 2 rows and 3  
columns is processed in about 10 minutes).
I intend to integrate it into a function, so it is obvious that time  
will be even longer.


If someone can advise me any possibility to modify which I have done  
or to the way I can do it, please give me a message.


King regards to all guys who develop and maintain R sources for such  
dummies as me

Alex Levitchi



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


David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] Re-execute previous command

On Feb 9, 2010, at 11:21 AM, mnstn wrote:

Hello All,
In bash, to re-execute a command, say, 'gnuplot plot.gnu', one can
type

!gnu. Is there a similar feature in the R console? For example:
plot(g1$mean,g2$mean)

In most system typeing up-arrow will prompt the console to echo the
last command (and repeating the process will scroll back through
history.)

There is also a .Last.value object but I don't know if it will replot.

--
David.

.
.
.
Lines, many lines of commands
-Here I want to repeat the previous plot command.
!plot and plot(+TAB) do not work.

This question was asked in this forum back in 2006 but the answer
seems to

have been lost somewhere in their discussions. This is the thread
http://n4.nabble.com/R-command-line-need-intelligent-command-history-recall-td793072.html#a793075
. Can some one comment if there is a way to do it?

Thank you,
MoonStone
--
View this message in context:
http://n4.nabble.com/Re-execute-previous-command-tp1474629p1474629.html
Sent from the R help mailing list archive at Nabble.com.

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

David Winsemius, MD
Heritage Laboratories
West Hartford, CT

Re: [R] Missing interaction effect in binomial GLMM with lmer

2010-02-09 Thread Ista Zahn

Hi Sam,
Good question. I originally guessed that the simple effect (I know
some people on this list don't seem to care for that term, but it's
always made sense to me) coefficients were in the same direction, such
that the effect if Origin at Treat=hot was significantly different
from zero, but not from the effect of Origin at Treat = cold. But a
quick look indicated that is not the case:

contrasts(hatch.frame$Treat) - contr.treatment(2, base=1)
model1-lmer(success~Origin*Treat+(1|Female),family=binomial,REML=TRUE,data=hatch.frame)
summary(model1)

Generalized linear mixed model fit by the Laplace approximation
Formula: success ~ Origin * Treat + (1 | Female)
  Data: hatch.frame
  AIC   BIC logLik deviance
 95.34 109.3 -42.6785.34
Random effects:
 Groups NameVariance Std.Dev.
 Female (Intercept) 0.54993  0.74157
Number of obs: 120, groups: Female, 20

Fixed effects:
   Estimate Std. Error z value Pr(|z|)
(Intercept) 3.609227   1.146844   3.147  0.00165 **
Origin2-0.004192   1.606214  -0.003  0.99792
Treat2 -5.401703   1.238911  -4.360  1.3e-05 ***
Origin2:Treat2  1.948242   1.697945   1.147  0.25121
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Correlation of Fixed Effects:
   (Intr) Orign2 Treat2
Origin2 -0.714
Treat2  -0.889  0.635
Orign2:Trt2  0.649 -0.907 -0.730

contrasts(hatch.frame$Treat) - contr.treatment(2, base=2)
model2-lmer(success~Origin*Treat+(1|Female),family=binomial,REML=TRUE,data=hatch.frame)
summary(model2)

Generalized linear mixed model fit by the Laplace approximation
Formula: success ~ Origin * Treat + (1 | Female)
  Data: hatch.frame
  AIC   BIC logLik deviance
 95.34 109.3 -42.6785.34
Random effects:
 Groups NameVariance Std.Dev.
 Female (Intercept) 0.54993  0.74157
Number of obs: 120, groups: Female, 20

Fixed effects:
  Estimate Std. Error z value Pr(|z|)
(Intercept) -1.7925 0.5683  -3.154  0.00161 **
Origin2  1.9441 0.7190   2.704  0.00686 **
Treat1   5.4017 1.2389   4.360  1.3e-05 ***
Origin2:Treat1  -1.9484 1.6979  -1.148  0.25116
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Correlation of Fixed Effects:
   (Intr) Orign2 Treat1
Origin2 -0.790
Treat1  -0.385  0.305
Orign2:Trt1  0.281 -0.336 -0.730


So I'm as stumped as you are. How can the effect of Origin at
treat=hot be significantly different from zero, but not significantly
different from -0.004? Clearly there is something here I'm not
understanding. I'm very curious to know the answer.

Best,
Ista

On Tue, Feb 9, 2010 at 12:22 PM, Weber, Sam sam.we...@exeter.ac.uk wrote:
 Dear all,

 I was wondering if anyone could help solve a problem of a missing interaction 
 effect!!

 I carried out a 2 x 2 factorial experiment to see if eggs from 2 different 
 locations (Origin =  1 or 2) had different hatching success under 2 different 
 incubation schedules (Treat = 1 or 2). Six eggs were taken from 10 females 
 (random = Female) at each location and split between the treatments, giving 
 30 eggs from each location in each treatment.

 Overall proportions hatching were as follows:

 Treat
                        1                     2
 Origin
 1                   29/30                 5/30
 2                   29/30               16/30


 I made a binomial response in which hatching was a success and not-hatching 
 was a failure, and analysed as a binomial GLMM. I'm particularly interested 
 in the interaction between the two factors. An expression reproducing the raw 
 data is attached at the end of the post in case it is helpful.

 hatch.frame$success-cbind(hatch.frame$Hatched,hatch.frame$Nothatched)
 model-lmer(success~Origin*Treat+(1|Female),family=binomial,method=ML,data=hatch.frame)
 model2-update(model,~.-Origin:Treat)
 anova(model,model2)

 Data:
 Models:
 model2: success ~ Origin + Treat + (1 | Female)
 model: success ~ Origin * Treat + (1 | Female)
            Df     AIC     BIC       logLik  Chisq  Chi Df   Pr(Chisq)
 model2  4  94.707 105.857 -43.353
 model   5  95.350 109.287 -42.675 1.3572      1      0.244

 model3-update(model2,~.-Origin)
 anova(model2,model3)

 Data:
 Models:
 model3: success ~ Treat + (1 | Female)
 model2: success ~ Origin + Treat + (1 | Female)
       Df     AIC     BIC  logLik  Chisq Chi Df Pr(Chisq)
 model3  3  98.863 107.225 -46.431
 model2  4  94.707 105.857 -43.353 6.1558      1    0.01310 *
 ---
 Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

 model4-update(model2,~.-Treat)
 anova(model2,model4)

 Data:
 Models:
 model4: success ~ Origin + (1 | Female)
 model2: success ~ Origin + Treat + (1 | Female)
       Df     AIC     BIC  logLik  Chisq Chi Df Pr(Chisq)
 model4  3 155.592 163.954 -74.796
 model2  4  94.707 105.857 -43.353 62.885      1  2.191e-15 ***
 ---
 Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
 So the model implies that there is a very significant effect of

[R] odfTable: table width and alignment

2010-02-09 Thread Aleksey Naumov

Dear R and odfWeave users,

I am trying to figure out how to control table width and alignment on the
page for a table generated by odfTable. Based on reading odfWeave
documentation (including formattingOut.odt), here is how I manipulate the
styles:

st = getStyleDefs()
# modify the table style
tab = getStyles()$table
st[[tab]]$align = center# seems to have no effect
st[[tab]]$marginLeft = 2.0 in # seems to have no effect
setStyleDefs(st)

My table always ends up fully justified (taking all page width). When I
check Table Format in the output .odf, Alignment is always Automatic. When
doing Column/Optimal Width on the table, the table shrinks but becomes left
aligned, not centered.

The table style is understood by odfWeave -- here is the output I get after
sourcing my style definition file:

 getStyles()$table
[1] RTable1
 getStyleDefs()[[getStyles()$table]]
$type
[1] Table

$marginLeft
[1] 2.0 in

$marginRight
[1] 0.05in

$marginTop
[1] 0.05in

$marginBottom
[1] 0.05in

$align
[1] center

I am not sure why these style options do not seem to have effect on the
output. My specific questions are:

(1) How do I get table alignment to work?

(2) Is the only way to control table width via setting the margins in the
$table style? What am I doing wrong in the above style code?

(3) What about column width -- is it possible to automatically size columns
to fit the content, like selecting Column/Optimal Width but programmatically
via odfWeave?


Thank you,

Aleksey

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[R] The default argument 'envir' of eval()

In ?eval, it says

    ...  If ‘envir’ is not
    specified, then the default is ‘parent.frame()’ (the environment
    where the call to ‘eval’ was made).

I tried the following example with eval(expr)  and eval(expr,
parent.frame()) in f(). The results are different, which are not
consistent with the help. Could somebody let know whether I
misunderstand the help? Or there is something with the help?


 expr=expression(print(paste('x =', x)))
 x=0
 f=function() {
+   x=2
+   eval(expr)
+   eval(expr, parent.frame())
+   eval(expr, parent.frame(2))
+   eval(expr, parent.frame(3))
+ }
 g=function() {
+   x=1
+   f()
+ }

 g()
[1] x = 2
[1] x = 1
[1] x = 0
[1] x = 0

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Re: [R] Re-execute previous command

On Feb 9, 2010, at 12:13 PM, David Winsemius wrote:

On Feb 9, 2010, at 11:21 AM, mnstn wrote:

Hello All,
In bash, to re-execute a command, say, 'gnuplot plot.gnu', one can
type

!gnu. Is there a similar feature in the R console? For example:
plot(g1$mean,g2$mean)

In most system typeing up-arrow will prompt the console to echo
the last command (and repeating the process will scroll back through
history.)

There is also a .Last.value object but I don't know if it will
replot.

It will if the plotting function returned a value as do the lattice
and ggplot functions, but not if it returns NULL as do the base
graphics functions.

--
David.

.
.
.
Lines, many lines of commands
-Here I want to repeat the previous plot command.
!plot and plot(+TAB) do not work.

This question was asked in this forum back in 2006 but the answer
seems to

Thank you,
MoonStone
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David Winsemius, MD
Heritage Laboratories
West Hartford, CT

[R] Kernel density / weights matrix?

2010-02-09 Thread Stephan Lindner

Dear everyone,


I'm coding the Horowitz-Spokoiny (2001) test [1], and I would be very
grateful or some advice regarding the Kernel density (apologies
beforehand if my terminology is not fully correct). I have looked into
ksmooth and npreg, but with no success. 

Given a (n x p) matrix of covariates X, I need to construct the
following matrix of Kernel densities or weights:

w(x_i, x_j) = 

K(x_i - x_j)
 -  
   sum_{k=1}^n K(x_i - x_k)


where x_i, x_j, x_k are (1 x p) vectors, and K is a multivariate normal
kernel. The resulting weighting matrix W has dimension (n x n). 


I have looked into npreg, but if I get this correctly, it does not
output this weighting matrix. I do need the weighting matrix itself
for the test statistic, and not just the kernel regression
estimates. I can construct it myself, but I thought I'd ask around
before doing so.

Best,


Stephan


[1] Horowitz Joel L. and Spokoiny Vladimir G. (2001): An Adaptive,
Rate-Optimal Test of a Parametric Mean-Regression Model against a
Nonparametric Alternative. Econometrica, Vol. 69, No. 3 (May, 2001),
pp. 599-631







-- 
---
Stephan Lindner
University of Michigan

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Re: [R] The default argument 'envir' of eval()

2010-02-09 Thread Duncan Murdoch


On 09/02/2010 12:18 PM, bluesky...@gmail.com wrote:

In ?eval, it says

...  If ‘envir’ is not
specified, then the default is ‘parent.frame()’ (the environment
where the call to ‘eval’ was made).

I tried the following example with eval(expr)  and eval(expr,
parent.frame()) in f(). The results are different, which are not
consistent with the help. Could somebody let know whether I
misunderstand the help? Or there is something with the help?
  


The help is correct, you misunderstood it.  Default values for arguments 
are evaluated in the evaluation frame of the function.  Explicit 
arguments are evaluated in the evaluation frame of the caller.  
parent.frame() has different meanings in those two contexts.


Duncan Murdoch


 expr=expression(print(paste('x =', x)))
 x=0
 f=function() {
+   x=2
+   eval(expr)
+   eval(expr, parent.frame())
+   eval(expr, parent.frame(2))
+   eval(expr, parent.frame(3))
+ }
 g=function() {
+   x=1
+   f()
+ }

 g()
[1] x = 2
[1] x = 1
[1] x = 0
[1] x = 0

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[R] Bar plot

2010-02-09 Thread Sunitap22


Hello 

(this might be a very simple question) 
My data is as follows (table name is student)
   YearStudentsPassed
1  2000300
2  2001360
3  2002450
4  2003450
5  2004270
6  2005280
7  2006400
8  2007270

I want to plot a barplot so for this 

If I use barplot(students) it says Error in barplot.default(students) :
'height' must be a vector or a matrix

so I used the following command
barplot(StudentsPassed)

But this doesnt give me the names for Xaxis, how can I get the names for
Xaxis, also how can I get labels for X and Y axis, and tittle, I tried 

title(main = Year Vs Number of Publications)
to get the title but its giving me an error

Thanks in advance
Sunita
I want to 
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Re: [R] Comparing means and trends in short time-series

2010-02-09 Thread Dieter Menne



Simone Vincenzi-2 wrote:
 
  I have a statistical problem with the comparison of short time-series,
 representing densities of fish in different streams. For each stream (6 in
 total, here below showed only part of the dataset) I have 8 years of
 density
 data, as follows:
 
  
year density stream
 
 1  2000 0.51  stream1
 
 2  2001 0.87  stream1
 Are mean densities different among streams from 2000 to
 2007? Is it sufficient and appropriate a test of correlation (using rcorr
 of
 the library Design, for example) to test if density trends in time are
 similar among streams? 
 
 

Have a look at function gls in in package nlme.

gls(density~year*stream,)

There are some similar examples in the book coming with nlme (chapter 5.4)

Dieter




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Re: [R] question about nlme...

2010-02-09 Thread Dieter Menne



Jeff Harring wrote:
 
 I am looking for R code to be able to fit a linear-linear piecewise 
 model with person-specific changepoint. I have searched the web, but 
 have not been able to locate any code.
 
 ..
 I would appreciate any advice on this code or guidance as to where to 
 search for example code to fit this nonlinear mixed effects model.
 
 

I had good success in fitting piecewise functions with continuous
derivatives, e.g. a logistic function with a linear continuation. Things did
not work well with nlme when there was a discontinuity. There is an
discussion from 2000 right from Douglas Bates' and Mary Lindstrom's mouth
for nls and the hockey-stick, so it could be possible to get is to work with
good data.

http://www.biostat.wustl.edu/archives/html/s-news/2000-04/msg00209.html

Also check

http://markmail.org/message/6p63i4cxfk6enlp5

Dieter


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Re: [R] Nonparametric alternative to LDA

2010-02-09 Thread Andrej Kastrin

Classification trees (rpart package), but know too little about your
problem...

Best, Andrej

On Feb 9, 5:50 pm, Robert Lonsinger rob.lonsin...@gmail.com wrote:
 Can anybody recommend a nonparametric alternative to linear discriminant
 function that may be available as a package in R?

 Cheers,
 Rob

         [[alternative HTML version deleted]]

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[R] model II major axis regression

2010-02-09 Thread garth


Hello all; My question is part statistical and part R. 

I have performed model II major axis regression in R using both the smatr()
and lmodel2() packages, but neither offers an option to statistically weight
my regression. I have a vector of weights which I would like to apply to
each of my regression points (x vs. y), and was hoping that someone out
there could provide me with some help on this issue. I have searched on the
R help pages and online with no luck.

Just to provide context, I am comparing 2 random variables which have been
geographically gridded so that each grid point has one x data and one y data
point. I would like to weight the regression of x vs. y such that grids with
larger geographic area (near the equator) get more weight than those with
smaller geographic area (near the poles). My weights are already calculated. 

Thanks in advance,
Dan.
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Re: [R] Bar plot

2010-02-09 Thread Peter Ehlers


Here is a simple 3-step solution:

1. type ?barplot
2. find the section labelled 'Arguments'
3. carefully read what each argument means/does

Voila, problem(s) solved.

(It was good to include some data, but not so good
to say that title(...) is giving an error. No need
to be secretive about the error; just say what it was.)

 -Peter Ehlers

Sunitap22 wrote:
Hello 

(this might be a very simple question) 
My data is as follows (table name is student)

   YearStudentsPassed
1  2000300
2  2001360
3  2002450
4  2003450
5  2004270
6  2005280
7  2006400
8  2007270

I want to plot a barplot so for this 


If I use barplot(students) it says Error in barplot.default(students) :
'height' must be a vector or a matrix

so I used the following command
barplot(StudentsPassed)

But this doesnt give me the names for Xaxis, how can I get the names for
Xaxis, also how can I get labels for X and Y axis, and tittle, I tried 


title(main = Year Vs Number of Publications)
to get the title but its giving me an error

Thanks in advance
Sunita
I want to 


--
Peter Ehlers
University of Calgary

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Re: [R] Missing interaction effect in binomial GLMM with lmer

2010-02-09 Thread Bert Gunter

You might do better posting this on the R-sig-mixed-models list. 

Bert Gunter
Genentech Nonclinical Statistics

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Ista Zahn
Sent: Tuesday, February 09, 2010 9:19 AM
To: Weber, Sam
Cc: r-help@R-project.org
Subject: Re: [R] Missing interaction effect in binomial GLMM with lmer

Hi Sam,
Good question. I originally guessed that the simple effect (I know
some people on this list don't seem to care for that term, but it's
always made sense to me) coefficients were in the same direction, such
that the effect if Origin at Treat=hot was significantly different
from zero, but not from the effect of Origin at Treat = cold. But a
quick look indicated that is not the case:

contrasts(hatch.frame$Treat) - contr.treatment(2, base=1)
model1-lmer(success~Origin*Treat+(1|Female),family=binomial,REML=TRUE,data=
hatch.frame)
summary(model1)

Generalized linear mixed model fit by the Laplace approximation
Formula: success ~ Origin * Treat + (1 | Female)
  Data: hatch.frame
  AIC   BIC logLik deviance
 95.34 109.3 -42.6785.34
Random effects:
 Groups NameVariance Std.Dev.
 Female (Intercept) 0.54993  0.74157
Number of obs: 120, groups: Female, 20

Fixed effects:
   Estimate Std. Error z value Pr(|z|)
(Intercept) 3.609227   1.146844   3.147  0.00165 **
Origin2-0.004192   1.606214  -0.003  0.99792
Treat2 -5.401703   1.238911  -4.360  1.3e-05 ***
Origin2:Treat2  1.948242   1.697945   1.147  0.25121
---
Signif. codes:  0 *** 0.001 ** 0.01 * 0.05 . 0.1   1

Correlation of Fixed Effects:
   (Intr) Orign2 Treat2
Origin2 -0.714
Treat2  -0.889  0.635
Orign2:Trt2  0.649 -0.907 -0.730

contrasts(hatch.frame$Treat) - contr.treatment(2, base=2)
model2-lmer(success~Origin*Treat+(1|Female),family=binomial,REML=TRUE,data=
hatch.frame)
summary(model2)

Generalized linear mixed model fit by the Laplace approximation
Formula: success ~ Origin * Treat + (1 | Female)
  Data: hatch.frame
  AIC   BIC logLik deviance
 95.34 109.3 -42.6785.34
Random effects:
 Groups NameVariance Std.Dev.
 Female (Intercept) 0.54993  0.74157
Number of obs: 120, groups: Female, 20

Fixed effects:
  Estimate Std. Error z value Pr(|z|)
(Intercept) -1.7925 0.5683  -3.154  0.00161 **
Origin2  1.9441 0.7190   2.704  0.00686 **
Treat1   5.4017 1.2389   4.360  1.3e-05 ***
Origin2:Treat1  -1.9484 1.6979  -1.148  0.25116
---
Signif. codes:  0 *** 0.001 ** 0.01 * 0.05 . 0.1   1

Correlation of Fixed Effects:
   (Intr) Orign2 Treat1
Origin2 -0.790
Treat1  -0.385  0.305
Orign2:Trt1  0.281 -0.336 -0.730

So I'm as stumped as you are. How can the effect of Origin at
treat=hot be significantly different from zero, but not significantly
different from -0.004? Clearly there is something here I'm not
understanding. I'm very curious to know the answer.

Best,
Ista

On Tue, Feb 9, 2010 at 12:22 PM, Weber, Sam sam.we...@exeter.ac.uk wrote:
 Dear all,

 I was wondering if anyone could help solve a problem of a missing
interaction effect!!

 I carried out a 2 x 2 factorial experiment to see if eggs from 2 different
locations (Origin =  1 or 2) had different hatching success under 2
different incubation schedules (Treat = 1 or 2). Six eggs were taken from 10
females (random = Female) at each location and split between the treatments,
giving 30 eggs from each location in each treatment.

 Overall proportions hatching were as follows:

 Treat
                        1                     2
 Origin
 1                   29/30                 5/30
 2                   29/30               16/30

 I made a binomial response in which hatching was a success and
not-hatching was a failure, and analysed as a binomial GLMM. I'm
particularly interested in the interaction between the two factors. An
expression reproducing the raw data is attached at the end of the post in
case it is helpful.

 hatch.frame$success-cbind(hatch.frame$Hatched,hatch.frame$Nothatched)

model-lmer(success~Origin*Treat+(1|Female),family=binomial,method=ML,data
=hatch.frame)
 model2-update(model,~.-Origin:Treat)
 anova(model,model2)

 Data:
 Models:
 model2: success ~ Origin + Treat + (1 | Female)
 model: success ~ Origin * Treat + (1 | Female)
            Df     AIC     BIC       logLik  Chisq  Chi Df   Pr(Chisq)
 model2  4  94.707 105.857 -43.353
 model   5  95.350 109.287 -42.675 1.3572      1      0.244

 model3-update(model2,~.-Origin)
 anova(model2,model3)

 Data:
 Models:
 model3: success ~ Treat + (1 | Female)
 model2: success ~ Origin + Treat + (1 | Female)
       Df     AIC     BIC  logLik  Chisq Chi Df Pr(Chisq)
 model3  3  98.863 107.225 -46.431
 model2  4  94.707 105.857 -43.353 6.1558      1    0.01310 *
 ---
 Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

 model4-update(model2,~.-Treat)
 anova(model2,model4)

 Data:
 Models:
 model4:

Re: [R] Interpretation of high order interaction terms.

2010-02-09 Thread RICHARD M. HEIBERGER

## Artificial data with all interactions significant.

## The interaction2wt plot shows all main effects and all pairwise
## interactions.  We see in the Y ~ A|B panel (or in the
## interaction.plot) that Y goes uphill for levels 1 and 2 of B and
## goes down and then up for level 3 of B.  This is the two-way A:B
## interaction.  At each level of B, the main effect of A differs.

## From the xyplot, we see that at level 1 of C, Y goes down for level
## 3 of B.  At level 2 of C, Y goes down a lot and then up for level 3
## of B.  This is the three-way A:B:C interaction.  At each level of C
## the two-way interaction of A and B differs.


set.seed(1)

require(HH)  ## needed for interaction2wt()

threeway -
data.frame(matrix(c(
1, 1, 1, 1,
2, 2, 1, 1,
3, 3, 1, 1,
1, 1, 2, 1,
2, 2, 2, 1,
3, 3, 2, 1,
3, 1, 3, 1,
2, 2, 3, 1,
1, 3, 3, 1,
1, 1, 1, 2,
2, 2, 1, 2,
3, 3, 1, 2,
1, 1, 2, 2,
2, 2, 2, 2,
3, 3, 2, 2,
3, 1, 3, 2,
-4, 2, 3, 2,
1, 3, 3, 2),
byrow=TRUE, 18, 4, dimnames=list(1:18,c(Y,A,B,C
for (i in 2:4) threeway[[i]] - factor(threeway[[i]])
threeway - rbind(threeway, threeway)
threeway$Y - threeway$Y + rnorm(36, s=.5)


anova(aov(Y ~ A * B * C, data=threeway))

with(threeway, interaction.plot(A, B, Y))  ## shows just the Y ~ A|B panel

## all two-way interactions and main effects
interaction2wt(Y ~ A + B + C, data=threeway)  ## library(HH) required

xyplot(x ~ A | C, groups=B,
   data=aggregate(threeway[,1], threeway[,-1], mean), type=l,
   auto.key=list(title=B, space=left,
 border=TRUE, lines=TRUE, points=FALSE),
   strip=strip.custom(strip.names=c(TRUE,TRUE)))

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Re: [R] Re-execute previous command

2010-02-09 Thread mnstn


Thanks David,
I guess I have to settle with the upper-arrow thing I have been doing all
along. My figures do not return any value so ggplot, lattice are out.
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Re: [R] Dickey Fuller test of a time series (problem)

2010-02-09 Thread sarathsv


i also having the problem with R .


R version 2.10.0 (2009-10-26)
Copyright (C) 2009 The R Foundation for Statistical Computing
ISBN 3-900051-07-0

R is free software and comes with ABSOLUTELY NO WARRANTY.
You are welcome to redistribute it under certain conditions.
Type 'license()' or 'licence()' for distribution details.

  Natural language support but running in an English locale

R is a collaborative project with many contributors.
Type 'contributors()' for more information and
'citation()' on how to cite R or R packages in publications.

Type 'demo()' for some demos, 'help()' for on-line help, or
'help.start()' for an HTML browser interface to help.
Type 'q()' to quit R.

[Previously saved workspace restored]

 z-rnorm(289)
 z
  [1]  2.330811003  1.504743043  1.768175890  0.629578885 -1.102466476 
1.022360003 -0.683596222  0.825797728  1.970304092
 [10] -0.434295690 -0.332111909  0.784754124  0.618029090 -0.323679397 
0.477139243  0.567437753 -0.082967056 -0.631445108
 [19] -0.599448056 -0.023043350  0.254275136  0.729704319 -0.557885077 
0.136863688  0.336083722 -0.574337438  0.489163710
 [28] -1.178968998 -0.500117153  0.194962840  1.077193868 -0.224323646 
0.830018849 -0.923133317  0.984637207 -0.967962617
 [37] -0.330364545  0.411477240  1.921856365 -0.402244385  0.287249079
-0.254981296  1.373816270  0.051474667  0.467294146
 [46] -0.084979045  1.010413164 -1.685650845 -0.225540927 -0.068473164 
0.076474205 -0.599026450  0.749528909 -0.005966977
 [55] -0.751531710 -0.566290715 -0.653128084  1.252402084 -0.475794837 
0.361830550  0.534634320  0.128657947 -0.998795343
 [64]  0.439908085 -0.616926920  0.616190407 -0.503356528 -1.571513634
-0.862492045 -0.858421322  0.37011  0.026994800
 [73] -0.072956095  0.750615120 -0.111619011  0.073884163 -0.962739147 
0.038873386 -0.343490682  0.752229210  0.965300992
 [82] -1.604473652  1.516414656 -0.938925299 -0.709151677  0.231854623
-0.218100137  1.543797726 -0.898485417 -0.383154763
 [91]  1.676661233 -0.050011605 -0.885938130 -0.759522611  0.825827497 
0.307787209 -1.582181323  1.548739934  0.874375153
[100] -1.929437918 -1.257364398 -1.789263683 -2.143550087 -0.299557119
-0.034432098  0.523581523  1.282782038 -0.627147412
[109]  1.379406957 -1.943450866 -0.109843039 -0.186738958  1.940752202
-0.399888402  0.889084467  0.347721291  0.221675595
[118]  1.014829243  0.815928201 -1.289821785  1.086385072 -0.299486086
-0.126573107  0.527587325 -1.988292033 -0.661784968
[127] -0.739984305  0.043729158  0.780508168 -0.329149317 -0.239561967 
0.479590885  0.483343301 -1.330296792 -1.347459911
[136] -0.588658836 -0.954908406 -0.215942979 -0.079013859 -0.386107147 
1.030946757  1.352902085  1.621552841  0.951461132
[145]  0.107764563 -0.143231089  0.837469606 -0.687510567  1.376661975 
0.837165093  0.063985577  1.381120151 -0.628153036
[154] -1.144266557  1.005945165  1.085202192  0.273625164  0.022451305
-0.372955252  0.283434935 -1.174292734  1.513095774
[163] -1.406613019 -0.954491994  0.275374664 -0.404428637 -0.791690077
-0.325169327  0.859515784 -1.605660446 -0.126481055
[172] -1.163844796  0.234669024  0.694055806 -0.503758028  0.747620192
-0.490237653 -0.673188493  0.597849322  0.508355060
[181]  0.220080232  1.254662349 -1.390019760  1.036511650  0.004499064
-0.171119971  0.341759420  0.883209195  0.196981832
[190]  1.93083 -0.472635901 -0.143698739  0.134008071  0.123061173
-0.847030934  0.478631946 -0.640264555  1.198188793
[199]  0.928747360 -1.223396075  0.042572331 -1.193181087 -0.284040641
-0.511904510 -0.591542425  0.763156940  0.649463947
[208] -0.762889670 -0.628481227  0.367263423 -0.286380274  0.354460516
-0.138982551 -1.184283483  0.489375881  1.005739798
[217]  0.267274437 -0.118960041 -1.207006456  0.461973434  0.325420233 
0.996123459  1.188010215 -2.000847324  1.043974965
[226]  0.356219212  0.620522833  3.800773985 -2.080091995  0.322824104
-1.793810727  0.49250 -0.092310467  2.893818046
[235] -0.846021981  0.402650410 -0.426312400  1.162983880  2.076818805
-0.260109032  0.725039578  0.361726563 -0.116508468
[244]  0.820778041  1.664995327  0.361725669  0.577447944 -1.008407703 
0.390553465  0.263064760  0.915409450  0.294579966
[253]  1.478028593  1.865177412 -1.050708541 -2.071520363  0.450842362 
0.530261688  0.226103509 -0.875767768  0.255960500
[262] -0.675455904 -1.436531605 -0.715532812  0.498582850  0.463511831 
0.568583368  0.337032963 -0.31489 -0.738982923
[271]  1.740242504  1.089138147 -0.128574330  1.834612295 -0.657243928 
0.701904808  1.550270461 -0.234375326 -0.420885683
[280]  1.845942481 -0.895760996 -0.593765244  0.586912606 -1.062899762 
0.277845308 -0.834841694 -0.614720628 -0.343506642
[289]  0.442537573
 x - data.frame(z) 
 x
   z
12.330811003
21.504743043
31.768175890
40.629578885
5   -1.102466476
61.022360003
7   -0.683596222
80.825797728
91.970304092
10  -0.434295690
11  -0.332111909
12   0.784754124
13   0.618029090
14  -0.323679397
15   0.477139243
16

[R] how to rest the variables

2010-02-09 Thread kayj


Hi ,

I have several commands that I want to run, I was wondering if there is an
easy way to reset my variables before the start of each run

for ( i in 1:200){
reset the variables 
run the commands
}

My solution right now is to set all my vectors to 0 and my data frames to
NULL after the start of the loop. Is there an easy way to do this?

Thanks,




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

Re: [R] how to rest the variables

2010-02-09 Thread jim holtman

What do you mean by 'reset'?  Why not put your processing in a
function that you call so that each time you invoke it, any variables
local to the function start in an 'initialized' state (depends on how
you are using them).  If you need them to have some specific value,
then put in the code necessary since R does not know which ones you
want initialized and which ones you want to keep the value for.  Are
you thinking about some other language that has this feature?  If so
how is it used there?

On Tue, Feb 9, 2010 at 2:02 PM, kayj kjaj...@yahoo.com wrote:

 Hi ,

 I have several commands that I want to run, I was wondering if there is an
 easy way to reset my variables before the start of each run

 for ( i in 1:200){
 reset the variables
 run the commands
 }

 My solution right now is to set all my vectors to 0 and my data frames to
 NULL after the start of the loop. Is there an easy way to do this?

 Thanks,




 --
 View this message in context: 
 http://n4.nabble.com/how-to-rest-the-variables-tp1474869p1474869.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.




-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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Re: [R] Installation require proxy settings (continued)

Timing can be an issue.  You need to set the environment variables before 
accessing the internet for the 1st time in each R session.  If you get the 
error, then try setting the variables, it will not work.

For my situation I have http_proxy and http_proxy_user set for the entire 
computer (in windows you do this through the control panel).  

If you want to go with the .Renviron route, just type: Sys.getenv('home')
At the command prompt and it will tell you what your Home folder is.  Just 
create the .Renviron file in that folder.



-- 
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-boun...@r-
 project.org] On Behalf Of Van Wyk, Jaap
 Sent: Monday, February 08, 2010 10:11 PM
 To: r-help@r-project.org
 Subject: Re: [R] Installation require proxy settings (continued)
 Importance: High
 
 Thanks to Greg and Michael for their advice.
 I have tried all these possibilities (and other); nothing seems to
 work; I persistently get the message:
 In open.connection(con, r) :
   cannot open: HTTP status was '407 Proxy Authentication Required'
 It seems as if proxy authentication required is causing the problem.
 It seems I have to set up an .Renviron file, but I am still learning R,
 and cannot find such a file.
 Would it help, and how do I create it (it that is a possible
 alternative to try).
 Thanks for any comments.
 Jacob
 
 
 Jacob L van Wyk
 Department of Statistics
 University of Johannesburg (APK)
 PO Box 524, Auckland Park, 2006
 Office ph: 011 559 3080
 Fax: 011 559 2499
 
 
 
 This email and all contents are subject to the following disclaimer:
 
 http://www.uj.ac.za/UJ_email_legal_disclaimer.htm
 
   [[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.

[R] subset in a matrix

2010-02-09 Thread DonDiego


Hi,

I have a matrix of data values like the example bellow. I would like to
extract a subset of the matrix for the values where the first column is
negative.  I am using the subset function. However, I am getting an error
message that the conditional variable doe snot exist.  For some reason, the
subset operation only works if I transform the matrix to a data set using
as.data.set(). The help indicates that the subset function can be applied to
matrixes and data sets. I am wondering if anyone has seen a similar problem
before. am I using the correct syntax?



n = 15
m = 5

cnames = paste(x,1:m,sep=)
rnames = 1:n

z = matrix(rnorm(n*m),n,m,dimnames =list(rnames,cnames))


Thanks,

Jorge




test = subset(z,x1  0, select = c(cnames))

-- 
View this message in context: 
http://n4.nabble.com/subset-in-a-matrix-tp1474958p1474958.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] Re-execute previous command

2010-02-09 Thread akea013


On Feb 10, 5:21 am, mnstn pavan.n...@gmail.com wrote:
 Hello All,
 In bash, to re-execute a command, say,  'gnuplot plot.gnu', one can type
 !gnu. Is there a similar feature in the R console? For example:
 plot(g1$mean,g2$mean)
 .
 .
 .
 Lines, many lines of commands
 -Here I want to repeat the previous plot command.
 !plot and plot(+TAB) do not work.

 This question was asked in this forum back in 2006 but the answer seems to
 have been lost somewhere in their discussions. This is the 
 threadhttp://n4.nabble.com/R-command-line-need-intelligent-command-history-...
 . Can some one comment if there is a way to do it?

 Thank you,
 MoonStone
 --

In bash, you can also use
Ctrl-R plot
to see the previous plot command -
then
Ctrl-R
repeatedly to cycle through them all.
Ctrl-S
to search forward on the same text.
Press enter to execute the one you want - or edit, etc.
This also works in the R console for me.
--
Anita

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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] Installation require proxy settings (continued)

2010-02-09 Thread Prof Brian Ripley


On Tue, 9 Feb 2010, Greg Snow wrote:


Timing can be an issue.  You need to set the environment variables before 
accessing the internet for the 1st time in each R session.  If you get the 
error, then try setting the variables, it will not work.

For my situation I have http_proxy and http_proxy_user set for the entire 
computer (in windows you do this through the control panel).

If you want to go with the .Renviron route, just type: Sys.getenv('home')
At the command prompt and it will tell you what your Home folder is.  Just 
create the .Renviron file in that folder.


I believe the OP specified internet2.dll on installation.  It might be 
better not to do so ... otherwise you are at the mercy of IE 
internals.  The rw-FAQ does say (Q2.19)


(a) Use the alternative internet2.dll by starting R with the flag 
--internet2 (see How do I install R for Windows?) or calling 
setInternet2(TRUE). These cause R to use the Internet Explorer 
internals.. Note that this does not work with proxies that need 
authentication.



--
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-boun...@r-
project.org] On Behalf Of Van Wyk, Jaap
Sent: Monday, February 08, 2010 10:11 PM
To: r-help@r-project.org
Subject: Re: [R] Installation require proxy settings (continued)
Importance: High

Thanks to Greg and Michael for their advice.
I have tried all these possibilities (and other); nothing seems to
work; I persistently get the message:
In open.connection(con, r) :
  cannot open: HTTP status was '407 Proxy Authentication Required'
It seems as if proxy authentication required is causing the problem.
It seems I have to set up an .Renviron file, but I am still learning R,
and cannot find such a file.
Would it help, and how do I create it (it that is a possible
alternative to try).
Thanks for any comments.
Jacob


Jacob L van Wyk
Department of Statistics
University of Johannesburg (APK)
PO Box 524, Auckland Park, 2006
Office ph: 011 559 3080
Fax: 011 559 2499



This email and all contents are subject to the following disclaimer:

http://www.uj.ac.za/UJ_email_legal_disclaimer.htm

[[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.


__
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.



--
Brian D. Ripley,  rip...@stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

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

Re: [R] subset in a matrix

2010-02-09 Thread Henrique Dallazuanna

Try this:

z[z[,1]  0,]

On Tue, Feb 9, 2010 at 6:12 PM, DonDiego jorge.nie...@moorecap.com wrote:

 Hi,

 I have a matrix of data values like the example bellow. I would like to
 extract a subset of the matrix for the values where the first column is
 negative.  I am using the subset function. However, I am getting an error
 message that the conditional variable doe snot exist.  For some reason, the
 subset operation only works if I transform the matrix to a data set using
 as.data.set(). The help indicates that the subset function can be applied to
 matrixes and data sets. I am wondering if anyone has seen a similar problem
 before. am I using the correct syntax?



 n = 15
 m = 5

 cnames = paste(x,1:m,sep=)
 rnames = 1:n

 z = matrix(rnorm(n*m),n,m,dimnames =list(rnames,cnames))


 Thanks,

 Jorge




 test = subset(z,x1  0, select = c(cnames))

 --
 View this message in context: 
 http://n4.nabble.com/subset-in-a-matrix-tp1474958p1474958.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.




-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

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

Re: [R] subset in a matrix

2010-02-09 Thread Dennis Murphy

Hi:

 z = matrix(rnorm(n*m),n,m,dimnames =list(rnames,cnames))
 z
   x1  x2   x3 x4  x5
1  -0.3942900 -0.61202639 -1.804958629 -0.1351786 -1.27659221
2  -0.0593134  0.34111969  1.465554862  1.1780870 -0.57326541
3   1.1000254 -1.12936310  0.153253338 -1.5235668 -1.22461261
4   0.7631757  1.43302370  2.172611670  0.5939462 -0.47340064
5  -0.1645236  1.98039990  0.475509529  0.3329504 -0.62036668
6  -0.2533617 -0.36722148 -0.709946431  1.0630998  0.04211587
7   0.6969634 -1.04413463  0.610726353 -0.3041839 -0.91092165
8   0.5566632  0.56971963 -0.934097632  0.3700188  0.15802877
9  -0.6887557 -0.13505460 -1.253633400  0.2670988 -0.65458464
10 -0.7074952  2.40161776  0.291446236 -0.5425200  1.76728727
11  0.3645820 -0.03924000 -0.443291873  1.2078678  0.71670748
12  0.7685329  0.68973936  0.001105352  1.1604026  0.91017423
13 -0.1123462  0.02800216  0.074341324  0.7002136  0.38418536
14  0.8811077 -0.74327321 -0.589520946  1.5868335  1.68217608
15  0.3981059  0.18879230 -0.568668733  0.5584864 -0.63573645
 z[z[, 1]  0, ]
   x1  x2  x3 x4  x5
1  -0.3942900 -0.61202639 -1.80495863 -0.1351786 -1.27659221
2  -0.0593134  0.34111969  1.46555486  1.1780870 -0.57326541
5  -0.1645236  1.98039990  0.47550953  0.3329504 -0.62036668
6  -0.2533617 -0.36722148 -0.70994643  1.0630998  0.04211587
9  -0.6887557 -0.13505460 -1.25363340  0.2670988 -0.65458464
10 -0.7074952  2.40161776  0.29144624 -0.5425200  1.76728727
13 -0.1123462  0.02800216  0.07434132  0.7002136  0.38418536


Is this what you were looking for?

HTH,
Dennis

On Tue, Feb 9, 2010 at 12:12 PM, DonDiego jorge.nie...@moorecap.com wrote:


 Hi,

 I have a matrix of data values like the example bellow. I would like to
 extract a subset of the matrix for the values where the first column is
 negative.  I am using the subset function. However, I am getting an error
 message that the conditional variable doe snot exist.  For some reason, the
 subset operation only works if I transform the matrix to a data set using
 as.data.set(). The help indicates that the subset function can be applied
 to
 matrixes and data sets. I am wondering if anyone has seen a similar problem
 before. am I using the correct syntax?



 n = 15
 m = 5

 cnames = paste(x,1:m,sep=)
 rnames = 1:n

 z = matrix(rnorm(n*m),n,m,dimnames =list(rnames,cnames))


 Thanks,

 Jorge




 test = subset(z,x1  0, select = c(cnames))

 --
 View this message in context:
 http://n4.nabble.com/subset-in-a-matrix-tp1474958p1474958.html
 Sent from the R help mailing list archive at Nabble.com.

 __
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 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.


[[alternative HTML version deleted]]

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

Re: [R] Interpretation of high order interaction terms.

This is useful. But it is not exactly what I am looking for. I can use
'contrast' to infer linear combination factor level means as discussed
in Statistical Inference by Casella and Berger. What is the equivalent
'contrast' for interaction terms? Is there a book that discuss this?

On Tue, Feb 9, 2010 at 12:29 PM, RICHARD M. HEIBERGER r...@temple.edu wrote:
 ## Artificial data with all interactions significant.

 ## The interaction2wt plot shows all main effects and all pairwise
 ## interactions.  We see in the Y ~ A|B panel (or in the
 ## interaction.plot) that Y goes uphill for levels 1 and 2 of B and
 ## goes down and then up for level 3 of B.  This is the two-way A:B
 ## interaction.  At each level of B, the main effect of A differs.

 ## From the xyplot, we see that at level 1 of C, Y goes down for level
 ## 3 of B.  At level 2 of C, Y goes down a lot and then up for level 3
 ## of B.  This is the three-way A:B:C interaction.  At each level of C
 ## the two-way interaction of A and B differs.


 set.seed(1)

 require(HH)  ## needed for interaction2wt()

 threeway -
 data.frame(matrix(c(
 1, 1, 1, 1,
 2, 2, 1, 1,
 3, 3, 1, 1,
 1, 1, 2, 1,
 2, 2, 2, 1,
 3, 3, 2, 1,
 3, 1, 3, 1,
 2, 2, 3, 1,
 1, 3, 3, 1,
 1, 1, 1, 2,
 2, 2, 1, 2,
 3, 3, 1, 2,
 1, 1, 2, 2,
 2, 2, 2, 2,
 3, 3, 2, 2,
 3, 1, 3, 2,
 -4, 2, 3, 2,
 1, 3, 3, 2),
 byrow=TRUE, 18, 4, dimnames=list(1:18,c(Y,A,B,C
 for (i in 2:4) threeway[[i]] - factor(threeway[[i]])
 threeway - rbind(threeway, threeway)
 threeway$Y - threeway$Y + rnorm(36, s=.5)


 anova(aov(Y ~ A * B * C, data=threeway))

 with(threeway, interaction.plot(A, B, Y))  ## shows just the Y ~ A|B panel

 ## all two-way interactions and main effects
 interaction2wt(Y ~ A + B + C, data=threeway)  ## library(HH) required

 xyplot(x ~ A | C, groups=B,
       data=aggregate(threeway[,1], threeway[,-1], mean), type=l,
       auto.key=list(title=B, space=left,
         border=TRUE, lines=TRUE, points=FALSE),
       strip=strip.custom(strip.names=c(TRUE,TRUE)))


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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] subset in a matrix

2010-02-09 Thread Peter Ehlers


As others have said, z[z[, 1]  0, ] does it. Just in case
you're wondering why your subset command won't work,
str() is your friend (as is so often the case):

str(z)
str(as.data.frame(z))  ## (I don't think that R has 'as.data.set')

So z is a matrix with column *names* x1, etc;
as.data.frame(z) is a data.frame with *variables* named x1 etc.

If you really want to use subset(), then

 subset(z, z[, x1]  0, select = ...)

will work, but I wouldn't use it.

 -Peter Ehlers

DonDiego wrote:

Hi,

I have a matrix of data values like the example bellow. I would like to
extract a subset of the matrix for the values where the first column is
negative.  I am using the subset function. However, I am getting an error
message that the conditional variable doe snot exist.  For some reason, the
subset operation only works if I transform the matrix to a data set using
as.data.set(). The help indicates that the subset function can be applied to
matrixes and data sets. I am wondering if anyone has seen a similar problem
before. am I using the correct syntax?



n = 15
m = 5

cnames = paste(x,1:m,sep=)
rnames = 1:n

z = matrix(rnorm(n*m),n,m,dimnames =list(rnames,cnames))


Thanks,

Jorge




test = subset(z,x1  0, select = c(cnames))



--
Peter Ehlers
University of Calgary

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

Re: [R] Error on start R in server

2010-02-09 Thread Martin Morgan

On 02/08/2010 11:05 PM, Jiiindo wrote:
 
 Thanks for your answer,
 I can run it with an application Java normal. In my web-application, when i
 need, i start R et calling R function (new REvaluator).eval(1+1) to
 calcule an expression 1+1example.
 And output of your code:
 
 Loading RInterpreter library
 Load the Java VM with .JavaInit()
 R version 2.10.1 (2009-12-14) 
 i686-pc-linux-gnu 
 
 locale:
  [1] LC_CTYPE=en_US.UTF-8   LC_NUMERIC=C  
  [3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
  [5] LC_MONETARY=C  LC_MESSAGES=en_US.UTF-8   
  [7] LC_PAPER=en_US.UTF-8   LC_NAME=C 
  [9] LC_ADDRESS=C   LC_TELEPHONE=C
 [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C   
 
 attached base packages:
 [1] stats graphics  grDevices utils datasets  methods   base 
 
 other attached packages:
 [1] SJava_0.72.0

Hi -- In your original email

Warning: stack imbalance in 'if', 153 then 154
Warning: stack imbalance in '{', 151 then 152

this is definitely coming from R.

If it does not occur in a normal Java application that uses SJava, but
does occur in the web application that uses SJava, then the web
application is configured differently from the normal application.

The web application might be accessing a different version of R or
SJava, or it might have the wrong java.library.path or LD_LIBRARY_PATH
or CLASSPATH.

The way you started the embedded R in the SJava web application (with
the --no-save) means that any .RData or other R startup file, including
those in the web application 'home' directory, will be loaded when R starts.

I hope this provides some help; if it works out of the web app, but not
in it, then it is a web app configuration issue.

Martin
-- 
Martin Morgan
Computational Biology / Fred Hutchinson Cancer Research Center
1100 Fairview Ave. N.
PO Box 19024 Seattle, WA 98109

Location: Arnold Building M1 B861
Phone: (206) 667-2793

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Re: [R] Interpretation of high order interaction terms.

2010-02-09 Thread Kevin Wright

Googling for: contrasts for interaction terms produces useful results.  For
example:
http://www.ats.ucla.edu/stat/spss/faq/test_subcommand_mixed_mod.htm

Rich's example gave you an interpretation of a three-way interaction using
contrasts.  Maybe a bit hidden, but they are there.  For example, excluded
levels of a factor have a contrast coefficient of 0.

Kevin


On Tue, Feb 9, 2010 at 12:29 PM, RICHARD M. HEIBERGER r...@temple.eduwrote:

 ## Artificial data with all interactions significant.

 ## The interaction2wt plot shows all main effects and all pairwise
 ## interactions.  We see in the Y ~ A|B panel (or in the
 ## interaction.plot) that Y goes uphill for levels 1 and 2 of B and
 ## goes down and then up for level 3 of B.  This is the two-way A:B
 ## interaction.  At each level of B, the main effect of A differs.

 ## From the xyplot, we see that at level 1 of C, Y goes down for level
 ## 3 of B.  At level 2 of C, Y goes down a lot and then up for level 3
 ## of B.  This is the three-way A:B:C interaction.  At each level of C
 ## the two-way interaction of A and B differs.


 set.seed(1)

 require(HH)  ## needed for interaction2wt()

 threeway -
 data.frame(matrix(c(
 1, 1, 1, 1,
 2, 2, 1, 1,
 3, 3, 1, 1,
 1, 1, 2, 1,
 2, 2, 2, 1,
 3, 3, 2, 1,
 3, 1, 3, 1,
 2, 2, 3, 1,
 1, 3, 3, 1,
 1, 1, 1, 2,
 2, 2, 1, 2,
 3, 3, 1, 2,
 1, 1, 2, 2,
 2, 2, 2, 2,
 3, 3, 2, 2,
 3, 1, 3, 2,
 -4, 2, 3, 2,
 1, 3, 3, 2),
 byrow=TRUE, 18, 4, dimnames=list(1:18,c(Y,A,B,C
 for (i in 2:4) threeway[[i]] - factor(threeway[[i]])
 threeway - rbind(threeway, threeway)
 threeway$Y - threeway$Y + rnorm(36, s=.5)


 anova(aov(Y ~ A * B * C, data=threeway))

 with(threeway, interaction.plot(A, B, Y))  ## shows just the Y ~ A|B
 panel

 ## all two-way interactions and main effects
 interaction2wt(Y ~ A + B + C, data=threeway)  ## library(HH) required

 xyplot(x ~ A | C, groups=B,
   data=aggregate(threeway[,1], threeway[,-1], mean), type=l,
   auto.key=list(title=B, space=left,
 border=TRUE, lines=TRUE, points=FALSE),
   strip=strip.custom(strip.names=c(TRUE,TRUE)))

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-- 
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Re: [R] lmer

2010-02-09 Thread Douglas Bates

On Tue, Feb 9, 2010 at 4:38 AM, Demirtas, Hakan demir...@uic.edu wrote:

 Does lmer do three-level mixed-effects models?

Yes.

 What forms of outcome
 variables can it handle (continuous, survival, binary)? I'd appreciate any
 help.

Continuous outcomes for lmer.  glmer can handle binary outcomes.

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Re: [R] Formula used to create new plot

It is not clear what exactly you want, but here are some ideas.

Lattice and ggplot2 allow you to save graph objects that can then be 
modified/printed/plotted, there is information in there on types of axes etc.

For basic graphics (and others) you can use the history command to see recent 
plotting commands issued.

The plot2script function in the TeachingDemos package will create a script that 
will recreate the current plot (but uses low level plotting commands, not the 
same as was typed in).

-- 
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-boun...@r-
 project.org] On Behalf Of Timothée Poisot
 Sent: Tuesday, February 09, 2010 9:24 AM
 To: r-h...@stat.math.ethz.ch
 Subject: [R] Formula used to create new plot
 
 Dear list users,
 
 I was wondering if there is any way to know, for an open graphical
 peripheric, the instruction that was used to call it. For example, if I
 create a plot using
 
 plot(runif(10))
 
 is there any way to have the call returned to me? That would be
 especially usefull to know it there are any log axes, for example
 
 thanks
 t
 
 
 
 ---
 Timothée POISOT
 -
 Institut des Sciences de l'Evolution
 Université Montpellier 2, CC 065
 Place Eugène Bataillon
 34095 Montpellier CEDEX 5
 -
 Phone :   (+33)4 67 14 40 61
 Fax   :   (+33)4 67 14 40 61
 E-mail:   tpoi...@um2.fr
 Web   :   http://www.timotheepoisot.fr/
 ---
 
 
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[R] [R-pkgs] survey 3.20

2010-02-09 Thread Thomas Lumley



Version 3.20 of the survey package is percolating through CRAN

The major additions since the last announcement on this list (3.18, in 
September) are
  
- database-backed designs can now use replicate weights
- some multivariate statistics: principal components, factor analysis.

The NEWS file has a more detailed list of additions and changes.

   -thomas

Thomas Lumley   Assoc. Professor, Biostatistics
tlum...@u.washington.eduUniversity of Washington, Seattle

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

Re: [R] Re-execute previous command

You can use the history function to see a list of previous commands (but it 
does not automatically run them).

Do

 history(pat='plot')

To get a list of all the commands that you ran that included the string 'plot' 
somewhere in it.  You can then copy and paste to rerun (under the windows GUI 
(and possibly others) you can highlight and hit a button to rerun).

Hope this 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-boun...@r-
 project.org] On Behalf Of mnstn
 Sent: Tuesday, February 09, 2010 9:21 AM
 To: r-help@r-project.org
 Subject: [R] Re-execute previous command
 
 
 Hello All,
 In bash, to re-execute a command, say,  'gnuplot plot.gnu', one can
 type
 !gnu. Is there a similar feature in the R console? For example:
 plot(g1$mean,g2$mean)
 .
 .
 .
 Lines, many lines of commands
 -Here I want to repeat the previous plot command.
 !plot and plot(+TAB) do not work.
 
 This question was asked in this forum back in 2006 but the answer seems
 to
 have been lost somewhere in their discussions. This is the thread
 http://n4.nabble.com/R-command-line-need-intelligent-command-history-
 recall-td793072.html#a793075
 . Can some one comment if there is a way to do it?
 
 Thank you,
 MoonStone
 --
 View this message in context: http://n4.nabble.com/Re-execute-previous-
 command-tp1474629p1474629.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.

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Re: [R] Installation require proxy settings (continued)

Thanks Prof. Ripley.  I had forgotten that the OP had specified internet2.  I 
don't use it, so my suggestions only apply when it is turned off.

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111


 -Original Message-
 From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
 Sent: Tuesday, February 09, 2010 1:18 PM
 To: Greg Snow
 Cc: Van Wyk, Jaap; r-help@r-project.org
 Subject: Re: [R] Installation require proxy settings (continued)
 
 On Tue, 9 Feb 2010, Greg Snow wrote:
 
  Timing can be an issue.  You need to set the environment variables
 before accessing the internet for the 1st time in each R session.  If
 you get the error, then try setting the variables, it will not work.
 
  For my situation I have http_proxy and http_proxy_user set for the
 entire computer (in windows you do this through the control panel).
 
  If you want to go with the .Renviron route, just type:
 Sys.getenv('home')
  At the command prompt and it will tell you what your Home folder
 is.  Just create the .Renviron file in that folder.
 
 I believe the OP specified internet2.dll on installation.  It might be
 better not to do so ... otherwise you are at the mercy of IE
 internals.  The rw-FAQ does say (Q2.19)
 
 (a) Use the alternative internet2.dll by starting R with the flag
 --internet2 (see How do I install R for Windows?) or calling
 setInternet2(TRUE). These cause R to use the Internet Explorer
 internals.. Note that this does not work with proxies that need
 authentication.
 
  --
  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-boun...@r-
  project.org] On Behalf Of Van Wyk, Jaap
  Sent: Monday, February 08, 2010 10:11 PM
  To: r-help@r-project.org
  Subject: Re: [R] Installation require proxy settings (continued)
  Importance: High
 
  Thanks to Greg and Michael for their advice.
  I have tried all these possibilities (and other); nothing seems to
  work; I persistently get the message:
  In open.connection(con, r) :
cannot open: HTTP status was '407 Proxy Authentication Required'
  It seems as if proxy authentication required is causing the
 problem.
  It seems I have to set up an .Renviron file, but I am still learning
 R,
  and cannot find such a file.
  Would it help, and how do I create it (it that is a possible
  alternative to try).
  Thanks for any comments.
  Jacob
 
 
  Jacob L van Wyk
  Department of Statistics
  University of Johannesburg (APK)
  PO Box 524, Auckland Park, 2006
  Office ph: 011 559 3080
  Fax: 011 559 2499
 
 
  
  This email and all contents are subject to the following disclaimer:
 
  http://www.uj.ac.za/UJ_email_legal_disclaimer.htm
 
 [[alternative HTML version deleted]]
 
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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.
 
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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.
 
 
 --
 Brian D. Ripley,  rip...@stats.ox.ac.uk
 Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
 University of Oxford, Tel:  +44 1865 272861 (self)
 1 South Parks Road, +44 1865 272866 (PA)
 Oxford OX1 3TG, UKFax:  +44 1865 272595

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Re: [R] Interpretation of high order interaction terms.

2010-02-09 Thread RICHARD M. HEIBERGER

## Continuing with the same example, I now show the full set of
contrasts and their
## associated regression coefficients.  Remember that regression coefficients in
## models with factors are meaningful only when the dummy variables
are also displayed.

## construct full set of dummy variables to display all contrasts
sapply(threeway[,2:4], contrasts)  ## default is treatment contrasts
contrasts(threeway$A) - cbind(Lin=c(-1,0,1), Quad=c(1,-2,1))
contrasts(threeway$B) - cbind(Lin=c(-1,0,1), Quad=c(1,-2,1))
contrasts(threeway$C) - cbind(Lin=c(-1,1))
sapply(threeway[,2:4], contrasts)  ## changed to orthogonal
   ## integer-valued polynomial contrasts

tw.aov - aov(Y ~ A * B * C, data=threeway, x=TRUE)
x - tw.aov$x## dummy variables x
t(x)

anova(tw.aov)
beta.hat - coef(tw.aov)  ## regression coefficients beta.hat

## construct predicted values from beta.hat
y.hat - x %*% beta.hat
y.hat
fitted(tw.aov)  ## compare to predicted values from fitted function

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Re: [R] ggplot2 stacked line plot

2010-02-09 Thread Liam Blanckenberg

Hadley,

Thanks for the pointing that error out to me. Unfortunately, revising
my code to:

p - ggplot2(~, aes(x = ~, y = ~, colour = Type)) +
geom_line(aes(colour = Type), position = 'stack')

still does not generate what I'm after. I'm essentially after a line
plot where each 'Type' ('series' in excel lingo) is stacked in an
analogous fashion to a stacked area chart (i.e. geom_area(aes(Fill =
Type), position = 'stack')) - again using lines rather than areas
though.

This chart should look just like a stacked area chart, however rather
than each 'layer' (representing a given 'Type') being filled with a
colour, each layer should have just a coloured line representing its
values. Summing each line vertically for a given value on the x-axis
across all the layers (i.e. 'Types') should then give the total y
value for that x value (just like vertically summing an area chart).
Really all I'm after is a stacked area chart where the fill for each
'Type' has been removed, leaving only a line for each (stacked) Type's
value...

I hope this is somewhat clear!

Many thanks,

Liam





On Wed, Feb 10, 2010 at 1:53 AM, hadley wickham h.wick...@gmail.com wrote:
 Hi Liam,

 Your syntax is a little off.  You want:

 p - ggplot2(~, aes(x = ~, y = ~, colour = Type)) +
  geom_area(aes(fill = Type), position = 'stack')

 Position isn't an aesthetic.

 Hadley

 On Sun, Feb 7, 2010 at 10:40 PM, Liam Blanckenberg
 liam.blanckenb...@gmail.com wrote:
 Hi all,

 I have been hunting around for hours trying to figure out how to
 generate a stacked line chart using ggplot2. This type of chart can be
 generated in excel 2007 by selecting: Chart type  Line  Stacked
 line. I can generate a stacked area chart using the following code:

   p - ggplot2(~, aes(x = ~, y = ~, colour = Type)) +
 geom_area(aes(position = 'stack', fill = Type))

 However, when I try and replicate this using the following code for 
 geom_line:

   p - ggplot(~, aes(x = ~, y = ~, colour = Type)) +
 geom_line(aes(position = 'stack'))

 the resulting plot is not stacked - i.e. each 'Type' is plotted at its
 actual value rather than cumulatively to form a stacked chart... I
 have poured through Hadley's ggplot2 book (ggplot2: elegant graphics
 for data analysis), the R help list and also done general google
 searching but cannot find a way to generate this type of plot.

 R version: 2.9.2
 ggplot2 version: 0.8.5
 OS: windows 7 (64-bit).

 Any suggestions or assistance would be greatly appreciated.

 Regards,

 Liam

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 --
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[R] Bad characters/fonts or something in R Help

2010-02-09 Thread Matthew Burton-Kelly

Hello,
I am experiencing problems viewing the help files in R.  The easiest  
way to describe this is to just post an image.  You can see what I'm  
talking about here: http://www.flickr.com/photos/matthewbk/4344722268/

If I copy and paste the weird characters, they show up like they  
should.  Am I missing a specific font?

Mac 10.6 Snow Leopard
R 2.10.1 GUI 1.31 Leopard build 32-bit (5537)

Thanks for any help.  I have uninstalled and reinstalled but to no  
avail.

Matt



-
Matthew Burton-Kelly, M.S.
Graduate Student
Department of Geology and Geological Engineering
University of North Dakota
(802) 922-3696
matthew.burton.ke...@und.nodak.edu
matthew.burtonke...@gmail.com
http://www.protichnoctem.com

About thirty years ago there was much talk that geologists ought only  
to observe and not theorize; and I well remember someone saying that  
at this rate a man might as well go into a gravelpit and count the  
pebbles and describe the colors.  How odd it is that anyone should not  
see that all observation must be for or against some view if it is to  
be of any service!
-Charles Darwin, in an 1861 letter to Henry Fawcett.


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[R] Double Integral Minimization Problem

2010-02-09 Thread MVika


Hello all,

I am trying to minimize a function which contains a double integral, using
nlminb for the minimization and adapt for the integral. The integral is
over two variables (thita and radiusb) 
and the 3 free parameters I want to derive from the minimization are
counts0, index and radius_eff.

I have used both tasks in the past successfully but this is the first time
I've tried using the functions where the free parameters are contained
within the integral. My code is copied below:

## Begin of the code
alpha=3.99
vita=4.4
ellip=0.2

majaxis_pix=c(3.0  6.5 10.0 13.5 17.0 20.5 24.0 27.5 31.0 34.5 38.0 41.5
45.0 48.5 52.0 55.5 59.0)
counts=c(2479.117 718.061 298.918 157.963 98.869  63.883 44.524 31.918
23.500 17.877 13.915 11.032 8.881  7.245  5.978  4.972  4.175112)
countserr=c(80.085 12.181  4.247  3.148  1.963 1.279  0.448  0.242 0.097 
0.055 0.034  0.022  0.015  0.011  0.0082 0.006  0.0043)


intensity=function(x,counts0,index,radius_eff){

thita=x[1]
radiusb=x[2]

counts2=(1-ellip)*(vita-1)/(pi*alpha^2)*counts0*radiusb*exp(-(2*index-0.324)*(radiusb/radius_eff)^(1/index))*(1+(((majaxis_pix^2)+(radiusb^2)-2*majaxis_pix*radiusb*cos(-thita)+(ellip^2-2*ellip)*((radiusb*sin(thita))^2))/(alpha^2)))^(-vita)
return(counts2)
}


sersic=function(p)
{

counts0=p[1]
index=p[2]
radius_eff=p[3]

value1=adapt(2,c(0,0),c(2*pi,200),functn=intensity,minpts=1000,maxpts=NULL,eps=0.01,counts0=counts0,index=index,radius_eff=radius_eff)

test=value1$value


f=sum(((counts-test)/countserr)^2)
return(f)
}


out-nlminb(c(16000.0, 3.0,10.0), sersic) 
##End of the code

Any suggestions are welcome,
Thank you,
Marina





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Re: [R] R ggplot2 legend text left justify

2010-02-09 Thread Paul Sutcliffe


Thanks


hadley wickham wrote:


Hi Paul,

That's a bug in the current version of ggplot.  I'm working on update
for later this week.

Hadley

On Mon, Feb 8, 2010 at 5:56 PM, Paul Sutcliffe psutc...@it.uts.edu.au wrote:
  

In ggplot2 how do you justify the legend text ?
In the example below the opts(legend.text = theme_text(size = 9,hjust=0))
 changes the size of the text OK but it remains right justified.




mydata=data.frame(RowID=c(A,B,C),Name=c(long long long long long
name,short name ,medium medium name),Speed=c(100,140,120))
mydata
  

 RowID  Name Speed
1 A long long long long long name   100
2 B   short name140
3 Cmedium medium name   120


ggplot(mydata, aes(RowID,Speed))+ opts(legend.text = theme_text(size =
9,hjust=0)) + geom_bar() + xlab(Name) + aes(colour = Name) +aes(fill=Name)

  

Thank in advance
Paul


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Re: [R] Comparing means and trends in short time-series

2010-02-09 Thread Simone Vincenzi

I received also this message that did not show up here:

This sounds like a problem in (linear) mixed effects models. Look into
packages
nlme (with the lme() function) and lme4 (function lmer()) for starters. What
you
have falls under the heading of longitudinal or repeated measures data. Both
packages allow flexibility in the choice of within-unit covariance
structure.

 

It seem that it is trickier than it seems. I cannot test the
gls(density~year*stream,..) as I do not have replicates within stream, but
only one measurement for each stream. I tried to compare via anova the two
models:

Mixed1 - lme(density~year, data=mydata, random=~1|sector)

Mixed2 - lme(density~year, data=mydata, random=~1|year/sector)

 

But I'm not sure I can fit the Mixed2 model.

Many thanks for the answers so far.

Any ideas?

 

 

Simone Vincenzi, PhD 
Department of Environmental Sciences
University of Parma
Viale G. P. Usberti, 33/A, 43100 Parma, Italy
Phone: +39 0521 905696
Fax: +39 0521 906611
e.mail:  mailto:svinc...@nemo.unipr.it svinc...@nemo.unipr.it

 


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Re: [R] ggplot2 stacked line plot

Hi Liam,

Yes, that's what that code should do.  Could you please send a small
reproducible example?

Hadley

On Tue, Feb 9, 2010 at 4:21 PM, Liam Blanckenberg
liam.blanckenb...@gmail.com wrote:
 Hadley,

 Thanks for the pointing that error out to me. Unfortunately, revising
 my code to:

 p - ggplot2(~, aes(x = ~, y = ~, colour = Type)) +
 geom_line(aes(colour = Type), position = 'stack')

 still does not generate what I'm after. I'm essentially after a line
 plot where each 'Type' ('series' in excel lingo) is stacked in an
 analogous fashion to a stacked area chart (i.e. geom_area(aes(Fill =
 Type), position = 'stack')) - again using lines rather than areas
 though.

 This chart should look just like a stacked area chart, however rather
 than each 'layer' (representing a given 'Type') being filled with a
 colour, each layer should have just a coloured line representing its
 values. Summing each line vertically for a given value on the x-axis
 across all the layers (i.e. 'Types') should then give the total y
 value for that x value (just like vertically summing an area chart).
 Really all I'm after is a stacked area chart where the fill for each
 'Type' has been removed, leaving only a line for each (stacked) Type's
 value...

 I hope this is somewhat clear!

 Many thanks,

 Liam





 On Wed, Feb 10, 2010 at 1:53 AM, hadley wickham h.wick...@gmail.com wrote:
 Hi Liam,

 Your syntax is a little off.  You want:

 p - ggplot2(~, aes(x = ~, y = ~, colour = Type)) +
  geom_area(aes(fill = Type), position = 'stack')

 Position isn't an aesthetic.

 Hadley

 On Sun, Feb 7, 2010 at 10:40 PM, Liam Blanckenberg
 liam.blanckenb...@gmail.com wrote:
 Hi all,

 I have been hunting around for hours trying to figure out how to
 generate a stacked line chart using ggplot2. This type of chart can be
 generated in excel 2007 by selecting: Chart type  Line  Stacked
 line. I can generate a stacked area chart using the following code:

   p - ggplot2(~, aes(x = ~, y = ~, colour = Type)) +
 geom_area(aes(position = 'stack', fill = Type))

 However, when I try and replicate this using the following code for 
 geom_line:

   p - ggplot(~, aes(x = ~, y = ~, colour = Type)) +
 geom_line(aes(position = 'stack'))

 the resulting plot is not stacked - i.e. each 'Type' is plotted at its
 actual value rather than cumulatively to form a stacked chart... I
 have poured through Hadley's ggplot2 book (ggplot2: elegant graphics
 for data analysis), the R help list and also done general google
 searching but cannot find a way to generate this type of plot.

 R version: 2.9.2
 ggplot2 version: 0.8.5
 OS: windows 7 (64-bit).

 Any suggestions or assistance would be greatly appreciated.

 Regards,

 Liam

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 --
 http://had.co.nz/





-- 
http://had.co.nz/

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[R] Model matrix using dummy regressors or deviation regressors

The model matrix for the code at the end the email is shown below.
Since the model matrix doesn't have -1, I think that it is made of
dummy regressors rather than deviation regressors. I'm wondering how
to make a model matrix using deviation regressors. Could somebody let
me know?

 model.matrix(aaov)
   (Intercept) A2 B2 B3 A2:B2 A2:B3
11  0  0  0 0 0
21  0  0  0 0 0
31  0  0  0 0 0
41  0  0  0 0 0
51  1  0  0 0 0
61  1  0  0 0 0
71  1  0  0 0 0
81  1  0  0 0 0
91  0  1  0 0 0
10   1  0  1  0 0 0
11   1  0  1  0 0 0
12   1  0  1  0 0 0
13   1  1  1  0 1 0
14   1  1  1  0 1 0
15   1  1  1  0 1 0
16   1  1  1  0 1 0
17   1  0  0  1 0 0
18   1  0  0  1 0 0
19   1  0  0  1 0 0
20   1  0  0  1 0 0
21   1  1  0  1 0 1
22   1  1  0  1 0 1
23   1  1  0  1 0 1
24   1  1  0  1 0 1
attr(,assign)
[1] 0 1 2 2 3 3
attr(,contrasts)
attr(,contrasts)$A
[1] contr.treatment

attr(,contrasts)$B
[1] contr.treatment



#
a=2
b=3
n=4
A = rep(sapply(1:a,function(x){rep(x,n)}),b)
B = as.vector(sapply(sapply(1:b, function(x){rep(x,n)}), function(x){rep(x,a)}))
Y = A + B + rnorm(a*b*n)

fr = data.frame(Y=Y,A=as.factor(A),B=as.factor(B))
aaov=aov(Y ~ A * B,fr)
summary(aaov)
model.matrix(aaov)

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[R] split strings in a vector and convert it to a data.frame

2010-02-09 Thread Martin Batholdy

hi,

I have a vector full of strings like;


xy_100_ab   xy_101_ab   xy_102_ab   xy_103_ab


I want to seperate each string in three pieces and the separator should be the 
_

at the end I want a data.frame like:

column1 column2 column3
xy  100 ab
xy  101 ab
xy  102 ab
xy  103 ab


I tried strsplit but I couldn't figure out how to convert the list I get into a 
data.frame.

I just need a function like read.table. But it seems that read.table only can 
handle files, not vectors...
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[R] Superimpose ksmooth() onto barplot

2010-02-09 Thread Nathan S. Watson-Haigh


I'd like to superimpose a ksmooth() onto a barplot().

My data is:
 d
2009-06-20 2009-06-21 2009-06-22 2009-06-23 2009-06-24 
2009-06-25 2009-06-26 2009-06-27 2009-06-28 2009-06-29 2009-06-30 
2009-07-01 2009-07-02
Same Breed (B)   12.64  21.08  13.52  12.51  
13.71   9.91  14.24   7.18  11.81   5.92  
12.04  17.96   2.99
Different Breed   3.70   3.43   5.43   4.43   
3.69   1.83   2.95   2.93   4.44   3.47   
9.90  10.63   2.92
2009-07-04 2009-07-05 2009-07-06 2009-07-07 2009-07-08 
2009-07-09 2009-07-10 2009-07-11 2009-07-12 2009-07-13 2009-07-14 
2009-07-15 2009-07-16
Same Breed (B)   27.81  26.79  24.93  23.39  
27.52  28.24  12.60   2.25   1.20   0.51   
0.18   1.31   0.40
Different Breed  12.16   4.53   8.51  11.77   
5.65  11.67   5.81   0.53   1.26   0.42   
1.10   0.20   0.29
2009-07-17 2009-07-18 2009-07-19 2009-07-21 2009-07-22 
2009-07-23 2009-07-24 2009-07-25 2009-07-26 2009-07-27 2009-07-28 
2009-07-29 2009-10-29
Same Breed (B)0.24   0.03   0.05  13.34   
7.70   9.50   9.12   8.62  10.60   7.46  
13.17  15.55   3.17
Different Breed   0.13   0.01   0.05   4.98   
2.75   4.38   2.17   3.51   1.97   2.68   
1.60   2.47   0.38
2009-10-30 2009-10-31 2009-11-01 2009-11-02 2009-11-03 
2009-11-04 2009-11-05 2009-11-06 2009-11-07 2009-11-08 2009-11-09 
2009-11-24 2009-11-25
Same Breed (B)1.39   3.57   3.53   1.30   
2.26   3.41   3.44   4.04   1.71   2.16   
5.01   4.34   3.58
Different Breed   0.51   0.85   1.49   1.99   
6.83   1.91   1.54   1.44   0.60   2.08   
1.55   2.13   1.44

2009-11-26 2009-11-27 2009-11-28 2009-11-29 2009-11-30
Same Breed (B)2.12   2.34   1.90   4.99   2.11
Different Breed   4.21   1.00   2.28   2.90   1.68

I thought the following might work ok, but the x-axis doesn't quite line 
up correctly for the lines because of the spacing between the bars. I 
could set the space to zero, but does anyone have any thoughts on how to 
maintain the spacing and stretch the lines data?
barplot(as.matrix(d), beside=FALSE, las=2, 
col=as.numeric(as.factor(rownames(d
lines(ksmooth(1:ncol(d), colSums(d), kernel=normal, 
bandwidth=bandwidth), type=l, lwd=2, col=blue)
lines(ksmooth(1:ncol(d), d[1,], kernel=normal, bandwidth=bandwidth), 
type=l, lwd=2, col=as.numeric(as.factor(rownames(d)))[1])
lines(ksmooth(1:ncol(d), d[2,], kernel=normal, bandwidth=bandwidth), 
type=l, lwd=2, col=as.numeric(as.factor(rownames(d)))[2])
legend(topright, c(rownames(d), Total), 
fill=c(as.numeric(as.factor(rownames(d))), blue))


Cheers,
Nathan

--

Dr. Nathan S. Watson-Haigh
OCE Post Doctoral Fellow
CSIRO Livestock Industries
University Drive
Townsville, QLD 4810
Australia

Tel: +61 (0)7 4753 8548
Fax: +61 (0)7 4753 8600
Web: http://www.csiro.au/people/Nathan.Watson-Haigh.html

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Re: [R] Model matrix using dummy regressors or deviation regressors



On Feb 9, 2010, at 6:33 PM, bluesky...@gmail.com wrote:


The model matrix for the code at the end the email is shown below.
Since the model matrix doesn't have -1, I think that it is made of
dummy regressors rather than deviation regressors. I'm wondering how
to make a model matrix using deviation regressors. Could somebody let
me know?


If I understand your terminology, then:

?contr.sum

--  
David.



model.matrix(aaov)

  (Intercept) A2 B2 B3 A2:B2 A2:B3
11  0  0  0 0 0
21  0  0  0 0 0
31  0  0  0 0 0
41  0  0  0 0 0
51  1  0  0 0 0
61  1  0  0 0 0
71  1  0  0 0 0
81  1  0  0 0 0
91  0  1  0 0 0
10   1  0  1  0 0 0
11   1  0  1  0 0 0
12   1  0  1  0 0 0
13   1  1  1  0 1 0
14   1  1  1  0 1 0
15   1  1  1  0 1 0
16   1  1  1  0 1 0
17   1  0  0  1 0 0
18   1  0  0  1 0 0
19   1  0  0  1 0 0
20   1  0  0  1 0 0
21   1  1  0  1 0 1
22   1  1  0  1 0 1
23   1  1  0  1 0 1
24   1  1  0  1 0 1
attr(,assign)
[1] 0 1 2 2 3 3
attr(,contrasts)
attr(,contrasts)$A
[1] contr.treatment

attr(,contrasts)$B
[1] contr.treatment



#
a=2
b=3
n=4
A = rep(sapply(1:a,function(x){rep(x,n)}),b)
B = as.vector(sapply(sapply(1:b, function(x){rep(x,n)}), function(x) 
{rep(x,a)}))

Y = A + B + rnorm(a*b*n)

fr = data.frame(Y=Y,A=as.factor(A),B=as.factor(B))
aaov=aov(Y ~ A * B,fr)
summary(aaov)
model.matrix(aaov)

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David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] split strings in a vector and convert it to a data.frame

2010-02-09 Thread Jorge Ivan Velez

Hi Martin,

Here is a sugestion:

string - c(xy_100_ab, xy_101_ab,xy_102_ab,xy_103_ab)
out - data.frame( do.call( rbind, strsplit( string, '_' ) ) )
names(out) - paste('column',1:3,sep=)
out

HTH,
Jorge

On Tue, Feb 9, 2010 at 6:46 PM, Martin Batholdy  wrote:

 hi,

 I have a vector full of strings like;


 xy_100_ab   xy_101_ab   xy_102_ab   xy_103_ab


 I want to seperate each string in three pieces and the separator should be
 the _

 at the end I want a data.frame like:

 column1 column2 column3
 xy  100 ab
 xy  101 ab
 xy  102 ab
 xy  103 ab


 I tried strsplit but I couldn't figure out how to convert the list I get
 into a data.frame.

 I just need a function like read.table. But it seems that read.table only
 can handle files, not vectors...
 __
 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.


[[alternative HTML version deleted]]

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Re: [R] split strings in a vector and convert it to a data.frame