date:20100726

On Jul 25, 2010, at 11:47 PM, Guillem R. wrote:

As far as I know, the predict command gives the predicted values (and
intervals) of y, but what I'm looking for is the conditional effects
(betas)

of x on y conditional on values of z.

If you want the betas, then simply print the object. Don't for get the
default parameterization is for treatment effects.

I'm trying to produce a plot similar to the first shown in this link:
http://homepages.nyu.edu/~mrg217/interaction.html#code

If you want to provide an example ...and a more complete description
of the desire output, I sure someone here can finish the job of
showing you how setting up a call to predict will get you to that goal.

--
David.

Thanks again

David Winsemius wrote:

On Jul 25, 2010, at 10:24 PM, Guillem R. wrote:

Dear all,

I'd like to plot the marginal effect of a variable in a
multiplicative

interaction regression, that is, the effect of a variable
conditional on the
values of another variable. As an illustration, given model lm1

lm1 - lm(y ~ x*z)

? predict

Perhaps:

predict(lm1, newdata=data.frame(x=1:10, z=5), interval=confidence)

I'd like to get the effects of x on y conditional on the values of
z, with
the corresponding confidence intervals if possible. Does anyone know
of any
package or simple way to do this?

Thanks

David Winsemius, MD
West Hartford, CT

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Re: [R] Manage several graphical devices in interactive mode

2010-07-26 Thread Sébastien Bihorel

Thanks Duncan and Ted,

That will be tremendously helpful to me. In combination with the
width/height and xpos/ypos arguments, I think I'll be able to create this
matrix of devices I am looking for.

Thanks again

Sebastien

On Sun, Jul 25, 2010 at 12:07 PM, Duncan Murdoch
murdoch.dun...@gmail.comwrote:

 On 25/07/2010 10:03 AM, (Ted Harding) wrote:

 On 25-Jul-10 13:22:11, Sébastien Bihorel wrote:

 Dear R-users,

 Does anybody know a good way to create and use several graphical
 devices at the same time in interactive mode? Ideally, I'd like
 to open 2 to 3 devices and assign names to them. This way, I could
 make any addition/modification/update to a particular device using
 its name.

 I did not see anything like a name argument in ?X11. Is there an
 alternative?

 Thanks in advance for your feeback.
 Sebastien


 You can certainly do this by hand. The basic R terminal is device
 number 1. If you do X11() (or invoke it implicitly by using a 'plot'
 command) this will be device number 2. With just this one open, any
 'plot' (or plot-related) commands will re-use it.

 However, having opened device #2, you can do X11() again and get a
 second graphics device which will be number 3. And so on, for as many
 additional devices as you like.

 By default, the device that the results of any 'plot' command are
 sent to will be the most recently opened one. However, you can select
 the device using the command dev.set(n) where n ( = 2,3,... ) is the
 device you want to use. The device which is active will be shown
 by the word (ACTIVE) attached to its display. All other devices
 will show (inactive).

 You could name your devices by indexing a vector (2:n) by names.

 Try the following, Starting with no graphics devices open:

 X11()  # dev 2
 X11()  # dev 3
 X11()  # dev 4  ## This one is currently active


 That's the right idea, but I wouldn't count on the numbers coming out 2, 3,
 4:  what if another device was already opened?  Better to use dev.cur() just
 after you open a new device to find out what its number was.  So do
 something like

 X11()
 plotA - dev.cur()

 X11()
 plotB - dev.cur()

 X11()
 plotC - dev.cur()


 dev.set(plotA)  # first one is active

 etc.

 Duncan Murdoch

## Arange these so that you can see all their top bars
 devN - c(plotA=2,plotB=3,plotC=4)

 dev.set(devN[plotA])  # dev 2 is active

 dev.set(devN[plotC])  # dev 4 is active

 dev.set(devN[plotB])  # dev 3 is active

 dev.set(devN[plotA])  # dev 2 is active

 dev.set(devN[plotC])  # dev 4 is active

 Hoping this helps,
 Ted.

 
 E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
 Fax-to-email: +44 (0)870 094 0861
 Date: 25-Jul-10   Time: 15:03:14
 -- XFMail --

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[R] Plot of a subset of a data.frame()

2010-07-26 Thread Steffen Uhlig


Hello,

my data.frame is sort of a collection of process values, i.e. huge 
run-chart. It consists of a time-stamp in the first column (date as 
string), factors in the following columns (used for subset-filtering), 
and some process-data columns.
Hereafter, two examples are listed, showing the problems that occour 
during print:


At first the example, that works fine:

~~
a = c(1:10) # create a vector of integers
b = rep(c(a,b),5)   # create a vector of chars, used
# as factor-levels  
d = rnorm(10)   # some random numbers
e = data.frame(a,b,d)   # connect to a data.frame

e.1 = subset(e, b==a)   # create two subsets
e.2 = subset(e, b==b)
plot(d~a, e.1, pch=3, col=2) # plot first data-subset
points(d~a, e.2, pch=4, col=3) # plot the 2nd one

~~
all looks fine in theses plots.


However, changing the content of vector a to a set of strings the 
following happens:


~~
a = c(a,b,c,d,e,f,g,h,i,j)
e = data.frame(a,b,d)   # re-build data.frame

e.1 = subset(e, b==a) # create two subsets
e.2 = subset(e, b==b)
plot(d~a, e.1, pch=3, col=2)
points(d~a, e.2, pch=4, col=3)
~~
The plot-command produces horizontal lines instead of dots. This seems 
to happen when the x-axis contains strings rather than numbers. is there 
a way out?


Best regards,
/Steffen
--
Steffen Uhlig, PhD
Mechatronik und Sensortechnik
HTW des Saarlandes
Goebenstraße 40
66117 Saarbrücken

Tel.: +49 (0) 681 58 67 274

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[R] what is a vignette?

2010-07-26 Thread Alaios

I am trying to find a simple R guide that explain what a vignette is but so far 
I didnt make any progress. I tried to search inside R's built in help.start() 
but it only returns results how to see vignettes.

So could you please tell me what a vignette is and if you can also could you 
give some simple guide that I can always use to read about these things?

Best Regards
Alex

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Re: [R] Plot of a subset of a data.frame()

2010-07-26 Thread Alain Guillet


 Hello,

It is completely normal. I advise you to read the manual An 
introduction to R on the CRAN website. For example you can find (part 
12.1.1) :



   12.1.1 The |plot()| function

 One of the most frequently used plotting functions in R is the 
|plot()| function. This is a /generic/ function: the type of plot 
produced is dependent on the type or /class/ of the first argument.


|plot(|x|, |y|)|
|plot(|xy|)|
   If x and y are vectors, |plot(|x|, |y|)| produces a scatterplot of y
   against x. The same effect can be produced by supplying one argument
   (second form) as either a list containing two elements x and y or a
   two-column matrix.
|plot(|x|)|
   If x is a time series, this produces a time-series plot. If x is a
   numeric vector, it produces a plot of the values in the vector
   against their index in the vector. If x is a complex vector, it
   produces a plot of imaginary versus real parts of the vector elements.
|plot(|f|)|
|plot(|f|, |y|)|
   f is a factor object, y is a numeric vector. The first form
   generates a bar plot of f; the second form produces boxplots of y
   for each level of f.
|plot(|df|)|
|plot(~ |expr|)|
|plot(|y| ~ |expr|)|
   df is a data frame, y is any object, expr is a list of object names
   separated by `|+|' (e.g., |a + b + c|). The first two forms produce
   distributional plots of the variables in a data frame (first form)
   or of a number of named objects (second form). The third form plots
   y against every object named in expr. 




Alain


On 26-Jul-10 13:38, Steffen Uhlig wrote:

Hello,

my data.frame is sort of a collection of process values, i.e. huge 
run-chart. It consists of a time-stamp in the first column (date as 
string), factors in the following columns (used for subset-filtering), 
and some process-data columns.
Hereafter, two examples are listed, showing the problems that occour 
during print:


At first the example, that works fine:

~~
a = c(1:10) # create a vector of integers
b = rep(c(a,b),5)# create a vector of chars, used
# as factor-levels
d = rnorm(10)# some random numbers
e = data.frame(a,b,d)# connect to a data.frame

e.1 = subset(e, b==a)# create two subsets
e.2 = subset(e, b==b)
plot(d~a, e.1, pch=3, col=2) # plot first data-subset
points(d~a, e.2, pch=4, col=3) # plot the 2nd one

~~
all looks fine in theses plots.


However, changing the content of vector a to a set of strings the 
following happens:


~~
a = c(a,b,c,d,e,f,g,h,i,j)
e = data.frame(a,b,d)   # re-build data.frame

e.1 = subset(e, b==a) # create two subsets
e.2 = subset(e, b==b)
plot(d~a, e.1, pch=3, col=2)
points(d~a, e.2, pch=4, col=3)
~~
The plot-command produces horizontal lines instead of dots. This seems 
to happen when the x-axis contains strings rather than numbers. is 
there a way out?


Best regards,
/Steffen


--
Alain Guillet
Statistician and Computer Scientist

SMCS - IMMAQ - Université catholique de Louvain
Bureau c.316
Voie du Roman Pays, 20
B-1348 Louvain-la-Neuve
Belgium

tel: +32 10 47 30 50

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[R] Odp: Plot of a subset of a data.frame()

2010-07-26 Thread Petr PIKAL

Hi

r-help-boun...@r-project.org napsal dne 26.07.2010 13:38:44:

 Hello,
 
 my data.frame is sort of a collection of process values, i.e. huge 
 run-chart. It consists of a time-stamp in the first column (date as 
 string), factors in the following columns (used for subset-filtering), 
 and some process-data columns.
 Hereafter, two examples are listed, showing the problems that occour 
 during print:
 
 At first the example, that works fine:
 
 ~~
 a = c(1:10)   # create a vector of integers
 b = rep(c(a,b),5)   # create a vector of chars, used
  # as factor-levels 
 d = rnorm(10)  # some random numbers
 e = data.frame(a,b,d)   # connect to a data.frame
 
 e.1 = subset(e, b==a)   # create two subsets
 e.2 = subset(e, b==b)
 plot(d~a, e.1, pch=3, col=2) # plot first data-subset

Rather strange plot call. I usually call

plot(a, d, pch=as.numeric(as.factor(b))+2, col=as.numeric(as.factor(b))+1)

as you could have problem when some point in second subset is outside a 
range of first subset.

 points(d~a, e.2, pch=4, col=3) # plot the 2nd one
 
 ~~
 all looks fine in theses plots.
 
 
 However, changing the content of vector a to a set of strings the 
 following happens:
 
 ~~
 a = c(a,b,c,d,e,f,g,h,i,j)
 e = data.frame(a,b,d)   # re-build data.frame
 
 e.1 = subset(e, b==a) # create two subsets
 e.2 = subset(e, b==b)
 plot(d~a, e.1, pch=3, col=2)
 points(d~a, e.2, pch=4, col=3)
 ~~
 The plot-command produces horizontal lines instead of dots. This seems 
 to happen when the x-axis contains strings rather than numbers. is there 

 a way out?

You actually called boxplots hence lines and labels under x axis. The way 
out depends on how do you want everything to be plotted. If a vector was 
a factor you could use conversion to numeric representation by

a.n-as.numeric(a)

and plot d against a.n with axis labels from a.

Try to go through
plot, plot.default, boxplot, factor

help pages

Regards
Petr

 
 Best regards,
 /Steffen
 -- 
 Steffen Uhlig, PhD
 Mechatronik und Sensortechnik
 HTW des Saarlandes
 Goebenstraße 40
 66117 Saarbrücken
 
 Tel.: +49 (0) 681 58 67 274
 
 __
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Re: [R] Plot of a subset of a data.frame()

2010-07-26 Thread ONKELINX, Thierry

You could have a look at the ggplot2 package to make such plots. The code for 
the plots is more readable than with base plots.

a = c(1:10) # create a vector of integers
b = rep(c(a,b),5)   # create a vector of chars, used
# as factor-levels  
d = rnorm(10)   # some random numbers
e = data.frame(a,b,d)

library(ggplot2)
ggplot(e, aes(x = a, y = d, colour = b, shape = b)) + geom_point()

a = c(a,b,c,d,e,f,g,h,i,j)
e = data.frame(a,b,d)   # re-build data.frame
ggplot(e, aes(x = a, y = d, colour = b, shape = b)) + geom_point()

HTH,

Thierry


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 Steffen Uhlig
 Verzonden: maandag 26 juli 2010 13:39
 Aan: r-help@r-project.org
 Onderwerp: [R] Plot of a subset of a data.frame()
 
 Hello,
 
 my data.frame is sort of a collection of process values, i.e. 
 huge run-chart. It consists of a time-stamp in the first 
 column (date as string), factors in the following columns 
 (used for subset-filtering), and some process-data columns.
 Hereafter, two examples are listed, showing the problems that 
 occour during print:
 
 At first the example, that works fine:
 
 ~~
 a = c(1:10)   # create a vector of integers
 b = rep(c(a,b),5) # create a vector of chars, used
   # as factor-levels  
 d = rnorm(10) # some random numbers
 e = data.frame(a,b,d) # connect to a data.frame
 
 e.1 = subset(e, b==a)   # create two subsets
 e.2 = subset(e, b==b)
 plot(d~a, e.1, pch=3, col=2) # plot first data-subset 
 points(d~a, e.2, pch=4, col=3) # plot the 2nd one
 
 ~~
 all looks fine in theses plots.
 
 
 However, changing the content of vector a to a set of 
 strings the following happens:
 
 ~~
 a = c(a,b,c,d,e,f,g,h,i,j)
 e = data.frame(a,b,d)   # re-build data.frame
 
 e.1 = subset(e, b==a) # create two subsets
 e.2 = subset(e, b==b)
 plot(d~a, e.1, pch=3, col=2)
 points(d~a, e.2, pch=4, col=3)
 ~~
 The plot-command produces horizontal lines instead of dots. 
 This seems to happen when the x-axis contains strings rather 
 than numbers. is there a way out?
 
 Best regards,
 /Steffen
 --
 Steffen Uhlig, PhD
 Mechatronik und Sensortechnik
 HTW des Saarlandes
 Goebenstraße 40
 66117 Saarbrücken
 
 Tel.: +49 (0) 681 58 67 274
 
 __
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Re: [R] Plot of a subset of a data.frame()



On Jul 26, 2010, at 7:38 AM, Steffen Uhlig wrote:


Hello,

my data.frame is sort of a collection of process values, i.e. huge  
run-chart. It consists of a time-stamp in the first column (date as  
string), factors in the following columns (used for subset- 
filtering), and some process-data columns.
Hereafter, two examples are listed, showing the problems that occour  
during print:


At first the example, that works fine:

~~
a = c(1:10) # create a vector of integers
b = rep(c(a,b),5)   # create a vector of chars, used
# as factor-levels  
d = rnorm(10)   # some random numbers
e = data.frame(a,b,d)   # connect to a data.frame


You've gotten several answers, but none have addressed an aspect of R  
behavior that took me longer to appreciate than it perhaps should  
have. The b column inside the e data.frame is now a factor column.  
I mention that because you later referred to it as a string which it  
is not. It is an integer with an associated  indexed level character  
vector. Many of the functions that you might think would work on  
strings will give either errors or unexpected results when applied  
to factors.





e.1 = subset(e, b==a)   # create two subsets
e.2 = subset(e, b==b)
plot(d~a, e.1, pch=3, col=2) # plot first data-subset
points(d~a, e.2, pch=4, col=3) # plot the 2nd one

~~
all looks fine in theses plots.


However, changing the content of vector a to a set of strings the  
following happens:


~~
a = c(a,b,c,d,e,f,g,h,i,j)
e = data.frame(a,b,d)   # re-build data.frame

e.1 = subset(e, b==a) # create two subsets
e.2 = subset(e, b==b)
plot(d~a, e.1, pch=3, col=2)
points(d~a, e.2, pch=4, col=3)
~~
The plot-command produces horizontal lines instead of dots. This  
seems to happen when the x-axis contains strings rather than  
numbers. is there a way out?


Best regards,
/Steffen

--

David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] vectorisation?

2010-07-26 Thread Raghu

Many thanks Dimitris.

On Mon, Jul 26, 2010 at 9:13 AM, Dimitris Rizopoulos 
d.rizopou...@erasmusmc.nl wrote:

 have a look at function rollapply() from package zoo.

 I hope it helps.

 Best,
 Dimitris


 On 7/26/2010 8:28 AM, Raghu wrote:

 Hi

 I have 3500 rows of data (say a single column) in a vector. If I want to
 compare every ith element with the simple average of the previous (i-5)
 elements, then I could write something like this:

 for(i in 6:NROW(data)){
if(data[i] =  mean(data[(i-5):i]) (counter[i]=1)}
 }

 Is it possible to replicate the above faster in R?



 --
 Dimitris Rizopoulos
 Assistant Professor
 Department of Biostatistics
 Erasmus University Medical Center

 Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
 Tel: +31/(0)10/7043478
 Fax: +31/(0)10/7043014




-- 
'Raghu'

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Re: [R] glm - prediction of a factor with several levels

2010-07-26 Thread blackscorpio


Thanks a lot for your anwers.
To Ben Bolker  : I am trying to perform an ordinal logistic regression to
predict an Y 3-class variable, having observed 3 continous predictors V1,
V2, V3.
With random data my code would be something like :  

# simulate 10 observations of 3 independant N(0,1) predictors
X=rmvnorm(10,rep(0,3))
# variable to predict
Y=c(1,1,1,2,2,2,3,3,3,3)
# create data frame
A=as.data.frame(cbind(X,Y))
# turn Y into class variable
A$Y=as.factor(A$Y)
# perform logisitic regression
glm(Y~V1+V2+V3,A,family=binomial)

As only one intercept is returned, it seems indeed that a 2-class model has
been performed instead, as said by zachmor.
Although I solved my problem by using polr instead of glm, I'd like to
understand what glm does in such a case since it gave me better
well-classification rates with the predict function.

Thanks a lot again !
-- 
View this message in context: 
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[R] Outlier detection in bimodal distribution

2010-07-26 Thread Tim Smith

Hi,

I was looking for a package that would help with outlier detection for bimodal 
distributions. I have tried 'outliers' and 'extremevalues' packages, but am not 
sure if they are ok for bimodal distribution.

Any help would be highly appreciated!

thanks,


  
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Re: [R] what is a vignette?

2010-07-26 Thread raghu

vignette is an R jargon for documentation. try ?vignette. Not all packages
come with vignettes. You may want to install a package and go to the
installed directory to see if any examples are given for the same.

HTH
Raghu

On Mon, Jul 26, 2010 at 12:56 PM, alaios [via R]
ml-node+2302242-1733300369-309...@n4.nabble.comml-node%2b2302242-1733300369-309...@n4.nabble.com
wrote:

I am trying to find a simple R guide that explain what a vignette is but so
far
I didnt make any progress. I tried to search inside R's built in
help.start()
but it only returns results how to see vignettes.

So could you please tell me what a vignette is and if you can also could
you
give some simple guide that I can always use to read about these things?

Best Regards
Alex

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Re: [R] Marginal effects from interaction regression model

2010-07-26 Thread Michael Friendly

The plots you are looking for are called 'effect plots'.  See the 
effects package by John Fox.


Guillem R. wrote:

As far as I know, the predict command gives the predicted values (and
intervals) of y, but what I'm looking for is the conditional effects (betas)
of x on y conditional on values of z.

I'm trying to produce a plot similar to the first shown in this link:
http://homepages.nyu.edu/~mrg217/interaction.html#code

Thanks again



David Winsemius wrote:


On Jul 25, 2010, at 10:24 PM, Guillem R. wrote:


Dear all,

I'd like to plot the marginal effect of a variable in a multiplicative
interaction regression, that is, the effect of a variable  
conditional on the

values of another variable. As an illustration, given model lm1

lm1 - lm(y ~ x*z)

? predict

Perhaps:

predict(lm1, newdata=data.frame(x=1:10, z=5), interval=confidence)


I'd like to get the effects of x on y conditional on the values of  
z, with
the corresponding confidence intervals if possible. Does anyone know  
of any

package or simple way to do this?

Thanks

--

David Winsemius, MD
West Hartford, CT

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Re: [R] what is a vignette?

2010-07-26 Thread Matt Shotwell

Alex, 

Vignettes are optional supplemental documentation. That is, they are in
addition to the required boilerplate documentation for R functions and
datasets. Vignettes are written in the spirit of sharing knowledge, and
assisting new users in learning the purpose and use of a package.

Maybe the best place to start is simply to read one, or a few. The `zoo`
package has a few, for example here:

http://cran.r-project.org/web/packages/zoo/index.html

The technical details of vignettes, and how to write one are contained
in the `Writing R Extensions` manual:

http://cran.r-project.org/manuals.html

-Matt

On Mon, 2010-07-26 at 07:55 -0400, Alaios wrote:
 I am trying to find a simple R guide that explain what a vignette is but so 
 far 
 I didnt make any progress. I tried to search inside R's built in help.start() 
 but it only returns results how to see vignettes.
 
 So could you please tell me what a vignette is and if you can also could you 
 give some simple guide that I can always use to read about these things?
 
 Best Regards
 Alex
 
 __
 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.

-- 
Matthew S. Shotwell
Graduate Student 
Division of Biostatistics and Epidemiology
Medical University of South Carolina

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Re: [R] glm - prediction of a factor with several levels

2010-07-26 Thread Marc Schwartz

On Jul 26, 2010, at 4:18 AM, blackscorpio wrote:

 
 Thanks a lot for your anwers.
 To Ben Bolker  : I am trying to perform an ordinal logistic regression to
 predict an Y 3-class variable, having observed 3 continous predictors V1,
 V2, V3.
 With random data my code would be something like :  
 
 # simulate 10 observations of 3 independant N(0,1) predictors
 X=rmvnorm(10,rep(0,3))
 # variable to predict
 Y=c(1,1,1,2,2,2,3,3,3,3)
 # create data frame
 A=as.data.frame(cbind(X,Y))
 # turn Y into class variable
 A$Y=as.factor(A$Y)
 # perform logisitic regression
 glm(Y~V1+V2+V3,A,family=binomial)
 
 As only one intercept is returned, it seems indeed that a 2-class model has
 been performed instead, as said by zachmor.
 Although I solved my problem by using polr instead of glm, I'd like to
 understand what glm does in such a case since it gave me better
 well-classification rates with the predict function.
 
 Thanks a lot again !


From the Details section of ?glm:

 For binomial and quasibinomial families the response can also be specified as 
a factor (when the first level denotes failure and all others success) 
...


So 1 is a failure and c(2, 3) denote success, thus a two level response.

BTW, if you are going to use a function (eg. rmvnorm()) from an external 
package, be sure to include the relevant library() call in your example code so 
folks don't need to guess which CRAN package(s) may be required to run it.

HTH,

Marc Schwartz

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Re: [R] does package QuantPsych function lm.beta can handle resultsof a regression with weights?

2010-07-26 Thread Tom Fletcher

The original function was created for a simple example. It never was
written to address weighted regression. A quick fix will work for you
situation.

### The original is:


lm.beta -
function (MOD) 
{
b - summary(MOD)$coef[-1, 1]
sx - sd(MOD$model[-1])
sy - sd(MOD$model[1])
beta - b * sx/sy
return(beta)
}

  A newer modification:

lm.betaW - 
function (MOD) 
{
b - summary(MOD)$coef[-1, 1]
sx - sd(MOD$model[-1][1])
sy - sd(MOD$model[1])
beta - b * sx/sy
return(beta)
}

The above should do the trick.

TF

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Dimitri Liakhovitski
Sent: Thursday, July 22, 2010 12:35 PM
To: r-help@r-project.org
Subject: [R] does package QuantPsych function lm.beta can handle
resultsof a regression with weights?

Hello, and sorry for not providing an example.
I run a regular linear regression (using lm) and use weights with it
(weights = ...).
I use QuantPsych package, its function lm.beta to extract standardized
regression weights from my lm regression object.

When I don't use weights, everything is fine.
But when I do use weights, I get an error that refers to lm.beta code:
In b * sx : longer object length is not a multiple of shorter object
length

This happens because there is an extra column in the object:
regr$model that lm.beta is using to get at the betas.
Is there some other package that just gives me the standardized
regression weights - even if I used weights for regression?

Thank you!

--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

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Re: [R] what is a vignette?

2010-07-26 Thread Max Kuhn

A good article is:

   http://www.statistik.lmu.de/~leisch/Sweave/Sweave-Rnews-2003-2.pdf

Max

On Mon, Jul 26, 2010 at 9:18 AM, Matt Shotwell shotw...@musc.edu wrote:
 Alex,

 Vignettes are optional supplemental documentation. That is, they are in
 addition to the required boilerplate documentation for R functions and
 datasets. Vignettes are written in the spirit of sharing knowledge, and
 assisting new users in learning the purpose and use of a package.

 Maybe the best place to start is simply to read one, or a few. The `zoo`
 package has a few, for example here:

 http://cran.r-project.org/web/packages/zoo/index.html

 The technical details of vignettes, and how to write one are contained
 in the `Writing R Extensions` manual:

 http://cran.r-project.org/manuals.html

 -Matt

 On Mon, 2010-07-26 at 07:55 -0400, Alaios wrote:
 I am trying to find a simple R guide that explain what a vignette is but so 
 far
 I didnt make any progress. I tried to search inside R's built in help.start()
 but it only returns results how to see vignettes.

 So could you please tell me what a vignette is and if you can also could you
 give some simple guide that I can always use to read about these things?

 Best Regards
 Alex

 __
 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.

 --
 Matthew S. Shotwell
 Graduate Student
 Division of Biostatistics and Epidemiology
 Medical University of South Carolina

 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
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-- 

Max

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[R] Concatenate a mix of numbers and letters to create a vector name

2010-07-26 Thread Panos Hadjinicolaou

Dear all,

I am trying to create a vector name, for example tmax.195012 from tmax., 1950 
and 12.  Obviously I don't wish to simply type it because the 3 name components 
 are changing in each iteration within a loop. Is there any way of  
concatenating those 3 components (which are a mixture of numbers and  letters)?

Thanks for reading,

Panos

-
Dr Panos Hadjinicolaou

Energy Environment  Water Research Center (EEWRC)
The Cyprus Institute
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Re: [R] Concatenate a mix of numbers and letters to create a vector name

2010-07-26 Thread Dimitris Rizopoulos


have a look at function paste(), i.e., ?paste


I hope it helps.

Best,
Dimitris


On 7/26/2010 3:44 PM, Panos Hadjinicolaou wrote:

Dear all,

I am trying to create a vector name, for example tmax.195012 from tmax., 1950 
and 12.  Obviously I don't wish to simply type it because the 3 name components 
 are changing in each iteration within a loop. Is there any way of  
concatenating those 3 components (which are a mixture of numbers and  letters)?

Thanks for reading,

Panos

-
Dr Panos Hadjinicolaou

Energy Environment  Water Research Center (EEWRC)
The Cyprus Institute
--
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--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center

Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014

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[R] = vs - operator

2010-07-26 Thread Alaios

Hello
I notice that in Linux the = operator works like the - operator
So a=3 is similar to a-3.
Could you please verify me that is correct? I would like to use = operator. 
Do 
you think that might be a problem in the future?

Best Regards
Alex

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[R] search and replace in a list of strings

2010-07-26 Thread Olivier Coupiac

Hi everybody,

I have a 4*4 matrix WD of wind data of the following form:

 WD
$date
[1] 07.07.2010 07.07.2010 07.07.2010 07.07.2010

$time
[1] 00:00:00 00:10:00 00:20:00 00:30:00

$CH1Avg
[1] 3.02 3.04 2.94 2.71

I would like to transform the date and time strings in usable numbers for 
plotting :

-  Transform 07.07.2010 in 07*24 + 07*31*24   for example

-  Transform the string of type 12:15:20 in a number equals to 
12+15/60+20/3600

I would hence need a function able to look for defined patterns in a list, like 
aa:bb:cc and evaluate aa+bb/60+cc/3600.

Anyone has a idea?

Thanks a lot!
Cheers

Olivier Coupiac

[[alternative HTML version deleted]]

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[R] Smalltalk with R

2010-07-26 Thread nero


Dear all,

i´m search for informations referring to R and Seaside respectively
VisualWorks or other cross-platforms for the implementation of Smalltalk.
Do you know a interface or a bridge,which enables work with such a
R-Smalltalk connection? 

Best Regards 
Nero
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Re: [R] search and replace in a list of strings

The most usable form is to convert it to POSIXct since most of the
plotting routines know how to handle date/time.  You could do:

myDate - as.POSIXct(paste(WD$date, WD$time), format='%m.%d.%Y %H:%M:%S')


e.g.,
 as.POSIXct('07.07.2010 12:34:15', format='%m.%d.%Y %H:%M:%S')
[1] 2010-07-07 12:34:15 EDT


On Mon, Jul 26, 2010 at 9:07 AM, Olivier Coupiac
olivier.coup...@renerco.com wrote:
 Hi everybody,

 I have a 4*4 matrix WD of wind data of the following form:

 WD
 $date
 [1] 07.07.2010 07.07.2010 07.07.2010 07.07.2010

 $time
 [1] 00:00:00 00:10:00 00:20:00 00:30:00

 $CH1Avg
 [1] 3.02 3.04 2.94 2.71

 I would like to transform the date and time strings in usable numbers for 
 plotting :

 -          Transform 07.07.2010 in 07*24 + 07*31*24   for example

 -          Transform the string of type 12:15:20 in a number equals to 
 12+15/60+20/3600

 I would hence need a function able to look for defined patterns in a list, 
 like aa:bb:cc and evaluate aa+bb/60+cc/3600.

 Anyone has a idea?

 Thanks a lot!
 Cheers

 Olivier Coupiac

        [[alternative HTML version deleted]]

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+1 513 646 9390

What is the problem that you are trying to solve?

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Re: [R] = vs - operator

Check the mail archieve on this;  there has been a long discussion.

To avoid trouble in the future, use - as the assignment operator.

On Mon, Jul 26, 2010 at 9:51 AM, Alaios ala...@yahoo.com wrote:
 Hello
 I notice that in Linux the = operator works like the - operator
 So a=3 is similar to a-3.
 Could you please verify me that is correct? I would like to use = operator. 
 Do
 you think that might be a problem in the future?

 Best Regards
 Alex

 __
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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
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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] = vs - operator

Hi,

Just as an example, here are three threads that discuss it.

http://www.mail-archive.com/r-help@r-project.org/msg16881.html

http://r.789695.n4.nabble.com/advice-opinion-on-vs-in-teaching-R-td1014502.html#a1014502

http://www.mail-archive.com/r-help@r-project.org/msg100034.html

Cheers,

Josh

On Mon, Jul 26, 2010 at 6:51 AM, Alaios ala...@yahoo.com wrote:
 Hello
 I notice that in Linux the = operator works like the - operator
 So a=3 is similar to a-3.
 Could you please verify me that is correct? I would like to use = operator. 
 Do
 you think that might be a problem in the future?

 Best Regards
 Alex

 __
 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.




-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

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[R] Optimization problem with nonlinear constraint

2010-07-26 Thread Uli Kleinwechter


Dear all,

I'm looking for a way to solve a simple optimization problem with a 
nonlinear constraint. An example would be


max x   s.t.   y = x * T ^(x-1)

where y and T are known values.

optim() and constrOptim() do only allow for box or linear constraints, 
so I did not succedd here. I also found hints to donlp2 but this does 
not seem to be available anymore.


Any hints are welcome,

Uli

--

Uli Kleinwechter
Agricultural and Food Policy Group (420a)
University of Hohenheim
D-70593 Stuttgart
E-mail: u.kleinwech...@uni-hohenheim.de

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Re: [R] Concatenate a mix of numbers and letters to create a vector name

2010-07-26 Thread Panos Hadjinicolaou

Thanks for the reply. Indeed the paste function results in  concatenation:
  paste(c(tmax., 1950, 12), collapse=)
[1] tmax.195012

but  I am looking for a way to subsequently get rid of the -  - in order to  
use tmax.195012 as an object (e.g. to define a vector with that name).  Any 
ideas?

Thanks,

Panos

  _  

From: Dimitris Rizopoulos [mailto:d.rizopou...@erasmusmc.nl]
To: Panos Hadjinicolaou [mailto:p.hadjinicol...@cyi.ac.cy]
Cc: r-help@r-project.org
Sent: Mon, 26 Jul 2010 16:48:31 +0300
Subject: Re: [R] Concatenate a mix of numbers and letters to create a vector 
name

have a look at function paste(), i.e., ?paste
  
  
  I hope it helps.
  
  Best,
  Dimitris
  
  
  On 7/26/2010 3:44 PM, Panos Hadjinicolaou wrote:
   Dear all,
  
   I am trying to create a vector name, for example tmax.195012 from  tmax., 
1950 and 12. Obviously I don't wish to simply type it because  the 3 name 
components are changing in each iteration within a loop. Is  there any way of 
concatenating those 3 components (which are a mixture  of numbers and letters)?
  
   Thanks for reading,
  
   Panos
  
   -
   Dr Panos Hadjinicolaou
  
   Energy Environment  Water Research Center (EEWRC)
   The Cyprus Institute
   --
[[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.
  
  
  -- 
  Dimitris Rizopoulos
  Assistant Professor
  Department of Biostatistics
  Erasmus University Medical Center
  
  Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
  Tel: +31/(0)10/7043478
  Fax: +31/(0)10/7043014
[[alternative HTML version deleted]]

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Re: [R] Concatenate a mix of numbers and letters to create a vector name

2010-07-26 Thread Dimitris Rizopoulos


have a look at assign() -- Best, Dimitris



On 7/26/2010 4:23 PM, Panos Hadjinicolaou wrote:

Thanks for the reply. Indeed the paste function results in  concatenation:
paste(c(tmax., 1950, 12), collapse=)
[1] tmax.195012

but  I am looking for a way to subsequently get rid of the -  - in order to  
use tmax.195012 as an object (e.g. to define a vector with that name).  Any ideas?

Thanks,

Panos

   _

From: Dimitris Rizopoulos [mailto:d.rizopou...@erasmusmc.nl]
To: Panos Hadjinicolaou [mailto:p.hadjinicol...@cyi.ac.cy]
Cc: r-help@r-project.org
Sent: Mon, 26 Jul 2010 16:48:31 +0300
Subject: Re: [R] Concatenate a mix of numbers and letters to create a vector 
name

have a look at function paste(), i.e., ?paste


   I hope it helps.

   Best,
   Dimitris


   On 7/26/2010 3:44 PM, Panos Hadjinicolaou wrote:
 Dear all,
   
 I am trying to create a vector name, for example tmax.195012 from  tmax., 
1950 and 12. Obviously I don't wish to simply type it because  the 3 name 
components are changing in each iteration within a loop. Is  there any way of 
concatenating those 3 components (which are a mixture  of numbers and letters)?
   
 Thanks for reading,
   
 Panos
   
 -
 Dr Panos Hadjinicolaou
   
 Energy Environment   Water Research Center (EEWRC)
 The Cyprus Institute
 --
  [[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.
   

   --
   Dimitris Rizopoulos
   Assistant Professor
   Department of Biostatistics
   Erasmus University Medical Center

   Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
   Tel: +31/(0)10/7043478
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--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center

Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
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Re: [R] Concatenate a mix of numbers and letters to create a vector name

2010-07-26 Thread Alain Guillet


 Hi,

assign(paste(c(tmax., 1950, 12), collapse=) ,1:10) does what you want.

Alain




On 26-Jul-10 16:23, Panos Hadjinicolaou wrote:

Thanks for the reply. Indeed the paste function results in  concatenation:
paste(c(tmax., 1950, 12), collapse=)
[1] tmax.195012

but  I am looking for a way to subsequently get rid of the -  - in order to  
use tmax.195012 as an object (e.g. to define a vector with that name).  Any ideas?

Thanks,

Panos

   _

From: Dimitris Rizopoulos [mailto:d.rizopou...@erasmusmc.nl]
To: Panos Hadjinicolaou [mailto:p.hadjinicol...@cyi.ac.cy]
Cc: r-help@r-project.org
Sent: Mon, 26 Jul 2010 16:48:31 +0300
Subject: Re: [R] Concatenate a mix of numbers and letters to create a vector 
name

have a look at function paste(), i.e., ?paste


   I hope it helps.

   Best,
   Dimitris


   On 7/26/2010 3:44 PM, Panos Hadjinicolaou wrote:
 Dear all,
   
 I am trying to create a vector name, for example tmax.195012 from  tmax., 
1950 and 12. Obviously I don't wish to simply type it because  the 3 name 
components are changing in each iteration within a loop. Is  there any way of 
concatenating those 3 components (which are a mixture  of numbers and letters)?
   
 Thanks for reading,
   
 Panos
   
 -
 Dr Panos Hadjinicolaou
   
 Energy Environment   Water Research Center (EEWRC)
 The Cyprus Institute
 --
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   -- 
   Dimitris Rizopoulos

   Assistant Professor
   Department of Biostatistics
   Erasmus University Medical Center

   Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
   Tel: +31/(0)10/7043478
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--
Alain Guillet
Statistician and Computer Scientist

SMCS - IMMAQ - Université catholique de Louvain
Bureau c.316
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[R] List to data frame

2010-07-26 Thread Johannes Graumann

Hi,

Any ideas on how to efficiently convert

 list(c(1,2,3),c(4,5,6))

to 

 data.frame(OriginalListIndex=c(1,1,1,2,2,2),Item=c(1,2,3,4,5,6))

Thanks for any hints,

Joh

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[R] Trouble using grid.layout in Sweave

2010-07-26 Thread Sebastian Weber

Hi!

I am troubled by Sweave which I want to use in order to plot graphics which
I build up successively by the use of grid.layout. Here is the code:

fig=TRUE,label=evolDist,height=6in,width=3in,pdf=FALSE=

## combined plot via grid viewports
grid.newpage()
pushViewport(viewport(layout=grid.layout(2,1)))
vlay - function(x,y) viewport(layout.pos.row=x, layout.pos.col=y)
print(pl$gscoreDist, vp=vlay(1,1))
print(pl$acceptDist, vp=vlay(2,1))
dev.off()

@

The error is the following from Sweave:

Error in grid.newpage() : Non-finite location and/or size for viewport
In addition: There were 24 warnings (use warnings() to see them)
Error in driver$runcode(drobj, chunk, chunkopts) :
  Error in grid.newpage() : Non-finite location and/or size for viewport
Calls: Sweave - Anonymous
Execution halted


Any hints? Of course, I can always wrap the code into a fig=FALSE, and
pdf()-call, but that is not how sweave is meant to be used, as I got it.

Many thanks in advance,

Sebastian Weber

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Re: [R] List to data frame

2010-07-26 Thread Henrique Dallazuanna

Try this:

stack(data.frame(list('A' = c(1,2,3), 'B' = c(4,5,6

On Mon, Jul 26, 2010 at 11:46 AM, Johannes Graumann 
johannes_graum...@web.de wrote:

 Hi,

 Any ideas on how to efficiently convert

  list(c(1,2,3),c(4,5,6))

 to

  data.frame(OriginalListIndex=c(1,1,1,2,2,2),Item=c(1,2,3,4,5,6))

 Thanks for any hints,

 Joh

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

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Re: [R] Plot of a subset of a data.frame()

2010-07-26 Thread Steffen Uhlig


Dear David, Petr, and Alain,

thank you very much for your fast responses. It's a typical 
handbook-not-read-error at my side. I will dig deeper into the 
plot-functions and the assignment of data. I was not aware of that the 
vector a is handled as a vector of factors with 10 levels. Thanks for 
your suggestions and hints!


Best regards,
/steffen


Am 26.07.2010 14:30, schrieb David Winsemius:


On Jul 26, 2010, at 7:38 AM, Steffen Uhlig wrote:


Hello,

my data.frame is sort of a collection of process values, i.e. huge
run-chart. It consists of a time-stamp in the first column (date as
string), factors in the following columns (used for subset-filtering),
and some process-data columns.
Hereafter, two examples are listed, showing the problems that occour
during print:

At first the example, that works fine:

~~
a = c(1:10) # create a vector of integers
b = rep(c(a,b),5) # create a vector of chars, used
# as factor-levels
d = rnorm(10) # some random numbers
e = data.frame(a,b,d) # connect to a data.frame


You've gotten several answers, but none have addressed an aspect of R
behavior that took me longer to appreciate than it perhaps should have.
The b column inside the e data.frame is now a factor column. I
mention that because you later referred to it as a string which it is
not. It is an integer with an associated indexed level character vector.
Many of the functions that you might think would work on strings
will give either errors or unexpected results when applied to factors.




e.1 = subset(e, b==a) # create two subsets
e.2 = subset(e, b==b)
plot(d~a, e.1, pch=3, col=2) # plot first data-subset
points(d~a, e.2, pch=4, col=3) # plot the 2nd one

~~
all looks fine in theses plots.


However, changing the content of vector a to a set of strings the
following happens:

~~
a = c(a,b,c,d,e,f,g,h,i,j)
e = data.frame(a,b,d) # re-build data.frame

e.1 = subset(e, b==a) # create two subsets
e.2 = subset(e, b==b)
plot(d~a, e.1, pch=3, col=2)
points(d~a, e.2, pch=4, col=3)
~~
The plot-command produces horizontal lines instead of dots. This seems
to happen when the x-axis contains strings rather than numbers. is
there a way out?

Best regards,
/Steffen



--
Steffen Uhlig, PhD
Mechatronik und Sensortechnik
HTW des Saarlandes
Goebenstraße 40
66117 Saarbrücken

Tel.: +49 (0) 681 58 67 274

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Re: [R] search and replace in a list of strings

2010-07-26 Thread Olivier Coupiac

It works, thanks a lot!

Cheers

-Original Message-
From: jim holtman [mailto:jholt...@gmail.com] 
Sent: Monday, July 26, 2010 4:04 PM
To: Olivier Coupiac
Cc: r-help@r-project.org
Subject: Re: [R] search and replace in a list of strings

The most usable form is to convert it to POSIXct since most of the
plotting routines know how to handle date/time.  You could do:

myDate - as.POSIXct(paste(WD$date, WD$time), format='%m.%d.%Y %H:%M:%S')

e.g.,
 as.POSIXct('07.07.2010 12:34:15', format='%m.%d.%Y %H:%M:%S')
[1] 2010-07-07 12:34:15 EDT

On Mon, Jul 26, 2010 at 9:07 AM, Olivier Coupiac
olivier.coup...@renerco.com wrote:
 Hi everybody,

 I have a 4*4 matrix WD of wind data of the following form:

 WD
 $date
 [1] 07.07.2010 07.07.2010 07.07.2010 07.07.2010

 $time
 [1] 00:00:00 00:10:00 00:20:00 00:30:00

 $CH1Avg
 [1] 3.02 3.04 2.94 2.71

 I would like to transform the date and time strings in usable numbers for 
 plotting :

 -          Transform 07.07.2010 in 07*24 + 07*31*24   for example

 -          Transform the string of type 12:15:20 in a number equals to 
 12+15/60+20/3600

 I would hence need a function able to look for defined patterns in a list, 
 like aa:bb:cc and evaluate aa+bb/60+cc/3600.

 Anyone has a idea?

 Thanks a lot!
 Cheers

 Olivier Coupiac

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What is the problem that you are trying to solve?

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Re: [R] hatching posibility in Panel.Polygon

2010-07-26 Thread HC


Thank you very much for your quick help.

In fact, I was trying to show confidence intervals from two different
methods (for various groups in panels) that had some overlaps and some
exclusive areas. And using fill hides the area of the one underneath.
Hatching may have been very good for this and that is why I was trying to
figure out a way for doing it.

Since the option is not there right now, I will use lines' to show the
comparison and be content with that for the time being!

Once again, thank you very much for your help.
HC
-- 
View this message in context: 
http://r.789695.n4.nabble.com/hatching-posibility-in-Panel-Polygon-tp2301863p2302461.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] List to data frame

Hi,

Here is another option if you already have a list you want to convert.
 This will handle different elements of the list being different
lengths.

#Using your example data
mydata - list(c(1,2,3),c(4,5,6))

data.frame(
 OriginalListIndex = rep(x = seq_along(mydata),
   times = unlist(lapply(mydata, length))),
 Item = unlist(mydata)
)

#Just to demonstrate that this method works generally
mydata - list(c(1,2,3), c(7,6), c(3,4,5,6,7,8,9))

data.frame(
 OriginalListIndex = rep(x = seq_along(mydata),
   times = unlist(lapply(mydata, length))),
 Item = unlist(mydata)
)


HTH,

Josh

On Mon, Jul 26, 2010 at 7:46 AM, Johannes Graumann
johannes_graum...@web.de wrote:
 Hi,

 Any ideas on how to efficiently convert

 list(c(1,2,3),c(4,5,6))

 to

 data.frame(OriginalListIndex=c(1,1,1,2,2,2),Item=c(1,2,3,4,5,6))

 Thanks for any hints,

 Joh

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-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

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Re: [R] Trouble using grid.layout in Sweave

2010-07-26 Thread Felipe Carrillo

I just run the code below with sweave and works fine
It looks like you might be missing the sequence of vplay


fig.R,echo=F,fig=T=
library(ggplot2)
vplay- function(x, y) 
viewport(layout.pos.row=x, layout.pos.col=y) 
grid.newpage() 
p - ggplot(diamonds, aes(x=carat, y=..density..)) + 
geom_histogram(binwidth=0.2)
p - p + facet_grid(. ~ cut)
pushViewport(viewport(layout=grid.layout(3,3))) 
print(p, vp=vplay(1,1)) 
print(p, vp=vplayt(2:3,2:3)) 
print(p, vp=vplay(1, 2:3)) 
print(p, vp=vplay(2:3, 1)) 
@
 
Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish  Wildlife Service
California, USA



- Original Message 
 From: Sebastian Weber sebastian.we...@physik.tu-darmstadt.de
 To: r-help@r-project.org
 Sent: Mon, July 26, 2010 7:51:06 AM
 Subject: [R] Trouble using grid.layout in Sweave
 
 Hi!
 
 I am troubled by Sweave which I want to use in order to plot graphics which
 I build up successively by the use of grid.layout. Here is the code:
 
 fig=TRUE,label=evolDist,height=6in,width=3in,pdf=FALSE=
 
 ## combined plot via grid viewports
 grid.newpage()
 pushViewport(viewport(layout=grid.layout(2,1)))
 vlay - function(x,y) viewport(layout.pos.row=x, layout.pos.col=y)
 print(pl$gscoreDist, vp=vlay(1,1))
 print(pl$acceptDist, vp=vlay(2,1))
 dev.off()
 
 @
 
 The error is the following from Sweave:
 
 Error in grid.newpage() : Non-finite location and/or size for viewport
 In addition: There were 24 warnings (use warnings() to see them)
 Error in driver$runcode(drobj, chunk, chunkopts) :
   Error in grid.newpage() : Non-finite location and/or size for viewport
 Calls: Sweave - Anonymous
 Execution halted
 
 
 Any hints? Of course, I can always wrap the code into a fig=FALSE, and
 pdf()-call, but that is not how sweave is meant to be used, as I got it.
 
 Many thanks in advance,
 
 Sebastian Weber
 
     [[alternative HTML version deleted]]
 
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Re: [R] How to generate a sequence of dates without hardcoding the year

2010-07-26 Thread Felipe Carrillo

Thanks Jim and Enrique, that should work since I am only trying to show
the month along my X axis it regardless of what year it is.
 
Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish  Wildlife Service
California, USA



- Original Message 
 From: jim holtman jholt...@gmail.com
 To: Felipe Carrillo mazatlanmex...@yahoo.com
 Cc: r-h...@stat.math.ethz.ch
 Sent: Sat, July 24, 2010 4:02:57 PM
 Subject: Re: [R] How to generate a sequence of dates without hardcoding the 
year
 
 Is this what you want if you want to assume that the date without a
 year is this year:
 
  seq(as.Date(7-1,%m-%d),by=week, length=52)
 [1] 2010-07-01 2010-07-08 2010-07-15 2010-07-22 2010-07-29
 2010-08-05 2010-08-12 2010-08-19
 [9] 2010-08-26 2010-09-02 2010-09-09 2010-09-16 2010-09-23
 2010-09-30 2010-10-07 2010-10-14
 [17] 2010-10-21 2010-10-28 2010-11-04 2010-11-11 2010-11-18
 2010-11-25 2010-12-02 2010-12-09
 [25] 2010-12-16 2010-12-23 2010-12-30 2011-01-06 2011-01-13
 2011-01-20 2011-01-27 2011-02-03
 [33] 2011-02-10 2011-02-17 2011-02-24 2011-03-03 2011-03-10
 2011-03-17 2011-03-24 2011-03-31
 [41] 2011-04-07 2011-04-14 2011-04-21 2011-04-28 2011-05-05
 2011-05-12 2011-05-19 2011-05-26
 [49] 2011-06-02 2011-06-09 2011-06-16 2011-06-23
 
 
 
 On Sat, Jul 24, 2010 at 5:07 PM, Felipe Carrillo
 mazatlanmex...@yahoo.com wrote:
  Hi:
  I have a dataframe named 'spring' and I am trying to add a new variable 
named
  'IdDate'
  This line of code works fine:
  spring$idDate - seq(as.Date(2008-07-01),as.Date(2009-06-30),by=week)
 
  But I don't want to hardcode the year because it will be used again the
  following year
  Is it possible to just generate dates with the month and day?
 
  I tried the code below:
  seq(as.Date(7-1,%B%d),as.Date(6-30,%B%d),by=week)
 
  and got this error message:
  Error in seq.int(0, to - from, by) : 'to' must be finite
  Thanks for any pointers
 
 
  Felipe D. Carrillo
  Supervisory Fishery Biologist
  Department of the Interior
  US Fish  Wildlife Service
  California, USA
 
 
 
 
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  R-help@r-project.org mailing list
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  PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
  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] Optimization problem with nonlinear constraint

2010-07-26 Thread Ravi Varadhan

Hi Uli,

I am not sure if this is the problem that you really want to solve.  The
answer is the solution to the equation y = x * T^(x-1), provided a solution
exists.  There is no optimization involved here.  What is the real problem
that you are trying to solve?

If you want to solve a more meaningful constrained optimization problem, you
may want to try the abalama package which I just put on CRAN.  It can
optimize smooth nonlinear functions subject to linear and nonlinear equality
and inequality constraints.

http://cran.r-project.org/web/packages/alabama/index.html

Let me know if you run into any problems using it.

Best,
Ravi.


-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Uli Kleinwechter
Sent: Monday, July 26, 2010 10:16 AM
To: r-help@r-project.org
Subject: [R] Optimization problem with nonlinear constraint

Dear all,

I'm looking for a way to solve a simple optimization problem with a 
nonlinear constraint. An example would be

max x   s.t.   y = x * T ^(x-1)

where y and T are known values.

optim() and constrOptim() do only allow for box or linear constraints, 
so I did not succedd here. I also found hints to donlp2 but this does 
not seem to be available anymore.

Any hints are welcome,

Uli

-- 

Uli Kleinwechter
Agricultural and Food Policy Group (420a)
University of Hohenheim
D-70593 Stuttgart
E-mail: u.kleinwech...@uni-hohenheim.de

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Re: [R] merge table rows (\multirow)

2010-07-26 Thread Fabian Scheipl

You can also automate it with this:

do.multirow-function(df, which=1:ncol(df)){
    for(c in which){
        runs - rle(as.character(df[,c]))
        if(all(runs$lengths1)){
            tmp - rep(, nrow(df))
            tmp[c(1, 1+head(cumsum(runs$lengths),-
1))] -
                    paste(\\multirow{,runs$lengths,}{*}{,df[c(1,
1+head(cumsum(runs$lengths),-1)),c],},sep=)
            df[,c] - tmp
        }
    }
    return(df)
}

This will replace the which-columns of data.frame df that have
only repeated entries  with the appropriate \multirow commands.

You then have to use print.xtable with sth like
print(xtable(do.multirow(df)),
        sanitize.text.function = function(x){
            return(x)
        }
)
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Re: [R] UseR! 2010 - my impressions

2010-07-26 Thread Tal Galili

Dear Ravi - I echo everything you wrote, useR2010 was an amazing experience
(for me, and for many others with whom I have spoken about it).
Many thanks should go to the wonderful people who put their efforts into
making this conference a reality (and Kate is certainly one of them).
Thank you for expressing feelings I had using your own words.

Best,
Tal


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




On Sat, Jul 24, 2010 at 2:50 AM, Ravi Varadhan rvarad...@jhmi.edu wrote:

 Dear UseRs!,

 Everything about UseR! 2010 was terrific!  I really mean everything - the
 tutorials, invited talks, kaleidoscope sessions, focus sessions, breakfast,
 snacks, lunch, conference dinner, shuttle services, and the participants.
 The organization was fabulous.  NIST were gracious hosts, and provided top
 notch facilities.  The rousing speech by Antonio Possolo, who is the chief
 of Statistical Engineering Division at NIST, set the tempo for the entire
 conference.  Excellent invited lectures by Luke Tierney, Frank Harrell, Mark
 Handcock, Diethelm Wurtz, Uwe Ligges, and Fritz Leisch.  All the sessions
 that I attended had many interesting ideas and useful contributions.  During
 the whole time that I was there, I could not help but get the feeling that I
 am a part of something great.

 Before I end, let me add a few words about a special person.  This
 conference would not have been as great as it was without the tireless
 efforts of Kate Mullen.  The great thing about Kate is that she did so much
 without ever hogging the limelight.  Thank you, Kate and thank you NIST!

 I cannot wait for UseR!2011!

 Best,
 Ravi.

 

 Ravi Varadhan, Ph.D.
 Assistant Professor,
 Division of Geriatric Medicine and Gerontology
 School of Medicine
 Johns Hopkins University

 Ph. (410) 502-2619
 email: rvarad...@jhmi.edu

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[R] OOP module

2010-07-26 Thread Albert-Jan Roskam

Hello,
 
Does anybody know if the OOP module (Chambers  Temple Lang) is going to 
replace the the S4 (and the S3) class system? 
http://www.omegahat.org/OOP/oop.pdf

Cheers!!
Albert-Jan

~~
All right, but apart from the sanitation, the medicine, education, wine, public 
order, irrigation, roads, a fresh water system, and public health, what have 
the Romans ever done for us?
~~


  
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[R] gapped sequence data summary

2010-07-26 Thread jd6688


   Id cat1 location item_values p-values sequence 
a111 1 3002737 100 0.01   1 
a112 1 3017821 102 0.05   2 
a113 2 3027730 103 0.02   3 
a114 2 3036220 104 0.04   4 
a115 1 3053984 105 0.03   5 

a118 1 3090500 106 0.02   8 
a119 1 3103304 107 0.03   9   
a120 2 3090500 106 0.02   10 
a121 2 3103304 107 0.03   11 

what I am trying to accomplish is:

for sequence 1:5
   cat1start of the location   end of the location,   peak value
of the item_values
  1  30027373053984 105
  2   30277303036220104

for sequence 8:11

   cat1start of the location   end of the location,   peak value
of the item_values
  1  3090500   3103304  107
  2  3090500   3103304  107

and so on...

I have been trying to find a way to accomplish this, however, I didn't find
one that worked as expected.
would you shed some light on this? Thanks,


 
-- 
View this message in context: 
http://r.789695.n4.nabble.com/gapped-sequence-data-summary-tp2302552p2302552.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] List to data frame

2010-07-26 Thread Johannes Graumann

Thanks a lot!

This solves my problem!

Joh

On Monday 26 July 2010 17:06:37 Joshua Wiley wrote:
 Hi,
 
 Here is another option if you already have a list you want to convert.
  This will handle different elements of the list being different
 lengths.
 
 #Using your example data
 mydata - list(c(1,2,3),c(4,5,6))
 
 data.frame(
  OriginalListIndex = rep(x = seq_along(mydata),
times = unlist(lapply(mydata, length))),
  Item = unlist(mydata)
 )
 
 #Just to demonstrate that this method works generally
 mydata - list(c(1,2,3), c(7,6), c(3,4,5,6,7,8,9))
 
 data.frame(
  OriginalListIndex = rep(x = seq_along(mydata),
times = unlist(lapply(mydata, length))),
  Item = unlist(mydata)
 )
 
 
 HTH,
 
 Josh
 
 On Mon, Jul 26, 2010 at 7:46 AM, Johannes Graumann
 
 johannes_graum...@web.de wrote:
  Hi,
  
  Any ideas on how to efficiently convert
  
  list(c(1,2,3),c(4,5,6))
  
  to
  
  data.frame(OriginalListIndex=c(1,1,1,2,2,2),Item=c(1,2,3,4,5,6))
  
  Thanks for any hints,
  
  Joh
  
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[R] Cluster analysis

2010-07-26 Thread Pablo Cerdeira

Hi all,

I have no idea if this question is to easy to be answered, but I´m starting
with R. So, here we go.

I have a large dataset with a lot of steps a judicial case. A sample is
attached.

I´d like to do a cluster analysis to try to understand with one is the most
usual path followed by this legal cases.

After that, I´d like to plot a cluster tree.

In the attached sample, the column:

- id_processo is the primary key of a legal case;
- number is the step number in the legal case;
- andamento is the description of the legal case step.

I have no idea on how to do it using R. Can someone help me?

Thanks in advanced

-- 
*Pablo de Camargo Cerdeira*
pa...@fgv.br
pablo.cerde...@gmail.com
+55 (21) 3799-6065
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Re: [R] gapped sequence data summary

2010-07-26 Thread William Dunlap

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of jd6688
 Sent: Monday, July 26, 2010 9:23 AM
 To: r-help@r-project.org
 Subject: [R] gapped sequence data summary

Id cat1 location item_values p-values sequence 
 a111 1 3002737 100 0.01   1 
 a112 1 3017821 102 0.05   2 
 a113 2 3027730 103 0.02   3 
 a114 2 3036220 104 0.04   4 
 a115 1 3053984 105 0.03   5 

 a118 1 3090500 106 0.02   8 
 a119 1 3103304 107 0.03   9   
 a120 2 3090500 106 0.02   10 
 a121 2 3103304 107 0.03   11 

 what I am trying to accomplish is:

 for sequence 1:5
cat1start of the location   end of the 
 location,   peak value
 of the item_values
   1  30027373053984   
   105
   2   30277303036220  
   104

 for sequence 8:11

cat1start of the location   end of the 
 location,   peak value
 of the item_values
   1  3090500   3103304
   107
   2  3090500   3103304
   107

 and so on...

To find which rows are the first and last row
of a run of numbers that differs by 1 you can use
the functions
first - function(x)c(TRUE, diff(x)!=1)
last - function(x)c(diff(x)!=1, TRUE)
on 'sequence'.

You can assign a group identifier to each run with
runNumber - cumsum(c(TRUE, diff(sequence)))
and use aggregate() or one of the functions in the
plyr package to apply a summary function to each
group.

If there might be NA's in the sequence variable you
will have to modify these a bit.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com  

 I have been trying to find a way to accomplish this, however, 
 I didn't find
 one that worked as expected.
 would you shed some light on this? Thanks,

 -- 
 View this message in context: 
 http://r.789695.n4.nabble.com/gapped-sequence-data-summary-tp2
302552p2302552.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
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Re: [R] Trouble using grid.layout in Sweave

2010-07-26 Thread Sebastian Weber

Hi!

I moved the definition of the vlay function before the grid.newpage call,
and now it works! This is weird, I don't get it what was wrong in the first
place, if someone can enlighten me, I would feel better.

Sebastian

To make it clear, this works:


evolDist,fig=TRUE=
vlay - function(x,y) viewport(layout.pos.row=x, layout.pos.col=y)

## combined plot via grid viewports
grid.newpage()
pushViewport(viewport(layout=grid.layout(2,1)))
print(pl$gscoreDist, vp=vlay(1,1))
print(pl$acceptDist, vp=vlay(2,1))
@

On Mon, Jul 26, 2010 at 5:32 PM, Felipe Carrillo
mazatlanmex...@yahoo.comwrote:

 I just run the code below with sweave and works fine
 It looks like you might be missing the sequence of vplay


 fig.R,echo=F,fig=T=
 library(ggplot2)
 vplay- function(x, y)
 viewport(layout.pos.row=x, layout.pos.col=y)
 grid.newpage()
 p - ggplot(diamonds, aes(x=carat, y=..density..)) +
 geom_histogram(binwidth=0.2)
 p - p + facet_grid(. ~ cut)
 pushViewport(viewport(layout=grid.layout(3,3)))
 print(p, vp=vplay(1,1))
 print(p, vp=vplayt(2:3,2:3))
 print(p, vp=vplay(1, 2:3))
 print(p, vp=vplay(2:3, 1))
 @

 Felipe D. Carrillo
 Supervisory Fishery Biologist
 Department of the Interior
 US Fish  Wildlife Service
 California, USA



 - Original Message 
  From: Sebastian Weber sebastian.we...@physik.tu-darmstadt.de
  To: r-help@r-project.org
  Sent: Mon, July 26, 2010 7:51:06 AM
  Subject: [R] Trouble using grid.layout in Sweave
 
  Hi!
 
  I am troubled by Sweave which I want to use in order to plot graphics
 which
  I build up successively by the use of grid.layout. Here is the code:
 
  fig=TRUE,label=evolDist,height=6in,width=3in,pdf=FALSE=
 
  ## combined plot via grid viewports
  grid.newpage()
  pushViewport(viewport(layout=grid.layout(2,1)))
  vlay - function(x,y) viewport(layout.pos.row=x, layout.pos.col=y)
  print(pl$gscoreDist, vp=vlay(1,1))
  print(pl$acceptDist, vp=vlay(2,1))
  dev.off()
 
  @
 
  The error is the following from Sweave:
 
  Error in grid.newpage() : Non-finite location and/or size for viewport
  In addition: There were 24 warnings (use warnings() to see them)
  Error in driver$runcode(drobj, chunk, chunkopts) :
Error in grid.newpage() : Non-finite location and/or size for viewport
  Calls: Sweave - Anonymous
  Execution halted
 
 
  Any hints? Of course, I can always wrap the code into a fig=FALSE, and
  pdf()-call, but that is not how sweave is meant to be used, as I got it.
 
  Many thanks in advance,
 
  Sebastian Weber
 
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[R] the real dimnames

2010-07-26 Thread Michael Lachmann


Hi,

R seems to have a feature that isn't used much, which I don't really  
now how to call. But, the dimnames function, can in addition to giving  
names to rows/columns/dim 3 rows/dim 4 rows... can also give labels to  
the dimensions themselves.


Thus, I can do:
A = matrix(1:9,3,3)

dimnames(A) = list(from=c(), to=c() )

and now, printing a prints these dimension labels nicely:
 A
  to
from   [,1] [,2] [,3]
  [1,]147
  [2,]258
  [3,]369

I don't know exactly what these labels are called, but they can be  
handy sometimes.

For example, they could be handy when using aperm. Instead of writing

aperm(A, c(2,1) )

It would be nice to be able to write:

aperm(A, c(from,to) )

This way, I don't have to remember what is the order of the dims of A.

It is easy to extend the functionality of aperm, so that the above  
code works:

---
my.aperm=function( A, d , ...) {
  n=names(dimnames(A))
  if( is.null(n) ) {
return( aperm(A,d, ...))
  }

  dimnum = seq(along=n)
  names(dimnum) = n

  return( aperm(A, dimnum[d], ... ))
}
--

It seems some functions support these dim labels (print, for example),  
and some don't (for example, matrix multiplication doesn't, aperm  
doesn't, apply doesn't).


Another cool way to use these, would be labeled arguments in array  
subsetting:

 a[to=1:2,from=2:3]
  to
from   [,1] [,2]
  [1,]47
  [2,]58
 a[from=1:2,to=2:3]
  to
from   [,1] [,2]
  [1,]47
  [2,]58

No, doesn't seem to work.

My questions are:
1. Do these have names, so that I can search the documentation for them?
2. How are they usually used?

thanks,
Michael

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[R] zoo objects and c

2010-07-26 Thread Erin Hodgess

Dear R People:

I would like to combine a zoo object with some observations at the end.

Here is the set up:

 xgh
2010-06-15 2010-06-16 2010-06-17 2010-06-18 2010-06-19 2010-06-20 2010-06-21
 5  6  1  5  0  0 13
2010-06-22 2010-06-23 2010-06-24 2010-06-25 2010-06-26 2010-06-27 2010-06-28
 9  6  4  6  0  0  9
2010-06-29 2010-06-30 2010-07-01 2010-07-02 2010-07-03 2010-07-04 2010-07-05
15 10  5  6  0  0  0
2010-07-06 2010-07-07 2010-07-08 2010-07-09 2010-07-10
12  8  3  6  0
 xs
 [1] 2010-06-15 2010-06-16 2010-06-17 2010-06-18 2010-06-19
 [6] 2010-06-20 2010-06-21 2010-06-22 2010-06-23 2010-06-24
[11] 2010-06-25 2010-06-26 2010-06-27 2010-06-28 2010-06-29
[16] 2010-06-30 2010-07-01 2010-07-02 2010-07-03 2010-07-04
[21] 2010-07-05 2010-07-06 2010-07-07 2010-07-08 2010-07-09
[26] 2010-07-10 2010-07-11 2010-07-12 2010-07-13 2010-07-14
[31] 2010-07-15
 zoo(c(xgh,-0.7,3.6,3.7,3.4,3.2),order=xs)
Error in rbind.zoo(...) : indexes overlap


What am I doing wrong, please?  I'm sure that it's something simple.

Thanks,
Erin

-- 
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com

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Re: [R] Outlier detection in bimodal distribution

2010-07-26 Thread Bert Gunter

I doubt that there are any.

For a bimodal distrbution, I think one would have to specify in
detail the nature of the distribution and then define what one means
by an outlier (a slippery, sinister notion, at best and a flimsy
cloak for skulduggery at worst) .

As has been said her frquently before -- what is the scientific
context? What is the scientific question?

I suspect you need to seek the help of a local statistician before you
sweep possibly important data under the outlier rug.

Bert Gunter
Genentech Nonclinical Biostatistics


On Mon, Jul 26, 2010 at 5:40 AM, Tim Smith tim_smith_...@yahoo.com wrote:

 Hi,

 I was looking for a package that would help with outlier detection for bimodal
 distributions. I have tried 'outliers' and 'extremevalues' packages, but am 
 not
 sure if they are ok for bimodal distribution.

 Any help would be highly appreciated!

 thanks,



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Re: [R] zoo objects and c

2010-07-26 Thread Gabor Grothendieck

On Mon, Jul 26, 2010 at 2:17 PM, Erin Hodgess erinm.hodg...@gmail.com wrote:
 Dear R People:

 I would like to combine a zoo object with some observations at the end.

 Here is the set up:

 xgh
 2010-06-15 2010-06-16 2010-06-17 2010-06-18 2010-06-19 2010-06-20 2010-06-21
         5          6          1          5          0          0         13
 2010-06-22 2010-06-23 2010-06-24 2010-06-25 2010-06-26 2010-06-27 2010-06-28
         9          6          4          6          0          0          9
 2010-06-29 2010-06-30 2010-07-01 2010-07-02 2010-07-03 2010-07-04 2010-07-05
        15         10          5          6          0          0          0
 2010-07-06 2010-07-07 2010-07-08 2010-07-09 2010-07-10
        12          8          3          6          0
 xs
  [1] 2010-06-15 2010-06-16 2010-06-17 2010-06-18 2010-06-19
  [6] 2010-06-20 2010-06-21 2010-06-22 2010-06-23 2010-06-24
 [11] 2010-06-25 2010-06-26 2010-06-27 2010-06-28 2010-06-29
 [16] 2010-06-30 2010-07-01 2010-07-02 2010-07-03 2010-07-04
 [21] 2010-07-05 2010-07-06 2010-07-07 2010-07-08 2010-07-09
 [26] 2010-07-10 2010-07-11 2010-07-12 2010-07-13 2010-07-14
 [31] 2010-07-15
 zoo(c(xgh,-0.7,3.6,3.7,3.4,3.2),order=xs)
 Error in rbind.zoo(...) : indexes overlap


 What am I doing wrong, please?  I'm sure that it's something simple.


please use dput when posting to r-help:

dput(xgh)

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Re: [R] zoo objects and c

2010-07-26 Thread Erin Hodgess

Whoops...sorry about that.  Here we go:

 dput(xgh)
structure(c(5, 6, 1, 5, 0, 0, 13, 9, 6, 4, 6, 0, 0, 9, 15, 10,
5, 6, 0, 0, 0, 12, 8, 3, 6, 0), index = structure(c(14775, 14776,
14777, 14778, 14779, 14780, 14781, 14782, 14783, 14784, 14785,
14786, 14787, 14788, 14789, 14790, 14791, 14792, 14793, 14794,
14795, 14796, 14797, 14798, 14799, 14800), class = Date), class = zoo)
 dput(xs)
structure(c(14775, 14776, 14777, 14778, 14779, 14780, 14781,
14782, 14783, 14784, 14785, 14786, 14787, 14788, 14789, 14790,
14791, 14792, 14793, 14794, 14795, 14796, 14797, 14798, 14799,
14800, 14801, 14802, 14803, 14804, 14805), class = Date)
 zoo(c(xgh,-0.7,3.6,3.7,3.4,3.2),order=xs)
Error in rbind.zoo(...) : indexes overlap



I just tried this:

zoo(c(as.numeric(xgh),-0.7,3.6,3.7,3.4,3.2),order=xs)

and it does work.  However, I'm not sure if that's the right way to go.

Thanks,
Erin


On Mon, Jul 26, 2010 at 1:20 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
 On Mon, Jul 26, 2010 at 2:17 PM, Erin Hodgess erinm.hodg...@gmail.com wrote:
 Dear R People:

 I would like to combine a zoo object with some observations at the end.

 Here is the set up:

 xgh
 2010-06-15 2010-06-16 2010-06-17 2010-06-18 2010-06-19 2010-06-20 2010-06-21
         5          6          1          5          0          0         13
 2010-06-22 2010-06-23 2010-06-24 2010-06-25 2010-06-26 2010-06-27 2010-06-28
         9          6          4          6          0          0          9
 2010-06-29 2010-06-30 2010-07-01 2010-07-02 2010-07-03 2010-07-04 2010-07-05
        15         10          5          6          0          0          0
 2010-07-06 2010-07-07 2010-07-08 2010-07-09 2010-07-10
        12          8          3          6          0
 xs
  [1] 2010-06-15 2010-06-16 2010-06-17 2010-06-18 2010-06-19
  [6] 2010-06-20 2010-06-21 2010-06-22 2010-06-23 2010-06-24
 [11] 2010-06-25 2010-06-26 2010-06-27 2010-06-28 2010-06-29
 [16] 2010-06-30 2010-07-01 2010-07-02 2010-07-03 2010-07-04
 [21] 2010-07-05 2010-07-06 2010-07-07 2010-07-08 2010-07-09
 [26] 2010-07-10 2010-07-11 2010-07-12 2010-07-13 2010-07-14
 [31] 2010-07-15
 zoo(c(xgh,-0.7,3.6,3.7,3.4,3.2),order=xs)
 Error in rbind.zoo(...) : indexes overlap


 What am I doing wrong, please?  I'm sure that it's something simple.


 please use dput when posting to r-help:

 dput(xgh)




-- 
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com

__
R-help@r-project.org mailing list
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[R] using string variable as order() function argument

2010-07-26 Thread mirek


Hello,

In my script I would like to use a loop, which sorts the dataframe 
according to different columns, pointed by the string variable.


id col1  col2  col3
1   1004  8
2   1112  2
3   1208  3
4   1305  5

Usually the order() function can be used like this:

sorted = mytable[order(column3) , ]

which results in properly sorted table:

id col1  col2  col3
2   1112  2
3   1208  3
4   1305  5
1   1004  8

But when trying to use a string variable instead of column3 name:

columnname = column3
sorted = mytable[order(columnname) , ]

this command is not properly evaluated and the effect is somewhat strange.

Would you suggest some solution?

Thanks a lot!

Mirek

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Re: [R] using string variable as order() function argument

2010-07-26 Thread Duncan Murdoch


On 26/07/2010 2:56 PM, mirek wrote:

Hello,

In my script I would like to use a loop, which sorts the dataframe 
according to different columns, pointed by the string variable.


 id col1  col2  col3
1   1004  8
2   1112  2
3   1208  3
4   1305  5

Usually the order() function can be used like this:

sorted = mytable[order(column3) , ]

which results in properly sorted table:

 id col1  col2  col3
2   1112  2
3   1208  3
4   1305  5
1   1004  8

But when trying to use a string variable instead of column3 name:

columnname = column3
sorted = mytable[order(columnname) , ]

this command is not properly evaluated and the effect is somewhat strange.


The argument to order() should be a vector whose sort order is to be 
returned.  So you just need to extract one column from the dataframe, e.g.


column - mytable[, columnname]
sorted - mytable[order(column) , ]

Duncan Murdoch

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Re: [R] zoo objects and c

2010-07-26 Thread Gabor Grothendieck

On Mon, Jul 26, 2010 at 2:24 PM, Erin Hodgess erinm.hodg...@gmail.com wrote:
 Whoops...sorry about that.  Here we go:

 dput(xgh)
 structure(c(5, 6, 1, 5, 0, 0, 13, 9, 6, 4, 6, 0, 0, 9, 15, 10,
 5, 6, 0, 0, 0, 12, 8, 3, 6, 0), index = structure(c(14775, 14776,
 14777, 14778, 14779, 14780, 14781, 14782, 14783, 14784, 14785,
 14786, 14787, 14788, 14789, 14790, 14791, 14792, 14793, 14794,
 14795, 14796, 14797, 14798, 14799, 14800), class = Date), class = zoo)
 dput(xs)
 structure(c(14775, 14776, 14777, 14778, 14779, 14780, 14781,
 14782, 14783, 14784, 14785, 14786, 14787, 14788, 14789, 14790,
 14791, 14792, 14793, 14794, 14795, 14796, 14797, 14798, 14799,
 14800, 14801, 14802, 14803, 14804, 14805), class = Date)
 zoo(c(xgh,-0.7,3.6,3.7,3.4,3.2),order=xs)
 Error in rbind.zoo(...) : indexes overlap



 I just tried this:

 zoo(c(as.numeric(xgh),-0.7,3.6,3.7,3.4,3.2),order=xs)

 and it does work.  However, I'm not sure if that's the right way to go.


It should be written like this:

   zoo(c(coredata(xgh), -0.7, 3.6, 3.7, 3.4, 3.2), xs)

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[R] HowTo get callback on destroy of grDevices::windows()

2010-07-26 Thread rrich...@fh-lausitz.de

 I'm build with the usage of the tcltk/tcltk2 package a gui which is 
started with rscript.exe. At this gui the user is able to open and close 
plot windows.


Now I'm looking for a way that a r function is called when a 
grDevices::windows window is closed.
Is there a way to attach a callback function or so, which would be 
triggert if the user click at close of the grDevices::window?


My workarround is to check continues the open windows with the tcl 
command after, but it's no good workarround.


thanks
roland

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[R] Time-dependent covarites in survreg function

2010-07-26 Thread Michael Haenlein

Dear all,

I'm doing a survival analysis with time-dependent covariates. Until now, I
have used a simple Cox model for this, specifically the coxph function from
the survival library. Now, I would like to try out an accelerated failure
time model with a parametric specification as implemented for example in the
survreg function.

Two questions: First, can survreg handle time-dependent covariates? The
description for this function does not make reference to them. And second,
in case survreg cannot deal with time-dependent covariates, is there a
similar function in some other package that can?

Thanks very much,

Michael

[[alternative HTML version deleted]]

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Re: [R] HowTo get callback on destroy of grDevices::windows()

2010-07-26 Thread Duncan Murdoch


On 26/07/2010 3:09 PM, rrich...@fh-lausitz.de wrote:
  I'm build with the usage of the tcltk/tcltk2 package a gui which is 
started with rscript.exe. At this gui the user is able to open and close 
plot windows.


Now I'm looking for a way that a r function is called when a 
grDevices::windows window is closed.
Is there a way to attach a callback function or so, which would be 
triggert if the user click at close of the grDevices::window?
  


I don't know if that's possible, but you'll be more likely to get a 
definitive answer on the R-devel list.


Duncan Murdoch
My workarround is to check continues the open windows with the tcl 
command after, but it's no good workarround.


thanks
roland

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Re: [R] the real dimnames

2010-07-26 Thread Allan Engelhardt


On 26/07/10 19:01, Michael Lachmann wrote:

Hi,

R seems to have a feature that isn't used much, which I don't really 
now how to call. But, the dimnames function, can in addition to giving 
names to rows/columns/dim 3 rows/dim 4 rows... can also give labels to 
the dimensions themselves.


Thus, I can do:
A = matrix(1:9,3,3)

dimnames(A) = list(from=c(), to=c() )


If you just want to set the names of the dimnames, just do

names(dimnames(A)) - c(from, to)

Remember that dimnames is a list and all lists have names.

Allan

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Re: [R] HowTo get callback on destroy of grDevices::windows()

2010-07-26 Thread rrich...@fh-lausitz.de


 okay, thanks I will try the R-devel list :)

BR
Roland

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[R] Sample size calculation for non-normal population with unknown mean and SD

2010-07-26 Thread Majonu


Basically, we have a population of 4,392 documents and we want to find out
the number of patents per document. We don’t want to go through all 4,392
documents, but want a reliable sample size from which to draw inferences. I
feel like this count data will not follow a normal distribution, but more
like a Poisson (skewed right.) The problem is we don’t have much similar
data to this data set, so mean and standard deviation are unknown. Is there
any way to derive a sample size based off the confidence interval, margin of
error, and population size for what I assume to be a non-normal population?
Any help would be greatly appreciated. 
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Re: [R] Sample size calculation for non-normal population with unknown mean and SD

2010-07-26 Thread Bert Gunter

The obvious:

Take a small sample, say 25-50. Get an estimate of your distribution
from that. Then use this to determine how many more (if any)
additional samples you need for desired precision. This latter can
probably easily be done via simulation/bootstrap if you don't want to
specify a parametric form.

My guess is that your distribution is right-skew but not Poisson --
probably more like a truncated Poisson. But of course I have no idea
what sorts of documents you've got, so how would I know?

Bert Gunter
Genentech Nonclinical Biostatistics

On Mon, Jul 26, 2010 at 1:28 PM, Majonu mnu...@andrew.cmu.edu wrote:

Basically, we have a population of 4,392 documents and we want to find out
the number of patents per document. We don’t want to go through all 4,392
documents, but want a reliable sample size from which to draw inferences. I
feel like this count data will not follow a normal distribution, but more
like a Poisson (skewed right.) The problem is we don’t have much similar
data to this data set, so mean and standard deviation are unknown. Is there
any way to derive a sample size based off the confidence interval, margin of
error, and population size for what I assume to be a non-normal population?
Any help would be greatly appreciated.
--
View this message in context:
http://r.789695.n4.nabble.com/Sample-size-calculation-for-non-normal-population-with-unknown-mean-and-SD-tp2302833p2302833.html
Sent from the R help mailing list archive at Nabble.com.

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[R] problem with building package on CRAN

2010-07-26 Thread William Revelle


Dear friends,
  I have just gotten a strange error message back from  Uwe saying 
that the most recent version of psych failed to pass R CMD check for 
Windows.


The error message was less than helpful, in that it seems to have 
failed when trying to include the Rcpp library, which I do not 
directly call.  (see below)



* using log directory 'd:/Rcompile/CRANpkg/local/2.11/psych.Rcheck'
* using R version 2.11.1 (2010-05-31)
* using session charset: ISO8859-1
* checking for file 'psych/DESCRIPTION' ... OK
* this is package 'psych' version '1.0-90'
* checking package name space information ... OK
* checking package dependencies ... OK
* checking if this is a source package ... OK
* checking whether package 'psych' can be installed ... ERROR
Installation failed.
The installation logfile:
-Id:/Rcompile/CRANpkg/lib/2.11/Rcpp/include



I do have several suggested packages  (polycor, GPArotation, MASS, 
graph, Rgraphviz, mvtnorm, Rcsdp),  but none of these are actually 
required.  My examples all ask if the suggested packages are 
available and then do not call them if they are not.


Any suggestions on what to do would be appreciated.

Thanks.

Bill



--
William Revelle http://personality-project.org/revelle.html
Professor   http://personality-project.org
Department of Psychology http://www.wcas.northwestern.edu/psych/
Northwestern University http://www.northwestern.edu/
Use R for psychology   http://personality-project.org/r
It is 6 minutes to midnight http://www.thebulletin.org

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[R] using string variable as order() function argument

2010-07-26 Thread mirek wyczesany


Hello,

In my script I would like to use a loop, which sorts the dataframe 
according to different columns, pointed by the string variable.


id col1  col2  col3
1   1004  8
2   1112  2
3   1208  3
4   1305  5

Usually the order() function can be used like this:

sorted = mytable**[order(column3) , ]

which results in properly sorted table:
**

id col1  col2  col3
2   1112  2
3   1208  3
4   1305  5
1   1004  8

**But when trying to use a string variable instead of column3 name:

columnname = column3
**sorted = mytable**[order(columnname) , ]**

this command is not properly evaluated and the effect is somewhat strange.

Would you suggest some solution?

Thanks a lot!

Mirek

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[R] easy debugging

2010-07-26 Thread ying_chen wang

I am new to R. Used to use FORTRAN. R is so different from FORTRAN. The
following codes would work in FOTRAN. I am trying to put an upper limit at
120. If the score is  120, it is assigned 120. Or else, keep the original
values.

version 1:

equated-11
result-11
equated-c(111.0,112.06, 112.9, 113.8, 115.0, 116.2, 117.0, 118.0, 120.5,
120.5, 120.5)

for (i in 1:11){
if (equated  120) result[i]-120
if (equated  120) result[i]-equated[i]
result-result
result
}
result

version2:

if (equated  120) result-120
if (equated  120) result-equated



If any of you can help, I would appreciate that.

G

[[alternative HTML version deleted]]

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[R] Model Fit Statistics in a Logit Model

2010-07-26 Thread Mathew, Abraham T

 

 

I'm running a logistic regression model in R. I've used both the Zelig and Car 
packages. However, I'm wondering if there is a simple way to get the model fit 
statistics for the model. (pseudo R-square, chi-square, log liklihood,etc)

 

Thanks

 

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Re: [R] easy debugging

Is this what you want:

 equated-c(111.0,112.06, 112.9, 113.8, 115.0, 116.2, 117.0, 118.0, 120.5,
+ 120.5, 120.5)
 equated[equated  120] - 120
 equated
 [1] 111.00 112.06 112.90 113.80 115.00 116.20 117.00 118.00 120.00
120.00 120.00



You should read up on 'indexing' in the R Intro paper.

On Mon, Jul 26, 2010 at 1:26 PM, ying_chen wang
gracedrop.w...@gmail.com wrote:
 I am new to R. Used to use FORTRAN. R is so different from FORTRAN. The
 following codes would work in FOTRAN. I am trying to put an upper limit at
 120. If the score is  120, it is assigned 120. Or else, keep the original
 values.

 version 1:

 equated-11
 result-11
 equated-c(111.0,112.06, 112.9, 113.8, 115.0, 116.2, 117.0, 118.0, 120.5,
 120.5, 120.5)

 for (i in 1:11){
 if (equated  120) result[i]-120
 if (equated  120) result[i]-equated[i]
 result-result
 result
 }
 result

 version2:

 if (equated  120) result-120
 if (equated  120) result-equated



 If any of you can help, I would appreciate that.

 G

        [[alternative HTML version deleted]]

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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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[R] Bug on r-bc?

2010-07-26 Thread Paul Smith

Dear All,

The following code should return 1, but it returns 0:

source(http://r-bc.googlecode.com/svn/trunk/R/bc.R;)
bc(9 % 2)

Do you confirm this bug?

Paul

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Re: [R] using string variable as order() function argument

Hi,

Is this what you want?

##
mytable - read.table(textConnection(
id col1  col2  col3
1004  8
1112  2
1208  3
1305  5
), header = TRUE)

mytable

columnname - col3

mytable[order(mytable[, columnname]), ]
###

Josh

On Mon, Jul 26, 2010 at 11:54 AM, mirek wyczesany
miroslaw.wyczes...@uj.edu.pl wrote:
 Hello,

 In my script I would like to use a loop, which sorts the dataframe according
 to different columns, pointed by the string variable.

    id col1  col2  col3
 1   10    0    4      8
 2   11    1    2      2
 3   12    0    8      3
 4   13    0    5      5

 Usually the order() function can be used like this:

 sorted = mytable**[order(column3) , ]

 which results in properly sorted table:
 **

    id col1  col2  col3
 2   11    1    2      2
 3   12    0    8      3
 4   13    0    5      5
 1   10    0    4      8

 **But when trying to use a string variable instead of column3 name:

 columnname = column3
 **sorted = mytable**[order(columnname) , ]**

 this command is not properly evaluated and the effect is somewhat strange.

 Would you suggest some solution?

 Thanks a lot!

 Mirek

 __
 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.




-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

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[R] how to generate a random data from a empirical distribition

2010-07-26 Thread xin wei


hi, this is more a statistical question than a R question. but I do want to
know how to implement this in R. 
I have 10,000 data points. Is there any way to generate a empirical
probablity distribution from it (the problem is that I do not know what
exactly this distribution follows, normal, beta?). My ultimate goal is to
generate addition 20,000 data point from this empirical distribution created
from the existing 10,000 data points. 
thank you all in advance.


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Re: [R] Your message to R-help awaits moderator approval

2010-07-26 Thread mirek wyczesany

verry sorry for posting the mwssage below from wrong email account. 
please reject it. sorry for the inconvenience...

mirek

On 07/26/2010 08:54 PM, r-help-boun...@r-project.org wrote:

Your mail to 'R-help' with the subject

 using string variable as order() function argument

Is being held until the list moderator can review it for approval.

The reason it is being held:

 Post by non-member to a members-only list

Either the message will get posted to the list, or you will receive
notification of the moderator's decision.  If you would like to cancel
this posting, please visit the following URL:

 
https://stat.ethz.ch/mailman/confirm/r-help/a091c4c24fa7c13e6e67722b19bbc43a0badf79f

   



--
Mirek Wyczesany
Jagiellonian University
Psychophysiology Lab
Kraków, PL

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[R] Accessing single element of data.frame

2010-07-26 Thread vacas


Hi I am new to R. 
I am having this problem
t1 - read.csv(myfile.csv)
t2 - data.frame(t1)
which have 10 row and 10 columns
t2[1,1] does not give the first element but it gives the levels, how can I
fix it. 

I will be thankful to community. 
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Re: [R] easy debugging

On Mon, Jul 26, 2010 at 2:06 PM, ying_chen wang
gracedrop.w...@gmail.com wrote:
 Thanks, it works. I found out the solution a moment ago. The 2nd one works.

 But, the weird thing is that if I use 'x', it works. If I use 'equated', it
 didn't work. Not sure why.

What is 'x' ?


 Thanks again.

 G

 On Mon, Jul 26, 2010 at 5:04 PM, Joshua Wiley jwiley.ps...@gmail.com
 wrote:

 What about these two options?

 #One way
 ifelse(equated  120, 120, equated)

 #Another way
 equated[equated  120] - 120

 HTH,

 Josh

 On Mon, Jul 26, 2010 at 10:26 AM, ying_chen wang
 gracedrop.w...@gmail.com wrote:
  I am new to R. Used to use FORTRAN. R is so different from FORTRAN. The
  following codes would work in FOTRAN. I am trying to put an upper limit
  at
  120. If the score is  120, it is assigned 120. Or else, keep the
  original
  values.
 
  version 1:
 
  equated-11
  result-11
  equated-c(111.0,112.06, 112.9, 113.8, 115.0, 116.2, 117.0, 118.0,
  120.5,
  120.5, 120.5)
 
  for (i in 1:11){
  if (equated  120) result[i]-120
  if (equated  120) result[i]-equated[i]
  result-result
  result
  }
  result
 
  version2:
 
  if (equated  120) result-120
  if (equated  120) result-equated
 
 
 
  If any of you can help, I would appreciate that.
 
  G
 
         [[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.
 



 --
 Joshua Wiley
 Ph.D. Student, Health Psychology
 University of California, Los Angeles
 http://www.joshuawiley.com/





-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

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Re: [R] easy debugging

What about these two options?

#One way
ifelse(equated  120, 120, equated)

#Another way
equated[equated  120] - 120

HTH,

Josh

On Mon, Jul 26, 2010 at 10:26 AM, ying_chen wang
gracedrop.w...@gmail.com wrote:
 I am new to R. Used to use FORTRAN. R is so different from FORTRAN. The
 following codes would work in FOTRAN. I am trying to put an upper limit at
 120. If the score is  120, it is assigned 120. Or else, keep the original
 values.

 version 1:

 equated-11
 result-11
 equated-c(111.0,112.06, 112.9, 113.8, 115.0, 116.2, 117.0, 118.0, 120.5,
 120.5, 120.5)

 for (i in 1:11){
 if (equated  120) result[i]-120
 if (equated  120) result[i]-equated[i]
 result-result
 result
 }
 result

 version2:

 if (equated  120) result-120
 if (equated  120) result-equated



 If any of you can help, I would appreciate that.

 G

        [[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.




-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

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Re: [R] using string variable as order() function argument

Try this:

 mytable
  id col1 col2 col3
1 10048
2 11122
3 12083
4 13055
 colName - 'col3'
 mytable[order(mytable[[colName]]),]
  id col1 col2 col3
2 11122
3 12083
4 13055
1 10048
 colName - 'id'
 mytable[order(mytable[[colName]]),]
  id col1 col2 col3
1 10048
2 11122
3 12083
4 13055




On Mon, Jul 26, 2010 at 2:54 PM, mirek wyczesany
miroslaw.wyczes...@uj.edu.pl wrote:
 Hello,

 In my script I would like to use a loop, which sorts the dataframe according
 to different columns, pointed by the string variable.

    id col1  col2  col3
 1   10    0    4      8
 2   11    1    2      2
 3   12    0    8      3
 4   13    0    5      5

 Usually the order() function can be used like this:

 sorted = mytable**[order(column3) , ]

 which results in properly sorted table:
 **

    id col1  col2  col3
 2   11    1    2      2
 3   12    0    8      3
 4   13    0    5      5
 1   10    0    4      8

 **But when trying to use a string variable instead of column3 name:

 columnname = column3
 **sorted = mytable**[order(columnname) , ]**

 this command is not properly evaluated and the effect is somewhat strange.

 Would you suggest some solution?

 Thanks a lot!

 Mirek

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What is the problem that you are trying to solve?

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Re: [R] problem with building package on CRAN

2010-07-26 Thread Romain Francois



Hello,

(ccing Rcpp-devel too because this is relevant)

This comes up every now and again on packages that are completely 
unrelated to Rcpp. We don't know yet why or what to do to fix the issue. 
I believe (but this might not be the case) that this is due to packages 
that do use Rcpp and fail to follow our guidelines and documentation and 
emails about using LinkingTo to pull in header files.
See 
http://lists.r-forge.r-project.org/pipermail/rcpp-devel/2010-July/000898.html


I'm really sorry you are caught in the middle of this, there is probably 
nothing wrong with your package, it just happens sort of randomly. I 
believe the chances of this happening would be lower if package 
developers (of package using Rcpp) would be so kind and follow our 
guidelines, but we can only offer to write the guidelines, we cannot 
force them to read or apply them.


Romain

Le 26/07/10 22:43, William Revelle a écrit :


Dear friends,
I have just gotten a strange error message back from Uwe saying that the
most recent version of psych failed to pass R CMD check for Windows.

The error message was less than helpful, in that it seems to have failed
when trying to include the Rcpp library, which I do not directly call.
(see below)


* using log directory 'd:/Rcompile/CRANpkg/local/2.11/psych.Rcheck'
* using R version 2.11.1 (2010-05-31)
* using session charset: ISO8859-1
* checking for file 'psych/DESCRIPTION' ... OK
* this is package 'psych' version '1.0-90'
* checking package name space information ... OK
* checking package dependencies ... OK
* checking if this is a source package ... OK
* checking whether package 'psych' can be installed ... ERROR
Installation failed.
The installation logfile:
-Id:/Rcompile/CRANpkg/lib/2.11/Rcpp/include



I do have several suggested packages (polycor, GPArotation, MASS, graph,
Rgraphviz, mvtnorm, Rcsdp), but none of these are actually required. My
examples all ask if the suggested packages are available and then do not
call them if they are not.

Any suggestions on what to do would be appreciated.

Thanks.

Bill




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[R] O/T good c/c++ code for LU decomposition

2010-07-26 Thread Erin Hodgess

Dear R People:

Could someone recommend a good c/c++ code (or Fortran) for LU
decomposition, please?

Sorry to bother about this.  I'm trying to do some non-R work that
requires a matrix inversion.

Thanks,
Erin


-- 
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com

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[R] [R-pkgs] version 2.0-0 of the car package

2010-07-26 Thread John Fox

Dear all,

Sandy Weisberg and I would like to announce version 2.0-0 of the car
package, now on CRAN. We've released this major revision of the package in
anticipation of the publication of An R Companion to Applied Regression,
Second Edition (Sage, in press), co-authored by us, which should be
available before the end of the year.

The new version of the car package has a number of new functions and data
sets, along with many changes to old functions, including a change in naming
conventions, using camel-case rather than period-separated names: for
example, av.plots() is now avPlots(). For the time-being, old function
names, such as av.plots(), are available as deprecated aliases. A few
functions from the alr3 package have also been rewritten, renamed, and are
now part of car. The alr3 package will be updated later in the year.


Regards,
John


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

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Re: [R] Accessing single element of data.frame

2010-07-26 Thread Jannis


The solution to (most R problems) is as follows:

1.if asking for help include reproducible examples including parts of 
your data otherwise we can just guess what kind of data you have.


2. In general, refer to the help pages of the functions you use ( 
help(read.csv),help(data.frame) ) (


i have to confess that a similar problem puzzled me for quite a while, 
so here comes suggestion 3:


3. As a start, have a look what kind of data you have:

class(t1)
class(t2)
class(t2[2,])

I would guess that t2 is a factor and not a vector and that the problem 
lies in read.csv converting whatever data you have to factors.


See

?read.csv


and the

stringsAsFactors


argument of that call. Adjusting its values might fix your problem. If 
not, go back to advice 1 ;-)



HTH
Jannis


vacas schrieb:
Hi I am new to R. 
I am having this problem

t1 - read.csv(myfile.csv)
t2 - data.frame(t1)
which have 10 row and 10 columns
t2[1,1] does not give the first element but it gives the levels, how can I
fix it. 

I will be thankful to community. 



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Re: [R] latent class analysis with mixed variable types

2010-07-26 Thread Ingmar Visser

Donald,
Mixed types are handled in flexmix and in depmixS4, not sure about ordinal
in flexmix (depmixS4 does not handle ordinal but does handle multinomial,
constraints may be an
option to deal with ordinal); both have glm distributions, ie gaussian,
binary and many
others.
Best, Ingmar

On Sat, Jul 24, 2010 at 3:52 AM, Donald Braman donald.bra...@gmail.comwrote:

 As an alternative to Latent GOLD, I'm wondering if anyone knows of and R
 package that can manage Latent Class Analysis with mixed variable types
 (continuous, ordinal, and nominal/binary).

[[alternative HTML version deleted]]

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[[alternative HTML version deleted]]

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Re: [R] O/T good c/c++ code for LU decomposition

2010-07-26 Thread Romain Francois



Hi,

Armadillo (http://arma.sourceforge.net/docs.html) has LU. Here is an 
example adapted from armadillo's documentation using Rcpp/RcppArmadillo 
and inline:


require( inline )
require( RcppArmadillo )

fx - cxxfunction( signature( A_ = matrix), '
using namespace arma ;
mat A = asmat(A_);

mat L;
mat U;
mat P;

lu(L, U, P, A);

mat B = trans(P)*L*U;
return List::create(
_[L] = L,
_[U] = U,
_[P] = P,
_[A] = A,
_[B] = B
) ;
', plugin = RcppArmadillo )
fx( matrix( runif(100), 10, 10) )


Romain

Le 26/07/10 23:46, Erin Hodgess a écrit :


Dear R People:

Could someone recommend a good c/c++ code (or Fortran) for LU
decomposition, please?

Sorry to bother about this.  I'm trying to do some non-R work that
requires a matrix inversion.

Thanks,
Erin





--
Romain Francois
Professional R Enthusiast
+33(0) 6 28 91 30 30
http://romainfrancois.blog.free.fr
|- http://bit.ly/bc8jNi : Rcpp 0.8.4
|- http://bit.ly/dz0RlX : bibtex 0.2-1
`- http://bit.ly/a5CK2h : Les estivales 2010

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[R] Switch Enter and Return in R.app?

2010-07-26 Thread Nick Matzke


Hi all,

I googled this but no luck.

I am using R.app 2.10.1 on Mac OSX 10.4.

Here's the problem:

When I type at the R.app command line and hit the carriage 
return Enter (right pinky, Return on some keyboards), it 
just adds a blank line.


To actually get the command to execute, I have to go all the 
way over to the number-keypad and hit Enter there.


Surely there must be a way to get R.app to synonymize these 
two keys?  All that moving over to the number pad is driving 
me nuts.


Thanks so much!!  Apologies if it is something obvious which 
I missed!!


Nick




--

Nicholas J. Matzke
Ph.D. Candidate, Graduate Student Researcher
Huelsenbeck Lab
Center for Theoretical Evolutionary Genomics
4151 VLSB (Valley Life Sciences Building)
Department of Integrative Biology
University of California, Berkeley

Graduate Student Instructor, IB200A
Principles of Phylogenetics: Systematics
http://ib.berkeley.edu/courses/ib200a/index.shtml

Lab websites:
http://ib.berkeley.edu/people/lab_detail.php?lab=54
http://fisher.berkeley.edu/cteg/hlab.html
Dept. personal page: 
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Lab personal page: 
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Lab phone: 510-643-6299
Dept. fax: 510-643-6264
Cell phone: 510-301-0179
Email: mat...@berkeley.edu

Mailing address:
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Berkeley, CA 94720-3140

-
[W]hen people thought the earth was flat, they were wrong. 
When people thought the earth was spherical, they were 
wrong. But if you think that thinking the earth is spherical 
is just as wrong as thinking the earth is flat, then your 
view is wronger than both of them put together.


Isaac Asimov (1989). The Relativity of Wrong. The 
Skeptical Inquirer, 14(1), 35-44. Fall 1989.

http://chem.tufts.edu/AnswersInScience/RelativityofWrong.htm

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Re: [R] Bug on r-bc?

2010-07-26 Thread Ben Bolker

Paul Smith phhs80 at gmail.com writes:

 
 Dear All,
 
 The following code should return 1, but it returns 0:
 
 source(http://r-bc.googlecode.com/svn/trunk/R/bc.R;)
 bc(9 % 2)
 
 Do you confirm this bug?
 

  It's not a bug in r-bc, it's a misfeature (?) in bc.
It has to do with the 'scale' parameter in bc (which gets
set to 100 explicitly by r-bc, but would be set to 20 in any
case by the use of the '-l' option to bc)

http://superuser.com/questions/31445/gnu-bc-modulo-with-scale-other-than-0
http://en.wikipedia.org/wiki/Bc_programming_language
http://www.linuxquestions.org/questions/programming-9/bc-using-l-messes-up-modulus-331003/

a workaround;

 source(http://r-bc.googlecode.com/svn/trunk/R/bc.R;)
 bc(scale=0; 9%2)
[1] 1

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Re: [R] Bug on r-bc?

2010-07-26 Thread Gabor Grothendieck

On Mon, Jul 26, 2010 at 1:44 PM, Paul Smith phh...@gmail.com wrote:
 Dear All,

 The following code should return 1, but it returns 0:

 source(http://r-bc.googlecode.com/svn/trunk/R/bc.R;)
 bc(9 % 2)


See FAQ 2 on the r-bc package home page:
http://r-bc.googlecode.com

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Re: [R] Plot of a subset of a data.frame()



On Jul 26, 2010, at 10:56 AM, Steffen Uhlig wrote:


Dear David, Petr, and Alain,

thank you very much for your fast responses. It's a typical  
handbook-not-read-error at my side. I will dig deeper into the  
plot-functions and the assignment of data. I was not aware of that  
the vector a is handled as a vector of factors with 10 levels.  
Thanks for your suggestions and hints!


You can prevent that behavior and instead get a character vector ...  
at least from functions that return such ... by using stringsAsFactors  
= FALSE within the data.frame call. You also have the option of  
setting that globally which at least one well known institution has  
adopted as the default policy for its work.


?data.frame
?options

--
David


Best regards,
/steffen


Am 26.07.2010 14:30, schrieb David Winsemius:


On Jul 26, 2010, at 7:38 AM, Steffen Uhlig wrote:


Hello,

my data.frame is sort of a collection of process values, i.e. huge
run-chart. It consists of a time-stamp in the first column (date as
string), factors in the following columns (used for subset- 
filtering),

and some process-data columns.
Hereafter, two examples are listed, showing the problems that occour
during print:

At first the example, that works fine:

~~
a = c(1:10) # create a vector of integers
b = rep(c(a,b),5) # create a vector of chars, used
# as factor-levels
d = rnorm(10) # some random numbers
e = data.frame(a,b,d) # connect to a data.frame


You've gotten several answers, but none have addressed an aspect of R
behavior that took me longer to appreciate than it perhaps should  
have.

The b column inside the e data.frame is now a factor column. I
mention that because you later referred to it as a string which  
it is
not. It is an integer with an associated indexed level character  
vector.

Many of the functions that you might think would work on strings
will give either errors or unexpected results when applied to  
factors.





e.1 = subset(e, b==a) # create two subsets
e.2 = subset(e, b==b)
plot(d~a, e.1, pch=3, col=2) # plot first data-subset
points(d~a, e.2, pch=4, col=3) # plot the 2nd one

~~
all looks fine in theses plots.


However, changing the content of vector a to a set of strings the
following happens:

~~
a = c(a,b,c,d,e,f,g,h,i,j)
e = data.frame(a,b,d) # re-build data.frame

e.1 = subset(e, b==a) # create two subsets
e.2 = subset(e, b==b)
plot(d~a, e.1, pch=3, col=2)
points(d~a, e.2, pch=4, col=3)
~~
The plot-command produces horizontal lines instead of dots. This  
seems

to happen when the x-axis contains strings rather than numbers. is
there a way out?

Best regards,
/Steffen



--
Steffen Uhlig, PhD
Mechatronik und Sensortechnik
HTW des Saarlandes
Goebenstraße 40
66117 Saarbrücken

Tel.: +49 (0) 681 58 67 274


David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] how to generate a random data from a empirical distribition