date:20121127

Hello, 

I have a data frame somewhat like this one: 

myframe - data.frame (Timestamp=c(24.09.2012 09:00, 24.09.2012 10:00,
24.09.2012 11:00, 
   25.09.2012 09:00, 25.09.2012 10:00,
25.09.2012 11:00), Hunger=c(1,1,1,2,2,1) )  
myframestime - as.POSIXct (strptime(as.character(myframe$Timestamp),
%d.%m.%Y %H:%M), tz=GMT)
myframe2 - cbind (myframe,myframestime)
myframe2$Timestamp - NULL  
myframe2

Now I want to get the sum of Hunger for each day. In the end I want
something which looks like the following dataframe: 

myoutcome - data.frame(Timestamp=c(24.09.2012, 25.09.2012),
sumHunger=c(3, 5))

Does anyone know how to do that?
That would be very helpful and to all people who are willing to help me:
Thank you in advance! 

Best regards, 
Tagmarie 



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Re: [R] Problem in Starting R Server - object of type 'closure' is not subsettable

2012-11-27 Thread Manoj G

Thank you so much Barry for pointing out my mistake. 

Manoj G



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Re: [R] Help R

2012-11-27 Thread Jim Lemon


On 11/26/2012 09:38 PM, alanaro...@sapo.pt wrote:

Goodmorning,
I'moneafazrtrbhoquhasa
variablefactorcomtwoNivesCandH.Queoaatravésdlinear
regressionrelationshipetrehaSaerifthe
variablespthiedoseogmerparawanttheHtothatanddouCrespctinteRaficthat
showsthe two curvesHandCLIRand
Selecionar tudo
Thank you
Ana C. Rocha Rua


Hi Ana,
I'm afraid that I cannot make much sense at all out of this. The best 
thing I can suggest is that you contact the Brazil R Users group.


http://leg.est.ufpr.br/doku.php/software:rbr

Jim

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Re: [R] sum per day


Hello,

Try


aggregate(Hunger ~ cut(myframe2$myframestime, day), data = myframe2, 
FUN = sum)


Hope this helps,

Rui Barradas
Em 27-11-2012 09:13, Tagmarie escreveu:

Hello,

I have a data frame somewhat like this one:

myframe - data.frame (Timestamp=c(24.09.2012 09:00, 24.09.2012 10:00,
24.09.2012 11:00,
25.09.2012 09:00, 25.09.2012 10:00,
25.09.2012 11:00), Hunger=c(1,1,1,2,2,1) )
myframestime - as.POSIXct (strptime(as.character(myframe$Timestamp),
%d.%m.%Y %H:%M), tz=GMT)
myframe2 - cbind (myframe,myframestime)
myframe2$Timestamp - NULL
myframe2

Now I want to get the sum of Hunger for each day. In the end I want
something which looks like the following dataframe:

myoutcome - data.frame(Timestamp=c(24.09.2012, 25.09.2012),
sumHunger=c(3, 5))

Does anyone know how to do that?
That would be very helpful and to all people who are willing to help me:
Thank you in advance!

Best regards,
Tagmarie



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[R] order.max specification problem in the ar.ols function

2012-11-27 Thread soham chakraborty

Hello
I am facing a curious problem.I have a time series data with which i want to
fit auto-regressive  model of order p, where p runs from 1:9.I am using a
for loop which will fit an AR(p) model for each value of p using the
*ar.ols* function.
I am using the following code
for ( p in 1:9){
a=ar.ols (x=data.ts, order.max=p, demean=T, intercept=T)
}
Specifying the *order.max* to be p, it gives me a fitted AR model of order
exactly p, though it may not be the best model.
The code works fine except for p=4,5, in which cases it fits an AR (3) model
to the data.
I don't seem to understand what is the problem.Also is there any alternative
way by which i can fit an AR(p) model where p is specified by the user?
Any help in this matter will be of immense help.
Thank you.
Soham Chakraborty



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Re: [R] R strange behaviour when building huge concatenation


Hello,

That instruction gave me no problems. I've tried it with and without the 
useless end semi-colon and all went well.
Something to do with your R session? Try it in a new session to see if 
it works.


Hope this helps,

Rui Barradas
Em 26-11-2012 21:24, angeloimm escreveu:

Any idea on the reason why the instruction
p1x-c(2, 1, 0, 0, 0, 7, 1, 0, 2, 0, 5, 0, 0, 0, 0, 0, 10, 1, 3, 0, 2, 0, 0,
0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 12, 0, 0,
1, 0, 0, 0, 0, 0, 0, 2, 1, 0, 1, 0, 4, 0, 1, 0, 0, 0, 2, 0, 0, 3, 0, 0, 1,
0, 1, 0, 0, 0, 3, 0, 0, 1, 0, 0, 0, 7, 0, 10, 0, 0, 1, 6, 0, 0, 0, 0, 0, 2,
4, 1, 5, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0,
0, 1, 0, 0, 0, 4, 3, 0, 1, 2, 0, 1, 2, 0, 2, 1, 1, 0, 5, 2, 7, 2, 0, 4, 13,
0, 4, 4, 0, 0, 1, 0, 1, 29, 0, 3, 0, 0, 1, 0, 10, 0, 0, 13, 0, 0, 1, 0, 0,
0, 0, 0, 0, 0, 6, 8, 0, 0, 0, 1, 0, 4, 0, 2, 3, 3, 0, 0, 0, 0, 0, 6, 0, 1,
0, 0, 2, 0, 2, 2, 0, 0, 1, 0, 1, 0, 0, 7, 0, 0, 0, 2, 0, 4, 0, 1, 1, 0, 0,
0, 0, 0, 2, 0, 1, 1, 2, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 1, 5, 0, 0, 2,
0, 1, 2, 0, 0, 1, 0, 1, 0, 5, 0, 0, 1, 2, 0, 1, 0, 0, 1, 2, 1, 0, 1, 0, 1,
0, 3, 1, 13, 0, 0, 0, 0, 0, 3, 3, 1, 0, 0, 3, 0, 5, 0, 2, 1, 0, 1, 0, 0, 1,
1, 0, 0, 0, 3, 1, 1, 0, 1, 0, 0, 0, 4, 0, 2, 0, 3, 0, 0, 1, 0, 1, 0, 2, 0,
1, 3, 25, 0, 0, 0, 0, 5, 0, 2, 0, 0, 0, 5, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0, 0,
1, 0, 2, 1, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 2, 4, 0, 0, 1, 3, 0, 1, 0, 0,
0, 0, 0, 1, 1, 1, 0, 0, 0, 11, 0, 0, 4, 0, 0, 1, 2, 0, 0, 0, 0, 1, 1, 0, 3,
1, 0, 0, 2, 0, 8, 1, 0, 2, 1, 0, 0, 1, 0, 2, 0, 0, 3, 0, 0, 0, 0, 0, 0, 2,
2, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 2, 4, 0, 0, 2, 0, 6, 0,
1, 0, 1, 1, 1, 2, 1, 0, 7, 4, 0, 0, 0, 7, 0, 1, 1, 0, 2, 1, 0, 4, 0, 1, 0,
0, 2, 0, 0, 4, 0, 0, 2, 1, 0, 1, 0, 11, 0, 4, 0, 0, 0, 4, 0, 3, 0, 1, 1, 0,
0, 6, 3, 0, 0, 0, 0, 2, 0, 2, 0, 18, 0, 1, 0, 1, 0, 0, 0, 5, 0, 1, 0, 6, 0,
2, 0, 0, 2, 0, 1, 1, 0, 0, 1, 0, 1, 2, 0, 0, 0, 1, 0, 2, 0, 2, 0, 0, 0, 0,
1, 10, 0, 1, 3, 3, 0, 2, 0, 0, 12, 0, 1, 2, 2, 0, 0, 0, 0, 5, 0, 0, 2, 0, 5,
0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 1, 7, 0, 0, 2, 0,
0, 2, 0, 4, 0, 0, 3, 0, 1, 0, 2, 2, 0, 1, 1, 2, 0, 0, 0, 0, 2, 0, 0, 1, 0,
1, 0, 3, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 3, 0, 4, 0, 0, 0, 0,
2, 1, 1, 0, 0, 0, 0, 5, 1, 0, 0, 1, 2, 1, 0, 0, 0, 0, 1, 0, 8, 1, 0, 6, 1,
2, 0, 3, 6, 0, 1, 0, 2, 5, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 13, 0, 0, 1, 3,
0, 5, 0, 2, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 4, 6, 1, 2, 0, 0, 2, 0, 7, 0, 0,
0, 0, 0, 3, 0, 5, 0, 2, 1, 3, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 1,
0, 0, 1, 10, 0, 0, 0, 0, 3, 3, 0, 0, 2, 5, 4, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0,
0, 0, 1, 1, 3, 1, 0, 0, 0, 0, 6, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 6, 0, 1, 0,
2, 0, 0, 7, 0, 0, 0, 2, 0, 0, 0, 0, 2, 0, 1, 13, 7, 0, 0, 3, 0, 1, 0, 1, 1,
2, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 2, 1, 0, 0, 1, 19, 2, 2, 0, 2, 2, 0,
0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 2, 0, 0, 0, 1, 0, 7, 0, 0, 0, 0, 0,
1, 0, 0, 1, 1, 0, 0, 2, 0, 1, 1, 0, 0, 4, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 5,
1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 7, 2, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, 0,
0, 0, 0, 0, 0, 1, 0, 4, 0, 1, 1, 0, 1, 0, 1, 2, 0, 1, 0, 0, 0, 2, 0, 4, 0,
1, 7, 1, 0, 1, 0, 0, 4, 0, 1, 0, 0, 0, 0, 1, 0, 7, 1, 0, 6, 0, 5, 0, 2, 0,
1, 0, 6, 0, 2, 0, 0, 0, 2, 2, 0, 0, 6, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 0, 0,
0, 1, 4, 0, 0, 1, 0, 1, 0, 0, 1, 0, 5, 2, 0, 0, 0, 3, 0, 12, 1, 0, 0, 2, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 1, 1, 0, 0, 3, 0, 1, 0, 1, 0,
0, 0, 0, 4, 0, 0, 2, 0, 5, 0, 3, 1, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0,
0, 0, 2, 3, 0, 0, 0, 0, 11, 0, 0, 2, 0, 1, 0, 7, 0, 0, 1, 0, 3, 0, 2, 0, 1,
0, 4, 2, 2, 1, 0, 0, 0, 0, 0, 1, 3, 0, 0, 1, 1, 6, 0, 0, 4, 0, 0, 1, 0, 2,
0, 1, 3, 7, 2, 0, 5, 0, 0, 0, 0, 5, 0, 12, 1, 0, 1, 0, 1, 0, 0, 0, 0, 3, 0,
0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 4, 0, 0, 0, 2, 0, 4, 0, 1, 0, 1, 0, 3,
0, 0, 0, 0, 0, 1, 0, 0, 3, 0, 3, 4, 1, 0, 2, 1, 2, 1, 0, 0, 2, 1, 0, 0, 0,
1, 7, 0, 6, 0, 6, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0,
0, 0, 0, 0, 3, 0, 0, 3, 0, 3, 0, 8, 0, 1, 0, 0, 2, 0, 0, 11, 0, 1, 8, 0, 1,
0, 1, 0, 0, 3, 0, 0, 0, 0, 0, 3, 1, 0, 1, 6, 0, 2, 1, 1, 0, 1, 1, 0, 3, 2,
0, 0, 0, 2, 1, 1, 0, 0, 0, 0, 3, 2, 0, 3, 0, 0, 1, 4, 0, 6, 0, 1, 1, 4, 0,
2, 0, 4, 0, 0, 0, 0, 2, 0, 1, 0, 1, 2, 0, 0, 1, 0, 0, 2, 1, 0, 0, 3, 0, 0,
0, 0, 0, 0, 1, 1, 3, 2, 0, 0, 0, 1, 0, 0, 0, 4, 0, 1, 3, 0, 0, 0, 0, 0, 0,
3, 1, 0, 1, 1, 2, 0, 2, 0, 0, 0, 3, 0, 1, 0, 1, 0, 9, 0, 0, 4, 0, 2, 0, 5,
0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 0, 0, 0, 4, 1, 0, 0, 1, 5, 0, 0, 0, 1, 0,
1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 3, 0, 0, 2, 0, 14, 4, 0, 0, 0, 0, 0, 0, 2, 0,
0, 2, 0, 3, 0, 2, 0, 0, 1, 1, 6, 1, 6, 0, 1, 2, 0, 0, 2, 1, 0, 2, 1, 0, 1,
0, 2, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 4, 0, 4, 0, 1, 0, 0, 5, 0, 2, 0, 1, 0,
5, 0, 1, 2, 0, 2, 2, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 3, 0, 4,
0, 1, 0, 0, 3, 1, 0, 2, 0, 2, 1, 0, 0, 0, 4, 0, 0, 0, 26, 1, 6, 0, 0, 1, 0,
0, 0, 2, 0, 0, 0, 2, 0, 0, 1, 2, 0, 1, 0, 0, 0, 14, 0, 1, 0, 2, 4, 0, 1, 6,

Re: [R] Problem installing knitr 0.5 or higher in Ubuntu

2012-11-27 Thread Pascal Oettli


Hello,

http://cran.at.r-project.org/bin/linux/ubuntu/README.html

Regards,
Pascal


Le 27/11/2012 19:27, ATANU a écrit :


Hi,

I am using Rstudio in Ubuntu . While installing R from the terminal I got
the version R 2.13.1 .
But the problem is that knitr 0.5 requires higher versions.
So is there any way to get the latest version of R in ubuntu (or 2.13.1 is
the maximum I  can get)?
In case the higher R version is not available can anybody please help me how
do I use the knitr Html option in R studio.

Thanks in advance.

-Atanu









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Re: [R] Problem installing knitr 0.5 or higher in Ubuntu

2012-11-27 Thread Berend Hasselman


On 27-11-2012, at 11:27, ATANU wrote:

 
 Hi,
 
 I am using Rstudio in Ubuntu . While installing R from the terminal I got
 the version R 2.13.1 .
 But the problem is that knitr 0.5 requires higher versions.
 So is there any way to get the latest version of R in ubuntu (or 2.13.1 is
 the maximum I  can get)?
 In case the higher R version is not available can anybody please help me how
 do I use the knitr Html option in R studio.

You haven't told us which version of Ubuntu you are using.

See http://cran.r-project.org/bin/linux/ubuntu/

You should first uninstall your current R.

Berend

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Re: [R] Problem installing knitr 0.5 or higher in Ubuntu


Hello,

How did you install R? I've installed it from the terminal and got the 
latest version, R 2.15.2 The commands I've used were


sudo apt-get update
sudo apt-get install r-base
sudo apt-get install rbase-dev


Hope this helps,

Rui Barradas
Em 27-11-2012 10:27, ATANU escreveu:

Hi,

I am using Rstudio in Ubuntu . While installing R from the terminal I got
the version R 2.13.1 .
But the problem is that knitr 0.5 requires higher versions.
So is there any way to get the latest version of R in ubuntu (or 2.13.1 is
the maximum I  can get)?
In case the higher R version is not available can anybody please help me how
do I use the knitr Html option in R studio.

Thanks in advance.

-Atanu









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Re: [R] Help with graphics in gamm4 library

2012-11-27 Thread Simon Wood

I would approach something like this using predict(fit41$gam,...) (see 
?predict.gam) to produce the data needed for the plots.


Another possibility is to use s(Time,trt,bs=fs) (but see 
?factor.smooth.interaction first).


Note that gamm4 does not have a correlation argument, because lme4 does 
not provide correlation structures, so corARH1() is just ignored (there 
is no error or warning produced because gamm4 does have a '...' argument!).


best,
Simon

On 27/11/12 00:07, MurphFL wrote:

My problem is relatively straight forward, but I cannot seem to find a way to
make it work.

I have a RCBD with repeated measurements over time. I have created a fit
using the gamm4 package. My model is:

fit4a - gamm4(Rate ~ s(Time,by=trt,bs=cr)+trt,data=qual.11.dat,
   random=~(1|block),correlation=corARH1())

What I would like to create is plots with the X-axis being time, the Y-axis
being the fitted Rates for each individual treatment with the smoothed
curves overlayed on the plot.

Every idea I have had to do this has resulted in some errors, and I have
reached my wits end. Can anyone steer me in the right direction?



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+44 (0)1225 386603   http://people.bath.ac.uk/sw283

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Re: [R] Error in Random Effect Panel

2012-11-27 Thread Hard Core

up



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[R] Effect of each term in the accuracy of Nonlinear multivariate regression fitting equation

2012-11-27 Thread dsfakianakis

Dear all,

I have a set of data with 4 inputs (independent variables) and one output
(dependent variable). I want to perform a regression analysis in order to
fit these data to a regression model, however due to the non-linearity of
the model I do not have a clue which equation to use. I am thinking of
starting with a very general equation including ^3 terms and interactions
between the variables however this will lead to a very long equation. Is
there a way to assess the effect of each term to the accuracy of the
regression model in order to discard the terms with the least importance?
Something like a sensitivity analysis of the effect of each term to the
accuracy regression model. I know one possible solution to my problem is
simply 'trial and error' however before going down that road I want to check
if there is an easier way.

e.g. Let's say I have four input variables A B C and D, one output 'JIM' and
let z1, z2, ... be the coefficients of the terms of the equation. The
regression will be something like that:

Result = nls(JIM ~ z1*A + z2*B + z3*A*B^2 + z4*C*D^3 + z5*A^2*B^2 ... )

Is there a way to assess the contribution of each term (z1*A, z3*A*B^3 etc)
to the accuracy of the regression model?

Thanks a lot

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Re: [R] how to input a string without quote

2012-11-27 Thread snawy ben


Wow, you are so lazy... But sometimes R is just designed for lazy guys...

##
f = function(a) {
s = substitute(a)
as.character(s)
}
##

Yihui

Cheers, Yihui, nearly 4 years after your post this still works like a charm.
ps: of course I am not lazy, I simply want to avoid error prone input! ;)



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Re: [R] question about A2R

2012-11-27 Thread dmatshalaga

DID YOU SOLVE THE PROBLEM?



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Re: [R] order.max specification problem in the ar.ols function

Hello,

Just set argument aic = FALSE. From the help page:

|aic|

Logical flag. If |TRUE| then the Akaike Information Criterion is used to 
choose the order of the autoregressive model. If |FALSE|, the model of 
order |order.max| is fitted.


Hope this helps,

Rui Barradas
Em 27-11-2012 10:36, soham chakraborty escreveu:
 Hello
 I am facing a curious problem.I have a time series data with which i want to
 fit auto-regressive  model of order p, where p runs from 1:9.I am using a
 for loop which will fit an AR(p) model for each value of p using the
 *ar.ols* function.
 I am using the following code
 for ( p in 1:9){
 a=ar.ols (x=data.ts, order.max=p, demean=T, intercept=T)
 }
 Specifying the *order.max* to be p, it gives me a fitted AR model of order
 exactly p, though it may not be the best model.
 The code works fine except for p=4,5, in which cases it fits an AR (3) model
 to the data.
 I don't seem to understand what is the problem.Also is there any alternative
 way by which i can fit an AR(p) model where p is specified by the user?
 Any help in this matter will be of immense help.
 Thank you.
 Soham Chakraborty



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Re: [R] Anova

2012-11-27 Thread Jose Iparraguirre

Hi, 

You can use the car package and choose Type-III test.
Also, have a look at the package easyanova.
Regards,
José

José Iparraguirre
Chief Economist
Age UK


-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of humming-bird
Sent: 27 November 2012 10:12
To: r-help@r-project.org
Subject: [R] Anova

Hi everyone,

I am new to this forum and also new to statistics and I would appreciated it
if someone would take some time to answer my question.

I am analyzing companies in regard to their leverage. I categorized the
companies into 3 groups: small, mid and large. For the group small, I have
55 debt multiples, for mid 42 and for large 72. (Unfortunately I can not
provide my data because it is confidential.)
I am now trying to find out whether the mean debt multiples are
significantly different for the 3 different groups. For this reason I
calculated an anova table with the aov function and to display the results
for each pair I did the tukey.hsd function.
Now my question: Am I allowed to use these functions given that my data is
unbalanced? Can use I read several times that aov is only valid for balanced
data? If not, is there another function that I can use?

Thank you very much for your answers.


Call:
aov(formula = Debt.Ebitdax ~ Company.Size, data = Anetdebtx2003)

Terms:
Company.Size Residuals
Sum of Squares 302.3089 926.2174
Deg. of Freedom 2 166

Residual standard error: 2.362123 
Estimated effects may be unbalanced

 summary(Anovanetdebtx2003)
Df Sum Sq Mean Sq F value Pr(F) 
Company.Size 2 302.3 151.15 27.09 6.58e-11 ***
Residuals 166 926.2 5.58 
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 


 TukeyHSD(Anovanetdebtx2003)
Tukey multiple comparisons of means
95% family-wise confidence level

Fit: aov(formula = Debt.Ebitdax ~ Company.Size, data = Anetdebtx2003)

$Company.Size
diff lwr upr p adj
Mid Market Buyout-Large Buyout -1.446292 -2.530922 -0.3616617 0.0054123
Small Buyout-Large Buyout -3.112143 -4.112545 -2.1117420 0.000
Small Buyout-Mid Market Buyout -1.665852 -2.810574 -0.5211300 0.0021037



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Re: [R] Help on function please

Andras,

What do you want your code to do?  Give us a little explanation with your 
code.

When I try to run your code, I get
Error: could not find function genoud.
Either supply the code for the functions included in your code, or tell us 
what packages we need to run it.

Jean



Andras Farkas motyoc...@yahoo.com wrote on 11/26/2012 02:43:25 PM:
 
 Dear All,
  
 I could use a bit of help here, this function is hard to figure out 
 (for me at least) I have the following so far:
  
 PKindex-data.frame(Subject=c(1),time=c(1,2,3,4,6,10,12),conc=c(32,
 28,25,22,18,14,11))
 Dose-200
 Tinf -0.5
  
 defun- function(time, y, parms) { 
  dCpdt - -parms[kel] * y[1] 
  list(dCpdt) 
  } 
 modfun - function(time,kel, Vd) {  
  out - lsoda(((Dose/Tinf)*(1/(kel*Vd)))*(1-exp(-kel*time)),c
 (0,time),defun,parms=c(kel=kel,Vd=Vd),rtol=1e-3,atol=1e-5) 
  out[-1,2]
  }
 objfun - function(par) {
  out - modfun(PKindex$time, par[1], par[2])
  gift - which( PKindex$conc != 0 )
  sum((PKindex$conc[gift]-out[gift])^2)
  }
 gen-genoud
 
(objfun,nvars=2,max=FALSE,pop.size=30,max.generations=100,wait.generations=100,starting.value=c
 (0.7,
 60),BFGS=FALSE,print.level=0,boundary.enforcement=2,Domains=matrix(c
 (0.01,0.01,100,100),2,2),MemoryMatrix=TRUE)
  
 but get the following:
  Error in lsoda(((Dose/Tinf) * (1/(kel * Vd))) * (1 - exp(-kel * time)), 
 : 
   The number of derivatives returned by func() (1must equal the 
 length of the initial conditions vector (7)
  
 i figured that having the time parameter in the equation screws 
 things up, but do not now how to fix it bc I do not understand the 
 warning message.
  
  
 your help is greatly apreciated,
  
 Sincerely,
  
 Andras

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[R] loop with date

Hello, 
I tried to construct my very first loop today and completly failed :-(
Maybe someone can help me? 
I have a dataframe somewhat like this one: 

myframe - data.frame (Timestamp=c(24.09.2012 09:00, 24.09.2012 10:00,
24.09.2012 11:00, 
   25.09.2012 09:00, 25.09.2012 10:00,
25.09.2012 11:00), 
Speed=c(1,1,2,5,1,6))  
myframestime - as.POSIXct (strptime(as.character(myframe$Timestamp),
%d.%m.%Y %H:%M), tz=GMT)
myframe2 - cbind (myframe,myframestime)
myframe2$Timestamp - NULL  
myframe2

I want to construct a loop for every day, i.e. for each day I want to do
some calculations. 
(I know in the example it would be easier to do it differently, my real data
are little more complex). 

And BTW: Thanks for helping me earlier today with that other problem :-) 




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Re: [R] error of runing R in R 2.15.2 w/o graphes generated

It works for me.  See  
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
 for some suggestions on how to report the problem in more detail.

Your sessionInfo() is probably an absolute necessity here.

Good luck

John Kane
Kingston ON Canada


 -Original Message-
 From: dtustud...@hotmail.com
 Sent: Mon, 26 Nov 2012 22:27:27 -0700
 To: r-help@r-project.org
 Subject: [R] error of runing R in R 2.15.2 w/o graphes generated
 
 
 Hi,
 
 I have installed R 2.15.2 on windows 7.
 
 http://cran.cnr.berkeley.edu/
 
 I tried to run some simple graph code:
 
 http://www.harding.edu/fmccown/r/
 
 But, no graphs are presented or poped up.
 
 Any help will be appreciated.
 
 Thanks
 
 
 
 
 
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desktop!

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Re: [R] loop with date

2012-11-27 Thread R. Michael Weylandt

On Tuesday, November 27, 2012, Tagmarie wrote:

 Hello,
 I tried to construct my very first loop today and completly failed :-(
 Maybe someone can help me?
 I have a dataframe somewhat like this one:

 myframe - data.frame (Timestamp=c(24.09.2012 09:00, 24.09.2012 10:00,
 24.09.2012 11:00,
25.09.2012 09:00, 25.09.2012 10:00,
 25.09.2012 11:00),
 Speed=c(1,1,2,5,1,6))
 myframestime - as.POSIXct (strptime(as.character(myframe$Timestamp),
 %d.%m.%Y %H:%M), tz=GMT)
 myframe2 - cbind (myframe,myframestime)
 myframe2$Timestamp - NULL
 myframe2


Where's the loopy bit that fails?

Cheers, MW




 I want to construct a loop for every day, i.e. for each day I want to do
 some calculations.
 (I know in the example it would be easier to do it differently, my real
 data
 are little more complex).

 And BTW: Thanks for helping me earlier today with that other problem :-)




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 http://r.789695.n4.nabble.com/loop-with-date-tp4650961.html
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 http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.


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Re: [R] Effect of each term in the accuracy of Nonlinear multivariate regression fitting equation

2012-11-27 Thread Keith Jewell

In this context, linear model means linear in the _coefficients_ not
(necessarily) linear in the predictors, so your model:

JIM ~ z1*A + z2*B + z3*A*B^2 + z4*C*D^3 + z5*A^2*B^2 ...
is a linear model (in z1, z2, ...).

So you don't need to use nls, lm is probably favourite. You can use all
the techniques around for evaluating linear models; anova.lm might give
you a start.

On 27/11/2012 11:40, dsfakianakis wrote:

Dear all,

e.g. Let's say I have four input variables A B C and D, one output 'JIM' and
let z1, z2, ... be the coefficients of the terms of the equation. The
regression will be something like that:

Result = nls(JIM ~ z1*A + z2*B + z3*A*B^2 + z4*C*D^3 + z5*A^2*B^2 ... )

Is there a way to assess the contribution of each term (z1*A, z3*A*B^3 etc)
to the accuracy of the regression model?

Thanks a lot

Re: [R] R strange behaviour when building huge concatenation

No idea. It seems fine to me.

Perhaps a bit more information would be useful expecially your sessionInfo() 

See  
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
 for some suggestions on forming questions.

John Kane
Kingston ON Canada


 -Original Message-
 From: angelo...@gmail.com
 Sent: Mon, 26 Nov 2012 13:24:28 -0800 (PST)
 To: r-help@r-project.org
 Subject: Re: [R] R strange behaviour when building huge concatenation
 
 Any idea on the reason why the instruction
 p1x-c(2, 1, 0, 0, 0, 7, 1, 0, 2, 0, 5, 0, 0, 0, 0, 0, 10, 1, 3, 0, 2, 0,
 0,
 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 12, 0,
 0,
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 1,
 0, 1, 0, 0, 0, 3, 0, 0, 1, 0, 0, 0, 7, 0, 10, 0, 0, 1, 6, 0, 0, 0, 0, 0,
 2,
 4, 1, 5, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0,
 0,
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 13,
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 0,
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 1,
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 0,
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 2,
 0, 1, 2, 0, 0, 1, 0, 1, 0, 5, 0, 0, 1, 2, 0, 1, 0, 0, 1, 2, 1, 0, 1, 0,
 1,
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 1,
 1, 0, 0, 0, 3, 1, 1, 0, 1, 0, 0, 0, 4, 0, 2, 0, 3, 0, 0, 1, 0, 1, 0, 2,
 0,
 1, 3, 25, 0, 0, 0, 0, 5, 0, 2, 0, 0, 0, 5, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0,
 0,
 1, 0, 2, 1, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 2, 4, 0, 0, 1, 3, 0, 1, 0,
 0,
 0, 0, 0, 1, 1, 1, 0, 0, 0, 11, 0, 0, 4, 0, 0, 1, 2, 0, 0, 0, 0, 1, 1, 0,
 3,
 1, 0, 0, 2, 0, 8, 1, 0, 2, 1, 0, 0, 1, 0, 2, 0, 0, 3, 0, 0, 0, 0, 0, 0,
 2,
 2, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 2, 4, 0, 0, 2, 0, 6,
 0,
 1, 0, 1, 1, 1, 2, 1, 0, 7, 4, 0, 0, 0, 7, 0, 1, 1, 0, 2, 1, 0, 4, 0, 1,
 0,
 0, 2, 0, 0, 4, 0, 0, 2, 1, 0, 1, 0, 11, 0, 4, 0, 0, 0, 4, 0, 3, 0, 1, 1,
 0,
 0, 6, 3, 0, 0, 0, 0, 2, 0, 2, 0, 18, 0, 1, 0, 1, 0, 0, 0, 5, 0, 1, 0, 6,
 0,
 2, 0, 0, 2, 0, 1, 1, 0, 0, 1, 0, 1, 2, 0, 0, 0, 1, 0, 2, 0, 2, 0, 0, 0,
 0,
 1, 10, 0, 1, 3, 3, 0, 2, 0, 0, 12, 0, 1, 2, 2, 0, 0, 0, 0, 5, 0, 0, 2, 0,
 5,
 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 1, 7, 0, 0, 2,
 0,
 0, 2, 0, 4, 0, 0, 3, 0, 1, 0, 2, 2, 0, 1, 1, 2, 0, 0, 0, 0, 2, 0, 0, 1,
 0,
 1, 0, 3, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 3, 0, 4, 0, 0, 0,
 0,
 2, 1, 1, 0, 0, 0, 0, 5, 1, 0, 0, 1, 2, 1, 0, 0, 0, 0, 1, 0, 8, 1, 0, 6,
 1,
 2, 0, 3, 6, 0, 1, 0, 2, 5, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 13, 0, 0, 1,
 3,
 0, 5, 0, 2, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 4, 6, 1, 2, 0, 0, 2, 0, 7, 0,
 0,
 0, 0, 0, 3, 0, 5, 0, 2, 1, 3, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0,
 1,
 0, 0, 1, 10, 0, 0, 0, 0, 3, 3, 0, 0, 2, 5, 4, 0, 0, 0, 0, 0, 0, 0, 1, 0,
 0,
 0, 0, 1, 1, 3, 1, 0, 0, 0, 0, 6, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 6, 0, 1,
 0,
 2, 0, 0, 7, 0, 0, 0, 2, 0, 0, 0, 0, 2, 0, 1, 13, 7, 0, 0, 3, 0, 1, 0, 1,
 1,
 2, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 2, 1, 0, 0, 1, 19, 2, 2, 0, 2, 2,
 0,
 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 2, 0, 0, 0, 1, 0, 7, 0, 0, 0, 0,
 0,
 1, 0, 0, 1, 1, 0, 0, 2, 0, 1, 1, 0, 0, 4, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0,
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 0,
 0, 0, 0, 0, 0, 1, 0, 4, 0, 1, 1, 0, 1, 0, 1, 2, 0, 1, 0, 0, 0, 2, 0, 4,
 0,
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 0,
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 1, 1, 0, 0, 3, 0, 1, 0, 1,
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 0, 0, 0, 4, 0, 0, 2, 0, 5, 0, 3, 1, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 2,
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 0, 4, 2, 2, 1, 0, 0, 0, 0, 0, 1, 3, 0, 0, 1, 1, 6, 0, 0, 4, 0, 0, 1, 0,
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 0,
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Re: [R] Effect of each term in the accuracy of Nonlinear multivariate regression fitting equation

2012-11-27 Thread David Winsemius



On Nov 27, 2012, at 7:44 AM, Keith Jewell wrote:

In this context, linear model means linear in the _coefficients_  
not (necessarily) linear in the predictors, so your model:

  JIM ~ z1*A + z2*B + z3*A*B^2 + z4*C*D^3 + z5*A^2*B^2 ...
is a linear model (in z1, z2, ...).

So you don't need to use nls, lm is probably favourite. You can use  
all the techniques around for evaluating linear models; anova.lm  
might give you a start.


The additional R coding tip would be the I() function,

lm(JIM ~  B + A*I(B^2) + I(C*D^3) + I(A^2*B^2) + ...

Note that A and I(B^2) would also get estimates because of the way *  
is interpreted in formulas. If the * is inside the I() function that  
interpretation is not expanded.


In the linear models context it might be wiser to forego this approach  
and instead use regression splines.


--
David.




KJ

On 27/11/2012 11:40, dsfakianakis wrote:

Dear all,

I have a set of data with 4 inputs (independent variables) and one  
output
(dependent variable). I want to perform a regression analysis in  
order to
fit these data to a regression model, however due to the non- 
linearity of
the model I do not have a clue which equation to use. I am thinking  
of
starting with a very general equation including ^3 terms and  
interactions
between the variables however this will lead to a very long  
equation. Is

there a way to assess the effect of each term to the accuracy of the
regression model in order to discard the terms with the least  
importance?
Something like a sensitivity analysis of the effect of each term to  
the
accuracy regression model. I know one possible solution to my  
problem is
simply 'trial and error' however before going down that road I want  
to check

if there is an easier way.

e.g. Let's say I have four input variables A B C and D, one output  
'JIM' and
let z1, z2, ...  be the coefficients of the terms of the equation.   
The

regression will be something like that:

Result = nls(JIM ~ z1*A + z2*B + z3*A*B^2 + z4*C*D^3 +  
z5*A^2*B^2 ... )


Is there a way to assess the contribution of each term (z1*A,  
z3*A*B^3 etc)

to the accuracy of the regression model?

Thanks a lot



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David Winsemius, MD
Alameda, CA, USA

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Re: [R] Anova

2012-11-27 Thread S Ellison

 Now my question: Am I allowed to use these functions given 
 that my data is unbalanced? 

Unusually, Yes, assuming all the other requisite assumptions are reasonably 
well satisfied. One-way ANOVA interpretation is not much affected by imbalance 
because (among other things) with only one factor there is no 'model order' 
dependence or hypothesis separation to worry about. You don't need to use 
alternate sum-of-squares calculations (eg type II and III as in car) because 
types I, II and III sums of squares are all the same in a one-way case.

In a two-factor or higher analysis it's a (very) different story. 

However, one-way anova is still adversely affected by failures in homogeneity 
of variance, non-normality etc. 

S Ellison

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Re: [R] loop with date

2012-11-27 Thread jim holtman

Here is an example of an approach:

 myframe - data.frame (Timestamp=c(24.09.2012 09:00, 24.09.2012 10:00,
+ 24.09.2012 11:00,
+25.09.2012 09:00, 25.09.2012 10:00,
+ 25.09.2012 11:00),
+ Speed=c(1,1,2,5,1,6))
 myframestime - as.POSIXct (strptime(as.character(myframe$Timestamp),
+ %d.%m.%Y %H:%M), tz=GMT)
 myframe2 - cbind (myframe,myframestime)
 myframe2$Timestamp - NULL
 myframe2
  Speedmyframestime
1 1 2012-09-24 09:00:00
2 1 2012-09-24 10:00:00
3 2 2012-09-24 11:00:00
4 5 2012-09-25 09:00:00
5 1 2012-09-25 10:00:00
6 6 2012-09-25 11:00:00

 # split the dataframe into days' and then find average of Speed (for example)
 tapply(myframe2$Speed, cut(myframe2$myframestime, 'day'), mean)
2012-09-24 2012-09-25
  1.33   4.00





On Tue, Nov 27, 2012 at 9:02 AM, Tagmarie ramga...@gmx.net wrote:
 Hello,
 I tried to construct my very first loop today and completly failed :-(
 Maybe someone can help me?
 I have a dataframe somewhat like this one:

 myframe - data.frame (Timestamp=c(24.09.2012 09:00, 24.09.2012 10:00,
 24.09.2012 11:00,
25.09.2012 09:00, 25.09.2012 10:00,
 25.09.2012 11:00),
 Speed=c(1,1,2,5,1,6))
 myframestime - as.POSIXct (strptime(as.character(myframe$Timestamp),
 %d.%m.%Y %H:%M), tz=GMT)
 myframe2 - cbind (myframe,myframestime)
 myframe2$Timestamp - NULL
 myframe2

 I want to construct a loop for every day, i.e. for each day I want to do
 some calculations.
 (I know in the example it would be easier to do it differently, my real data
 are little more complex).

 And BTW: Thanks for helping me earlier today with that other problem :-)




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-- 
Jim Holtman
Data Munger Guru

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

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Re: [R] order.max specification problem in the ar.ols function

2012-11-27 Thread soham chakraborty

Thank you very much for the advice.It works that way.
Soham Chakraborty



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Re: [R] sum per day

HI,
Try this:
library(plyr)
res-ddply(myframe2,.(Timestamp=as.Date(myframe2$myframestime)),function(x) 
sum(x$Hunger))
 names(res)[2]-sumHunger
 res[,1]-format(res[,1],%d.%m.%Y)
 res
#   Timestamp sumHunger
#1 24.09.2012 3
#2 25.09.2012 5
A.K.




- Original Message -
From: Tagmarie ramga...@gmx.net
To: r-help@r-project.org
Cc: 
Sent: Tuesday, November 27, 2012 4:13 AM
Subject: [R] sum per day

Hello, 

I have a data frame somewhat like this one: 

myframe - data.frame (Timestamp=c(24.09.2012 09:00, 24.09.2012 10:00,
24.09.2012 11:00, 
                                   25.09.2012 09:00, 25.09.2012 10:00,
25.09.2012 11:00), Hunger=c(1,1,1,2,2,1) )                                  
myframestime - as.POSIXct (strptime(as.character(myframe$Timestamp),
%d.%m.%Y %H:%M), tz=GMT)
myframe2 - cbind (myframe,myframestime)
myframe2$Timestamp - NULL  
myframe2

Now I want to get the sum of Hunger for each day. In the end I want
something which looks like the following dataframe: 

myoutcome - data.frame(Timestamp=c(24.09.2012, 25.09.2012),
sumHunger=c(3, 5))

Does anyone know how to do that?
That would be very helpful and to all people who are willing to help me:
Thank you in advance! 

Best regards, 
Tagmarie 



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Re: [R] loop with date

HI,
If you want to do daily mean, sum etc.
you could try ?tapply(), ?ave(), ?aggregate(), ?ddply() etc.
For e.g.
 ave(myframe2$Speed,as.Date(myframe2$myframestime),FUN=sum)
#[1]  4  4  4 12 12 12
tapply(myframe2$Speed,as.Date(myframe2$myframestime),FUN=mean)
#2012-09-24 2012-09-25 
 # 1.33   4.00 

It is better to show the complex data as an example using dput()
A.K.




- Original Message -
From: Tagmarie ramga...@gmx.net
To: r-help@r-project.org
Cc: 
Sent: Tuesday, November 27, 2012 9:02 AM
Subject: [R] loop with date

Hello, 
I tried to construct my very first loop today and completly failed :-(
Maybe someone can help me? 
I have a dataframe somewhat like this one: 

myframe - data.frame (Timestamp=c(24.09.2012 09:00, 24.09.2012 10:00,
24.09.2012 11:00, 
                                   25.09.2012 09:00, 25.09.2012 10:00,
25.09.2012 11:00), 
                        Speed=c(1,1,2,5,1,6))                                  
myframestime - as.POSIXct (strptime(as.character(myframe$Timestamp),
%d.%m.%Y %H:%M), tz=GMT)
myframe2 - cbind (myframe,myframestime)
myframe2$Timestamp - NULL  
myframe2

I want to construct a loop for every day, i.e. for each day I want to do
some calculations. 
(I know in the example it would be easier to do it differently, my real data
are little more complex). 

And BTW: Thanks for helping me earlier today with that other problem :-) 




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[R] Variance component estimation in glmmPQL

2012-11-27 Thread nblarson

Hi all,

I've been attempting to fit a logistic glmm using glmmPQL in order to
estimate variance components for a score test, where the model is of the
form logit(mu) = X*a+ Z1*b1 + Z2*b2.  Z1 and Z2 are actually reduced rank
square root matrices of the assumed covariance structure (up to a constant)
of random effects c1 and c2, respectively, such that b1 ~ N(0,sig.1^2*I) and
c1 ~ N(0,sig.1^2*K1) , where K1 = Z1*t(Z1), and  c1 = Z1*b1.

The model form I've been using is just the following:

m-glmmPQL(y~1,random=list(f=pdBlocked(list(pdIdnot(~Z.1-1),pdIdnot(~Z.2-1
,family=binomial(link=logit))

I've been extracting the variance components using VarCorr(), but I've
noticed that the reported variances associated with my random effects are
not even close to the values I get if I evaluate their variances empirically
(eg var(random.effects(m.12)).  I know that's not how they're actually
estimated, but there may be a whole order of magnitude difference in the
values.

Below is an example under R 2.14 on a linux machine:

library(MASS)
library(mgcv)
library(boot)

set.seed(1234)

G.1-matrix(rnorm(5000,0,0.25),nrow=100)
G.2-matrix(rnorm(5000,0,0.25),nrow=100)

K.1-G.1%*%t(G.1)
K.2-G.2%*%t(G.2)

Z.1-mroot(K.1,method=svd)
Z.2-mroot(K.2,method=svd)

b.1-matrix(rnorm(ncol(Z.1),0,0.25),ncol=1)
b.2-matrix(rnorm(ncol(Z.2),0,0.50),ncol=1)

p-inv.logit(Z.1%*%b.1+Z.2%*%b.2)

y-rbinom(100,1,p)
f-rep(1,100)

m.fit-glmmPQL(y~1,random=list(f=pdBlocked(list(pdIdent(~Z.1-1),pdIdent(~Z.2-1,
  family=binomial(link=logit))

VarCorr(m.fit)
var(as.numeric(random.effects(m.fit))[1:ncol(Z.1)])
var(as.numeric(random.effects(m.fit))[-c(1:ncol(Z.1))])

From the above, VarCorr() results in variance component estimates of sig.1^2
= 0.444 and sig.2^2  = 0.2778, whereas the empirical estimates are sig.1^2 =
0.2060 and sig.1^2 = 0.097.  I know variance component estimation in general
is a little shaky, but my simulations suggest that VarCorr() is extracting
values that are way too large on a consistent basis.

I'm largely assuming I'm misinterpreting something here, just can't figure
out what.

-Nick



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Re: [R] R strange behaviour when building huge concatenation

2012-11-27 Thread angeloimm

Hello John
It seems correct to me too but in my R console it seems to not be working
Here there is what I did:
i copied the statement on one row (leaving and removing the final useless
semi colomn)
i tried to execute it in the R console (in order to open my R console I
simply opened a terminal window on my ubuntu machine and I typed R)
when I click enter the inserted statement doesn't not run...I simply see
the cursor on a new line and this new line starts with +

Here there is a little stack of what I see on my terminal:
, 1, 1, 2, 2, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 17, 2, 2, 0,
2, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 1, 0, 0,
1, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 7, 0, 0, 1, 0, 0, 3, 0, 0, 2, 4,
0, 1, 0, 0, 0, 0, 2, 1, 0, 4, 2, 0, 0, 1, 3, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0,
0, 0, 0, 0, 0, 0, 0, 0, 2, 3, 0, 1, 0, 1, 1, 3, 3, 0, 0, 0, 0, 1, 0, 2, 0,
1, 0, 0, 0, 1, 1, 0, 2, 0, 2, 0, 3, 0, 4, 0, 1, 1, 0, 0, 1, 0, 3, 0, 0, 0,
1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 3, 0, 6, 0, 0, 0, 0, 3, 0, 0, 0, 2, 5, 0,
1, 0, 0, 1, 0, 1, 0, 0, 2, 0, 1, 2, 0, 0, 7, 0, 0, 2, 0, 2, 2, 0, 0, 0, 0,
0, 2, 0, 8, 0, 0, 5, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 3, 0, 2, 0, 0, 0, 2, 1,
0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 2, 0, 0, 1, 0, 1, 2, 0, 0, 1, 0, 2, 0, 0,
3, 0, 0, 0, 1, 0, 0, 0, 1, 0, 9, 3, 0, 0, 0, 0, 0, 0, 3, 0, 0, 2, 0, 1, 1,
1, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0,
1, 4, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 2, 0, 1, 0, 0,
6, 0, 0, 0, 0, 1, 0, 0, 2, 6, 0, 1, 0, 0, 0, 0, 2, 0, 7, 0, 1, 2, 1, 1, 1,
0, 0, 1, 5, 0, 1, 0, 0, 2, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 5, 1, 0,
1, 2, 0, 0, 1, 2, 0, 1, 0, 2, 0, 1, 2, 7, 1, 2, 0, 0, 0, 5, 0, 4, 0))
+ 
+ 
+ 
+ 
+ 

Each time i click enter i go on a new line starting with +
Then if i start in tying some character I get an error like this:

+ 
+ 
+ q
+ 
+ q
Errore: unexpected symbol in:

q

Sometimes this error (Errore: unexpected symbol in:) appears in the
statement execution

Here there are my sessionInfo() result:
 sessionInfo()
R version 2.14.1 (2011-12-22)
Platform: i686-pc-linux-gnu (32-bit)

locale:
 [1] LC_CTYPE=it_IT.UTF-8   LC_NUMERIC=C  
 [3] LC_TIME=it_IT.UTF-8LC_COLLATE=it_IT.UTF-8
 [5] LC_MONETARY=it_IT.UTF-8LC_MESSAGES=it_IT.UTF-8   
 [7] LC_PAPER=C LC_NAME=C 
 [9] LC_ADDRESS=C   LC_TELEPHONE=C
[11] LC_MEASUREMENT=it_IT.UTF-8 LC_IDENTIFICATION=C   

attached base packages:
[1] stats graphics  grDevices utils datasets  methods   base 


I really don't know why this happensabove all because it seems to me a
very very simple statement...
Thank to all you for the support




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Re: [R] Problems with nls

2012-11-27 Thread Yifen

Hi all,

I am encountering the same problem when I use nls to estimate the Bass
diffusion model. I used similar code in the previous replies and it works
with the data in this original post. However, I got an error when I use my
own data.

Error in nls(formula = Bass.Model, start = c(p = 0.03, q = 0.4, m =
max(CuSale)),  : 
  step factor 0.000488281 reduced below 'minFactor' of 0.000976563


Here is my code:

Laptop_sale - c(1405,
1863,2027,2669,2938,5275,6595,6943,8621,10905,12420,22400,32380,31600,34900,43163,47838,47592)
CuSale - Laptop_sale
time - seq_along(Laptop_sale) 
Bass.Model - Laptop_sale ~ m * ((p + q)^2/p) * (exp(-(p + q) * time)/((q /
p) * exp(-(p + q) * time) + 1)^2) 
Bass.Fit - nls(formula = Bass.Model, start = c(p = 0.03, q = 0.4, m =
max(CuSale)), trace = TRUE) 


Can someone help me to figure out what the problem is? Thank you very much.



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Re: [R] confidence intervals with glmmPQL

2012-11-27 Thread Sally_roman

Hi - Thanks for the reply.  I provided answers below.
Sally


Sally_roman sroman at umassd.edu writes:

  Hi - I am using R version 2.13.0.  I have run several GLMMs using
 the glmmPQL function to model the proportion of fish caught in one
 net to the total caught in both nets by length.  I started with a
 polynomial regression full model with three length terms: l, l^2,
 and l^3 (l=length).  The length terms and intercept were the fixed
 effects and the random effect was a paired haul (n=18).
 
 m1-glmmPQL(fixed=Proportion~1+Length+second+third,random=~1|Pair,
 family=binomial,data=species,verbose=T,niter=2,
 weight=(Experimental+Control))

  Why did you set niter=2?  That seems like a bad idea
(the default is 10; I can imagine increasing it if there
 are warnings that the fit hasn't converged, but I don't
 see why you would decrease it).

I set niter=2 after I ran the models with the default.  All models converged
within 2 iterations.  
 
 For the majority of the models, I ended up with a constant model with no
 length effect.

This isn't quite clear: did you do some kind of stepwise model
reduction or AIC-based model selection?

I used a step wise backward selection for length terms and terms were
removed if they were not significant.  

 The issue I am having is with the confidence intervals that
 were calculated.  For two models the CIs are not symmetrical around the
 mean
 proportion from the model.  The CIs for the other constant models are
 symmetrical around the mean.  I was wondering if anyone has an idea why
 this
 would be or if anyone has any suggestions.  
 Thanks Sally

  There's not enough detail here to answer.  How did you get your
confidence intervals?  A reproducible example would be helpful.
If m1 is the result of a glmmPQL fit, then confint(m1) gives odd
results (because it calls confint.default, which doesn't really
know what to do with the results of the fit); intervals(m1) might
be more what you're looking for.

#subsample of data
PairSpecies Length  ExperimentalControl Proportion  second  third
2   Monkfish23  0   1   0   529 12167
2   Monkfish27  0   1   0   729 19683
7   Monkfish27  0   2   0   729 19683
8   Monkfish27  0   1   0   729 19683
14  Monkfish28  1   0   1   784 21952
13  Monkfish29  0   1   0   841 24389
14  Monkfish29  0   1   0   841 24389
1   Monkfish30  0   1   0   900 27000
5   Monkfish30  0   2   0   900 27000
18  Monkfish30  1   0   1   900 27000
4   Monkfish31  1   0   1   961 29791
5   Monkfish31  0   1   0   961 29791
11  Monkfish31  0   1   0   961 29791
2   Monkfish32  0   1   0   102432768
11  Monkfish32  0   1   0   102432768
12  Monkfish32  0   1   0   102432768
18  Monkfish32  1   1   0.5 102432768
4   Monkfish34  0   1   0   115639304
17  Monkfish34  0   1   0   115639304
18  Monkfish34  0   1   0   115639304
18  Monkfish35  0   1   0   122542875
1   Monkfish37  0   1   0   136950653
2   Monkfish40  1   0   1   160064000
14  Monkfish40  0   1   0   160064000
18  Monkfish40  1   0   1   160064000


This is my model code
#Models
#Select trip
t-t1
#select species
species-subset(t,Species==Monkfish) 
#define polynomials
#2nd order poly
second-species$Length^2
#3rd order poly
third-species$Length^3
#combine species,second  third
species-cbind(species,second,third)

#backward selection
#full model
m1-glmmPQL(fixed=Proportion~1+Length+second+third,random=~1|Pair,family=binomial,data=species,verbose=T,niter=2,weight=(Experimental+Control))
summary(m1)
#2nd order
m2-update(m1, . ~. -third)
summary(m2)
#linear
m3-update(m2, . ~. -second)
summary(m3)
#constant 
m4-update(m3, . ~. -Length)
summary(m4)

I used some code that was passed to me from a colleague who got it from
someone else.  I tried to go through the code to understand it, but got
stuck on a couple of areas.  I did not hear back from the original person
who wrote the code.   

Here is the code for a constant model - 

#CIs
get.sel.and.conf.band.catch=function(parm,varm,l.min,l.max,n=200){
  lgt-seq(l.min,l.max,length=n)
  X=sapply(1:length(parm),function(i){lgt^(i-1)})
  reg.line=X%*%parm
  se=apply(X,1,function(x,varm){sqrt(t(x)%*%varm%*%x)},varm)

[R] in par(mfrow=c(1, 2)), how to keep one half plot static and the other half changing

2012-11-27 Thread Baoqiang Cao

Hi,

I'm trying to plot something in the following way and would like if
you could help:

I'd like in a same plot window, two plots are shown, the left one is a
bird-view plot of the whole data, the right half keep changing, i.e.,
different plots will be shown up on request, so that when I
select/click on some where in the left plot, the right plot will be
the corresponding plot.

What I did is:

par(mfrow=c(1,2))
plot(x, y)

while(1) {
...
pxy - locator(1, type=p)

#select data point (dx,dy) based on pxy for a new plot
..

plot(dx,dy)
}

I ended up with the left plot is overwritten by plot(dx,dy). Is there
anyway to keep the left side intact while changing plots on the right
side?


Thanks a lot!

Baoqiang

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Re: [R] binning by frequency


Hello,

I'm not sure I ubderstand but,

h - hist(b1,brea=100)

which.max(h$counts)   # max frequency
findInterval(b1, h$breaks)

Hope this helps,

Rui Barradas

Em 27-11-2012 14:59, Santosh escreveu:

Dear Rxperts,

is there way to identify intervals from continuous data (having some kind
of a pattern) and then pick the value of most frequency?


a1 - round(rnorm(50,mean=0,0.1),2)
a2 - round(rnorm(50,mean=1,0.2),1)
a3 - round(rnorm(50,mean=5,1),0)
a4 - round(rnorm(50,mean=14,4),0)
a5 - round(rnorm(50,mean=30,8),0)

b1 - rbind(a1,a2,a3,a4,a5)

hist(b1,brea=100) # shows intervals and values with varying frequency.

unlike the mean values of a1 a5 above, I don't know the nominal values.
I would like an algorithm to identify intervals and pick the value with
most frequency.

I tried cut, split and was not successful.
Any suggestions/tips are highly welcome.
Thanks and regards,
Santosh

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[R] best HDF5 package: h5r or rhdf5?

2012-11-27 Thread Johann Hibschman

What is the current best package for manipulating HDF5 data files?

I tried hdf5 a long time ago, but I ran into memory problems. h5r is on
CRAN now, and rhdf5 is part of bioconductor.

Ideally, I'd like to read simple vectors or tables, either the entire thing
or a subset of rows. I don't need much writing support, but it would be
nice. Compression is a must, though.

Thanks,
Johann

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Re: [R] glm convergence warning

It could be that for some levels of your independent factor variables (WS, 
SS), the response is either all zeroes or all ones.  Or, for your 
continuous independent variables (DV, DS), there is a clean break between 
the zeroes and ones.  For example, if all the CIDs are one when DS = 18 
but all of the CIDs are zero when DS =20, then there is no single best 
fit for a logistic model to that relation, the curve could be straight up 
steep, or gradual and shallow.

Do you get convergence if you fit these subset models?

glm(CID ~ WS, data=kimu, family=binomial)
glm(CID ~ SS, data=kimu, family=binomial)
glm(CID ~ DV, data=kimu, family=binomial)
glm(CID ~ DS, data=kimu, family=binomial)

Can you see the problem when you plot the data?
attach(kimu)
plot(WS, CID)
plot(SS, CID)
plot(DV, CID)
plot(DS, CID)

Jean



Anne Schaefer annelsch...@gmail.com wrote on 11/26/2012 08:46:26 PM:
 
 Hello,
 
 When I run the following glm model:
 
 modelresult=glm(CID~WS+SS+DV+DS, data=kimu, family=binomial)
 
 I get the following warning messages:
 
 1: glm.fit: algorithm did not converge
 2: glm.fit: fitted probabilities numerically 0 or 1 occurred
 
 What I am trying to do is model my response variable (CID: correct bird
 identification) as a function of the predictor variables weather state
 (WS), sea state (SS), distance from the vessel (DV) and duration of the
 sighting (DS). I defined both sea state and weather state as factors 
with
 three levels (0, 1, or 2). Distance of the vessel values are 100, 80, 
60,
 40, and 20. Duration of the sighting ranges from 0 to 58 seconds.
 
 The output R is giving me is:
 
 Deviance Residuals:
Min  1Q  Median  3Q Max
 -3.562e-05  -2.100e-08   2.100e-08   2.100e-08   3.632e-05
 
 Coefficients:
   Estimate Std. Error z value Pr(|z|)
 (Intercept) -2.000e+02  1.067e+06   0.0001.000
 WSf1 7.744e+01  9.086e+04   0.0010.999
 WSf2 1.285e+01  6.199e+04   0.0001.000
 SSf1-1.042e+02  1.683e+05  -0.0011.000
 SSf2-1.859e+02  1.432e+05  -0.0010.999
 DV   6.770e-01  9.394e+03   0.0001.000
 DS   9.822e+00  1.884e+04   0.0011.000
 
 
 What do the warning messages mean? Can I still use coefficient estimates
 and standard error values?
 
 Thank you!

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Re: [R] Accumulate objects in list after try()

Create an empty list called result before you run the loop.  Then store 
the results of the loop in the list.  For example:

result - vector(mode=list, length=1000)
for(i in 1:1000){ 
result[[i]] - try(harvest(i))
}

Jean



mdvaan mathijsdev...@gmail.com wrote on 11/27/2012 12:09:38 AM:
 
 Hi,
 
 I have written a function harvest and I would like to run the function 
for
 each value in a vector c(1:1000). The function returns 4 list objects
 (obj_1, obj_3, obj_3, obj_4) using the following code at the end of the
 function: return(list(obj_1 = obj_1, obj_2 = obj_2, obj_3 = obj_3, obj_4 
=
 obj_4)).
 
 Since I am connecting with the web in the function and the connection
 sometimes fails causing errors to occur, I invoke the function as 
follows:
 
 for(i in 1:1000){
   result - try(harvest(i));
   if(class(result) == try-error) next;
   }
 
 Everything works well accept for the fact that result only stores 
obj_1, 
 obj_2, obj_3,  obj_4 for the last i in the loop. How do I store obj_1, 
 obj_2, obj_3,  obj_4 for the first i in the first 4 elements of result, 
the
 objects for the second i in the next 4 elements, etc?
 
 Thank you very much. 

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Re: [R] Help with optimization problem

2012-11-27 Thread Berend Hasselman

Jorge,

First of all, I really do think that questions such as these should be posted 
directly to R-help.
More people will then see it with a greater chance of getting useful replies.
I am subscribed to R-help so will see posts.

I have cc'ed  this reply to R-help so that you may get more answers than I can 
give you.

I know absolutely nothing about transportation problems. 
For comments see inline.

On 27-11-2012, at 00:24, Jorge I Velez wrote:

 Dear Dr. Hasselman,
 
 Please receive my apologies in advance for contacting you directly, but I 
 have seen your replies in R-help and just thought that, perhaps, you could 
 give me a hand. I am facing an optimization problem and unfortunately can not 
 find a way to get around.  I will try to explain myself as good as I can.
 
 Let us consider an optimization problem in which it is of interest to 
 distribute resources from 2 to 3 points. The costs associated are as follows:
 
Destination
 From  12 3 Total 
  1   13 4  300
   2  32 3  200
 Total150  250  100 500
 
 If X_i (i = 1, 2, ..., 6) is the cost of shipping X units from/to the ith 
 combination (i.e., i = 1 means from shipping from point 1 to destination 1; i 
 = 2 from point 1 to destination 2 and so on), the formulation in R would be 
 as follows:
 

This description  is confusing. So x4 means shipping from point to destination 
4 (you only have 3 destinations)?

 require(lpSolve)
 f - matrix(c(1, 3, 4, 3, 2, 3), ncol = 3, byrow = TRUE)
 row.rhs - c(300, 200)
 col.rhs - c(150, 250, 100)
 row.signs - rep(==, length(row.rhs))
 col.signs - rep(==, length(col.rhs))
 D - lp.transport(f, min, row.signs, row.rhs, col.signs, col.rhs)
 D$solution
 
 So far so good until this point.  However, in addition to spend *all* the 
 resources as above, in my application I would like to set up a couple of more 
 constraints as follows.
 
 # row and column constraints
 x1 + x2 + x3 == 300# row 1
  x4 + x5 + x6== 200# row 2
 x1  + x4   == 150# col 1
  x2 + x5   == 250# col 2
   x3   + x6== 100# col 3
 x1 + x2 + x3 + x4 + x5 + x6  == 500# all available
 

But these constraints are already satisfied in the solution.

 Now, if w is a vector representing some constants (the length of w is the 
 number of destinations from which we ship stuff), I would like to include the 
 following three constraints, where r is fixed and known:
 

I don't understand the number of destinations from which we ship stuff.
Shouldn't from be to?


 w1*x1   + w2*x4   == r# col 1
w1*x2 + w2*x5  == r# col 2
  w3*x3   + w6*x6  == r# col 3
 

vector w is length 6 but you only have three destinations (from the initial 
description)
I don't understand what you are doing here.

 Another constraint that I would like to impose is that, by column, the number 
 of X_i's greater than zero should be at least two.  Furthermore, the 
 maximization function should not be the coefficients in the f matrix, but the 
 standard deviation of g = c(l1, l2, l3) where
 
 l1 = w1*x1 + w2*x4
 l2 = w1*x2 + w2*x5 
 l3 = w3*x3 + w6*x6 
 

How does this relate to your previous formula?

 In my application I have up to 100 columns and 50 rows and I am able to 
 include all the constraints but not the number of zeros neither the new 
 function to optimize. Could you please give some advice me on how to do both? 
  I have been reading a bit on quadratic programing as the function to 
 optimize is the standard deviation, but I do not really know how to set up 
 the constraints, let alone the complete problem. 
 

From this description I gather that you want to optimize (minimize?) the 
standard deviations, perhaps a weighted sum?

 Thank you very much in advance for any help you can provide me.
 

There is far too much unclear about your problem to give any sensible advice, 
assuming that I have any.

Berend

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Re: [R] loop with date

Ok, sorry, I thought the more complex details might be confusing and nobody
might answer. 
Here is something which looks more like my real dataframe and also what I
want to do with it: 

That's my data frame: 

myframe - data.frame (Timestamp=c(24.09.2012 09:00, 24.09.2012 10:00,
24.09.2012 11:00, 
   25.09.2012 09:00, 25.09.2012 10:00,
25.09.2012 11:00), 
Hunger=c(1,1,2,5,1,6) ,
Longitude=c(8.91617, 8.92700, 8.92711, 8.92722,
8.92733, 8.92744),
Latitude=c(54.5485, 54.5410, 54.5412, 54.5413,
54.5414, 54.5424) ,
AnimalID= c(rep(Ernie)))  
  
head(myframe)
myframestime - as.POSIXct (strptime(as.character(myframe$Timestamp),
%d.%m.%Y %H:%M), tz=GMT)
myframe2 - cbind (myframe,myframestime)
myframe2$Timestamp - NULL  
myframesxy - project(cbind(myframe2$Longitude,myframe2$Latitude),+proj=utm
+zone=32 +ellps=WGS84)
colnames(myframesxy) - c(Long, Lat)
myframe3 - cbind(myframe2, myframesxy)
myframe3$Longitude - NULL
myframe3$Latitude - NULL
myframe3

And here is what I want to do with it (make an ltraj element, calculate the
brownian bridge homerange and get the kernel area of it - I take the 95
level): 

library(adehabitatHR)
myframe3ltraj - as.ltraj(myframesxy,myframestime,  id=myframe3$AnimalID)
myframeLiker - liker (myframe3ltraj, sig2=18, rangesig1=c(1,10) )
myframeLiker
MyframeBB - kernelbb(myframe3ltraj, sig1=4.6036, sig2=18)
kernel.area(MyframeBB, unout=c(km2)) 

With the only difference that I don't want to calculate the homerange for
the complete time but for each day. 
 I could use subset and do it for every day by hand. 
But I'd have to do it for three month and then again for several animals. So
I thought using a loop for each animal to get the results by date would be
much faster. 
Does anybody have any idea? 





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Re: [R] aggregate() runs out of memory

2012-11-27 Thread Sam Steingold

 * Steve Lianoglou znvyvatyvfg.ubarl...@tznvy.pbz [2012-11-26 19:47:25 
 -0500]:

 On Monday, November 26, 2012, Sam Steingold wrote:
 [snip]


 there is precisely one country for each id.
 i.e., unique(country) is the same as country[1].
 thanks a lot for the suggestion!

  R result - f[, list(min=min(delay), max=max(delay),
  count=.N,country=country[1L]), by=share.id]


 And is it performant?

acceptable.

 It just occurred to me that this is even better:

 R setkeyv(f, c(share.id, delay))
 R result - f[,  list(min=delay[1L], max=delay[.N], count=.N,
 country=country[1L]), by=share.id]


this assumes that delays are sorted (like in my example)
which, in reality, they are not.
thanks for your help!

-- 
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http://www.childpsy.net/ http://honestreporting.com
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Illiterate?  Write today, for free help!

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Re: [R] install the ggplot2 package

2012-11-27 Thread Jonsson

I did upgrade but that did not solve the problem



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[R] calling c function from R

2012-11-27 Thread Li Shangru

Hello

I want to call C function from R. I follow the instruction below using the
example of foo.c
http://www.stat.umn.edu/~charlie/rc/

I saved the foo.c in my work directory. Then I entered the command

R CMD SHLIB foo.c , I got the following message



Error: unexpected symbol in R CMD


How did it go wrong? Thanks!

Regards

Shangru Li

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[R] error of installing a R package on Win 7

2012-11-27 Thread Jack Bryan


Hi, 
I am trying to run R 2.15.2 on Win 7. 
I am trying to run some example R code of the book :Portfolio Optimization with 
R/Rmetrics
I was told that :
To install all packages required for the examples of this ebook we 
recommendthat you install the bundle package ebookPortfolio. 
This can be done with the following commands in the R environment. If there is 
no binary package for your operating system, you can install the package from 
source with the argument type = source.
 install.packages(ebookPortfolio,repos = 
 c(http://pkg.rmetrics.org,http://cran.r-project.org;),type = 
 getOption(pkgType))
I got :
Warning: unable to access index for repository 
http://pkg.rmetrics.org/src/contribWarning: package ebookPortfolio is not 
available (for R version 2.15.2)Warning: unable to access index for repository 
http://pkg.rmetrics.org/bin/windows/contrib/2.15
Is this an error ? 
I am a new user of R.
How to handle it ? How to install ebookPortfolio package ? 
Any help will be appreciated. 
Thanks 
  
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[R] R.dll problem during loading the package lme4

2012-11-27 Thread Prasenjit

Greetings,
I am using R 2.14.1 and while loading package lme4 I am getting the error
message: the procedure entry point R_Check_class_etc could not be located in
the dynamic link library R.dll. Could anybody help me on this?

Thanks and Regards,
Roy



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[R] Stuck trying to modify a function

2012-11-27 Thread Benjamin Ward (ENV)

Hi,

I have the following data:

Path_Number - 5
ID.Path - c(1:Path_Number) # Make vector of ID's.
No_of_X - sample(50:550, length(ID.Path), replace=TRUE) #
X - split(sample(1:1, sum(No_of_X), replace=TRUE), rep(ID.Path, No_of_X))
Y - lapply(X,function(x) sample(x, round(runif(1, min=10, max=50

X and Y are both lists, and I've made the following function to work on that 
data as part of a simulation I'm building:

Mutate-function(x){
  l-0
  for(i in x){
l2-0
l-l+1
for(i in x[[l]]){
  l2-l2+1
  if(runif(1)  0.9) ifelse(runif(1) 0.5, x[[l]][l2] - x[[l]][l2]+1, 
x[[l]][l2] - x[[l]][l2]-1)
}
  }
  return(x)
}

I call this with Effectors-Mutate(X)
The function is designed to alter the values of each element in X by either + 
or - 1 (50:50 chance wether + or -). However Y, elements of which are a subset 
of the corresponding elements of X, need to be consistent i.e. if a value in X 
is changed, and that value is part of the Y subset, then the value in Y also 
needs to be changed. however, since Y is a smaller subset it will not be 
indexed the same. My idea was to include in the function an if statement that 
checks if Y contains the value to be changed, removes it, and then after the 
value in X is changed, put the new value in Y. I attempted this with:


Mutate-function(x,y){
  l-0
  for(i in x){
l2-0
l-l+1
for(i in x[[l]]){
  l2-l2+1
  if(runif(1)  0.9){
if(x[[l]][l2] %in% y[[l]] == TRUE){
  y[[l]]-[which(y[[l]]!=x[[l]][l2])]
  if(runif(1) 0.5){
x[[l]][l2] - x[[l]][l2]+1
y[[l]]-append(x[[l]][l2])
  }else{
x[[l]][l2] - x[[l]][l2]-1
y[[l]]-append(x[[l]][l2])
  }
}
ifelse(runif(1) 0.5, x[[l]][l2] - x[[l]][l2]+1, x[[l]][l2] - 
x[[l]][l2]-1)
  }
}
  }
  return(list(x,y))
}

Bit of an eyesore so I've put the altered stuff in bold. I've basically taken 
what the ifelse statement does in the first function, (which is still there and 
run if Y does not contain the X value being altered) and broken it down into an 
if and an else segment with multiple operations in curly braces to accommodate 
the extra actions needed to alter Y as well as X. This was all I could think of 
to keep changes between the two in sync, however this does not work when I 
try to load the function into workspace:

Error: unexpected '}' in }

I hope someone can point out what it is I've done that isn't working, or a 
better way to do this.

Best Wishes,

Ben W.

UEA (ENV)  The Sainsbury Laboratory.

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[R] interactions in GAMs

2012-11-27 Thread chirleu

Hi all, 
I wonder if it's possible to include a double interaction in a GAM formula.
Example:

If I do this:
mod=gam(energy~s(size, *by=color, by=sex*, k=5) + temperature, ...)

I get the interaction betwen size*color and size*sex.

But I need size*color*sex, being size a smoother.
I've created a new variable (colorsex) which combines all the level of both
color (2 levels) and sex (2 level), so that I have a new variable with 4
level. In this case I can do:

mod=gam(energy~s(size, *by=colorsex*, k=5) + temperature, ...)

What do you think of this approach?

In this case, should I also include colorsex (or color*sex) in the
parametric term *even if it's not significant*(as it's the case)?

Many thanks

David








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Re: [R] binning by frequency

2012-11-27 Thread Mark Lamias

You might find the binning function in the sm package helpful here.

--Mark Lamias

 From: Santosh santosh2...@gmail.com
To: r-help r-help@r-project.org 
Sent: Tuesday, November 27, 2012 9:59 AM
Subject: [R] binning by frequency

Dear Rxperts,

is there way to identify intervals from continuous data (having some kind
of a pattern) and then pick the value of most frequency?

a1 - round(rnorm(50,mean=0,0.1),2)
a2 - round(rnorm(50,mean=1,0.2),1)
a3 - round(rnorm(50,mean=5,1),0)
a4 - round(rnorm(50,mean=14,4),0)
a5 - round(rnorm(50,mean=30,8),0)

b1 - rbind(a1,a2,a3,a4,a5)

hist(b1,brea=100) # shows intervals and values with varying frequency.

unlike the mean values of a1 a5 above, I don't know the nominal values.
I would like an algorithm to identify intervals and pick the value with
most frequency.

I tried cut, split and was not successful.
Any suggestions/tips are highly welcome.
Thanks and regards,
Santosh

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Re: [R] interactions in GAMs

2012-11-27 Thread Simon Wood


David,

I think that the colorsex approach is the right one, and colorsex should 
initially be included as a main effect, because the smooths are centred 
for factor by variables (see e.g. ?gam.models). Whether you then choose 
to drop this main effect, as it appears to be non-significant, is a 
matter of taste (I would tend to leave it in).


best,
Simon

On 27/11/12 16:42, chirleu wrote:

Hi all,
I wonder if it's possible to include a double interaction in a GAM formula.
Example:

If I do this:
mod=gam(energy~s(size, *by=color, by=sex*, k=5) + temperature, ...)

I get the interaction betwen size*color and size*sex.

But I need size*color*sex, being size a smoother.
I've created a new variable (colorsex) which combines all the level of both
color (2 levels) and sex (2 level), so that I have a new variable with 4
level. In this case I can do:

mod=gam(energy~s(size, *by=colorsex*, k=5) + temperature, ...)

What do you think of this approach?

In this case, should I also include colorsex (or color*sex) in the
parametric term *even if it's not significant*(as it's the case)?

Many thanks

David








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Re: [R] Using cumsum with 'group by' ?

2012-11-27 Thread peter dalgaard


On Nov 27, 2012, at 14:30 , TheRealJimShady wrote:

 Thanks everyone for your help. All good now. Once I'd got the data I
 ended up doing this (replaced with real column names rather than the
 fakes before).
 
 microaeth_data$date-as.Date(microaeth_data$date_time,format=%Y-%m-%d
 %H:%M:%S)
 
 dat2-microaeth_data[order(microaeth_data[,13],microaeth_data[,14]),]
 
 dat2$cumsum-ave(dat2$bc,list(dat2$person_id,dat2$date),FUN=cumsum)
 
 If someone can be bothered I'd appreciate a breakdown of the final
 line of code so that I understood what I did, but no problem if not.
 

OK. ave() is originally designed to replace observations by their per-group 
average. It works by splitting the input vector into groups according to one or 
more factors, performing FUN on each group, and putting the result back in the 
original positions. The default for FUN is mean, but it doesn't need to be a 
scalar function; it also works with a FUN that returns a vector the same length 
as the input, such as FUN=cumsum.

The specification does not seem to allow a list() of grouping factors. It does 
work, but as it is unauthorised, it might not keep working.

Elaborating on the help page example:

 attach(warpbreaks)
 ave(breaks,wool,tension)
 [1] 44.6 44.6 44.6 44.6 44.6 44.6 44.6 44.6
 [9] 44.6 24.0 24.0 24.0 24.0 24.0 24.0 24.0
[17] 24.0 24.0 24.6 24.6 24.6 24.6 24.6 24.6
[25] 24.6 24.6 24.6 28.2 28.2 28.2 28.2 28.2
[33] 28.2 28.2 28.2 28.2 28.8 28.8 28.8 28.8
[41] 28.8 28.8 28.8 28.8 28.8 18.8 18.8 18.8
[49] 18.8 18.8 18.8 18.8 18.8 18.8
 ave(breaks,wool,tension, FUN=sum)
 [1] 401 401 401 401 401 401 401 401 401 216 216 216 216 216 216 216 216 216 221
[20] 221 221 221 221 221 221 221 221 254 254 254 254 254 254 254 254 254 259 259
[39] 259 259 259 259 259 259 259 169 169 169 169 169 169 169 169 169
 ave(breaks,wool,tension, FUN=cumsum)
 [1]  26  56 110 135 205 257 308 334 401  18  39  68  85  97 115 150 180 216  36
[20]  57  81  99 109 152 180 195 221  27  41  70  89 118 149 190 210 254  42  68
[39]  87 103 142 170 191 230 259  20  41  65  82  95 110 125 141 169
 ave(breaks,list(wool,tension), FUN=cumsum) # not actually supposed to work
 [1]  26  56 110 135 205 257 308 334 401  18  39  68  85  97 115 150 180 216  36
[20]  57  81  99 109 152 180 195 221  27  41  70  89 118 149 190 210 254  42  68
[39]  87 103 142 170 191 230 259  20  41  65  82  95 110 125 141 169

 


 Thanks
 
 James
 
 On 23 November 2012 20:06, arun kirshna [via R]
 ml-node+s789695n4650584...@n4.nabble.com wrote:
 HI,
 
 If that is the case, this should work:
 dat1-read.table(text=
 id,  x,  date
 1,  5,  2012-06-05 12:01
 1,  10,2012-06-05 12:02
 1,  45,2012-06-05 12:03
 2,  5,  2012-06-05 12:01
 2,  3,  2012-06-05 12:03
 2,  2,  2012-06-05 12:05
 3,  5,  2012-06-05 12:03
 3,  5,  2012-06-05 12:04
 3,  8,  2012-06-05 12:05
 1,  5,  2012-06-08 13:01
 1,  9,  2012-06-08 13:02
 1,  3,  2012-06-08 13:03
 2,  0,  2012-06-08 13:15
 2,  1,  2012-06-08 13:18
 2,  8,  2012-06-08 13:20
 2,  4,  2012-06-08 13:21
 3,  6,  2012-06-08 13:15
 3,  2,  2012-06-08 13:16
 3,  7,  2012-06-08 13:17
 3,  2,  2012-06-08 13:18
 ,sep=,,header=TRUE,stringsAsFactors=FALSE)
 dat1$date-as.Date(dat1$date,format=%Y-%m-%d %H:%M)
 dat2-dat1[order(dat1[,1],dat1[,3]),]
 dat2$Cumsum-ave(dat2$x,list(dat2$id,dat2$date),FUN=cumsum)
 
 head(dat2)
 #   id  x   date Cumsum
 #1   1  5 2012-06-05  5
 #2   1 10 2012-06-05 15
 #3   1 45 2012-06-05 60
 #10  1  5 2012-06-08  5
 #11  1  9 2012-06-08 14
 #12  1  3 2012-06-08 17
 #or
 with(dat2,aggregate(x,by=list(id=id,date=date),cumsum))
 #  id   datex
 #1  1 2012-06-055, 15, 60
 #2  2 2012-06-05 5, 8, 10
 #3  3 2012-06-055, 10, 18
 #4  1 2012-06-085, 14, 17
 #5  2 2012-06-08  0, 1, 9, 13
 #6  3 2012-06-08 6, 8, 15, 17
 A.K.
 
 
 
 - Original Message -
 From: TheRealJimShady [hidden email]
 To: [hidden email]
 Cc:
 Sent: Friday, November 23, 2012 6:04 AM
 Subject: Re: [R] Using cumsum with 'group by' ?
 
 Hi Arun  everyone,
 
 Thank you very much for your helpful suggestions. I've been working
 through them, but have realised that my data is a little more
 complicated than I said and that the solutions you've kindly provided
 don't work. The problem is that there is more than one day of data for
 each person. It looks like this:
 
 id  x  date
 1  5  2012-06-05 12:01
 1  102012-06-05 12:02
 1

[R] Concordant und Discordant Paars of Logistic Regression

2012-11-27 Thread Marcus Tullius

Hallo there,

 can anymore show me how to get results about konkordant und diskordant paars 
(Sommer-D, Gudman-Krustal-Gama, Kendall-Tau-a) within a logistic regression) in 
R?

 Thanks a lot.
 MT

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Re: [R] R strange behaviour when building huge concatenation

2012-11-27 Thread Jagat.K.Sheth

You probably hit a buffer limit in X11/xterm on your ubuntu machine with copy 
and paste. I get that behavior with Putty using your vector or when pasting 
(long) commands into Putty. 

If you really prefer copy and paste for this vector then try something like

 eval(parse(text=scan(clipboard, what=)))
Read 2357 items
 head(p1x)
[1] 2 1 0 0 0 7
 tail(p1x)
[1] 0 0 5 0 4 0

Otherwise you can save your code and source() the file. There is also ESS 
(Emacs Speaks Statistics) and friends if you want to avoid copy/paste approach. 
 

BTW you can use Ctrl-C to get back to command prompt in your R console rather 
than typing q in your output below. 
 

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of angeloimm
 Sent: Tuesday, November 27, 2012 9:04 AM
 To: r-help@r-project.org
 Subject: Re: [R] R strange behaviour when building huge concatenation
 
 Hello John
 It seems correct to me too but in my R console it seems to not be
 working
 Here there is what I did:
 i copied the statement on one row (leaving and removing the final
 useless
 semi colomn)
 i tried to execute it in the R console (in order to open my R console I
 simply opened a terminal window on my ubuntu machine and I typed R)
 when I click enter the inserted statement doesn't not run...I simply
 see
 the cursor on a new line and this new line starts with +
 
 Here there is a little stack of what I see on my terminal:
 , 1, 1, 2, 2, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 17, 2, 2,
 0,
 2, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 1, 0,
 0,
 1, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 7, 0, 0, 1, 0, 0, 3, 0, 0, 2,
 4,
 0, 1, 0, 0, 0, 0, 2, 1, 0, 4, 2, 0, 0, 1, 3, 0, 0, 1, 0, 0, 1, 0, 1, 1,
 0,
 0, 0, 0, 0, 0, 0, 0, 0, 2, 3, 0, 1, 0, 1, 1, 3, 3, 0, 0, 0, 0, 1, 0, 2,
 0,
 1, 0, 0, 0, 1, 1, 0, 2, 0, 2, 0, 3, 0, 4, 0, 1, 1, 0, 0, 1, 0, 3, 0, 0,
 0,
 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 3, 0, 6, 0, 0, 0, 0, 3, 0, 0, 0, 2, 5,
 0,
 1, 0, 0, 1, 0, 1, 0, 0, 2, 0, 1, 2, 0, 0, 7, 0, 0, 2, 0, 2, 2, 0, 0, 0,
 0,
 0, 2, 0, 8, 0, 0, 5, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 3, 0, 2, 0, 0, 0, 2,
 1,
 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 2, 0, 0, 1, 0, 1, 2, 0, 0, 1, 0, 2, 0,
 0,
 3, 0, 0, 0, 1, 0, 0, 0, 1, 0, 9, 3, 0, 0, 0, 0, 0, 0, 3, 0, 0, 2, 0, 1,
 1,
 1, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0,
 0,
 1, 4, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 2, 0, 1, 0,
 0,
 6, 0, 0, 0, 0, 1, 0, 0, 2, 6, 0, 1, 0, 0, 0, 0, 2, 0, 7, 0, 1, 2, 1, 1,
 1,
 0, 0, 1, 5, 0, 1, 0, 0, 2, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 5, 1,
 0,
 1, 2, 0, 0, 1, 2, 0, 1, 0, 2, 0, 1, 2, 7, 1, 2, 0, 0, 0, 5, 0, 4, 0))
 +
 +
 +
 +
 +
 
 Each time i click enter i go on a new line starting with +
 Then if i start in tying some character I get an error like this:
 
 +
 +
 + q
 +
 + q
 Errore: unexpected symbol in:
 
 q
 
 Sometimes this error (Errore: unexpected symbol in:) appears in the
 statement execution
 
 Here there are my sessionInfo() result:
  sessionInfo()
 R version 2.14.1 (2011-12-22)
 Platform: i686-pc-linux-gnu (32-bit)
 
 locale:
  [1] LC_CTYPE=it_IT.UTF-8   LC_NUMERIC=C
  [3] LC_TIME=it_IT.UTF-8LC_COLLATE=it_IT.UTF-8
  [5] LC_MONETARY=it_IT.UTF-8LC_MESSAGES=it_IT.UTF-8
  [7] LC_PAPER=C LC_NAME=C
  [9] LC_ADDRESS=C   LC_TELEPHONE=C
 [11] LC_MEASUREMENT=it_IT.UTF-8 LC_IDENTIFICATION=C
 
 attached base packages:
 [1] stats graphics  grDevices utils datasets  methods   base
 
 
 I really don't know why this happensabove all because it seems to
 me a
 very very simple statement...
 Thank to all you for the support
 
 
 
 
 --
 View this message in context: http://r.789695.n4.nabble.com/R-strange-
 behaviour-when-building-huge-concatenation-tp4650817p4650970.html
 Sent from the R help mailing list archive at Nabble.com.
 
 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-
 guide.html
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Re: [R] Help with graphics in gamm4 library

2012-11-27 Thread MurphFL

Thank you so much! I really appreciate the rapid response, and I was able to
solve my issues with this advice!



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[R] Fitting and plotting a coxph with survfit, package(surv)

2012-11-27 Thread Maziar Mohaddes

Hi Dear R-users

I have a database with 18000 observations and 20 variables. I am running
cox regression on five variables and trying to use survfit to plot the
survival based on a specific variable without success.

Lets say I have the following coxph:
library(survival)
fit - coxph(Surv(futime, fustat) ~ age + rx, data = ovarian)
fit
what I am trying to do is plot a survival comparing objects based on rx.
Using this
plot(survfit(fit, newdata=data.frame(rx =c(1:2), age=c(60)),
 xscale=365.25, xlab = Years, ylab=Survival))
I get the survival for patients at 60, but is there an option to get a
survfit for the patients regardless of the value in variable age?

Thanks in advance
Maziar Mohaddes
M.D.
Gothenburg, Sweden

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Re: [R] calling c function from R

2012-11-27 Thread Berend Hasselman


On 27-11-2012, at 17:21, Li Shangru wrote:

 Hello
 
 I want to call C function from R. I follow the instruction below using the
 example of foo.c
 http://www.stat.umn.edu/~charlie/rc/
 
 I saved the foo.c in my work directory. Then I entered the command
 
 R CMD SHLIB foo.c , I got the following message
 
 
 
 Error: unexpected symbol in R CMD
 
 
 How did it go wrong? Thanks!

Not how but why?
The message is given by R when  the syntax is wrong.

You didn't do what you were supposed to do according to the instructions.
The command

R CMD SHLIB foo.c 

should be executed outside of R.

Berend

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[R] loop command to matrix

2012-11-27 Thread eliza botto


Dear UseRs,Extremely sorry for a basic question. I have a matrix of 19 rows and 
365 columns. what i want to do is the following...First i want to leave out 
column number 1 and want to calculate the row wise mean of the remaining 
columns, which will obviously give me 365 values in one column, and then 
subtracting these values from the column i left out i.e. col=1 then i want to 
leave out column 2 and calculate the row wise mean of the remaining columns 
which includes column 1 too and then subtracting these values from the column i 
left out i.e. col=2.and then continuing this process the last column. i know a 
kind of manual way of doing things but its extremely long and laborious.Is 
there any loop command or shorter way??
thanks in advanceregardseliza 
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Re: [R] R strange behaviour when building huge concatenation

2012-11-27 Thread Milan Bouchet-Valat

Le mardi 27 novembre 2012 à 07:04 -0800, angeloimm a écrit :
 Hello John
 It seems correct to me too but in my R console it seems to not be working
 Here there is what I did:
 i copied the statement on one row (leaving and removing the final useless
 semi colomn)
 i tried to execute it in the R console (in order to open my R console I
 simply opened a terminal window on my ubuntu machine and I typed R)
 when I click enter the inserted statement doesn't not run...I simply see
 the cursor on a new line and this new line starts with +
Are you aware that this usually means R is waiting for the end of the
command, e.g. because a closing parenthesis is missing?

 Here there is a little stack of what I see on my terminal:
 , 1, 1, 2, 2, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 17, 2, 2, 0,
 2, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 1, 0, 0,
 1, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 7, 0, 0, 1, 0, 0, 3, 0, 0, 2, 4,
 0, 1, 0, 0, 0, 0, 2, 1, 0, 4, 2, 0, 0, 1, 3, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0,
 0, 0, 0, 0, 0, 0, 0, 0, 2, 3, 0, 1, 0, 1, 1, 3, 3, 0, 0, 0, 0, 1, 0, 2, 0,
 1, 0, 0, 0, 1, 1, 0, 2, 0, 2, 0, 3, 0, 4, 0, 1, 1, 0, 0, 1, 0, 3, 0, 0, 0,
 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 3, 0, 6, 0, 0, 0, 0, 3, 0, 0, 0, 2, 5, 0,
 1, 0, 0, 1, 0, 1, 0, 0, 2, 0, 1, 2, 0, 0, 7, 0, 0, 2, 0, 2, 2, 0, 0, 0, 0,
 0, 2, 0, 8, 0, 0, 5, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 3, 0, 2, 0, 0, 0, 2, 1,
 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 2, 0, 0, 1, 0, 1, 2, 0, 0, 1, 0, 2, 0, 0,
 3, 0, 0, 0, 1, 0, 0, 0, 1, 0, 9, 3, 0, 0, 0, 0, 0, 0, 3, 0, 0, 2, 0, 1, 1,
 1, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0,
 1, 4, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 2, 0, 1, 0, 0,
 6, 0, 0, 0, 0, 1, 0, 0, 2, 6, 0, 1, 0, 0, 0, 0, 2, 0, 7, 0, 1, 2, 1, 1, 1,
 0, 0, 1, 5, 0, 1, 0, 0, 2, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 5, 1, 0,
 1, 2, 0, 0, 1, 2, 0, 1, 0, 2, 0, 1, 2, 7, 1, 2, 0, 0, 0, 5, 0, 4, 0))
This command is not the same as the one you posted, it ends with two
parentheses... Could you post a whole copy/paste of the output,
including the beginning of the command? Have you tried copying the exact
contents of the command from your e-mail and check it still produces the
errors on your machine?

I think you should split the command in two parts, check if they work
separately. If one of them does not work, split it again, until you
identify either a problematic, possibly invisible char, or a length
limit...


My two cents

 + 
 + 
 + 
 + 
 + 
 
 Each time i click enter i go on a new line starting with +
 Then if i start in tying some character I get an error like this:
 
 + 
 + 
 + q
 + 
 + q
 Errore: unexpected symbol in:
 
 q
 
 Sometimes this error (Errore: unexpected symbol in:) appears in the
 statement execution
 
 Here there are my sessionInfo() result:
  sessionInfo()
 R version 2.14.1 (2011-12-22)
 Platform: i686-pc-linux-gnu (32-bit)
 
 locale:
  [1] LC_CTYPE=it_IT.UTF-8   LC_NUMERIC=C  
  [3] LC_TIME=it_IT.UTF-8LC_COLLATE=it_IT.UTF-8
  [5] LC_MONETARY=it_IT.UTF-8LC_MESSAGES=it_IT.UTF-8   
  [7] LC_PAPER=C LC_NAME=C 
  [9] LC_ADDRESS=C   LC_TELEPHONE=C
 [11] LC_MEASUREMENT=it_IT.UTF-8 LC_IDENTIFICATION=C   
 
 attached base packages:
 [1] stats graphics  grDevices utils datasets  methods   base 
 
 
 I really don't know why this happensabove all because it seems to me a
 very very simple statement...
 Thank to all you for the support
 
 
 
 
 --
 View this message in context: 
 http://r.789695.n4.nabble.com/R-strange-behaviour-when-building-huge-concatenation-tp4650817p4650970.html
 Sent from the R help mailing list archive at Nabble.com.
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.

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Re: [R] aggregate() runs out of memory

2012-11-27 Thread Steve Lianoglou

Hi,

On Tue, Nov 27, 2012 at 11:29 AM, Sam Steingold s...@gnu.org wrote:
 * Steve Lianoglou znvyvatyvfg.ubarl...@tznvy.pbz [2012-11-26 19:47:25 
 -0500]:
[snip]
 It just occurred to me that this is even better:

 R setkeyv(f, c(share.id, delay))
 R result - f[,  list(min=delay[1L], max=delay[.N], count=.N,
 country=country[1L]), by=share.id]


 this assumes that delays are sorted (like in my example)
 which, in reality, they are not.
 thanks for your help!

When you include delay in the call to `setkeyv` as I did above, it
sorts low to high w/in each share.id group.

-steve

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

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[R] Some questions about chron package..

2012-11-27 Thread 박상규

Hello,


I have questions while reviewing chron package(e.g.,chron.R).


1. What is the differences between 3 kinds of function definition ? 
 1) name lt;- function(...
 2) 'name' lt;- function(...
 3) name lt;- function(...
Do you know Why  author used various kinds of  definitions ? 
Is there no functional differences between them ? 


2. I don't understand the meaning of the below code:
 
as.chron lt;- function(x, ...) UseMethod(as.chron)
as.chron.default lt;- function (x, format, ...)
{ 


Could you let me know what is role of UseMethod(as.chron) ? 


3. In the following code,



chron lt;-
function(dates. = NULL, times. = NULL,
 format = c(dates = m/d/y, times = h:m:s),
 out.format, origin.)


'dates', 'origin' and 'times' has dot(.) as at the end of word.
Do you know the meaning of dots ? 


4. Could you introduce me some books on S/R language, especially for 
understanding of generic functions ? 
Some interesting packages such as quantmod, xts, zoo, etc., are seems to be 
coded using generic functions(UseMethod(..)).
But I can't easily understand the code.


Thanks in advance,




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Re: [R] loop command to matrix


Hello,

Try

x - matrix(1:18, ncol = 6)
sapply(seq_len(ncol(x)), function(i) x[, i] - rowMeans(x[, -i]))


Hope this helps,

Rui Barradas
Em 27-11-2012 17:51, eliza botto escreveu:

Dear UseRs,Extremely sorry for a basic question. I have a matrix of 19 rows and 365 
columns. what i want to do is the following...First i want to leave out column number 1 
and want to calculate the row wise mean of the remaining columns, which will obviously 
give me 365 values in one column, and then subtracting these values from the column i 
left out i.e. col=1 then i want to leave out column 2 and calculate the row wise mean of 
the remaining columns which includes column 1 too and then subtracting these values from 
the column i left out i.e. col=2.and then continuing this process the last column. i know 
a kind of manual way of doing things but its extremely long and laborious.Is 
there any loop command or shorter way??
thanks in advanceregardseliza   
[[alternative HTML version deleted]]

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Re: [R] loop command to matrix

2012-11-27 Thread eliza botto

thanks rui!! how can it be that you advise something and it doesn't work :)

eliza

 Date: Tue, 27 Nov 2012 18:14:41 +
 From: ruipbarra...@sapo.pt
 To: eliza_bo...@hotmail.com
 CC: r-help@r-project.org
 Subject: Re: [R] loop command to matrix

 Hello,

 Try

 x - matrix(1:18, ncol = 6)
 sapply(seq_len(ncol(x)), function(i) x[, i] - rowMeans(x[, -i]))

 Hope this helps,

 Rui Barradas
 Em 27-11-2012 17:51, eliza botto escreveu:
  Dear UseRs,Extremely sorry for a basic question. I have a matrix of 19 rows 
  and 365 columns. what i want to do is the following...First i want to leave 
  out column number 1 and want to calculate the row wise mean of the 
  remaining columns, which will obviously give me 365 values in one column, 
  and then subtracting these values from the column i left out i.e. col=1 
  then i want to leave out column 2 and calculate the row wise mean of the 
  remaining columns which includes column 1 too and then subtracting these 
  values from the column i left out i.e. col=2.and then continuing this 
  process the last column. i know a kind of manual way of doing things but 
  its extremely long and laborious.Is there any loop command or shorter way??
  thanks in advanceregardseliza   
  [[alternative HTML version deleted]]

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Re: [R] loop with date

Sorry again, 
project is part of the rgdal package.
Tagmarie



--
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http://r.789695.n4.nabble.com/loop-with-date-tp4650961p4651005.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] loop command to matrix

HI,
May be this helps you:
set.seed(5)
mat1-matrix(sample(1:400,80,replace=TRUE),ncol=8,nrow=10)
 split(mat1,col(mat1))
t(do.call(rbind,lapply(lapply(split(mat1,col(mat1)),function(x) 
cbind(matrix(x,ncol=1),mat1)),function(x){ 
res1-rowMeans(t(apply(x,1,function(x) 
x[!(duplicated(x)|duplicated(x,fromLast=TRUE))])))
  res-x[,1]-res1
  res})))

#   1    2  3  4   5  6
# [1,] -163.0 -129.8571429  152.42857  -52.14286  118.142857  -79.57143
# [2,]  105.14286   16.000  121.14286 -181.71429 -100.571429   50.28571
# [3,]  162.14286 -111.000 -160.14286 -130.42857  121.00  164.42857
# [4,] -123.71429    2.000 -150.0 -239.14286   -9.428571  192.85714
 [5,] -146.42857  -73.2857143 -130.42857 -187.57143  230.714286  233.0
# [6,]   57.28571 -171.2857143  -44.42857  -41.0  -12.428571  -89.0
# [7,]   19.14286  -44.8571429  -23.14286   50.0  125.428571 -105.42857
# [8,]  115.71429  152.2857143  187.71429   19.71429 -222.571429 -136.85714
# [9,]  184.85714    0.8571429 -187.71429  -70.0  110.571429 -162.57143
#[10,] -189.85714  143.8571429  195.28571  -59.57143   49.00 -178.42857
   7  8
# [1,]   30.14286  123.85714
# [2,] -129.14286  118.85714
# [3,]  139.28571 -185.28571
# [4,]  198.57143  128.85714
# [5,]  257.0 -183.0
# [6,]  144.14286  156.71429
## [7,]  183.71429 -204.85714
# [8,] -182.57143   66.57143
# [9,]  173.42857  -49.42857
#[10,]  137.0  -97.28571

A.K.



- Original Message -
From: eliza botto eliza_bo...@hotmail.com
To: r-help@r-project.org r-help@r-project.org
Cc: 
Sent: Tuesday, November 27, 2012 12:51 PM
Subject: [R] loop command to matrix


Dear UseRs,Extremely sorry for a basic question. I have a matrix of 19 rows and 
365 columns. what i want to do is the following...First i want to leave out 
column number 1 and want to calculate the row wise mean of the remaining 
columns, which will obviously give me 365 values in one column, and then 
subtracting these values from the column i left out i.e. col=1 then i want to 
leave out column 2 and calculate the row wise mean of the remaining columns 
which includes column 1 too and then subtracting these values from the column i 
left out i.e. col=2.and then continuing this process the last column. i know a 
kind of manual way of doing things but its extremely long and laborious.Is 
there any loop command or shorter way??
thanks in advanceregardseliza                            
    [[alternative HTML version deleted]]

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Re: [R] aggregate() runs out of memory

2012-11-27 Thread Sam Steingold

 * Steve Lianoglou znvyvatyvfg.ubarl...@tznvy.pbz [2012-11-27 12:53:23 
 -0500]:
 On Tue, Nov 27, 2012 at 11:29 AM, Sam Steingold s...@gnu.org wrote:
 * Steve Lianoglou znvyvatyvfg.ubarl...@tznvy.pbz [2012-11-26 19:47:25 
 -0500]:
 [snip]
 It just occurred to me that this is even better:

 R setkeyv(f, c(share.id, delay))
 R result - f[,  list(min=delay[1L], max=delay[.N], count=.N,
 country=country[1L]), by=share.id]


 this assumes that delays are sorted (like in my example)
 which, in reality, they are not.

 When you include delay in the call to `setkeyv` as I did above, it
 sorts low to high w/in each share.id group.

Ah, but then I would have to _sort_ (~n*log(n)) by delay within each ID
group, while all I care about is min/max (~n).

thanks again!

-- 
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
http://www.childpsy.net/ http://think-israel.org http://truepeace.org
http://thereligionofpeace.com http://mideasttruth.com http://www.memritv.org
If You Want Breakfast In Bed, Sleep In the Kitchen.

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[R] GLM Coding Issue

2012-11-27 Thread Craig P O'Connell



Dear all, 

Â Â  I am having a recurring problem when I attempt to conduct a GLM.Â  Here is 
what I am attempting (with fake data): 
First, I created a txt file, changed the directory in R (to the proper folder 
containing the file) and loaded the file: 

#avoid-read.table(avoid.txt,header=TRUE);avoid 
#Â  treatment feeding avoid noavoid 
#1Â Â  controlÂ  nofeedÂ Â Â Â  1Â Â Â Â  357 
#2Â Â  controlÂ Â Â  feedÂ Â Â Â  2Â Â Â Â  292 
#3Â Â  controlÂ Â Â Â  satÂ Â Â Â  4Â Â Â Â  186 
#4Â Â Â Â Â  procÂ  nofeedÂ Â Â  15Â Â Â Â  291 
#5Â Â Â Â Â  procÂ Â Â  feedÂ Â Â  25Â Â Â Â  288 
#6Â Â Â Â Â  procÂ Â Â Â  satÂ Â Â  17Â Â Â Â  140 
#7Â Â Â Â Â Â  magÂ  nofeedÂ Â Â  87Â Â Â Â  224 
#8Â Â Â Â Â Â  magÂ Â Â  feedÂ Â Â  34Â Â Â Â  229 
#9Â Â Â Â Â Â  magÂ Â Â Â  satÂ Â Â  46Â Â Â Â  151 

I then try to attach(avoid) the data, but continue to get an error message ( 
The following object(s) are masked _by_ .GlobalEnv :), so to fix this, I do the 
following: 

#newavoid-avoid 
#newavoidÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (does this do anything?) 

Lastly, I have several GLM's I wanted to conduct.Â  Please see the following: 

#model1-glm(cbind(avoid, noavoid)~treatment,data=,family=binomial) 

#model2=glm(cbind(avoid, noavoid)~feeding, familiy=binomial) 

#model3=glm(cbind(avoid, noavoid)~treatment+feeding, familiy=binomial) 

After running model1, I receive the error message Error in 
model.frame.default(formula = cbind(avoid, noavoid) ~ treatment,Â  : 
Â  invalid type (list) for variable 'cbind(avoid, noavoid)'.Â  

It would be greatly appreciated if somebody can help me with my coding, as you 
can see I amÂ a novice but doing my best to learn.Â  I figured if I can get 
model1 to run, I should be able to figure out the rest of my models.Â  

Kind Regards! 



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[R] Error: R could not find listDescription

2012-11-27 Thread Jack Bryan


Hi,

I am running R on Win 7. 

I got error for  listDescription(fPortfolio)

Error: could not find function listDescription

What do I need to install for solving this ? 

Any help will be appreciated. 

Thanks
  
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[R] Books for fully understanding internal logics on some packages(quantmod, xts, zoo and chron)

2012-11-27 Thread 박상규

Hello,


I'm very interested in using financial time series data, but I'm a beginner of 
R programming.
I'd like to fully understand internal logics on several time-series related 
packages such as quantmod, xts, zoo, chron, etc.
So, I read some books, 'R Cookbook' and 'Art of R Programming' and another 
simple tutorials.
But I still can't understand grammars of the packages codes.


Could you recommend some other books or educational materials about S/R 
language ? 


Thanks in advance,





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Re: [R] Error: R could not find listDescription


Hello,

There are two packages with a function listDescription, package fBasics 
and package fUtilities. You probably need to install one of them, which 
I can tell.


install.packages('fUtilities')  # for instance, run this only once
library(fUtilities)  # load it into R session


Hope this helps,

Rui Barradas
Em 27-11-2012 19:37, Jack Bryan escreveu:

Hi,

I am running R on Win 7.

I got error for  listDescription(fPortfolio)

Error: could not find function listDescription

What do I need to install for solving this ?

Any help will be appreciated.

Thanks

[[alternative HTML version deleted]]

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


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Re: [R] Error: R could not find listDescription

2012-11-27 Thread Sarah Goslee

Where did you get the document that tells you to use that code? Does
it also tell you to load particular packages?

www.rseek.org turns up a listDescription() function in the fBasics
package, but that isn't necessarily the one you need for whatever
application you're pursuing.

Sarah

On Tue, Nov 27, 2012 at 2:37 PM, Jack Bryan dtustud...@hotmail.com wrote:

 Hi,

 I am running R on Win 7.

 I got error for  listDescription(fPortfolio)

 Error: could not find function listDescription

 What do I need to install for solving this ?

 Any help will be appreciated.

 Thanks


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

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Re: [R] Error: R could not find listDescription

2012-11-27 Thread Jack Bryan


Hi, 

Thanks for your reply.  
I am trying to run R 2.15.2 on Win 7. 

I am trying to run some example R code of the book :Portfolio Optimization with 
R/Rmetrics

fPortfolio package is included in ebookPortfolio. 

From the book, I was told that :

To install all packages required for the examples of this ebook we 
recommendthat you install the bundle package ebookPortfolio. 
This can be done with the following commands in the R environment. If there is 
no binary package for your operating system, you can install the package from 
source with the argument type = source.

 install.packages(ebookPortfolio,repos = 
 c(http://pkg.rmetrics.org,http://cran.r-project.org;),type = 
 getOption(pkgType))

I got :
Warning: unable to access index for repository 
http://pkg.rmetrics.org/src/contribWarning: package ebookPortfolio is not 
available (for R version 2.15.2)Warning: unable to access index for repository 
http://pkg.rmetrics.org/bin/windows/contrib/2.15
Is this an error ? 

I am a new user of R.

How to handle it ? How to install ebookPortfolio package ? 

Any help will be appreciated. 

Thanks 

 Date: Tue, 27 Nov 2012 14:46:53 -0500
 Subject: Re: [R] Error: R could not find listDescription
 From: sarah.gos...@gmail.com
 To: dtustud...@hotmail.com
 CC: r-help@r-project.org
 
 Where did you get the document that tells you to use that code? Does
 it also tell you to load particular packages?
 
 www.rseek.org turns up a listDescription() function in the fBasics
 package, but that isn't necessarily the one you need for whatever
 application you're pursuing.
 
 Sarah
 
 On Tue, Nov 27, 2012 at 2:37 PM, Jack Bryan dtustud...@hotmail.com wrote:
 
  Hi,
 
  I am running R on Win 7.
 
  I got error for  listDescription(fPortfolio)
 
  Error: could not find function listDescription
 
  What do I need to install for solving this ?
 
  Any help will be appreciated.
 
  Thanks
 
 
 --
 Sarah Goslee
 http://www.functionaldiversity.org
  
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Re: [R] binning by frequency

2012-11-27 Thread Santosh

Thanks for your response.
Was wondering if there are any R functions/packages to perform optimal
binning of continuous data.

Thanks, again.
Santosh.

On Tue, Nov 27, 2012 at 9:09 AM, Mark Lamias mlam...@yahoo.com wrote:

 You might find the binning function in the sm package helpful here.

 --Mark Lamias

   --
 *From:* Santosh santosh2...@gmail.com
 *To:* r-help r-help@r-project.org
 *Sent:* Tuesday, November 27, 2012 9:59 AM
 *Subject:* [R] binning by frequency

 Dear Rxperts,

 is there way to identify intervals from continuous data (having some kind
 of a pattern) and then pick the value of most frequency?


 a1 - round(rnorm(50,mean=0,0.1),2)
 a2 - round(rnorm(50,mean=1,0.2),1)
 a3 - round(rnorm(50,mean=5,1),0)
 a4 - round(rnorm(50,mean=14,4),0)
 a5 - round(rnorm(50,mean=30,8),0)

 b1 - rbind(a1,a2,a3,a4,a5)

 hist(b1,brea=100) # shows intervals and values with varying frequency.

 unlike the mean values of a1 a5 above, I don't know the nominal values.
 I would like an algorithm to identify intervals and pick the value with
 most frequency.

 I tried cut, split and was not successful.
 Any suggestions/tips are highly welcome.
 Thanks and regards,
 Santosh

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Re: [R] Some questions about chron package..


Hello,

Inline.
Em 27-11-2012 18:06, 박상규 escreveu:

Hello,


I have questions while reviewing chron package(e.g.,chron.R).


1. What is the differences between 3 kinds of function definition ?
  1) name lt;- function(...
  2) 'name' lt;- function(...
  3) name lt;- function(...
Do you know Why  author used various kinds of  definitions ?
Is there no functional differences between them ?

Not that I know of.




2. I don't understand the meaning of the below code:
  
as.chron lt;- function(x, ...) UseMethod(as.chron)

as.chron.default lt;- function (x, format, ...)
{ 


Could you let me know what is role of UseMethod(as.chron) ?
It creates a generic function named as.chron. The next instruction 
creates the default method for that function.



3. In the following code,



chron lt;-
function(dates. = NULL, times. = NULL,
  format = c(dates = m/d/y, times = h:m:s),
  out.format, origin.)


'dates', 'origin' and 'times' has dot(.) as at the end of word.
Do you know the meaning of dots ?


I am not sure but I believe the programmer is trying to have arguments 
with names not conflicting with classes dates and times, or other 
objects or arguments.





4. Could you introduce me some books on S/R language, especially for 
understanding of generic functions ?
Some interesting packages such as quantmod, xts, zoo, etc., are seems to be 
coded using generic functions(UseMethod(..)).
But I can't easily understand the code.


Try this.

f - function(x) UseMethod(f)
f.default - function(x) print('x' is not of class 'matrix')
f.matrix - function(x) print('x' is of class 'matrix')

f(1)
f(a)
f(matrix(1, ncol=1))
f(list(1, 2, 3))


The S3 object oriented programming system is documented in many places, 
it dispatches based on the class of the first argument so in the third 
case it's f.matrix that is called and in the others, f.default. But you 
need only to call f(args) (no suffix). Common examples are print(), 
summary() and plot(). Any of these functions is generic and methods for 
each class can and are written to them. You can see which methods there 
are by running the command


methods(print)  # 174 in R 2.15.2, new session

The syntax for these methods is function.class. Then you can call 
function that the S3 system will find the appropriate method, if any, 
or will default to function.default. (Like in as.chron.default)

There are free books on CRAN, read them. And google S3 R Programming.

Hope this helps,

Rui Barradas



Thanks in advance,




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Re: [R] Error: R could not find listDescription

2012-11-27 Thread Sarah Goslee

The website given no longer exists.

The usual first approach is to contact the author of the books to ask
if there's a new location.

You could also try installing fPortfolio directly from CRAN:
install.packages(fPortfolio)

Sarah

On Tue, Nov 27, 2012 at 2:57 PM, Jack Bryan dtustud...@hotmail.com wrote:
 Hi,

 Thanks for your reply.

 I am trying to run R 2.15.2 on Win 7.

 I am trying to run some example R code of the book :Portfolio Optimization
 with R/Rmetrics

 fPortfolio package is included in ebookPortfolio.

 From the book, I was told that :

 To install all packages required for the examples of this ebook we
 recommendthat you install the bundle package ebookPortfolio.
 This can be done with the following commands in the R environment. If there
 is no binary package for your operating system, you can install the package
 from source with the argument type = source.

 install.packages(ebookPortfolio,repos =
 c(http://pkg.rmetrics.org,http://cran.r-project.org;),type =
 getOption(pkgType))

 I got :
 Warning: unable to access index for repository
 http://pkg.rmetrics.org/src/contribWarning: package ‘ebookPortfolio’ is not
 available (for R version 2.15.2)Warning: unable to access index for
 repository http://pkg.rmetrics.org/bin/windows/contrib/2.15
 Is this an error ?

 I am a new user of R.

 How to handle it ? How to install ebookPortfolio package ?

 Any help will be appreciated.

 Thanks



 Date: Tue, 27 Nov 2012 14:46:53 -0500
 Subject: Re: [R] Error: R could not find listDescription
 From: sarah.gos...@gmail.com
 To: dtustud...@hotmail.com
 CC: r-help@r-project.org

 Where did you get the document that tells you to use that code? Does
 it also tell you to load particular packages?

 www.rseek.org turns up a listDescription() function in the fBasics
 package, but that isn't necessarily the one you need for whatever
 application you're pursuing.

 Sarah

 On Tue, Nov 27, 2012 at 2:37 PM, Jack Bryan dtustud...@hotmail.com
 wrote:
 
  Hi,
 
  I am running R on Win 7.
 
  I got error for  listDescription(fPortfolio)
 
  Error: could not find function listDescription
 
  What do I need to install for solving this ?
 
  Any help will be appreciated.
 
  Thanks
 


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

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Error: R could not find listDescription


Hello,

Inline.
Em 27-11-2012 19:44, Rui Barradas escreveu:

Hello,

There are two packages with a function listDescription, package 
fBasics and package fUtilities. You probably need to install one of 
them, which I can tell.


Sorry, it's obviously can't

Rui Barradas


install.packages('fUtilities')  # for instance, run this only once
library(fUtilities)  # load it into R session


Hope this helps,

Rui Barradas
Em 27-11-2012 19:37, Jack Bryan escreveu:

Hi,

I am running R on Win 7.

I got error for  listDescription(fPortfolio)

Error: could not find function listDescription

What do I need to install for solving this ?

Any help will be appreciated.

Thanks

[[alternative HTML version deleted]]

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Re: [R] binning by frequency

2012-11-27 Thread William Dunlap

You might look at the 'mdlp' function in the 'discretization' package.
(SPSS has a procedure called 'optimal binning' that uses the 'minimum
description length principle' to do the binning.)

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf
 Of Santosh
 Sent: Tuesday, November 27, 2012 11:59 AM
 To: r-help
 Subject: Re: [R] binning by frequency
 
 Thanks for your response.
 Was wondering if there are any R functions/packages to perform optimal
 binning of continuous data.
 
 Thanks, again.
 Santosh.
 
 On Tue, Nov 27, 2012 at 9:09 AM, Mark Lamias mlam...@yahoo.com wrote:
 
  You might find the binning function in the sm package helpful here.
 
  --Mark Lamias
 
--
  *From:* Santosh santosh2...@gmail.com
  *To:* r-help r-help@r-project.org
  *Sent:* Tuesday, November 27, 2012 9:59 AM
  *Subject:* [R] binning by frequency
 
  Dear Rxperts,
 
  is there way to identify intervals from continuous data (having some kind
  of a pattern) and then pick the value of most frequency?
 
 
  a1 - round(rnorm(50,mean=0,0.1),2)
  a2 - round(rnorm(50,mean=1,0.2),1)
  a3 - round(rnorm(50,mean=5,1),0)
  a4 - round(rnorm(50,mean=14,4),0)
  a5 - round(rnorm(50,mean=30,8),0)
 
  b1 - rbind(a1,a2,a3,a4,a5)
 
  hist(b1,brea=100) # shows intervals and values with varying frequency.
 
  unlike the mean values of a1 a5 above, I don't know the nominal values.
  I would like an algorithm to identify intervals and pick the value with
  most frequency.
 
  I tried cut, split and was not successful.
  Any suggestions/tips are highly welcome.
  Thanks and regards,
  Santosh
 
  [[alternative HTML version deleted]]
 
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  PLEASE do read the posting guide
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  and provide commented, minimal, self-contained, reproducible code.
 
 
 
 
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Re: [R] Error: R could not find listDescription

2012-11-27 Thread Nordlund, Dan (DSHS/RDA)

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of Sarah Goslee
 Sent: Tuesday, November 27, 2012 12:10 PM
 To: Jack Bryan
 Cc: r help
 Subject: Re: [R] Error: R could not find listDescription

 The website given no longer exists.

 The usual first approach is to contact the author of the books to ask
 if there's a new location.

 You could also try installing fPortfolio directly from CRAN:
 install.packages(fPortfolio)

 Sarah

 On Tue, Nov 27, 2012 at 2:57 PM, Jack Bryan dtustud...@hotmail.com
 wrote:
  Hi,

  Thanks for your reply.

  I am trying to run R 2.15.2 on Win 7.

  I am trying to run some example R code of the book :Portfolio
 Optimization
  with R/Rmetrics

  fPortfolio package is included in ebookPortfolio.

  From the book, I was told that :

  To install all packages required for the examples of this ebook we
  recommendthat you install the bundle package ebookPortfolio.
  This can be done with the following commands in the R environment. If
 there
  is no binary package for your operating system, you can install the
 package
  from source with the argument type = source.

  install.packages(ebookPortfolio,repos =
  c(http://pkg.rmetrics.org,http://cran.r-project.org;),type =
  getOption(pkgType))

  I got :
  Warning: unable to access index for repository
  http://pkg.rmetrics.org/src/contribWarning: package ‘ebookPortfolio’
 is not
  available (for R version 2.15.2)Warning: unable to access index for
  repository http://pkg.rmetrics.org/bin/windows/contrib/2.15
  Is this an error ?

  I am a new user of R.

  How to handle it ? How to install ebookPortfolio package ?

  Any help will be appreciated.

  Thanks

  Date: Tue, 27 Nov 2012 14:46:53 -0500
  Subject: Re: [R] Error: R could not find listDescription
  From: sarah.gos...@gmail.com
  To: dtustud...@hotmail.com
  CC: r-help@r-project.org

  Where did you get the document that tells you to use that code? Does
  it also tell you to load particular packages?

  www.rseek.org turns up a listDescription() function in the fBasics
  package, but that isn't necessarily the one you need for whatever
  application you're pursuing.

  Sarah

  On Tue, Nov 27, 2012 at 2:37 PM, Jack Bryan dtustud...@hotmail.com
  wrote:

   Hi,

   I am running R on Win 7.

   I got error for  listDescription(fPortfolio)

   Error: could not find function listDescription

   What do I need to install for solving this ?

   Any help will be appreciated.

   Thanks

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

The website/URL below explains how to install the Rmetrics software from your 
local CRAN repository.

https://wiki.rmetrics.org/install_rmetrics

It seemed to work ok for me.

Hope this is helpful,

Dan

Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204

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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.

Re: [R] Books for fully understanding internal logics on some packages(quantmod, xts, zoo and chron)

2012-11-27 Thread Patrick Burns


One place to look would be the
archives of the r-sig-finance list.

A blog post with suggestions on how to
achieve that is:

http://www.portfolioprobe.com/2012/01/19/how-to-search-the-r-sig-finance-archives/

Pat


On 27/11/2012 19:41, 박상규 wrote:

Hello,


I'm very interested in using financial time series data, but I'm a beginner of 
R programming.
I'd like to fully understand internal logics on several time-series related 
packages such as quantmod, xts, zoo, chron, etc.
So, I read some books, 'R Cookbook' and 'Art of R Programming' and another 
simple tutorials.
But I still can't understand grammars of the packages codes.


Could you recommend some other books or educational materials about S/R 
language ?


Thanks in advance,





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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.



--
Patrick Burns
pbu...@pburns.seanet.com
twitter: @portfolioprobe
http://www.portfolioprobe.com/blog
http://www.burns-stat.com
(home of 'Some hints for the R beginner'
and 'The R Inferno')

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Re: [R] problem with svyglm

2012-11-27 Thread Pablo Menese

I colud not, even without attach the dataset.

The thing is, when I use this





On Fri, Nov 23, 2012 at 5:56 PM, David Winsemius dwinsem...@comcast.netwrote:


 On Nov 23, 2012, at 12:08 PM, Pablo Menese wrote:

  I have this problem.

 test - svydesign(id=~1,weights=~peso)

 logit - svyglm(bach ~ job2 + mujer + egp4 + programa + delay + mdeo + str
 + evprivate, family=binomial,design=test)

 then appear:

 Error in svyglm.survey.design(bach ~ job2 + mujer + egp4 + programa +  :
  all variables must be in design= argument

 I don't know what this mean...


 I suspect you have attach()-ed your dataset and are expecting regression
 functions to be aware of your column names. That expectation doesn't
 always get fulfilled since the authrs of regression packages are expecting
 dataframe arguments to be supplied. You may want to detach the dataset and
 use data= arguments in svydesign().

 You should forget that you ever heard about the function attach().

 --

 David Winsemius, MD
 Alameda, CA, USA



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Re: [R] problem with svyglm

2012-11-27 Thread Pablo Menese

Sorry, it send it alone...

When I use it:

logit - glm(bach ~ egp4 + programa, weight=wst7,
family=quasibinomial(linklogit))

I reach the same betas that in STATA, but the hypothesis test, the t value,
and the std. error is different.

I think that the solution can't be so far from this...

[[alternative HTML version deleted]]

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Re: [R] Problems with weight

2012-11-27 Thread Pablo Menese

I can't ... I don't know why but I can't

When I use it:

logit - glm(bach ~ egp4 + programa, weight=wst7,
family=quasibinomial(linklogit))

I reach the same betas that in STATA, but the hypothesis test, the t value,
and the std. error is different.

I think that the solution can't be so far from this...


On Fri, Nov 23, 2012 at 9:49 PM, Anthony Damico ajdam...@gmail.com wrote:

 from your stata output, it looks like you need to use the survey package
 in R

 for step-by-step instructions about how to do this (and comparisons to
 stata), see

 http://journal.r-project.org/archive/2009-2/RJournal_2009-2_Damico.pdf

 once you're ready to run the regression, use svyglm() instead of glm() and
 drop the weights argument (since it will already be part of the survey
 design)   :)



 On Fri, Nov 23, 2012 at 3:13 PM, Pablo Menese pmen...@gmail.com wrote:

 Until a weeks ago I used stata for everything.
 Now I'm learning R and trying to move. But, in this stage I'm testing R
 trying to do the same things than I used to do in stata whit the same
 outputs.
 I have a problem with the logit, applying weights.

 in stata I have this output
 . svy: logit bach job2 mujer i.egp4 programa delay mdeo i.str evprivate
 (running logit on estimation sample)

 Survey: Logistic regression

 Number of strata   = 1  Number of obs  =
 248
 Number of PSUs =   248  Population size=
 5290.1639
 Design df  =   247
 F(  11,237)=  4.39
 Prob  F   =0.


 Linearized
 bach   Coef.   Std. Err.  tPt [95% Conf. Interval]

 job2   -.4437446   .4385934-1.01   0.313-1.307605.4201154
 mujer1.070595   .4169919 2.57   0.011 .24928121.891908

 egp4
 2-.4839342.539808-0.90   0.371-1.547148.5792796
 3-1.288947   .5347344-2.41   0.017-2.342168   -.2357263
 4-.8569793   .5106425-1.68   0.095-1.862748.1487898

 programa.9694352   .5677642 1.71   0.089-.14884152.087712
 delay   -1.552582   .5714967-2.72   0.007-2.678211-.426954
 mdeo   -.7938904   .3727571-2.13   0.034-1.528078   -.0597025

 str
 2-1.122691   .5731879-1.96   0.051 -2.25165.0062682
 3-2.056682   .6350485-3.24   0.001-3.307483   -.8058812

 evprivate   -1.962431   .5674143-3.46   0.001-3.080018   -.8448431
 _cons2.308699   .7274924 3.17   0.002 .87581873.741578


 the best that i get in R was:

 glm(formula = bach ~ job2 + mujer + egp4 + programa + delay +
 mdeo + str + evprivate, family = quasibinomial(link = logit),
 weights = wst7)

 Deviance Residuals:
  Min1QMedian3Q   Max
 -12.5951   -3.9034   -0.94123.8268   11.2750

 Coefficients:
Estimate Std. Error t value Pr(|t|)
 (Intercept)  2.3087 0.7173   3.218  0.00147 **
 job2-0.4437 0.4355  -1.019  0.30926
 mujer1.0706 0.3558   3.009  0.00290 **
 egp4intermediate (iii, iv)  -0.4839 0.4946  -0.978  0.32890
 egp4skilled manual workers  -1.2889 0.5268  -2.447  0.01514 *
 egp4working class   -0.8570 0.4625  -1.853  0.06514 .
 programa 0.9694 0.4951   1.958  0.05141 .
 delay   -1.5526 0.4878  -3.183  0.00166 **
 mdeo-0.7939 0.4207  -1.887  0.06037 .
 strest. ii  -1.1227 0.4809  -2.334  0.02042 *
 strestr. iii-2.0567 0.5134  -4.006 8.28e-05 ***
 evprivate   -1.9624 0.6490  -3.024  0.00277 **
 ---
 Signif. codes:  0 *** 0.001 ** 0.01 * 0.05 . 0.1   1

 (Dispersion parameter for quasibinomial family taken to be 23.14436)

 Null deviance: 7318.5  on 246  degrees of freedom
 Residual deviance: 5692.8  on 235  degrees of freedom
   (103 observations deleted due to missingness)
 AIC: NA

 Number of Fisher Scoring iterations: 6

 Warning message:
 In summary.glm(logit) :
   observations with zero weight not used for calculating dispersion

 this has the same betas but the hypothesis test has differents values...


 HELP

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Re: [R] Stuck trying to modify a function

2012-11-27 Thread Milan Bouchet-Valat

Le mardi 27 novembre 2012 à 16:45 +, Benjamin Ward (ENV) a écrit :
 Hi,
 
 I have the following data:
 
 Path_Number - 5
 ID.Path - c(1:Path_Number) # Make vector of ID's.
 No_of_X - sample(50:550, length(ID.Path), replace=TRUE) #
 X - split(sample(1:1, sum(No_of_X), replace=TRUE), rep(ID.Path,
 No_of_X))
 Y - lapply(X,function(x) sample(x, round(runif(1, min=10, max=50
 
 X and Y are both lists, and I've made the following function to work
 on that data as part of a simulation I'm building:
 
 Mutate-function(x){
   l-0
   for(i in x){
 l2-0
 l-l+1
 for(i in x[[l]]){
   l2-l2+1
   if(runif(1)  0.9) ifelse(runif(1) 0.5, x[[l]][l2] -
 x[[l]][l2]+1, x[[l]][l2] - x[[l]][l2]-1)
 }
   }
   return(x)
 }
 
 I call this with Effectors-Mutate(X)
 The function is designed to alter the values of each element in X by
 either + or - 1 (50:50 chance wether + or -). However Y, elements of
 which are a subset of the corresponding elements of X, need to be
 consistent i.e. if a value in X is changed, and that value is part of
 the Y subset, then the value in Y also needs to be changed. however,
 since Y is a smaller subset it will not be indexed the same. My idea
 was to include in the function an if statement that checks if Y
 contains the value to be changed, removes it, and then after the value
 in X is changed, put the new value in Y. I attempted this with:
You should really read about vectorizing operations: you can most likely
get the same results in R using only a few lines, with a much better
performance. runif(length(x)) will directly give you a vector of the
needed length, and you can add or subtract 1 from x in one line:

new.x - ifelse(runif(length(x))  .5, x + 1, x - 1)
y[match(x, y, nomatch=0)] - new.x
y - ifelse(y %in% x, new.x[match(y, x)], y)
x - new.x

This is of course a proof of concept, I'm not sure this is really what
you asked for. See below for some debugging of your code.

 
 
 Mutate-function(x,y){
   l-0
   for(i in x){
 l2-0
 l-l+1
 for(i in x[[l]]){
   l2-l2+1
   if(runif(1)  0.9){
 if(x[[l]][l2] %in% y[[l]] == TRUE){
   y[[l]]-[which(y[[l]]!=x[[l]][l2])]
Bug is here:   ^

You should specify an object to index.

   if(runif(1) 0.5){
 x[[l]][l2] - x[[l]][l2]+1
 y[[l]]-append(x[[l]][l2])
   }else{
 x[[l]][l2] - x[[l]][l2]-1
 y[[l]]-append(x[[l]][l2])
   }
 }
 ifelse(runif(1) 0.5, x[[l]][l2] - x[[l]][l2]+1, x[[l]][l2]
 - x[[l]][l2]-1)
   }
 }
   }
   return(list(x,y))
 }
 
 Bit of an eyesore so I've put the altered stuff in bold. I've
 basically taken what the ifelse statement does in the first function,
 (which is still there and run if Y does not contain the X value being
 altered) and broken it down into an if and an else segment with
 multiple operations in curly braces to accommodate the extra actions
 needed to alter Y as well as X. This was all I could think of to keep
 changes between the two in sync, however this does not work when I
 try to load the function into workspace:
 
 Error: unexpected '}' in }
You should have copied the full output. This is only the last error
message: once an error happens, the whole syntax is broken and every
bracket can trigger an error. Only by looking at the first error you can
understand what's the problem.


Regards

 I hope someone can point out what it is I've done that isn't working,
 or a better way to do this.
 
 Best Wishes,
 
 Ben W.
 
 UEA (ENV)  The Sainsbury Laboratory.
 
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Re: [R] Problems with weight

2012-11-27 Thread Milan Bouchet-Valat

Le mardi 27 novembre 2012 à 18:33 -0300, Pablo Menese a écrit :
 I can't ... I don't know why but I can't
 
 When I use it:
 
 logit - glm(bach ~ egp4 + programa, weight=wst7,
 family=quasibinomial(linklogit))
You were advised to use svyglm(), not glm(). It's usually considered
polite to read carefully the anwsers you get to your questions...


Regards

 I reach the same betas that in STATA, but the hypothesis test, the t value,
 and the std. error is different.
 
 I think that the solution can't be so far from this...
 
 
 On Fri, Nov 23, 2012 at 9:49 PM, Anthony Damico ajdam...@gmail.com wrote:
 
  from your stata output, it looks like you need to use the survey package
  in R
 
  for step-by-step instructions about how to do this (and comparisons to
  stata), see
 
  http://journal.r-project.org/archive/2009-2/RJournal_2009-2_Damico.pdf
 
  once you're ready to run the regression, use svyglm() instead of glm() and
  drop the weights argument (since it will already be part of the survey
  design)   :)
 
 
 
  On Fri, Nov 23, 2012 at 3:13 PM, Pablo Menese pmen...@gmail.com wrote:
 
  Until a weeks ago I used stata for everything.
  Now I'm learning R and trying to move. But, in this stage I'm testing R
  trying to do the same things than I used to do in stata whit the same
  outputs.
  I have a problem with the logit, applying weights.
 
  in stata I have this output
  . svy: logit bach job2 mujer i.egp4 programa delay mdeo i.str evprivate
  (running logit on estimation sample)
 
  Survey: Logistic regression
 
  Number of strata   = 1  Number of obs  =
  248
  Number of PSUs =   248  Population size=
  5290.1639
  Design df  =   247
  F(  11,237)=  4.39
  Prob  F   =0.
 
 
  Linearized
  bach   Coef.   Std. Err.  tPt [95% Conf. Interval]
 
  job2   -.4437446   .4385934-1.01   0.313-1.307605.4201154
  mujer1.070595   .4169919 2.57   0.011 .24928121.891908
 
  egp4
  2-.4839342.539808-0.90   0.371-1.547148.5792796
  3-1.288947   .5347344-2.41   0.017-2.342168   -.2357263
  4-.8569793   .5106425-1.68   0.095-1.862748.1487898
 
  programa.9694352   .5677642 1.71   0.089-.14884152.087712
  delay   -1.552582   .5714967-2.72   0.007-2.678211-.426954
  mdeo   -.7938904   .3727571-2.13   0.034-1.528078   -.0597025
 
  str
  2-1.122691   .5731879-1.96   0.051 -2.25165.0062682
  3-2.056682   .6350485-3.24   0.001-3.307483   -.8058812
 
  evprivate   -1.962431   .5674143-3.46   0.001-3.080018   -.8448431
  _cons2.308699   .7274924 3.17   0.002 .87581873.741578
 
 
  the best that i get in R was:
 
  glm(formula = bach ~ job2 + mujer + egp4 + programa + delay +
  mdeo + str + evprivate, family = quasibinomial(link = logit),
  weights = wst7)
 
  Deviance Residuals:
   Min1QMedian3Q   Max
  -12.5951   -3.9034   -0.94123.8268   11.2750
 
  Coefficients:
 Estimate Std. Error t value Pr(|t|)
  (Intercept)  2.3087 0.7173   3.218  0.00147 **
  job2-0.4437 0.4355  -1.019  0.30926
  mujer1.0706 0.3558   3.009  0.00290 **
  egp4intermediate (iii, iv)  -0.4839 0.4946  -0.978  0.32890
  egp4skilled manual workers  -1.2889 0.5268  -2.447  0.01514 *
  egp4working class   -0.8570 0.4625  -1.853  0.06514 .
  programa 0.9694 0.4951   1.958  0.05141 .
  delay   -1.5526 0.4878  -3.183  0.00166 **
  mdeo-0.7939 0.4207  -1.887  0.06037 .
  strest. ii  -1.1227 0.4809  -2.334  0.02042 *
  strestr. iii-2.0567 0.5134  -4.006 8.28e-05 ***
  evprivate   -1.9624 0.6490  -3.024  0.00277 **
  ---
  Signif. codes:  0 *** 0.001 ** 0.01 * 0.05 . 0.1   1
 
  (Dispersion parameter for quasibinomial family taken to be 23.14436)
 
  Null deviance: 7318.5  on 246  degrees of freedom
  Residual deviance: 5692.8  on 235  degrees of freedom
(103 observations deleted due to missingness)
  AIC: NA
 
  Number of Fisher Scoring iterations: 6
 
  Warning message:
  In summary.glm(logit) :
observations with zero weight not used for calculating dispersion
 
  this has the same betas but the hypothesis test has differents values...
 
 
  HELP
 
  [[alternative HTML version deleted]]
 
 
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Re: [R] Stuck trying to modify a function

Ben,

You can use the sample() function to randomly add -1, 0, or 1 to each 
observation, and control for the probability of mutation at the same time. 
 Then you can use the match() function to make sure that any mutations in 
X are carried through to Y in the same way.  I wrote the function to do 
each list element separately.  So a gene in X[[1] and Y[[1]] will be 
mutated in the same way, but the same gene in X[[2]] and Y[[2]] may be 
mutated in a different way.  Not sure if that is what you want.

Mutate - function(x, y, prob.mutate=0.9) {
ux - unique(c(x, y))
new.ux - ux + sample(c(-1, 0, 1), size=length(ux), replace=TRUE, 
prob=c(prob.mutate/2, 1-prob.mutate, prob.mutate/2))
new.x - new.ux[match(x, ux)]
new.y - new.ux[match(y, ux)]
list(xm=new.x, ym=new.y)
}

Effectors - lapply(seq(X), function(i) Mutate(X[[i]], Y[[i]]))

Jean



Benjamin Ward (ENV) b.w...@uea.ac.uk wrote on 11/27/2012 10:45:23 AM:
 
 Hi,
 
 I have the following data:
 
 Path_Number - 5
 ID.Path - c(1:Path_Number) # Make vector of ID's.
 No_of_X - sample(50:550, length(ID.Path), replace=TRUE) #
 X - split(sample(1:1, sum(No_of_X), replace=TRUE), 
rep(ID.Path,No_of_X))
 Y - lapply(X,function(x) sample(x, round(runif(1, min=10, max=50
 
 X and Y are both lists, and I've made the following function to work
 on that data as part of a simulation I'm building:
 
 Mutate-function(x){
   l-0
   for(i in x){
 l2-0
 l-l+1
 for(i in x[[l]]){
   l2-l2+1
   if(runif(1)  0.9) ifelse(runif(1) 0.5, x[[l]][l2] - x[[l]]
 [l2]+1, x[[l]][l2] - x[[l]][l2]-1)
 }
   }
   return(x)
 }
 
 I call this with Effectors-Mutate(X)
 The function is designed to alter the values of each element in X by
 either + or - 1 (50:50 chance wether + or -). However Y, elements of
 which are a subset of the corresponding elements of X, need to be 
 consistent i.e. if a value in X is changed, and that value is part 
 of the Y subset, then the value in Y also needs to be changed. 
 however, since Y is a smaller subset it will not be indexed the 
 same. My idea was to include in the function an if statement that 
 checks if Y contains the value to be changed, removes it, and then 
 after the value in X is changed, put the new value in Y. I attemptedthis 
with:
 
 
 Mutate-function(x,y){
   l-0
   for(i in x){
 l2-0
 l-l+1
 for(i in x[[l]]){
   l2-l2+1
   if(runif(1)  0.9){
 if(x[[l]][l2] %in% y[[l]] == TRUE){
   y[[l]]-[which(y[[l]]!=x[[l]][l2])]
   if(runif(1) 0.5){
 x[[l]][l2] - x[[l]][l2]+1
 y[[l]]-append(x[[l]][l2])
   }else{
 x[[l]][l2] - x[[l]][l2]-1
 y[[l]]-append(x[[l]][l2])
   }
 }
 ifelse(runif(1) 0.5, x[[l]][l2] - x[[l]][l2]+1, x[[l]][l2]
 - x[[l]][l2]-1)
   }
 }
   }
   return(list(x,y))
 }
 
 Bit of an eyesore so I've put the altered stuff in bold. I've 
 basically taken what the ifelse statement does in the first 
 function, (which is still there and run if Y does not contain the X 
 value being altered) and broken it down into an if and an else 
 segment with multiple operations in curly braces to accommodate the 
 extra actions needed to alter Y as well as X. This was all I could 
 think of to keep changes between the two in sync, however this 
 does not work when I try to load the function into workspace:
 
 Error: unexpected '}' in }
 
 I hope someone can point out what it is I've done that isn't 
 working, or a better way to do this.
 
 Best Wishes,
 
 Ben W.
 
 UEA (ENV)  The Sainsbury Laboratory.

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Re: [R] Some questions about chron package..

2012-11-27 Thread Jeff Newmiller

1. I am not aware of a difference, and don't know why the various forms were 
used.

2. That handles identifying the correct function to call based on the types of 
arguments supplied when the function was called. Read about the S4 
object-oriented programming features to learn about method dispatch.

3. It has no syntactic effect. Using periods in parameter names may make it 
less likely to be the same as names defined by the user, but only if the user 
cooperates by avoiding that naming convention.

4. There is some quite good documentation supplied with R. You can also look on 
CRAN (http://www.r-project.org/doc/bib/R-publications.html).

Please read the posting guide, and post in plain text.
---
Jeff NewmillerThe .   .  Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#.   ##.#.  Live Go...
  Live:   OO#.. Dead: OO#..  Playing
Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
/Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
--- 
Sent from my phone. Please excuse my brevity.

박상규 birdfir...@naver.com wrote:

Hello,


I have questions while reviewing chron package(e.g.,chron.R).


1. What is the differences between 3 kinds of function definition ? 
 1) name lt;- function(...
 2) 'name' lt;- function(...
 3) name lt;- function(...
Do you know Why  author used various kinds of  definitions ? 
Is there no functional differences between them ? 


2. I don't understand the meaning of the below code:
 
as.chron lt;- function(x, ...) UseMethod(as.chron)
as.chron.default lt;- function (x, format, ...)
{ 


Could you let me know what is role of UseMethod(as.chron) ? 


3. In the following code,



chron lt;-
function(dates. = NULL, times. = NULL,
 format = c(dates = m/d/y, times = h:m:s),
 out.format, origin.)


'dates', 'origin' and 'times' has dot(.) as at the end of word.
Do you know the meaning of dots ? 


4. Could you introduce me some books on S/R language, especially for
understanding of generic functions ? 
Some interesting packages such as quantmod, xts, zoo, etc., are seems
to be coded using generic functions(UseMethod(..)).
But I can't easily understand the code.


Thanks in advance,




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[R] Finding values in one column and

2012-11-27 Thread Steven Ranney

All -

I have a data frame

data.a
ID  valueA  valueB
6   12  12
17  15  14
58  18  16
98  11  12
73  19  20
84  19  14
58  20  14
24  11  12
81  15  16
21  15  14
62  14  12
67  13  14
78  13  17
35  10  13
13  11  15
14  17  18
85  16  15
35  13  9
18  15  16

and a data frame

data.b
ID  valueA  valueB
6   
84  
21  
78  
14  

I'd like to have R find the data.b$ID in data.a$ID and insert the
corresponding data.a$valueA and data.a$valueB into the appropriate
columns in data.b.

How can I do this?

Thanks for you help.

SR
Steven H. Ranney

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Re: [R] R strange behaviour when building huge concatenation

I am currently getting the very strange results that if I paste your orginal 
data from your first message into my R terminal I get the same errors you do. 
By the way that ; is not needed in R. 

If I paste the same data into Rstudo, either into the editor or the console it 
works fine.

If I paste the data into gedit and send it to the console it works fine but if 
I paste it into the gedita console I get your error messages again. A quick 
try, pasting it into an R buffer in EMACS seems to work just fine.

I have no idea what is happening unless jagat.k.sheth is correct and we are 
hitting a buffer limit of some kind.

I'd suggest getting a decent editor and going with it.  Working directly in an 
R terminal is enough to drive most people crazy.  

Any of the editiors/ides mentioned above are good with different strengtsts 
etc.  At a guess, Rstudio is the easiest to install and get running on Ubuntu, 
gedit is pretty easy but you need to install the r gedit plug-in. EMACS as far 
as I can tell is very good and very powerful but I have not used it enough to 
really comment on it.

In any case one way or the other they can read in the data.  I'd also suggest 
not even 'thinking' about writing a vector statement that long. It would be 
much easier to do something like write the numbers in a column in a spreadsheet 
and import from there. There are various ways to do this but the simplist would 
be to just save the file as a csv file and import it using read.table or 
read.csv.  

I'm  sorry that I cannot be of more help. Perhaps one of the R gurus can 
comment on the possible buffer problem.

John Kane
Kingston ON Canada


 -Original Message-
 From: angelo...@gmail.com
 Sent: Tue, 27 Nov 2012 07:04:13 -0800 (PST)
 To: r-help@r-project.org
 Subject: Re: [R] R strange behaviour when building huge concatenation
 
 Hello John
 It seems correct to me too but in my R console it seems to not be working
 Here there is what I did:
 i copied the statement on one row (leaving and removing the final useless
 semi colomn)
 i tried to execute it in the R console (in order to open my R console I
 simply opened a terminal window on my ubuntu machine and I typed R)
 when I click enter the inserted statement doesn't not run...I simply
 see
 the cursor on a new line and this new line starts with +
 
 Here there is a little stack of what I see on my terminal:
 , 1, 1, 2, 2, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 17, 2, 2,
 0,
 2, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 1, 0,
 0,
 1, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 7, 0, 0, 1, 0, 0, 3, 0, 0, 2,
 4,
 0, 1, 0, 0, 0, 0, 2, 1, 0, 4, 2, 0, 0, 1, 3, 0, 0, 1, 0, 0, 1, 0, 1, 1,
 0,
 0, 0, 0, 0, 0, 0, 0, 0, 2, 3, 0, 1, 0, 1, 1, 3, 3, 0, 0, 0, 0, 1, 0, 2,
 0,
 1, 0, 0, 0, 1, 1, 0, 2, 0, 2, 0, 3, 0, 4, 0, 1, 1, 0, 0, 1, 0, 3, 0, 0,
 0,
 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 3, 0, 6, 0, 0, 0, 0, 3, 0, 0, 0, 2, 5,
 0,
 1, 0, 0, 1, 0, 1, 0, 0, 2, 0, 1, 2, 0, 0, 7, 0, 0, 2, 0, 2, 2, 0, 0, 0,
 0,
 0, 2, 0, 8, 0, 0, 5, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 3, 0, 2, 0, 0, 0, 2,
 1,
 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 2, 0, 0, 1, 0, 1, 2, 0, 0, 1, 0, 2, 0,
 0,
 3, 0, 0, 0, 1, 0, 0, 0, 1, 0, 9, 3, 0, 0, 0, 0, 0, 0, 3, 0, 0, 2, 0, 1,
 1,
 1, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0,
 0,
 1, 4, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 2, 0, 1, 0,
 0,
 6, 0, 0, 0, 0, 1, 0, 0, 2, 6, 0, 1, 0, 0, 0, 0, 2, 0, 7, 0, 1, 2, 1, 1,
 1,
 0, 0, 1, 5, 0, 1, 0, 0, 2, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 5, 1,
 0,
 1, 2, 0, 0, 1, 2, 0, 1, 0, 2, 0, 1, 2, 7, 1, 2, 0, 0, 0, 5, 0, 4, 0))
 +
 +
 +
 +
 +
 
 Each time i click enter i go on a new line starting with +
 Then if i start in tying some character I get an error like this:
 
 +
 +
 + q
 +
 + q
 Errore: unexpected symbol in:
 
 q
 
 Sometimes this error (Errore: unexpected symbol in:) appears in the
 statement execution
 
 Here there are my sessionInfo() result:
 sessionInfo()
 R version 2.14.1 (2011-12-22)
 Platform: i686-pc-linux-gnu (32-bit)
 
 locale:
  [1] LC_CTYPE=it_IT.UTF-8   LC_NUMERIC=C
  [3] LC_TIME=it_IT.UTF-8LC_COLLATE=it_IT.UTF-8
  [5] LC_MONETARY=it_IT.UTF-8LC_MESSAGES=it_IT.UTF-8
  [7] LC_PAPER=C LC_NAME=C
  [9] LC_ADDRESS=C   LC_TELEPHONE=C
 [11] LC_MEASUREMENT=it_IT.UTF-8 LC_IDENTIFICATION=C
 
 attached base packages:
 [1] stats graphics  grDevices utils datasets  methods   base
 
 
 I really don't know why this happensabove all because it seems to me
 a
 very very simple statement...
 Thank to all you for the support
 
 
 
 
 --
 View this message in context:
 http://r.789695.n4.nabble.com/R-strange-behaviour-when-building-huge-concatenation-tp4650817p4650970.html
 Sent from the R help mailing list archive at Nabble.com.
 
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 PLEASE do read the posting guide

Re: [R] Finding values in one column and

Have a look at ?merge

John Kane
Kingston ON Canada

 -Original Message-
 From: steven.ran...@gmail.com
 Sent: Tue, 27 Nov 2012 15:17:58 -0700
 To: r-help@r-project.org
 Subject: [R] Finding values in one column and

 All -

 I have a data frame

 data.a
 IDvalueA  valueB
 6 12  12
 1715  14
 5818  16
 9811  12
 7319  20
 8419  14
 5820  14
 2411  12
 8115  16
 2115  14
 6214  12
 6713  14
 7813  17
 3510  13
 1311  15
 1417  18
 8516  15
 3513  9
 1815  16

 and a data frame

 data.b
 IDvalueA  valueB
 6
 84
 21
 78
 14

 I'd like to have R find the data.b$ID in data.a$ID and insert the
 corresponding data.a$valueA and data.a$valueB into the appropriate
 columns in data.b.

 How can I do this?

 Thanks for you help.

 SR
 Steven H. Ranney

 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide
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Publish your photos in seconds for FREE
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Re: [R] Finding values in one column and

2012-11-27 Thread Steven Ranney

Thanks.

Soon after I posted this question, I discovered merge().

Steven H. Ranney


On Tue, Nov 27, 2012 at 3:26 PM, John Kane jrkrid...@inbox.com wrote:
 Have a look at ?merge

 John Kane
 Kingston ON Canada


 -Original Message-
 From: steven.ran...@gmail.com
 Sent: Tue, 27 Nov 2012 15:17:58 -0700
 To: r-help@r-project.org
 Subject: [R] Finding values in one column and

 All -

 I have a data frame

 data.a
 IDvalueA  valueB
 6 12  12
 1715  14
 5818  16
 9811  12
 7319  20
 8419  14
 5820  14
 2411  12
 8115  16
 2115  14
 6214  12
 6713  14
 7813  17
 3510  13
 1311  15
 1417  18
 8516  15
 3513  9
 1815  16

 and a data frame

 data.b
 IDvalueA  valueB
 6
 84
 21
 78
 14

 I'd like to have R find the data.b$ID in data.a$ID and insert the
 corresponding data.a$valueA and data.a$valueB into the appropriate
 columns in data.b.

 How can I do this?

 Thanks for you help.

 SR
 Steven H. Ranney

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

 
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Re: [R] GLM Coding Issue

2012-11-27 Thread Steve Lianoglou

Hi,

Comments inline:

On Tue, Nov 27, 2012 at 1:00 PM, Craig P O'Connell
coconne...@umassd.edu wrote:


 Dear all,

I am having a recurring problem when I attempt to conduct a GLM.  Here is 
 what I am attempting (with fake data):
 First, I created a txt file, changed the directory in R (to the proper folder 
 containing the file) and loaded the file:

 #avoid-read.table(avoid.txt,header=TRUE);avoid
 #  treatment feeding avoid noavoid
 #1   control  nofeed 1 357
 #2   controlfeed 2 292
 #3   control sat 4 186
 #4  proc  nofeed15 291
 #5  procfeed25 288
 #6  proc sat17 140
 #7   mag  nofeed87 224
 #8   magfeed34 229
 #9   mag sat46 151

 I then try to attach(avoid) the data, but continue to get an error message 
 ( The following object(s) are masked _by_ .GlobalEnv :), so to fix this, I do 
 the following:

 #newavoid-avoid
 #newavoid(does this do anything?)

It essentially makes a copy of `avoid` to `newavoid` -- what did you
want it to do?

That having been said, a good rule of thumb is to never use `attach`,
so let's avoid it for now.

 Lastly, I have several GLM's I wanted to conduct.  Please see the following:

 #model1-glm(cbind(avoid, noavoid)~treatment,data=,family=binomial)

 #model2=glm(cbind(avoid, noavoid)~feeding, familiy=binomial)

 #model3=glm(cbind(avoid, noavoid)~treatment+feeding, familiy=binomial)

`cbind`-ing doesn't make much sense here. What is your target (y)
variable here? are you trying to predict `avoid` or `noavoid` status?

Let's assume you were predicting `noavoid` from just `treatment` and
`feeding` (I guess you have more data (rows) than you show), you would
build a model like so:

R model - glm(noavoid ~ treatment + feeding, binomial, avoid)

Or to be explicit about the parameters:

R model - glm(noavoid ~ treatment + feeding, family=binomial, data=avoid)


 It would be greatly appreciated if somebody can help me with my coding, as 
 you can see I am a novice but doing my best to learn.  I figured if I can get 
 model1 to run, I should be able to figure out the rest of my models.

Since you're just getting started, maybe it would be helpful for
people writing documentation/tutorials/whatever what needs to be
explained better.

For instance, I'm curious why you thought to `cbind` in your first glm
call, which was:

model1-glm(cbind(avoid, noavoid)~treatment,data=,family=binomial)

What did you think `cbind`-ing was accomplishing for you? Is there an
example somewhere that's doing that as the first parameter to a `glm`
call?

Also, why just have `data=nothing`?

I'm not criticizing, just trying to better understand.

-steve

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

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

Re: [R] Finding values in one column and

Hi,
Try ?merge(), ?join() from library(plyr)
data.a-read.table(text=
ID    valueA    valueB
6    12    12
17    15    14
58    18    16
98    11    12
73    19    20
84    19    14
58    20    14
24    11    12
81    15    16
21    15    14
62    14    12
67    13    14
78    13    17
35    10    13
13    11    15
14    17    18
85    16    15
35    13    9
18    15    16
,sep=,header=TRUE)
data.b-read.table(text=
ID   
6   
84   
21   
78   
14    
,sep=,header=TRUE)
library(plyr)
join(data.a,data.b,by=ID,type=inner)
#  ID valueA valueB
#1  6 12 12
#2 84 19 14
#3 21 15 14
#4 78 13 17
#5 14 17 18
A.K.




- Original Message -
From: Steven Ranney steven.ran...@gmail.com
To: r-help@r-project.org
Cc: 
Sent: Tuesday, November 27, 2012 5:17 PM
Subject: [R] Finding values in one column and

All -

I have a data frame

data.a
ID    valueA    valueB
6    12    12
17    15    14
58    18    16
98    11    12
73    19    20
84    19    14
58    20    14
24    11    12
81    15    16
21    15    14
62    14    12
67    13    14
78    13    17
35    10    13
13    11    15
14    17    18
85    16    15
35    13    9
18    15    16

and a data frame

data.b
ID    valueA    valueB
6        
84        
21        
78        
14        

I'd like to have R find the data.b$ID in data.a$ID and insert the
corresponding data.a$valueA and data.a$valueB into the appropriate
columns in data.b.

How can I do this?

Thanks for you help.

SR
Steven H. Ranney

__
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Re: [R] Finding values in one column and