date:20100721



On Jul 21, 2010, at 5:33 AM, Mangalani Peter Makananisa wrote:


Dear R Gurus,


I saw no reason to copy Rob Hyndman. I did not see that this involves  
any of the packages he maintains.


I am having two dummy csv data sets A and B containing 19 and 15
cases/observations respectively. From the two data set 13 cases are
intersection. From one of the two (any) data set, How do I then  
retrieve

the unmerged data ? let's take A for example, six cases must appear in
our results. See the R codes below.


?setdiff

Perhaps:

setdiff( (NAME(A), NAME(B) )

You can also do a merge that is an outer join that includes all the  
NAME information and then extract the records with SALARY  
and .B.SALARY data. Untested in absence of working example:


?merge
mer - merge(A,B, all=TRUE)
mer[ mer$NAME %in% setdiff(NAME(A), NAME(B) ),  ]

--
David.





A = read.csv(C:/Documents and Settings/S1033067/Desktop/A.csv,

header = TRUE, dec =,, sep = ,)


names(A)


[1] NAME   SALARY


dim(A)[1]


[1] 19


B = read.csv(C:/Documents and Settings/S1033067/Desktop/B.csv,

header = TRUE, dec =,, sep = ,)


names(B)


[1] NAME B.SALARY


dim(B)[1]


[1] 15


common = merge(A,B)



names(common)


[1] NAME SALARY   B.SALARY


dim(common)[1]


[1] 13





David Winsemius, MD
West Hartford, CT

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[R] Validation in R for GAMM

2010-07-21 Thread Karen

My GAMM model is to find drivers of species richness in forests is

gamm1- gamm(Total Species Richness ~ fROT + s(PH) + s(LOI) + ASP +
s(SQRT_ELEV) + CANCOV + s(SQRT_TOTCWD) + s(WELLF) + s(WELLN) +
s(OLDWDLD) + s(DISTWOOD) + s(Annprec), random=list (fSITE =~1), family
= poisson, data = BIOFOR3)

My issues are that the validation graphs are using methods I'm
unfamiliar with e.g. square rooting values and Pearson's residuals.
I'm following as best I can recommendations in Wood 2006 and Zuur 2009
but have some (I think) basic questions on graphical validation:

1. plot sqrt transformed fitted values vs square root transformed
observed values - should be straight line.
I presume the observed values mentioned above refer only to observed
RESPONSE variable

2. Pearson residuals vs. sqrt fitted values (should form band with no
patterns)
I'm totally unfamilar with using Pearson residuals.  Until I can see
example of a band with no patterns I can't really say if my output is
OK! Could this band also be described as a cloud of points with no
pattern?

I have 2 factors and 2 unsmoothed explanatory variables in my model –
what residual graphs are suitable for these? Is it the traditional
ones of normalized residuals vs raw explanatory variables?

Also for a lot of other model checking residuals are plotted against
explanatory variables..what’s appropriate residuals to plot
vs.smoothed explanatory variables for a GAMM?

3.   Raw residuals (observed vs fitted) vs. sqrt fitted values (should
show clear cone)
 But my output doesn't show clear cone and I don't know what action to
take to rectify this issue
 Infact graphs 2. and 3. are the exact same.

Some help with this would be much appreciated.

Thanks,

Karen

The only examples I have for graph output is Wood 2006 Fig 6.11

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Re: [R] Chi-square distribution probability density function:



On Jul 21, 2010, at 6:52 AM, Knut Krueger wrote:


Hi to all I found
an formular of an **
***p-Value Calculator for the Chi-Square test*
*http://www.danielsoper.com/statcalc/calc11.aspx*
*with the formula*
*http://www.danielsoper.com/statkb/topic11.aspx*
*what's the gamma function of this formula in r?*
*df=5*
*ch2=25.50878*
*the following code does not give the result  0.001 for the values  
above *
*p= ((0.5^(df/2))/gamma(df))*(ch2^((df/2)-1))* (2.718281828459^(- 
ch2/2))

or is there any other error?


Check your implementation of that formula. You made an error in the  
gamma argument.


See also:

?Chisquare
?gamma
--

David Winsemius, MD
West Hartford, CT

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Re: [R] xtable with ifelse statement

2010-07-21 Thread Jannis

What exactly do you want to do? To me it looks more like a problem with the 
arrangement of your lists as with xtable (or ifelse) (so you are better next 
time of with a more precise subject line).

Please read the posting guide and provide some reproducable code so we can help 
you (and include your lists etc. via  dump('x',file=stdout())).

As first suggestion I would rearrange your lists in a way that the data you 
want to have together ends up in one list, but there are several alternatives, 
depending on what you want to achieve.


Jannis

--- vivi84 cipullu...@hotmail.com schrieb am Mi, 21.7.2010:

 Von: vivi84 cipullu...@hotmail.com
 Betreff: [R] xtable with ifelse statement
 An: r-help@r-project.org
 Datum: Mittwoch, 21. Juli, 2010 07:23 Uhr
 
 Hi there,
 I'm very new on R and I hope someone can help me to solve
 the problem in
 using the ifelse statement with the xtable function(library
 xtable). 
 I'm trying to get the printing of the elements of two lists
 in a sorted way.
 These two list have in common the their names.
 I will try to give an example:
 the first list looks like this:
 
 $code1
 
 Code              code1
 Nation         
    Usa
 
 $code2
 
 Code              code2
 Nation         
    Spain
 
 
 the second one is like this
 
 $code 1
 var1               
     87 
 var2               
    73
 
 $code2  
 var1               
     34
 var2               
     23
 
 
 As you can see the two list have the same names.I tried to
 implement the
 following ifelse statement but it doesn't work
 orde2=function(x,y){
 ifelse(names(x)==names(y),list(print(xtable(x)),print(xtable(y))),NULL)
 
 What I want is the xtable of this first:
 $code1
 
 Code              code1
 Nation         
    Usa
 
 
 and after this:
 
 $code 1
 var1               
     87 
 var2               
    73
 
 
 And as the first case, the xtable of this;
 $code2
 
 Code              code2
 Nation         
    Spain
 
 
 and after this:
 
 $code2  
 var1               
     34
 var2               
     23
 
 Someone Knows if it possible to do this and how??
 Thanks a lot!
 
 -- 
 View this message in context: 
 http://r.789695.n4.nabble.com/xtable-with-ifelse-statement-tp2296694p2296694.html
 Sent from the R help mailing list archive at Nabble.com.
 
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 mailing list
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[R] Contingency Table Analysis - specific cell to specific cell comparisons

2010-07-21 Thread Tsunhin John Wong

Dear R users,

I have a question of how to do some specific cell to cell comparisons
on a R x C contingency table.
The table is a 3 x 5 table with frequency / count data.

 langcons.table - table(lang, cons)
 langcons.table[cbind(lang,cons)] - freq
 langcons.table

  Adj Int Oth Pas Tra
C  69 221  17   3 198
E  56 214  33  31 174
J  36 291   8   9 164

I know how to do an independent model test using Poisson in glm
 glm.out1 - glm(freq~lang+cons, family=poisson, data=langcons.data)
 summary(glm.out1)

And then fit the saturated model
 glm.out2 - glm(freq~lang*cons, family=poisson, data=langcons.data)
 summary(glm.out2)

However, the results are difficult to interpret:
C and Adj are used to as a baseline.
And I can only see main effects and interactions and *always according
to the baseline*.
Coefficients:
 Estimate Std. Error z value Pr(|z|)
(Intercept)
lang1
lang2
cons1
cons2
cons3
cons4
lang1:cons1
lang2:cons1
lang1:cons2
lang2:cons2
lang1:cons3
lang2:cons3
lang1:cons4
lang2:cons4

If anyone know, please suggest me some way to do specific cell to cell
comparison on such a contingency table.
Say, to compare pairs of cells:
along a column: 3 vs 31, 9 vs 31, 3 vs 9
along a row: 36 vs 9
or even across column and row: 36 vs 31, and 36 vs 3

Thanks for your help in advance.

Best,

John

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Re: [R] Contingency Table Analysis - specific cell to specific cell comparisons



On Jul 21, 2010, at 8:07 AM, Tsunhin John Wong wrote:


Dear R users,

I have a question of how to do some specific cell to cell comparisons
on a R x C contingency table.
The table is a 3 x 5 table with frequency / count data.


langcons.table - table(lang, cons)
langcons.table[cbind(lang,cons)] - freq
langcons.table


 Adj Int Oth Pas Tra
C  69 221  17   3 198
E  56 214  33  31 174
J  36 291   8   9 164

I know how to do an independent model test using Poisson in glm

glm.out1 - glm(freq~lang+cons, family=poisson, data=langcons.data)
summary(glm.out1)


And then fit the saturated model

glm.out2 - glm(freq~lang*cons, family=poisson, data=langcons.data)
summary(glm.out2)


However, the results are difficult to interpret:
C and Adj are used to as a baseline.
And I can only see main effects and interactions and *always according
to the baseline*.
Coefficients:
Estimate Std. Error z value Pr(|z|)
(Intercept)
lang1
lang2
cons1
cons2
cons3
cons4
lang1:cons1
lang2:cons1
lang1:cons2
lang2:cons2
lang1:cons3
lang2:cons3
lang1:cons4
lang2:cons4

If anyone know, please suggest me some way to do specific cell to cell
comparison on such a contingency table.


Even if you are daunted by the task of plugging the covariates into  
the formula, exp(intercept+sum(beta_N*var_n)), you can always use the  
predict function to create an estimate for all (or a specific set) of  
the covariates. They come out on the log(rate) scale so would need to  
be exponentiated. Consult your instructor for further help.




Say, to compare pairs of cells:
along a column: 3 vs 31, 9 vs 31, 3 vs 9
along a row: 36 vs 9
or even across column and row: 36 vs 31, and 36 vs 3



David Winsemius, MD
West Hartford, CT

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Re: [R] how to unsubscribe

2010-07-21 Thread Jannis

As in nearly every e-mail list, you just need to click on one of the links at 
the end of each mail (its the first link in this case) and follow the 
instructions.


Jannis

--- barbara horta e costa barbarahco...@gmail.com schrieb am Mi, 21.7.2010:

 Von: barbara horta e costa barbarahco...@gmail.com
 Betreff: [R] how to unsubscribe
 An: r-help@r-project.org
 Datum: Mittwoch, 21. Juli, 2010 09:50 Uhr
 Hello,
 I need to unsubcribe this email to the R help mailing list
 because I need to
 create an email only for this mailing list. My email
 account doesn't have
 that much space anymore and I don't have access to this
 every day.
 How can I do that?
 Thanks a lot
 
 -- 
 Bárbara H. Costa
 Marine Biologist Researcher
 SCIAENA - Marine Sciences and Cooperation
 www.sciaena.org
 
 ISPA | http://www.ispa.pt/ui/uie/index.asp
 BIOMARES | http://www.ccmar.ualg.pt/biomares/
 MSI-UCSB | http://www.msi.ucsb.edu/
 
 bco...@ispa.pt
 barbarahco...@gmail.com
 
     [[alternative HTML version deleted]]
 
 
 -Integrierter Anhang folgt-
 
 __
 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.
 



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Re: [R] Random Forest - Strata

2010-07-21 Thread mxkuhn

If you use the index argument of the trainControl() function in the caret 
package, the train() function can be used for this type of resampling (and 
you'll get some decent summaries and visualizations to boot)

Max

On Jul 21, 2010, at 7:11 AM, Tim Howard tghow...@gw.dec.state.ny.us wrote:

 Coll,
 
 An alternative approach is to do that subsetting yourself before sending it 
 to RF and treat each group as an external validation group, as follows:
 - extract Site A, build a RF model (Model 1) on sites B and C
 - validate this model by running a predict on site A using the model, use 
 ROCR or other evaluation metrics to look at the effectiveness of Model 1. 
 - extract Site B, build a RF model (Model 2) on sites A and C.
 - validate this model by trying to predict presence in site B using model 2.
 - continue through all your sites.
 
 This is called 'leave-one-out' and is used in some fields for model 
 validation.  You final accuracy estimates of your model could be based on the 
 averages of values obtained for each model. 
 
 Hope that Helps. 
 Tim
 
 
 
 --
 
 Message: 44
 Date: Tue, 20 Jul 2010 08:48:04 -0700 (PDT)
 From: Coll gbco...@gmail.com
 To: r-help@r-project.org 
 Subject: [R] Random Forest - Strata
 Message-ID: 1279640884553-2295731.p...@n4.nabble.com
 Content-Type: text/plain; charset=us-ascii
 
 
 Hi all,
 
 Had struggled in getting Strata in randomForest to work on this. 
 
 Can I get randomForest for each of its TREE, to get ALL sample from some
 strata to build tree, while leaving some strata TOTALLY untouched as oob?
 
 e.g. in below, how I can tell RF to, 
 - for tree 1 in the forest, to use only Site A and B to build the tree,
 while using the WHOLE Site C data for the oob error rate,
 - for tree 2, use only site A and C to build tree, while using whole site B
 data for oob
 - for tree 3, use Site B and C, A as oob...?
 
 My command does not work as it would use some sample in all of the sites:
 rforest.obj - randomForest(Presence.f ~., data=dataset.subset, strata =
 site.factor)
 
 while 
 the setting the corresponding sampsize argument seems would only screen
 out the Site in all tree building...
 
 SitePresence  Length  Sulphur
 AYes   3.5019.42
 ANo3.9051.09
 ANo3.6026.75
 BYes   2.609.71
 BNo2.209.77
 BNo2.608.60
 BNo3.0035.59
 CYes   3.5016.07
 CNo3.4049.96
 CNo3.1035.35
 
 Any idea / comments are welcomed.
 
 Thanks in advance.
 
 Coll
 -- 
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[R] Saving R plots as image in oracle database tables

2010-07-21 Thread vikrant


Dear All,
I want to know how to save R plots as image/object in a oracle database??
I am using RODBC package for connecting R to oracle
Please explain by giving example..
Thanks in advance


Cheers,
Vikrant
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View this message in context: 
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[R] fix()ing an S4 method

2010-07-21 Thread Albert-Jan Roskam

Hi R experts,
 
The fix() function canbe used to edit normal functions. I would like to know 
whether it's also possible to use something similar to edit a method of an S4 
class. In other words, is there a fix-like function that allows me to edit 
method definitions without having to go back to the source code?
 
setGeneric (name=doStuff,def = function(object){standardGeneric(doStuff)})
setClass(SomeClass, representation(text = character))
setMethod(f = doStuff, signature (SomeClass), definition = function(object){
    return(print(paste(***, obj...@text)))  })
fix(doStuff) # Does NOT give the intended result. How can I make R show the 
method definition?

Cheers!!
Albert-Jan

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


  
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Re: [R] how to unsubscribe

2010-07-21 Thread barbara horta e costa

I already changed my email account.
Thanks a lot

On 21 July 2010 12:50, Jannis bt_jan...@yahoo.de wrote:

 As in nearly every e-mail list, you just need to click on one of the links
 at the end of each mail (its the first link in this case) and follow the
 instructions.


 Jannis

 --- barbara horta e costa barbarahco...@gmail.com schrieb am Mi,
 21.7.2010:

  Von: barbara horta e costa barbarahco...@gmail.com
  Betreff: [R] how to unsubscribe
  An: r-help@r-project.org
  Datum: Mittwoch, 21. Juli, 2010 09:50 Uhr
   Hello,
  I need to unsubcribe this email to the R help mailing list
  because I need to
  create an email only for this mailing list. My email
  account doesn't have
  that much space anymore and I don't have access to this
  every day.
  How can I do that?
  Thanks a lot
 
  --
  Bárbara H. Costa
  Marine Biologist Researcher
  SCIAENA - Marine Sciences and Cooperation
  www.sciaena.org
 
  ISPA | http://www.ispa.pt/ui/uie/index.asp
  BIOMARES | http://www.ccmar.ualg.pt/biomares/
  MSI-UCSB | http://www.msi.ucsb.edu/
 
  bco...@ispa.pt
  barbarahco...@gmail.com
 
  [[alternative HTML version deleted]]
 
 
  -Integrierter Anhang folgt-
 
  __
  R-help@r-project.org
  mailing list
  https://stat.ethz.ch/mailman/listinfo/r-help
   PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html
  and provide commented, minimal, self-contained,
  reproducible code.
 






-- 
Bárbara H. Costa
Marine Biologist Researcher
SCIAENA - Marine Sciences and Cooperation
www.sciaena.org

ISPA | http://www.ispa.pt/ui/uie/index.asp
BIOMARES | http://www.ccmar.ualg.pt/biomares/
MSI-UCSB | http://www.msi.ucsb.edu/

bco...@ispa.pt
barbarahco...@gmail.com

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[R] DIF Analysis starting from a gpcm class object

2010-07-21 Thread Dr. Federico Andreis


Dear useRs,

does any of you have suggestions on how to conduct a proper DIF analysis
starting from a model of
class gpcm (from the wonderful package ltm by prof. Rizopoulos)? 

difR will handle only dichotomous items, and I have a mix of dicho- and
polytomous ones (that's why I chose the partial credit model). 

I also found the package lordif, but I'm not really sure if that's what I
need.

I basically want to regress item difficulties on some other covariates (sex,
age class,...)..

Is there a package you know of that handle such models for a DIF analysis or
should I code such a thing by myself? and if so, how can I extract what I
need from the model object?

thanks to everybody

/federico

-
Dr. Federico Andreis
Università degli Studi di Milano-Bicocca, PhD Student
MEB Department, Karolinska Institutet, Stockholm, Visiting PhD Student
-
Problem: To Catch a Lion in the Sahara Desert - The Dirac Method 

We observe that wild lions are, ipso facto, not observable in the Sahara
Desert. Consequently, if there are any lions in the Sahara, they are tame.
The capture of a tame lion may be left as an exercise for the reader.
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[R] Fractional Polynomials - Hazard Ratios and Relative Hazard Plots

2010-07-21 Thread Laura Bonnett

Dear All,

I am using Windows and R version 2.9.2 with libraries cmprsk, mfp and
Design.

I have a dataset with approximately 1700 patients (1 row per patient) and I
have 12 outcomes, three of which are continuous.  I have performed
univariate analyses to see if any factors are associated with a higher
likelihood of the event of interest (achieving 12 month remission from
epileptic seizures) and also an analysis adjusted for multiple variables.

I have tried log and fractional polynomial (FP) transformations of each
continuous variable.  In the case of age, used for the example below, the FP
transformation led to the best model fit according to the AIC.  I have
therefore applied this transformation for all analyses.

To begin with I have fitted a Cox model stratified by randomisation period,
rpa, (either before or after a certain date).

fit1 - mfp(Surv(Remtime,Rcens) ~ fp(age) + strata(rpa), family=cox,
data=nearma, select=0.05, verbose=TRUE)

I would like two things from this model, hazard ratios and and an associated
hazard ratio plot.  I am aware that the hazard ratios produced from a
fractional polynomial transformation are not to be used directly (i.e. those
obtained from summary(coxfitf1)).  Instead the derived functional form of
the variable should be used to estimate hazard ratios post hoc.  I have
attached a word document explaining how hazard ratios and confidence
interval can be derived and given a worked example for the variable, age.
The univariate results are:

Age (years)

≤10

(10 to 25]

(25 to 37]

(37 to 50]

(50 to 71]

71

1.00

1.00 (1.00 to 1.00)

0.99 (0.99 to 1.00)

0.99 (0.99 to 0.99)

0.99 (0.98 to 0.99)

0.98 (0.97 to 0.99)

To create a plot of the relative hazard I have used the code:
plot(nearma$age,predict(fit1,type=risk),xlab=Age,ylab=Relative Hazard)
The produced plot is attached.
As you can clearly see, the hazard ratios above and the relative hazard plot
do not agree.
This is also the case for the other two continuous variables that have been
transformed via the FP approach.

The hazard ratios for age using the model adjusted for multiple variables
are as follows, which do coincide with the plot:

Age (years)

≤10

(10 to 25]

(25 to 37]

(37 to 50]

(50 to 71]

71

1.00

0.86 (0.77 to 0.95)

0.78 (0.65 to 0.94)

0.80 (0.64 to 1.00)

1.01 (0.81 to 1.25)

1.61 (1.27 to 2.05)

Can anyone therefore explain why the univariate hazard ratios do not
coincide with the relative hazard plot and yet the hazard ratios from the
multivariable model do?  I know that the calculations are correct for both
sets of hazard ratios.

Thank you for any help you can provide as I am at a loss to explain the
difference in the plot fo the calculations - they should, after all, be
saying the same thing!

Thank you,
Laura
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[R] R Cheat Sheets and Reference Cards

2010-07-21 Thread Tim Church

FYI - There is a new collection of free cheat sheets and reference cards for
R on DevCheatSheet - http://devcheatsheet.com/tag/r/

If anyone knows of any other R cheat sheets that are not listed, please use
the 'Suggest a Missing Cheat Sheet' link on the site.

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[R] Piecewise regression using lme()

2010-07-21 Thread goulvenS


Hi everyone,

I'm trying to fit a of piecewise regression model on a time series. The idea
is to divide the series into segments and then to apply linear regression
models on each segment but in a global way and considering
heteroskedasticity between the segments. For example, I build a time series
y with 3 segments: 

segment1=1:20+rnorm(20,0,2)
segment2=20-2*1:30+rnorm(30,0,5)
segment3=-40+0.5*1:15+rnorm(15,0,1)

group=c(rep(1,20),rep(2,30),rep(3,15))
y=c(segment1,segment2,segment3)

Data=data.frame(y,t=1:65,group=as.factor(group))

the model I'd like to fit is:

y_t=
(beta_01+beta_11*t+error_1)*(group==1)+(beta_02+beta_12*t+error_2)*(group==2)+(beta_03+beta_13*t+error_3)*(group==3)

It looks like a mixed effects model were the fixed effect are the piecewise
linear regression parts (beta_0i+beta_1i*t) and the random effects are the
variance components error_i. 

The problem is that I can't find the way the write this model correctly
using the lme() function of the package nlme. I can't remove the global
intercept and the global variance component. Here's what I tried:

#1 Using a prior piecewise linear regression

lm.list=lmList(y~t|group,Data)

model.lme=lme(lm.list,weights=varIdent(form=~1|group))

#2 Trying to estimate the whole model directly and considering the different
lines as random effects

model.lme=lme(y~1,random=~t|group,data=Data)

but the intercept remains... 

I read a lot of R-help messages before posting this one (my first!) and I'm
not getting any closer. It looks like no one tried to estimate the exact
same model. I'll be very grateful if someone could help me on this. 

Thanks

Goulven




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[R] Error message: attempt to set rownames on object with no dimensions

2010-07-21 Thread Igor Blanco

Hi there,

I am trying to analyze some data using the lme package.
In my case there is a variable called Newmarker that can only take 2 values
(number 1 or number 2).
I have used the as.factor to remark this fact but an error message appears
stating: attempt to set rownames on object with no dimensions.

Here is the code I have used:

*Dataset -   read.table(Data2.txt, header=TRUE, sep=, na.strings=NA,
dec=., strip.white=TRUE)

Dataset$Newmarker - as.factor(Dataset$Newmarker)

attach(Dataset)

lme.1 - lme(Newmarker ~ Age, data=Dataset,  random= ~1|ID)

summary(lme.1)*

If I don't use the as.factor I have no problems but I don't think this is
the correct way to proceed... The other variables I use can have any value.

Can anybody help me please?

Thanking you in advance,

Regards,

Igor Blanco

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Re: [R] Error message: attempt to set rownames on object with no dimensions

2010-07-21 Thread Prof Brian Ripley

?lme (at least the one in package nlme) seems not to mention that the 
response in 'fixed' must be a numeric variable, but it must (and a 
factor is not numeric).


If you want to use a binary response, you need to use other methods, 
and it would be better to ask on R-sig-mixed-models making clear what 
your aims actually are.


On Wed, 21 Jul 2010, Igor Blanco wrote:


Hi there,

I am trying to analyze some data using the lme package.


Perhaps you mean the nlme package?


In my case there is a variable called Newmarker that can only take 2 values
(number 1 or number 2).
I have used the as.factor to remark this fact but an error message appears
stating: attempt to set rownames on object with no dimensions.

Here is the code I have used:

*Dataset -   read.table(Data2.txt, header=TRUE, sep=, na.strings=NA,
dec=., strip.white=TRUE)

Dataset$Newmarker - as.factor(Dataset$Newmarker)

attach(Dataset)

lme.1 - lme(Newmarker ~ Age, data=Dataset,  random= ~1|ID)

summary(lme.1)*

If I don't use the as.factor I have no problems but I don't think this is
the correct way to proceed... The other variables I use can have any value.

Can anybody help me please?

Thanking you in advance,

Regards,

Igor Blanco

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

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Re: [R] Error message: attempt to set rownames on object with no dimensions



On Jul 21, 2010, at 9:39 AM, Igor Blanco wrote:


Hi there,

I am trying to analyze some data using the lme package.
In my case there is a variable called Newmarker that can only take 2  
values

(number 1 or number 2).
I have used the as.factor to remark this fact but an error message  
appears

stating: attempt to set rownames on object with no dimensions.

Here is the code I have used:

*Dataset -   read.table(Data2.txt, header=TRUE, sep=,  
na.strings=NA,

dec=., strip.white=TRUE)

Dataset$Newmarker - as.factor(Dataset$Newmarker)


Why do you use as.factor? It's probably already a factor. What does  
str(Dataset) show?




attach(Dataset)


You do not need to do this, and it creates potential for confusion.  
The data= argument will allow the column names to be accessed by the  
lme() function.




lme.1 - lme(Newmarker ~ Age, data=Dataset,  random= ~1|ID)

summary(lme.1)*

If I don't use the as.factor I have no problems but I don't think  
this is
the correct way to proceed... The other variables I use can have any  
value.



--

David Winsemius, MD
West Hartford, CT

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[R] The opposite of lag

Hello!

I have a data frame A (below) with a grouping factor (group). I take
my DV and create the new, lagged DV by applying the function lag.it
(below). It works fine.

A - data.frame(year=rep(c(1980:1984),3), group=
factor(sort(rep(1:3,5))), DV=c(rnorm(15)))
lag.it - function(x) {
  DV - ts(x$DV, start = x$year[1])
  idx - seq(length = length(DV))
  DVs - cbind(DV, lag(DV, -1))[idx,]
  out-cbind(x, DVs[,2])  # wages[,2]
   names(out)[length(out)]-DV.lag
   return(out)
}
A
A.lagged - do.call(rbind, by(A, A$group, lag.it))
A.lagged


Now, I am trying to create the oppostive of lag for DV (should I call
it lead?)
I tried exactly the same as above, but with a different number under
lag function (below), but it's not working. I am clearly doing
something wrong. Any advice?
Thanks a lot!


lead.it - function(x) {
  DV - ts(x$DV, start = x$year[1])
  idx - seq(length = length(DV))
  DVs - cbind(DV, lag(DV, 2))[idx,]
 out-cbind(x, DVs[,2])
   names(out)[length(out)]-DV.lead
   return(out)
}
A
A.lead - do.call(rbind, by(A, A$group, lead.it))
A.lead

-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

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[R] How do you calculate DIC from coda files (R2WinBUGS)?

2010-07-21 Thread Patrick McKann

 I tried calculating pD as 1/2 variance of the deviance, but I got hugely
inflated numbers.

Does anybody know how to calculate pD from the coda files output from
R2WinBUGS?

By the way, 'set DIC' is greyed out for some reason within WinBUGS, so I
can't monitor DIC.

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Re: [R] best way to apply a list of functions to a dataset ?

2010-07-21 Thread Peter Ehlers


Dennis' ddply solution would be my choice. Here is
a small variation that makes it easy to modify what
list of functions is applied:

#
ma- melt(attitude)

f - function(x,v) summarise(x,
 mean = mean(v),
 sd = sd(v),
 skewness = skewness(v),
 mean.gt.med = mean.gt.med(v)
)

ddply(ma, .(variable), function(x) f(x, v = x[[value]]))
#

Another option is to use data.frame in place of summarise:

#
f - function(x,v) data.frame(
 mean = mean(v),
 sd = sd(v),
 skewness = skewness(v),
 mean.gt.med = mean.gt.med(v)
)
#

  -Peter Ehlers


On 2010-07-21 0:41, Dennis Murphy wrote:

Hi:

On Tue, Jul 20, 2010 at 5:37 PM, Glen Barnettglnbr...@gmail.com  wrote:


Hi Dennis,

Thanks for the reply.

Yes, that's easier, but the conversion to a matrix with rbind has
converted the output of that final function to a numeric.



If you look at the output of lapply(attitude, f), you'll see that the
conversion from logical to numeric has already taken place. Different
components of lists can have different types, but within a component, all of
the elements must have the same class.
You can patch up the result as follows:

x- data.frame(do.call(rbind, lapply(attitude, f)))
x[, 5]- as.logical(x[, 5])

x

meansdskewness median mean.gt.med
rating 64.6 12.172562 -0.35792491   65.5   FALSE
complaints 66.6 13.314757 -0.21541749   65.0TRUE
privileges 53.1 12.235430  0.37912287   51.5TRUE
learning   56.36667 11.737013 -0.05403354   56.5   FALSE
raises 64.6 10.397226  0.19754317   63.5TRUE
critical   74.76667  9.894908 -0.86577893   77.5   FALSE
advance42.9 10.288706  0.85039799   41.0TRUE

but if you're doing this sort of thing over a large data frame such fixes
may be impractical.


I included that last function in the example secifically to preclude

people assuming that functions would always return the same type.




There is a plyr solution, although it's a little more long-winded than I'd
prefer in the end:

library(ggplot2)
# melt the data frame so that the variables become factor levels
  ma- melt(attitude)
Using  as id variables

dim(ma)

[1] 210   2
# Use ddply to get the set of summaries by variable:

ddply(ma, .(variable), summarise, mean = mean(value), sd = sd(value),
skewness = skewness(value), median = median(value),
mean.gt.med = mean.gt.med(value))
 variable meansdskewness median mean.gt.med
1 rating 64.6 12.172562 -0.35792491   65.5   FALSE
2 complaints 66.6 13.314757 -0.21541749   65.0TRUE
3 privileges 53.1 12.235430  0.37912287   51.5TRUE
4   learning 56.36667 11.737013 -0.05403354   56.5   FALSE
5 raises 64.6 10.397226  0.19754317   63.5TRUE
6   critical 74.76667  9.894908 -0.86577893   77.5   FALSE
7advance 42.9 10.288706  0.85039799   41.0TRUE

Notice that now the logical class of mean.gt.med is preserved.  The trick
with ddply() in package plyr is that, in this case, it is convenient to melt
the data frame first before doing the summarizations. This is because
ddply() requires a grouping variable - in this example, the groups are the
variables themselves.

HTH,
Dennis

I guess this doesn't matter too much for a logical, but what if

instead the function returned a character (say mean, median, or
equal - indicating which one was larger, or equal which could
easily happen with discrete data). This precludes using rbind (which I
also used at first, before I noticed that sometimes I could have
functions that don't return numerics).

Glen


On Tue, Jul 20, 2010 at 6:55 PM, Dennis Murphydjmu...@gmail.com  wrote:

Hi:

This might be a little easier (?):

library(datasets)
skewness- function(x) mean(scale(x)^3)
mean.gt.med- function(x) mean(x)median(x)

# --
# construct the function to apply to each variable in the data frame
f- function(x) c(mean = mean(x), sd = sd(x), skewness = skewness(x),
  median = median(x), mean.gt.med = mean.gt.med(x))

# map function to each variable with lapply and combine with do.call():
do.call(rbind, lapply(attitude, f))
meansdskewness median mean.gt.med
rating 64.6 12.172562 -0.35792491   65.5   0
complaints 66.6 13.314757 -0.21541749   65.0   1
privileges 53.1 12.235430  0.37912287   51.5   1
learning   56.36667 11.737013 -0.05403354   56.5   0
raises 64.6 10.397226  0.19754317   63.5   1
critical   74.76667  9.894908 -0.86577893   77.5   0
advance42.9 10.288706  0.85039799   41.0   1

HTH,
Dennis


On Mon, Jul 19, 2010 at 10:51 PM, Glen Barnettglnbr...@gmail.com

wrote:


Assuming I have a matrix of data (or under some restrictions that will
become obvious, possibly a data frame), I want to be able to apply a
list of functions (initially producing a single number from a vector)
to the data and produce a data

Re: [R] RGoogleDocs ability to write to spreadsheets broken as of yesterday

2010-07-21 Thread Ben Bolker

Harlan Harris harlan at harris.name writes:

 
 Hi,
 
 I'm using RGoogleDocs/RCurl to update a Google Spreadsheet. Everything
 worked OK until this morning, when my ability to write into spreadsheet
 cells went away. I get the following weird error:
 
 Error in els[[type + 1]] : subscript out of bounds
 
 Looking at the Google Docs API changelog, I see the following:
 
 http://code.google.com/apis/spreadsheets/changelog.html
 Release 2010-01 (July 14, 2010)
 
 This is an advanced notice about an upcoming change.
 
- Starting July 19, 2010, all links returned by all Spreadsheets API
feeds will use HTTPS. This is being done in the interests of increased
security. If you require the use of HTTP, we recommend that you remove the
replace https with http in these links. Another announcement will be made
on July 19, 2010, when this change goes to production.
 
 I suspect this is the problem. Fixing it is above my head, I'm afraid. Could
 anyone help? This is urgent. Thank you,


   This is an Omegahat package (took me a little while to find it). Perhaps
you should write to the package maintainer?

library(RGoogleDocs)
help(package=RGoogleDocs)

or, more obscurely:

help(package=RGoogleDocs)$info[[1]][9]

 (there may be a better way to deal with objects of type packageInfo
but I can't figure it out right at the moment).
  It looks as though one might be able to fix this by hacking the 
hard-coded URLs in the code, but as you suggest that might be above
your head.

  good luck ...
Ben Bolker

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Re: [R] RGoogleDocs ability to write to spreadsheets broken as of yesterday

2010-07-21 Thread Henrique Dallazuanna

On Wed, Jul 21, 2010 at 11:24 AM, Ben Bolker bbol...@gmail.com wrote:

 Harlan Harris harlan at harris.name writes:

 
  Hi,
 
  I'm using RGoogleDocs/RCurl to update a Google Spreadsheet. Everything
  worked OK until this morning, when my ability to write into spreadsheet
  cells went away. I get the following weird error:
 
  Error in els[[type + 1]] : subscript out of bounds
 
  Looking at the Google Docs API changelog, I see the following:
 
  http://code.google.com/apis/spreadsheets/changelog.html
  Release 2010-01 (July 14, 2010)
 
  This is an advanced notice about an upcoming change.
 
 - Starting July 19, 2010, all links returned by all Spreadsheets API
 feeds will use HTTPS. This is being done in the interests of increased
 security. If you require the use of HTTP, we recommend that you remove
 the
 replace https with http in these links. Another announcement will be
 made
 on July 19, 2010, when this change goes to production.
 
  I suspect this is the problem. Fixing it is above my head, I'm afraid.
 Could
  anyone help? This is urgent. Thank you,


This is an Omegahat package (took me a little while to find it). Perhaps
 you should write to the package maintainer?

 library(RGoogleDocs)
 help(package=RGoogleDocs)

 or, more obscurely:

 help(package=RGoogleDocs)$info[[1]][9]

  (there may be a better way to deal with objects of type packageInfo
 but I can't figure it out right at the moment).


Maybe:
packageDescription('RGoogleDocs', fields = 'Author')


  It looks as though one might be able to fix this by hacking the
 hard-coded URLs in the code, but as you suggest that might be above
 your head.

  good luck ...
 Ben Bolker

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-- 
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[R] Clustering groups

2010-07-21 Thread syrvn


Hi,

is there a way in R to identify those cluster methods / distance measures
which best reflect predefined cluster groups.

Given 10 observations O1...O10. Optimally, these 10 observations cluster as
follows:
cluster1: O1, O2, O3, O4
cluster2: O5, O6
cluster3: O7, O8, O9, O10.

What I want is a method which identifies that cluster method / distance
measure which best reflect my predefined groups.

Is that somehow possible to do?
Cheers
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Re: [R] The opposite of lag

On Wed, Jul 21, 2010 at 10:14 AM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
 Hello!

 I have a data frame A (below) with a grouping factor (group). I take
 my DV and create the new, lagged DV by applying the function lag.it
 (below). It works fine.

 A - data.frame(year=rep(c(1980:1984),3), group=
 factor(sort(rep(1:3,5))), DV=c(rnorm(15)))
 lag.it - function(x) {
  DV - ts(x$DV, start = x$year[1])
  idx - seq(length = length(DV))
  DVs - cbind(DV, lag(DV, -1))[idx,]
  out-cbind(x, DVs[,2])  # wages[,2]
   names(out)[length(out)]-DV.lag
   return(out)
 }
 A
 A.lagged - do.call(rbind, by(A, A$group, lag.it))
 A.lagged


 Now, I am trying to create the oppostive of lag for DV (should I call
 it lead?)
 I tried exactly the same as above, but with a different number under
 lag function (below), but it's not working. I am clearly doing
 something wrong. Any advice?
 Thanks a lot!


 lead.it - function(x) {
  DV - ts(x$DV, start = x$year[1])
  idx - seq(length = length(DV))
  DVs - cbind(DV, lag(DV, 2))[idx,]
  out-cbind(x, DVs[,2])
   names(out)[length(out)]-DV.lead
   return(out)
 }
 A
 A.lead - do.call(rbind, by(A, A$group, lead.it))
 A.lead


Try this:

library(zoo)
z - read.zoo(A, index = 1, split = group, frequency = 1)
lag(z, c(-1, 0, 1))

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Re: [R] best way to apply a list of functions to a dataset ?

2010-07-21 Thread Hadley Wickham

 ddply(ma, .(variable), summarise, mean = mean(value), sd = sd(value),
       skewness = skewness(value), median = median(value),
       mean.gt.med = mean.gt.med(value))

In principle, you should be able to do:

ddply(ma, .(variable), colwise(each(mean, sd, skewness, median, mean.gt.med)))

but currently colwise and each don't work together that well.

Hadley

-- 
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/

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Re: [R] Help with time in R

2010-07-21 Thread Aaditya Nanduri

Ms. Chisholm,

If you could tell us how you plan to use the variables, we will have a
better understanding of what you are looking for and will be able to help
you.
Are you looking for the time in seconds? In that case, do as Mr. Holfman
says. He just skipped the part about converting the factors to characters.
You can do that by:
y - as.character(x) where x is the vector of factors.

Are you looking to have a list of hours, minutes and seconds? That can be
done too...Although it would be much easier to just have hours and min.sec

On Tue, Jul 20, 2010 at 7:33 AM, Sarah Chisholm sarah.chisholm...@ucl.ac.uk
 wrote:

 Hi,

 I have a problem with the time formatting in R. I have entered time in the
 format MM:SS.xyz and R has automatically classified this as a factor, but
 I need it numerically. However when I use as.numeric() it gives me totally
 different numbers. Is there any way I can tell R to read thes input as a
 number?

 Thank you very much

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-- 
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aaditya.nand...@gmail.com

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[R] Interactions in GAMMs

2010-07-21 Thread Karen Moore

Hi,

I've an issue adding an interaction to a GAMM:

My model was of form:

gamm1 - gamm(TOTSR ~ fROT + s(PH) + s(LOI) + s(ASP) + s(SQRT_ELEV) + CANCOV
+ s(SQRT_TOTCWD) + s(WELLF) + s(WELLN) + s(OLDWDLD) + s(DISTWOOD) +
s(Annprec) + s(OLDWDLD:DISTWOOD) + (1|fSITE),  family = poisson, data =
BIOFOR2)

with interaction of s(OLDWDLD:DISTWOOD).

However I got this error message below but don't know what it means? (my
data is composed of info for 150 plots)

#I Warning messages:
#2: In OLDWDLD:DISTWOOD :
 # numerical expression has 150 elements: only the first used
#3: In OLDWDLD:DISTWOOD :
#  numerical expression has 150 elements: only the first used

Can anyone offer advice on how to include this interaction in GAMM model?
Thanks

Karen Moore

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Re: [R] RGoogleDocs ability to write to spreadsheets broken as of yesterday - CAN PAY FOR FIX

2010-07-21 Thread Harlan Harris

I unfortunately haven't received any responses about this problem. We (the
company I work for) are willing to discuss payment to someone who is willing
to quickly contribute a fix to the RGoogleDocs/RCurl toolchain that will
restore write access. Please contact me directly if you're interested. Thank
you,

 -Harlan Harris

On Tue, Jul 20, 2010 at 10:19 AM, Harlan Harris har...@harris.name wrote:

 Hi,

 I'm using RGoogleDocs/RCurl to update a Google Spreadsheet. Everything
 worked OK until this morning, when my ability to write into spreadsheet
 cells went away. I get the following weird error:

 Error in els[[type + 1]] : subscript out of bounds

 Looking at the Google Docs API changelog, I see the following:

 http://code.google.com/apis/spreadsheets/changelog.html
 Release 2010-01 (July 14, 2010)

 This is an advanced notice about an upcoming change.

- Starting July 19, 2010, all links returned by all Spreadsheets API
feeds will use HTTPS. This is being done in the interests of increased
security. If you require the use of HTTP, we recommend that you remove the
replace https with http in these links. Another announcement will be
made on July 19, 2010, when this change goes to production.


 I suspect this is the problem. Fixing it is above my head, I'm afraid.
 Could anyone help? This is urgent. Thank you,

  -Harlan Harris



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Re: [R] Interactions in GAMMs

2010-07-21 Thread Vito Muggeo (UniPa)


Hi Karen,
I think you should decide what you mean for interaction. s(x:y) is 
meaningless


If you want to fit a surface you should use s(x,y).

If you want to fit a varying coefficient model (interaction between a 
linear and a smooth term) you should use the argument by in s().


The help files of the mgcv package by S. Wood  are quite clear and they 
also include a lot of examples.


Hope this helps you

vito



Karen Moore ha scritto:

Hi,

I've an issue adding an interaction to a GAMM:

My model was of form:

gamm1 - gamm(TOTSR ~ fROT + s(PH) + s(LOI) + s(ASP) + s(SQRT_ELEV) + CANCOV
+ s(SQRT_TOTCWD) + s(WELLF) + s(WELLN) + s(OLDWDLD) + s(DISTWOOD) +
s(Annprec) + s(OLDWDLD:DISTWOOD) + (1|fSITE),  family = poisson, data =
BIOFOR2)

with interaction of s(OLDWDLD:DISTWOOD).

However I got this error message below but don't know what it means? (my
data is composed of info for 150 plots)

#I Warning messages:
#2: In OLDWDLD:DISTWOOD :
 # numerical expression has 150 elements: only the first used
#3: In OLDWDLD:DISTWOOD :
#  numerical expression has 150 elements: only the first used

Can anyone offer advice on how to include this interaction in GAMM model?
Thanks

Karen Moore

[[alternative HTML version deleted]]

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--

Vito M.R. Muggeo
Dip.to Sc Statist e Matem `Vianelli'
Università di Palermo
viale delle Scienze, edificio 13
90128 Palermo - ITALY
tel: 091 23895240
fax: 091 485726/485612
http://dssm.unipa.it/vmuggeo

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[R] lm: order of dropped columns

2010-07-21 Thread Anirban Mukherjee

Hi all,

If presented with a singular design matrix, lm drops columns to make the
design matrix non-singular. What algorithm is used to select which (and how
many) column(s) to drop? Particularly, given a factor, how does lm choose
levels of the factor to discard?

Thanks for the help.

Best,
Anirban

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Re: [R] Interactions in GAMMs



On Jul 21, 2010, at 11:17 AM, Karen Moore wrote:


Hi,

I've an issue adding an interaction to a GAMM:

My model was of form:


# Package? Probably:
require(mgcv)



gamm1 - gamm(TOTSR ~ fROT + s(PH) + s(LOI) + s(ASP) + s(SQRT_ELEV)  
+ CANCOV

+ s(SQRT_TOTCWD) + s(WELLF) + s(WELLN) + s(OLDWDLD) + s(DISTWOOD) +
s(Annprec) + s(OLDWDLD:DISTWOOD) + (1|fSITE),  family = poisson,  
data =

BIOFOR2)

with interaction of s(OLDWDLD:DISTWOOD).

However I got this error message below but don't know what it means?


The documentation for s() says the unnamed arguments must be a list of  
covariates and I do not think OLDWDLD:DISTWOOD constitutes a list. (On  
the other hand it appears that the term list is being used a bit  
loosely here, since the examples do not explicitly enclose the  
covariates in the list() function and when I did so with the example  
on the te help page I get an error.)



(my
data is composed of info for 150 plots)

#I Warning messages:
#2: In OLDWDLD:DISTWOOD :
# numerical expression has 150 elements: only the first used
#3: In OLDWDLD:DISTWOOD :
#  numerical expression has 150 elements: only the first used

Can anyone offer advice on how to include this interaction in GAMM  
model?


Do you have Wood's book? Ch/sect 6.7 has worked examples using the  
te() function which I believe is the GAM(M) counterpart to linear  
interactions. He also makes the point that random arguments to gamm()  
need to be in the list form. (The documentation suggests this has not  
changes since the book was published.)


--
David Winsemius, MD
West Hartford, CT

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[R] Default For list.len Argument (list output truncated)

2010-07-21 Thread Soeren . Vogel

Hello

How can I set a default for the 'list.len' argument in 'str'?

x - as.data.frame(matrix(1:1000, ncol=1000))
str(x)
formals(str) - alist(object=, ...=, list.len=1000)
args(str); formals(str)
str(x)

Does not display errors, but does not work either.

Thanks for help

Sören

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Re: [R] The opposite of lag

On Wed, Jul 21, 2010 at 10:50 AM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
 On Wed, Jul 21, 2010 at 10:14 AM, Dimitri Liakhovitski
 dimitri.liakhovit...@gmail.com wrote:
 Hello!

 I have a data frame A (below) with a grouping factor (group). I take
 my DV and create the new, lagged DV by applying the function lag.it
 (below). It works fine.

 A - data.frame(year=rep(c(1980:1984),3), group=
 factor(sort(rep(1:3,5))), DV=c(rnorm(15)))
 lag.it - function(x) {
  DV - ts(x$DV, start = x$year[1])
  idx - seq(length = length(DV))
  DVs - cbind(DV, lag(DV, -1))[idx,]
  out-cbind(x, DVs[,2])  # wages[,2]
   names(out)[length(out)]-DV.lag
   return(out)
 }
 A
 A.lagged - do.call(rbind, by(A, A$group, lag.it))
 A.lagged


 Now, I am trying to create the oppostive of lag for DV (should I call
 it lead?)
 I tried exactly the same as above, but with a different number under
 lag function (below), but it's not working. I am clearly doing
 something wrong. Any advice?
 Thanks a lot!


 lead.it - function(x) {
  DV - ts(x$DV, start = x$year[1])
  idx - seq(length = length(DV))
  DVs - cbind(DV, lag(DV, 2))[idx,]
  out-cbind(x, DVs[,2])
   names(out)[length(out)]-DV.lead
   return(out)
 }
 A
 A.lead - do.call(rbind, by(A, A$group, lead.it))
 A.lead


 Try this:

 library(zoo)
 z - read.zoo(A, index = 1, split = group, frequency = 1)
 lag(z, c(-1, 0, 1))


The ### line was missing:

library(zoo)
z - read.zoo(A, index = 1, split = group, frequency = 1)
z - as.zooreg(z) ###
lag(z, c(-1, 0, 1))

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Re: [R] The opposite of lag

Never mind- I figured it out:

A
library(zoo)
z - read.zoo(A, index = 1, split = group, frequency = 1)
z - as.zooreg(z) ###
result-lag(z, c(-1, 0, 1))
A
result

Thank you very much!

On Wed, Jul 21, 2010 at 12:08 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
 Thank you, Gabor, but sorry - what is the exact order of those rows again?
 Thank you!
 Dimitri

 On Wed, Jul 21, 2010 at 11:59 AM, Gabor Grothendieck
 ggrothendi...@gmail.com wrote:
 On Wed, Jul 21, 2010 at 10:50 AM, Gabor Grothendieck
 ggrothendi...@gmail.com wrote:
 On Wed, Jul 21, 2010 at 10:14 AM, Dimitri Liakhovitski
 dimitri.liakhovit...@gmail.com wrote:
 Hello!

 I have a data frame A (below) with a grouping factor (group). I take
 my DV and create the new, lagged DV by applying the function lag.it
 (below). It works fine.

 A - data.frame(year=rep(c(1980:1984),3), group=
 factor(sort(rep(1:3,5))), DV=c(rnorm(15)))
 lag.it - function(x) {
  DV - ts(x$DV, start = x$year[1])
  idx - seq(length = length(DV))
  DVs - cbind(DV, lag(DV, -1))[idx,]
  out-cbind(x, DVs[,2])  # wages[,2]
   names(out)[length(out)]-DV.lag
   return(out)
 }
 A
 A.lagged - do.call(rbind, by(A, A$group, lag.it))
 A.lagged


 Now, I am trying to create the oppostive of lag for DV (should I call
 it lead?)
 I tried exactly the same as above, but with a different number under
 lag function (below), but it's not working. I am clearly doing
 something wrong. Any advice?
 Thanks a lot!


 lead.it - function(x) {
  DV - ts(x$DV, start = x$year[1])
  idx - seq(length = length(DV))
  DVs - cbind(DV, lag(DV, 2))[idx,]
  out-cbind(x, DVs[,2])
   names(out)[length(out)]-DV.lead
   return(out)
 }
 A
 A.lead - do.call(rbind, by(A, A$group, lead.it))
 A.lead


 Try this:

 library(zoo)
 z - read.zoo(A, index = 1, split = group, frequency = 1)
 lag(z, c(-1, 0, 1))


 The ### line was missing:

 library(zoo)
 z - read.zoo(A, index = 1, split = group, frequency = 1)
 z - as.zooreg(z) ###
 lag(z, c(-1, 0, 1))




 --
 Dimitri Liakhovitski
 Ninah Consulting
 www.ninah.com




-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

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R-help@r-project.org mailing list
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] The opposite of lag

Sorry, I don't think it's working.
the last 3 columns (on the right) of result contain the original data
of each group.
But there is no shift at all. I am trying to reach the following
result for each group: The first number disappears and at the bottom
an NA appears.
Is it possible?

On Wed, Jul 21, 2010 at 12:12 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
 Never mind- I figured it out:

 A
 library(zoo)
 z - read.zoo(A, index = 1, split = group, frequency = 1)
 z - as.zooreg(z) ###
 result-lag(z, c(-1, 0, 1))
 A
 result

 Thank you very much!

 On Wed, Jul 21, 2010 at 12:08 PM, Dimitri Liakhovitski
 dimitri.liakhovit...@gmail.com wrote:
 Thank you, Gabor, but sorry - what is the exact order of those rows again?
 Thank you!
 Dimitri

 On Wed, Jul 21, 2010 at 11:59 AM, Gabor Grothendieck
 ggrothendi...@gmail.com wrote:
 On Wed, Jul 21, 2010 at 10:50 AM, Gabor Grothendieck
 ggrothendi...@gmail.com wrote:
 On Wed, Jul 21, 2010 at 10:14 AM, Dimitri Liakhovitski
 dimitri.liakhovit...@gmail.com wrote:
 Hello!

 I have a data frame A (below) with a grouping factor (group). I take
 my DV and create the new, lagged DV by applying the function lag.it
 (below). It works fine.

 A - data.frame(year=rep(c(1980:1984),3), group=
 factor(sort(rep(1:3,5))), DV=c(rnorm(15)))
 lag.it - function(x) {
  DV - ts(x$DV, start = x$year[1])
  idx - seq(length = length(DV))
  DVs - cbind(DV, lag(DV, -1))[idx,]
  out-cbind(x, DVs[,2])  # wages[,2]
   names(out)[length(out)]-DV.lag
   return(out)
 }
 A
 A.lagged - do.call(rbind, by(A, A$group, lag.it))
 A.lagged


 Now, I am trying to create the oppostive of lag for DV (should I call
 it lead?)
 I tried exactly the same as above, but with a different number under
 lag function (below), but it's not working. I am clearly doing
 something wrong. Any advice?
 Thanks a lot!


 lead.it - function(x) {
  DV - ts(x$DV, start = x$year[1])
  idx - seq(length = length(DV))
  DVs - cbind(DV, lag(DV, 2))[idx,]
  out-cbind(x, DVs[,2])
   names(out)[length(out)]-DV.lead
   return(out)
 }
 A
 A.lead - do.call(rbind, by(A, A$group, lead.it))
 A.lead


 Try this:

 library(zoo)
 z - read.zoo(A, index = 1, split = group, frequency = 1)
 lag(z, c(-1, 0, 1))


 The ### line was missing:

 library(zoo)
 z - read.zoo(A, index = 1, split = group, frequency = 1)
 z - as.zooreg(z) ###
 lag(z, c(-1, 0, 1))




 --
 Dimitri Liakhovitski
 Ninah Consulting
 www.ninah.com




 --
 Dimitri Liakhovitski
 Ninah Consulting
 www.ninah.com




-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

__
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] RGoogleDocs ability to write to spreadsheets broken as of yesterday

2010-07-21 Thread Ben Bolker

On Wed, Jul 21, 2010 at 10:38 AM, Henrique Dallazuanna www...@gmail.com wrote:


 Maybe:
 packageDescription('RGoogleDocs', fields = 'Author')



or rather

packageDescription('RgoogleDocs', fields='Maintainer')

 (usually but not always the same person)

  thanks
Ben

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

Re: [R] The opposite of lag

Basically, if you look at the A lagged below, then you see how in each
group the values of A move downwards: and NA is added on top and
the last value in each group disappears.
I am trying to get the opposite: the first value in each group
disappears and NA is added at the bottom.
D.

set.seed(1234)
A - data.frame(year=rep(c(1980:1984),3), group=
factor(sort(rep(1:3,5))), DV=c(rnorm(15)))
lag.it - function(x) {
  DV - ts(x$DV, start = x$year[1])
idx - seq(length = length(DV))
DVs - cbind(DV, lag(DV, -1))[idx,]
out-cbind(x, DVs[,2])  # wages[,2]
   names(out)[length(out)]-DV.lag
   return(out)
}
A
A.lagged - do.call(rbind, by(A, A$group, lag.it))
A.lagged





On Wed, Jul 21, 2010 at 12:18 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
 Sorry, I don't think it's working.
 the last 3 columns (on the right) of result contain the original data
 of each group.
 But there is no shift at all. I am trying to reach the following
 result for each group: The first number disappears and at the bottom
 an NA appears.
 Is it possible?

 On Wed, Jul 21, 2010 at 12:12 PM, Dimitri Liakhovitski
 dimitri.liakhovit...@gmail.com wrote:
 Never mind- I figured it out:

 A
 library(zoo)
 z - read.zoo(A, index = 1, split = group, frequency = 1)
 z - as.zooreg(z) ###
 result-lag(z, c(-1, 0, 1))
 A
 result

 Thank you very much!

 On Wed, Jul 21, 2010 at 12:08 PM, Dimitri Liakhovitski
 dimitri.liakhovit...@gmail.com wrote:
 Thank you, Gabor, but sorry - what is the exact order of those rows again?
 Thank you!
 Dimitri

 On Wed, Jul 21, 2010 at 11:59 AM, Gabor Grothendieck
 ggrothendi...@gmail.com wrote:
 On Wed, Jul 21, 2010 at 10:50 AM, Gabor Grothendieck
 ggrothendi...@gmail.com wrote:
 On Wed, Jul 21, 2010 at 10:14 AM, Dimitri Liakhovitski
 dimitri.liakhovit...@gmail.com wrote:
 Hello!

 I have a data frame A (below) with a grouping factor (group). I take
 my DV and create the new, lagged DV by applying the function lag.it
 (below). It works fine.

 A - data.frame(year=rep(c(1980:1984),3), group=
 factor(sort(rep(1:3,5))), DV=c(rnorm(15)))
 lag.it - function(x) {
  DV - ts(x$DV, start = x$year[1])
  idx - seq(length = length(DV))
  DVs - cbind(DV, lag(DV, -1))[idx,]
  out-cbind(x, DVs[,2])  # wages[,2]
   names(out)[length(out)]-DV.lag
   return(out)
 }
 A
 A.lagged - do.call(rbind, by(A, A$group, lag.it))
 A.lagged


 Now, I am trying to create the oppostive of lag for DV (should I call
 it lead?)
 I tried exactly the same as above, but with a different number under
 lag function (below), but it's not working. I am clearly doing
 something wrong. Any advice?
 Thanks a lot!


 lead.it - function(x) {
  DV - ts(x$DV, start = x$year[1])
  idx - seq(length = length(DV))
  DVs - cbind(DV, lag(DV, 2))[idx,]
  out-cbind(x, DVs[,2])
   names(out)[length(out)]-DV.lead
   return(out)
 }
 A
 A.lead - do.call(rbind, by(A, A$group, lead.it))
 A.lead


 Try this:

 library(zoo)
 z - read.zoo(A, index = 1, split = group, frequency = 1)
 lag(z, c(-1, 0, 1))


 The ### line was missing:

 library(zoo)
 z - read.zoo(A, index = 1, split = group, frequency = 1)
 z - as.zooreg(z) ###
 lag(z, c(-1, 0, 1))




 --
 Dimitri Liakhovitski
 Ninah Consulting
 www.ninah.com




 --
 Dimitri Liakhovitski
 Ninah Consulting
 www.ninah.com




 --
 Dimitri Liakhovitski
 Ninah Consulting
 www.ninah.com




-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.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.

Re: [R] Constrain density to 0 at 0?

2010-07-21 Thread Ravi Varadhan

This is the boundary problem in density estimation.  One simple trick is to 
use the idea of reflection.  If your data is (y1, y2, ...,yn), you create a 
reflected data by appending `n' negative values to your data, call this y*.  
Estimate the kernel density for this as fhat(y*).  Redefine this estimate so 
that it is zero for all y  0, and is equal to 2 * fhat(y*).

Here is an example:

y - rexp(100, rate=10)

yref - c(-y, y)

fref - density(yref)

sel - fref$x = 0

fref$y[sel] - 2 * fref$y[sel]

par(mfrow=c(2,1))

plot(density(y))

plot(fref$x[sel], fref$y[sel], type=l)

This approach assumes that the density has 0 gradient at the boundary (gradient 
defined from the right hand side).

There are other (better) approaches as well.  Look at a book on density 
estimation.

Ravi.  


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

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


- Original Message -
From: Farley, Robert farl...@metro.net
Date: Tuesday, July 20, 2010 2:47 pm
Subject: Re: [R] Constrain density to 0 at 0?
To: r-help@r-project.org r-help@r-project.org


 Read the help page for density more carefully. Especially the bw and  
 
  from arguments.
  
  
  Yes,  it was my inability to make sense of the help page that 
 motivated my email.
  
  My distances range from 0.4 to 7.6 but the density plot ranges from 
 -2 to 10.  
  
  from=0 seems to hide the negative portion of the density without 
 setting it to 0.
  
  
  I've tried adjust, but it requires a value of 0.2, which results in a 
 very, very spiky plot.
  
  
  I was hoping for a process that could forbid impossible (negative) 
 values, yet allow the density plot to blend the individual 
 measurements. Is there a variable bw that could be set small at the 
 extrema, and larger in the range of the data?
  
  
  
  
   
  Robert Farley
  Metro
  www.Metro.net 
   
  -Original Message-
  From: David Winsemius [ 
  Sent: Monday, July 19, 2010 19:31
  To: Farley, Robert
  Cc: r-help@r-project.org
  Subject: Re: [R] Constrain density to 0 at 0?
  
  
  On Jul 19, 2010, at 9:57 PM, Farley, Robert wrote:
  
   I'm plotting some trip length frequencies using the following code:
  
   plot( density(zTestData$Distance, weights=zTestData$Actual),
  xlim=c(0,10),
  main=Test TLFD,
  xlab=Distance,
 col=6  )
   lines(density(zTestData$Distance, weights=zTestData$FlatWeight),  
   col=2)
   lines(density(zTestData$Distance, weights=zTestData$BrdWeight ),  
   col=3)
  
   which works fine except the distances are all positive, but the  
   densities don't drop to 0 until around -2 or -3.
  
   Is there a way for me to force the density plot to 0 at 0?
  
  Yes. (Assuming it can be zero, given the data.)
  
  Read the help page for density more carefully. Especially the bw and  
 
  from arguments.
  
  -- 
  David.
  
  __
  R-help@r-project.org mailing list
  
  PLEASE do read the posting guide 
  and provide commented, minimal, self-contained, reproducible code.

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Objects within environment

2010-07-21 Thread Duncan Murdoch


On 21/07/2010 12:27 PM, Megh Dal wrote:

Hi Duncan, thanks for your clarification. However I do not think I could really 
understand the envir argument in objects() function.

It is saying an alternative argument for name Is the alternative means the 
alternative of, let say, package:graphics (which is the name of an environment?). Can you give me 
an example of an alternative argument of that particular environment?
  


as.environment(package:graphics) is an environment.  
package:graphics is its name.

This is specifying the environment evaluation environment. What does the phase 
environment evaluation environment mean? Can you give me an example?
  


That looks like a typo to me.  Environment evaluation environment is 
meaningless.

Mostly there for back compatibility: again totally in dark, what does it mean for 
back compatibility? An example would definitely be great.
  


Presumably some older release of R used envir, and we still have it so 
that old code will still work.  But that's a signal that new code should 
never need to use it.

Toy said you need to give an environment, not the name of one. If I call 
someone I need to call with his name, right? Then if I need to give an environment then, 
without its name how can I do so?
  


This is a computer language, not a conversation.  Words have technical 
meanings that aren't always perfect matches to English meanings of the 
same words.  Here the name of an environment is what you get when you 
ask for the name attribute of it.  There are lots of different ways to 
refer to objects other than by the name in that technical sense.  For 
example, you could say


x - as.environment(package:graphics) # uses the environment's name
ls(envir=x) # refers to it by a variable holding it.

Duncan Murdoch

Can you please explain me in simple english? I think R help file should use 
more non-technical simple english language so that student like me can 
understand R in more easier way!

--- On Wed, 7/21/10, Duncan Murdoch murdoch.dun...@gmail.com wrote:

 From: Duncan Murdoch murdoch.dun...@gmail.com
 Subject: Re: [R] Objects within environment
 To: Megh Dal megh700...@yahoo.com
 Cc: r-h...@stat.math.ethz.ch
 Date: Wednesday, July 21, 2010, 4:48 PM
 On 21/07/2010 5:57 AM, Megh Dal
 wrote:
  Hi all, I have following environments loaded with my
 current R session:
  
  search()
  [1] .GlobalEnv   
 package:stats 
package:graphics 
 package:grDevices
  [5] package:utils 
package:datasets 
 package:methods   Autoloads   
 [9] package:base
  How can I find the objects under a specific

 environment? Here I tries following:
  
  objects(envir=package:base)

  Error in objects(envir = package:base) : invalid
 'envir' argument
  
  It would be great if somebody would point me the

 correct arguments for object() function to find the onjects
 associated with it. In help file it is written that:
   envir: an alternative argument to ‘name’ for
 specifying the

environment evaluation environment. Mostly

 there for back

compatibility

  What is the wrong in my code?
 
 The easiest way to pick an item from the search list is by

 number:
 
 objects(3)

  or
 ls(3)
 
 will give you the objects in the graphics package, with the

 search list as above.  You can also specify the name as
 the name argument, e.g.
 
 objects(package:base)
 
 If you want to use the envir argument (why?), you need to

 give an environment, not the name of one.
 
 Duncan Murdoch
 
 



  



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Re: [R] The opposite of lag

On Wed, Jul 21, 2010 at 12:18 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
 Sorry, I don't think it's working.
 the last 3 columns (on the right) of result contain the original data
 of each group.
 But there is no shift at all. I am trying to reach the following
 result for each group: The first number disappears and at the bottom
 an NA appears.
 Is it possible?


It works for me.  Here is the result.  As we see

- the first 3 columns are lagged ahead so that they start at 1981
  which is one year later than the original columns start,
- the second set of 3 columns are the originals so they start at 1980
- the last set of 3 columns are lagged so they start at 1979 which is
  one year before the original columns start.

 set.seed(123)
 A - data.frame(year=rep(c(1980:1984),3), group=
+ factor(sort(rep(1:3,5))), DV=c(rnorm(15)))
 library(zoo)
 z - read.zoo(A, index = 1, split = group, frequency = 1)
 z - as.zooreg(z) ###
 lag(z, c(-1, 0, 1))
X1.lag.1   X2.lag.1   X3.lag.1 X1.lag0X2.lag0X3.lag0
1979  NA NA NA  NA NA NA
1980  NA NA NA -0.56047565  1.7150650  1.2240818
1981 -0.56047565  1.7150650  1.2240818 -0.23017749  0.4609162  0.3598138
1982 -0.23017749  0.4609162  0.3598138  1.55870831 -1.2650612  0.4007715
1983  1.55870831 -1.2650612  0.4007715  0.07050839 -0.6868529  0.1106827
1984  0.07050839 -0.6868529  0.1106827  0.12928774 -0.4456620 -0.5558411
1985  0.12928774 -0.4456620 -0.5558411  NA NA NA
 X1.lag1X2.lag1X3.lag1
1979 -0.56047565  1.7150650  1.2240818
1980 -0.23017749  0.4609162  0.3598138
1981  1.55870831 -1.2650612  0.4007715
1982  0.07050839 -0.6868529  0.1106827
1983  0.12928774 -0.4456620 -0.5558411
1984  NA NA NA
1985  NA NA NA

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Re: [R] RGoogleDocs ability to write to spreadsheets broken as of yesterday - CAN PAY FOR FIX

2010-07-21 Thread Duncan Temple Lang

Hi Harlan

  Can you send some code so that we can reproduce the problem.
That will enable me to fix the problem quicker.

  D.

On 7/21/10 8:26 AM, Harlan Harris wrote:
 I unfortunately haven't received any responses about this problem. We
 (the company I work for) are willing to discuss payment to someone who
 is willing to quickly contribute a fix to the RGoogleDocs/RCurl
 toolchain that will restore write access. Please contact me directly if
 you're interested. Thank you,
 
  -Harlan Harris
 
 On Tue, Jul 20, 2010 at 10:19 AM, Harlan Harris har...@harris.name
 mailto:har...@harris.name wrote:
 
 Hi,
 
 I'm using RGoogleDocs/RCurl to update a Google Spreadsheet.
 Everything worked OK until this morning, when my ability to write
 into spreadsheet cells went away. I get the following weird error:
 
 Error in els[[type + 1]] : subscript out of bounds
 
 Looking at the Google Docs API changelog, I see the following:
 
 http://code.google.com/apis/spreadsheets/changelog.html
 
 
 Release 2010-01 (July 14, 2010)
 
 This is an advanced notice about an upcoming change.
 
 * Starting July 19, 2010, all links returned by all Spreadsheets
   API feeds will use HTTPS. This is being done in the interests
   of increased security. If you require the use of HTTP, we
   recommend that you remove the replace |https| with |http| in
   these links. Another announcement will be made on July 19,
   2010, when this change goes to production.
 
 
 I suspect this is the problem. Fixing it is above my head, I'm
 afraid. Could anyone help? This is urgent. Thank you,
 
  -Harlan Harris
 


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[R] Problem with invoking R from the command line (Windows Vista)

2010-07-21 Thread Camster

Dear all

I am running in a problem when trying to run R from Windows command line.
I am runnning Windows Vista, R-2.10.1.

I have a script I would like to run remotely from another program. As it did
not work, I thought I would test the script from the Windows command line
which highlighted a problem.
When I open the command line and type R at the prompt (or any other variant
(R CMD BATCH, Rgui.exe or Rterm.exe), I get the error message:
'Rscript' is not recognized as an internal or external command,
operable program or batch file.

if I move to the R directory (cd usr\R\R-2.10.1\bin), and try running the
script ( R myscript.R --vanilla)
it works fine.
I thought the problem may come from the environment variables (tried
specifying explicit location of R.exe), but this did not solve the problem.
I also re-install R but that did not help either.

I cannot think of anything else.
I have read the R installation and rwfaq notes, and look through the posts
but I cannot find any suggestion on how to fix the bug.

If anyone can suggest anything, that will be much appreciated.

Many thanks

Camille

PS: I am also running R on another machine (Windows XP) and I don't have a
similar problem!

--
View this message in context:
http://r.789695.n4.nabble.com/Problem-with-invoking-R-from-the-command-line-Windows-Vista-tp2297429p2297429.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] [SPAM](Aktien) Re: Chi-square distribution probability density function:



On Jul 21, 2010, at 12:36 PM, Knut Krueger wrote:


David Winsemius schrieb:



On Jul 21, 2010, at 6:52 AM, Knut Krueger wrote:


Hi to all I found
an formular of an **
***p-Value Calculator for the Chi-Square test*
*http://www.danielsoper.com/statcalc/calc11.aspx*
*with the formula*
http://www.danielsoper.com/statkb/topic11.aspx
*what's the gamma function of this formula in r?*
*df=5*
*ch2=25.50878*
*the following code does not give the result  0.001 for the  
values above *
*p= ((0.5^(df/2))/gamma(df))*(ch2^((df/2)-1))* (2.718281828459^(- 
ch2/2))

or is there any other error?


Check your implementation of that formula. You made an error in the  
gamma argument.


Sorry a copy and paste error but gamma(df/2) does not solve the  
problem

I am not able to get the implementation of the function working :-(


See also:

?Chisquare
?gamma


there is also a different result using this function from the helpf  
files and the function

/f_n(x) = 1 / (2^(n/2) Gamma(n/2)) x^(n/2-1) e^(-x/2)


That is (perhaps) because you are confusing a density function with a  
cumulative probability function. The expression above should give the  
same result as a (proper) R transcription of the formula on the page  
you offered above and should be what dchisq referenced on the ? 
Chisquare page returns as well.


Sure enough...
 dchisq(25.50878, 5)
[1] 4.951e-05
 p= ((0.5^(df/2))/gamma(df/2))*(ch2^((df/2)-1))* (2.718281828459^(- 
ch2/2))

 p
[1] 4.951e-05

 Those formulae only differ in how they use grouping of the constant  
terms. (In R the function is gamma, not Gamma.) You should be able to  
get the P(X^2|x25.508, df=5) from an integration of that function.




Knut

/


David Winsemius, MD
West Hartford, CT

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[R] String processing - is there a better way

2010-07-21 Thread Davis, Brian

I have a two part question

Part 1) 
I am trying to remove characters in a string based on the position of a key 
character in another string.  I have a solution that works but it requires a 
for-loop.  A vectorized way of doing this has alluded me.  

CleanRead-function(x,y) {

  if (!is.character(x)) 
    x - as.character(x)
  if (!is.character(y)) 
    y - as.character(y)

  idx-grep(\\*, x, value=FALSE)
  starpos-gregexpr(\\*, x[idx])
  
  ysplit-strsplit(y[idx], '')
  n-length(idx)
  for(i in 1:n) {
    ysplit[[i]][starpos[[i]]] = 
  }

  y[idx]-unlist(lapply(ysplit, paste, sep='', collapse=''))
  return(y)
}

x-c(AA*.*A **a.a*,,,A, C*c.., **aA) 
y-c(abcdefghi, abcdefghij, abcde, abcd)

CleanRead(x,y)
[1] abdfghi cdeghij acde    cd


Is there a better way to do this?

Part 2) 
My next step in the string processing is to take the characters in the output 
of CleanRead and subtract 33 from the ascii value of the character to obtain an 
integer. Again I have a solution that works, involving splitting the string 
into characters then converting them to factors (starting at ascii 34) and 
using unclass to get the integer value. (kindof a atoi(x)-33 all in one step)

I looked for the C equivalent of atoi, but the only help I could find (R-help 
2003) suggested using as.numeric.  However, the help file (and testing) shows 
you get 'NA'.   

Am I missing an easier way to do this?



Thanks in advance,

Brian

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Re: [R] Problem with invoking R from the command line (Windows Vista)

2010-07-21 Thread Duncan Murdoch


On 21/07/2010 12:55 PM, Camster wrote:

Dear all

I am running in a problem when trying to run R from Windows command line.
I am runnning Windows Vista, R-2.10.1. 


I have a script I would like to run remotely from another program. As it did
not work, I thought I would test the script from the Windows command line
which highlighted a problem. 
When I open the command line and type R at the prompt (or any other variant
(R CMD BATCH, Rgui.exe or Rterm.exe), I get the error message: 
'Rscript' is not recognized as an internal or external command,

operable program or batch file.
  


This is a sign that your PATH environment variable doesn't include the 
directory RHOME\bin. 




if I move to the R directory (cd usr\R\R-2.10.1\bin), and try running the
script ( R  myscript.R --vanilla) 
 it works fine. 
I thought the problem may come from the environment variables (tried

specifying explicit location of R.exe), but this did not solve the problem.
I also re-install R but that did not help either.
  


You don't say exactly what you tried, but if you tried changing your 
PATH, you should try again, and verify that the change is in place in 
the session where you're trying to run R.


Duncan Murdoch

I cannot think of anything else. 
I have read the R installation and rwfaq notes, and look through the posts

but I cannot find any suggestion on how to fix the bug.

If anyone can suggest anything, that will be much appreciated.

Many thanks

Camille

PS: I am also running R on another machine (Windows XP) and I don't have a
similar problem!






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Re: [R] The opposite of lag

I reinstalled zoo and now I can see the years on left (raw names).
Before - I could not see them.
Thank you!

On Wed, Jul 21, 2010 at 12:51 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
 On Wed, Jul 21, 2010 at 12:18 PM, Dimitri Liakhovitski
 dimitri.liakhovit...@gmail.com wrote:
 Sorry, I don't think it's working.
 the last 3 columns (on the right) of result contain the original data
 of each group.
 But there is no shift at all. I am trying to reach the following
 result for each group: The first number disappears and at the bottom
 an NA appears.
 Is it possible?


 It works for me.  Here is the result.  As we see

 - the first 3 columns are lagged ahead so that they start at 1981
  which is one year later than the original columns start,
 - the second set of 3 columns are the originals so they start at 1980
 - the last set of 3 columns are lagged so they start at 1979 which is
  one year before the original columns start.

 set.seed(123)
 A - data.frame(year=rep(c(1980:1984),3), group=
 + factor(sort(rep(1:3,5))), DV=c(rnorm(15)))
 library(zoo)
 z - read.zoo(A, index = 1, split = group, frequency = 1)
 z - as.zooreg(z) ###
 lag(z, c(-1, 0, 1))
        X1.lag.1   X2.lag.1   X3.lag.1     X1.lag0    X2.lag0    X3.lag0
 1979          NA         NA         NA          NA         NA         NA
 1980          NA         NA         NA -0.56047565  1.7150650  1.2240818
 1981 -0.56047565  1.7150650  1.2240818 -0.23017749  0.4609162  0.3598138
 1982 -0.23017749  0.4609162  0.3598138  1.55870831 -1.2650612  0.4007715
 1983  1.55870831 -1.2650612  0.4007715  0.07050839 -0.6868529  0.1106827
 1984  0.07050839 -0.6868529  0.1106827  0.12928774 -0.4456620 -0.5558411
 1985  0.12928774 -0.4456620 -0.5558411          NA         NA         NA
         X1.lag1    X2.lag1    X3.lag1
 1979 -0.56047565  1.7150650  1.2240818
 1980 -0.23017749  0.4609162  0.3598138
 1981  1.55870831 -1.2650612  0.4007715
 1982  0.07050839 -0.6868529  0.1106827
 1983  0.12928774 -0.4456620 -0.5558411
 1984          NA         NA         NA
 1985          NA         NA         NA




-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

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Re: [R] Problem with invoking R from the command line (Windows Vista)

On Wed, Jul 21, 2010 at 12:55 PM, Camster c.szmar...@googlemail.com wrote:

 Dear all

 I am running in a problem when trying to run R from Windows command line.
 I am runnning Windows Vista, R-2.10.1.

 I have a script I would like to run remotely from another program. As it did
 not work, I thought I would test the script from the Windows command line
 which highlighted a problem.
 When I open the command line and type R at the prompt (or any other variant
 (R CMD BATCH, Rgui.exe or Rterm.exe), I get the error message:
 'Rscript' is not recognized as an internal or external command,
 operable program or batch file.

 if I move to the R directory (cd usr\R\R-2.10.1\bin), and try running the
 script ( R  myscript.R --vanilla)
  it works fine.
 I thought the problem may come from the environment variables (tried
 specifying explicit location of R.exe), but this did not solve the problem.
 I also re-install R but that did not help either.

 I cannot think of anything else.
 I have read the R installation and rwfaq notes, and look through the posts
 but I cannot find any suggestion on how to fix the bug.

 If anyone can suggest anything, that will be much appreciated.

 Many thanks

 Camille

 PS: I am also running R on another machine (Windows XP) and I don't have a
 similar problem!


If you don't want to change your path you can use Rgui.bat, Rterm.bat,
etc. in the batchfiles distribution at
   http://batchfiles.googlecode.com
Just put any of them that you wish to use on your path (they are self
contained Windows batch files, i.e. with no dependencies) and then
when you invoke any of them they will look up R in the registry and
run the corresponding command.

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Re: [R] The opposite of lag

On Wed, Jul 21, 2010 at 1:16 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
 I reinstalled zoo and now I can see the years on left (raw names).
 Before - I could not see them.
 Thank you!

Note that the result is a zoo object.  A zoo object consists of data
(which is everything except
the years, here) and the time index which is shown along the left hand side.

If lg is the zoo object then coredata(lg) and index(lg) give the two
components while
as.data.frame(lg) does give a data frame with the years as rownames.

lg - lag(z, c(-1, 0, 1))
coredata(lg)
index(lg)
as.data.frame(lg)

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Re: [R] Chi-square distribution probability density function:



On Jul 21, 2010, at 1:18 PM, Knut Krueger wrote:


David Winsemius schrieb:



And exactly why did you think I offered

?Chisquare
I was completely on the wrong way, and tried to find a solution with  
the formula instead to substitute the formula.


So I tried to implement pchisq into the formula - and of course I  
got wrong values ...


 p - function(x) ((0.5^(5/2))/gamma(5/2))*(x^((5/2)-1))*  
(2.718281828459^(-x/2))

 integrate(p, 25.508, Inf)
0.000 with absolute error  2.7e-05

Check result

 pchisq(25.508,5)
[1] 0.
 1-pchisq(25.508,5)
[1] 0.000



Thank's Knut


David Winsemius, MD
West Hartford, CT

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Re: [R] The opposite of lag

Thanks a lot, it's very helpful!

On Wed, Jul 21, 2010 at 1:21 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
 On Wed, Jul 21, 2010 at 1:16 PM, Dimitri Liakhovitski
 dimitri.liakhovit...@gmail.com wrote:
 I reinstalled zoo and now I can see the years on left (raw names).
 Before - I could not see them.
 Thank you!

 Note that the result is a zoo object.  A zoo object consists of data
 (which is everything except
 the years, here) and the time index which is shown along the left hand side.

 If lg is the zoo object then coredata(lg) and index(lg) give the two
 components while
 as.data.frame(lg) does give a data frame with the years as rownames.

 lg - lag(z, c(-1, 0, 1))
 coredata(lg)
 index(lg)
 as.data.frame(lg)




-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

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Re: [R] String processing - is there a better way