[R] Re : View code of function
Hi, About code read this document by Mr Uwe Ligges http://www.statistik.uni-dortmund.de/~ligges/R_Help_Desk_preview.pdf Justin BEM Elève Ingénieur Statisticien Economiste BP 294 Yaoundé. Tél (00237)9597295. - Message d'origine De : Wee-Jin Goh <[EMAIL PROTECTED]> À : R-Help Envoyé le : Vendredi, 24 Novembre 2006, 7h57mn 31s Objet : Re: [R] View code of function Hi, Just type mean.default without the () and the code of the function will be displayed. Wee-Jin On 24 Nov 2006, at 06:31, Lize van der Merwe wrote: > Dear list > > I need to see the code behind a function. I used to be able to > see the > code behind a function by typing e.g. mean(). Now I get the error > message: > Error in mean.default() : argument "x" is missing, with no > default. Please > advise. > > Regards > > Lize van der Merwe > > > > > > > > > -- > This e-mail and its contents are subject to the > South African Medical Research Council > e-mail legal notice available at http://www.mrc.ac.za/about/ > EmailLegalNotice.htm > > > [[alternative HTML version deleted]] > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting- > guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch 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. ___ Découvrez une nouvelle façon d'obtenir des réponses à toutes vos questions ! Profitez des connaissances, des opinions et des expériences des internaut [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Code of plot.TukeyHSD
Dear list I need to see the code of plot.TukeyHSD. There is no .default for it. Please help. Lize -- This e-mail and its contents are subject to the South African Medical Research Council e-mail legal notice available at http://www.mrc.ac.za/about/EmailLegalNotice.htm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] barplot help needed
hello, I would like to create the following barplot: I have 4 different data sets (same length + stddev for each data point) data1 sd1 data2 sd2 data3 sd3 data4 sd4 now, I'd like to plot in the following way: data1[1],data2[1],data3[1],data4[1] with it's sd-values side-by-side at one x-axis label (named "position 1") and each bar in different colors. data1[2],data2[2],data3[2],data4[2] at the next x-axis label (named "position 2") with the same color scheme and so on over the whole length. I managed to plot one set in the following way: par(mai=c(1.5,1,1,0.6)) plotInfo <- barplot(data1, las=2, ylim = c(0,plotMax+1), ylab = "Percentage") arrows(plotInfo,data1,plotInfo, data1 + sd1, length=0.1, angle=90) arrows(plotInfo,data1,plotInfo, data1 - sd1, length=0.1, angle=90) could anybody give me a help on this? Antje __ R-help@stat.math.ethz.ch 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] Re : View code of function
justin bem wrote: > Hi, > About code read this document by Mr Uwe Ligges > http://www.statistik.uni-dortmund.de/~ligges/R_Help_Desk_preview.pdf I should remove it, because the updated version has been published in the most recent R Newsletter. Please check out that one. Thanks, Uwe Ligges > Justin BEM > Elève Ingénieur Statisticien Economiste > BP 294 Yaoundé. > Tél (00237)9597295. > > > > - Message d'origine > De : Wee-Jin Goh <[EMAIL PROTECTED]> > À : R-Help > Envoyé le : Vendredi, 24 Novembre 2006, 7h57mn 31s > Objet : Re: [R] View code of function > > > Hi, > > Just type mean.default without the () and the code of the function > will be displayed. > > Wee-Jin > > On 24 Nov 2006, at 06:31, Lize van der Merwe wrote: > >> Dear list >> >> I need to see the code behind a function. I used to be able to >> see the >> code behind a function by typing e.g. mean(). Now I get the error >> message: >> Error in mean.default() : argument "x" is missing, with no >> default. Please >> advise. >> >> Regards >> >> Lize van der Merwe >> >> >> >> >> >> >> >> >> -- >> This e-mail and its contents are subject to the >> South African Medical Research Council >> e-mail legal notice available at http://www.mrc.ac.za/about/ >> EmailLegalNotice.htm >> >> >> [[alternative HTML version deleted]] >> >> __ >> R-help@stat.math.ethz.ch mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posting- >> guide.html >> and provide commented, minimal, self-contained, reproducible code. > > __ > R-help@stat.math.ethz.ch 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. > > > > > > > ___ > Découvrez une nouvelle façon d'obtenir des réponses à toutes vos questions ! > Profitez des connaissances, des opinions et des expériences des internaut > > [[alternative HTML version deleted]] > > > > > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch 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] Code of plot.TukeyHSD
getAnywhere("plot.TukeyHSD")
On Fri, 24 Nov 2006, Lize van der Merwe wrote:
> Dear list
> I need to see the code of plot.TukeyHSD. There is no .default for it.
> Please help.
> Lize
--
Brian D. Ripley, [EMAIL PROTECTED]
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] barplot help needed
tab <- do.call(rbind, list(data1, data2, data3, data4)) etype <- rep(c(sd1, sd2, sd3, sd4), length(data1)) b <- barplot(tab, beside=T) arrows(unlist(b), unlist(tab) - etype, unlist(b), unlist(tab) + etype, code=3) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Antje a écrit : > hello, > > I would like to create the following barplot: > > I have 4 different data sets (same length + stddev for each data point) > > data1 > sd1 > data2 > sd2 > data3 > sd3 > data4 > sd4 > > now, I'd like to plot in the following way: > > data1[1],data2[1],data3[1],data4[1] with it's sd-values side-by-side at > one x-axis label (named "position 1") and each bar in different colors. > > data1[2],data2[2],data3[2],data4[2] at the next x-axis label (named > "position 2") with the same color scheme > > and so on over the whole length. > > I managed to plot one set in the following way: > > par(mai=c(1.5,1,1,0.6)) > plotInfo <- barplot(data1, las=2, ylim = c(0,plotMax+1), ylab = > "Percentage") > arrows(plotInfo,data1,plotInfo, data1 + sd1, length=0.1, angle=90) > arrows(plotInfo,data1,plotInfo, data1 - sd1, length=0.1, angle=90) > > could anybody give me a help on this? > > Antje > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > __ R-help@stat.math.ethz.ch 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] t.test()
On 23 Nov 2006, at 13:46, Peter Dalgaard wrote:
> Robin Hankin <[EMAIL PROTECTED]> writes:
>
>> Hi
>>
>> I have a vector x of length n. I am interested in x[1]
>> being different from the other observations (ie x[-1]).
>>
[snip]
>>
>> What arguments do I need to send to t.test() to test my null?
>
>
[snip]
> Alternatively, just write up the formula for the t statistic:
>
>> x <- c(23,25,29,27,30,30)
>> (x[1]-mean(x[-1]))/sqrt(var(x[-1])*(1+1/(length(x)-1)))
>
The "gotcha" in Peter Dalgaard's formula is that the maximum
likelihood estimate for the
variance, with a sample size of one, is zero. This is why var[x[1]]
doesn't appear.
[R reports var(1) as NA because it uses the unbiased formula with
(n-1) on the denominator,
as documented]
Last night I derived the likelihood test for testing my null. Consider
H1: x~N(mu_x,s^2); y~N(mu_y,s^2)
H2: x,y~N(mu,s^2)
The support gained by allowing the two means to differ [ie compare H2
to H1] is:
\[
E=
\frac{n}{2}\ln\left(
\frac{\sum(z_i-\overline{z})^2}{
\sum(x_i-\overline{x})^2+
\sum(y_i-\overline{y})^2
}\right)
\]
where z=c(x,y) is both sets of observations taken together. This
formula supposes one
uses the appropriate maximum likelihood estimate for the (common)
variance. Note that
the MLE for the variance is different on H1 and H2.
Thus if E > 2 we can reject H2 and if !(E <2) we can accept (sic) H2.
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
tel 023-8059-7743
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Re: [R] statistics help
Dear Rohan,
Why would you want to simulate these probabilities? As far as I can tell by
your description these are all solvable analytically, see eg Kemeney,
Mirkil, Snell & Thompson, 1958, Finite Mathematical Structures. There are
undoubtedly more recent publications that cover first passage times in
Markov models, which is what you are looking for.
Hth, Ingmar
> From: Rohan Saldanha <[EMAIL PROTECTED]>
> Date: Thu, 23 Nov 2006 19:07:54 +
> To:
> Subject: [R] statistics help
>
> hi
>
> im a bioinformatics student as i have never had any previous programming
> experience
> i need help
>
> this is the question i need to answer:
>
> Random walk model
>
> we want to model a random walk where you take a step to the left with
> probability p
> and one to the right with probability 1-p. Now assume that there is a line
> of 11 squares.
> once you are in square 0 or in square 10 the walk ends. The aim of this
> problem set is
> to write a simulation for the random walk and analyse its dynamics
>
> 1. write a simulation for a random walk which allows you to calculate the
> probability of ending
> up in square 0 starting from any other square.
>
> 2. analyse the probability of ending up in square 0 starting from any other
> square. Also calculate the mean time until you have reached square 1 for p=
> 0.1,0.2,0.3,0.4 and 0.5. What is the probability of reaching square 10 for
> these parameters
> hint simulate each scenario 1000 times and plot on histogram
>
>
> this is the code that i have come up with but its not working very well.
>
> rw<-function(sw,p,nrep){
> Z=0
> T=0
> count=0
>for (i in 1:nrep)
>{
> n<-0
> s=sw
>
> while (s>0 & s<10)
> {
> x<-runif(1, min=0, max=1)
> if (x else{s<-s+1}
> print (s)
> n<-n+1
> }
> count<-count+n
> print (count)
>
> if (s==0) {Z<-Z+1}
> else {T<-T+1}
>}
> PrZ=Z/nrep
> print(c("PrZ",PrZ),quote=FALSE)
> }
>
> if you could shed some light on this it woul be really helpful. please let
> me know how much
> you would like as payment aswell.
>
> thanks
> rohan
>
> _
> Eat well and eat right. Get tips on nutrition from Naini Setalvad
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] how to loop this?[Solved]
Petr Pikal escribió:
> Hi
>
> when working with lists, considerable option is to look on lapply,
> sapply, mapply or other *apply functions. If I am not mistaken you
> can do
>
> ttt<-lapply(lasker, function(x)
> data.frame(table(substr(names(subset(x, x >=4)), 1, 7
>
> to get a list with desired output.
>
> HTH
> Petr
>
Dear Petr,
It works great!!
Many Thanks,
Antonio
>
> On 23 Nov 2006 at 21:06, antonio rodriguez wrote:
>
> Date sent:Thu, 23 Nov 2006 21:06:18 +0100
> From: antonio rodriguez <[EMAIL PROTECTED]>
> To: R-Help
> Subject: [R] how to loop this?
>
>
>> Hi,
>>
>> I have the next procedure:
>>
>> t1<-data.frame(table(substr(names(subset(lasker[[1]], lasker[[1]] >=
>> 4)), 1, 7)))
>>
>> t1[1:5,]
>>
>> Var1 Freq
>> 1 1988-023
>> 2 1988-031
>> 3 1988-041
>> 4 1988-052
>> 5 1988-063
>>
>> How to make a new list?, dataframe? having 189 elements in the
>> 'lasker' list:
>>
>> > str(lasker[[1]])
>> 'table' int [, 1:1274] 1 1 3 2 1 5 4 1 1 4 ...
>> - attr(*, "dimnames")=List of 1
>> ..$ : chr [1:1274] "1988-01-13" "1988-01-16" "1988-01-20"
>> "1988-01-25" ...
>> .
>> .
>> > str(lasker[[189]])
>> 'table' int [, 1:464] 1 1 4 1 4 4 6 2 3 3 ...
>> - attr(*, "dimnames")=List of 1
>> ..$ : chr [1:464] "1999-07-21" "1999-07-23" "1999-07-25"
>> "1999-07-31" ...
>>
>>
>> I've tried:
>>
>> windows<-vector("list",189)
>>
>> for (i in 1:189){
>> windows[[i]]<-data.frame(table(substr(names(subset(lasker[[i]],
>> lasker[[i]] >= 4)), 1, 7)))
>> }
>>
>> Erro en rep.int(rep.int(seq_len(nx), rep.int(rep.fac, nx)), orep) :
>> invalid number of copies in rep.int()
>>
>> I think this is due because each new element (like t1) has 2 columns.
>> But what kind of structure is needed to accomodate the new elements?
>>
>> Thanks
>>
>> Antonio
>>
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Re: [R] command option for R CMD BATCH
Thanks.
R.
On Thursday 23 November 2006 16:32, Prof Brian Ripley wrote:
> On Thu, 23 Nov 2006, Ramon Diaz-Uriarte wrote:
> > On Thursday 23 November 2006 15:44, Prof Brian Ripley wrote:
> >> Try this:
> >>
> >> gannet% cat month.R
> >> x <- commandArgs()
> >> print(x[length(x)])
> >>
> >> gannet% R --slave --args January < month.R
> >> [1] "January"
> >
> > Is the above
> > "R --slave --args January < month.R"
> > the preferred way of using it?
>
> Yes it is. That's exactly what --args was added to allow.
>
> > I tend to use
> >
> > "R --slave < month.R January"
> >
> > instead (as a consequence of reconverting former scripts that used R CMD
> > BATCH). The second call produces a "ARGUMENT 'January' __ignored__" but
> > otherwise seems to do the same thing.
--
Ramón Díaz-Uriarte
Bioinformatics
Centro Nacional de Investigaciones Oncológicas (CNIO)
(Spanish National Cancer Center)
Melchor Fernández Almagro, 3
28029 Madrid (Spain)
Fax: +-34-91-224-6972
Phone: +-34-91-224-6900
http://ligarto.org/rdiaz
PGP KeyID: 0xE89B3462
(http://ligarto.org/rdiaz/0xE89B3462.asc)
**NOTA DE CONFIDENCIALIDAD** Este correo electrónico, y en s...{{dropped}}
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[R] VB linked to R
help I'd like to develop an VB 6 application and use R as the "backgroud calculator". How do I call R with its functions and link it to the VB interface? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Advice on Grid Data
Dear All, I would like to automate the analysis and plotting of data taken from a grid. Typically I deal with 2 spatial coordinates and a scalar f(x,y), but the spatial grid is not evenly spaced at all and usually given in this form: x y f(x,y) 0.0 0.048979383 2.7659438106975056 0.0 0.044986665 2.603891585041688 0.0023807306 0.04787451 2.715949356768243 0.0 0.040993948 2.469223979694342 0.0023807306 0.043881793 2.5625191444824265 0.004761461 0.046769638 2.6629703119429022 0.0 0.03700123 2.361940994655468 0.0023807306 0.039889075 2.436480700580665 0.004761461 0.04277692 2.517958884562618 0.0071421918 0.045664765 2.606844303834078 0.0 0.060880877 3.470808435449538 0.0 0.05691371 3.1907723461238686 0.0020467786 0.059650626 3.3672237912200016 0.0 0.05294655 2.9558174712065237 0.0020467786 0.055683464 3.1075221272152054 0.004093557 0.05842038 3.268965886866726 0.0020467786 0.051716298 2.8929170062920653 0.004093557 0.054453213 3.029154848759894 0.006140336 0.057190128 3.1754081073235088 0.0 0.02473231 2.138648866573983 0.0 0.020556964 2.092324627395541 I tried the image plot and lattice but unsuccessfully. Now I am reading about the sp package ( http://cran.r-project.org/src/contrib/Descriptions/sp.html ), but I mainly would like a piece of advice about what tools to use and how to read and plot these data (I suppose it must be common e.g. in geography to deal with this kind of problems). Kind Regards Lorenzo __ R-help@stat.math.ethz.ch 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] VB linked to R
See the rw-FAQ Q2.18, since this must surely be under Windows (unstated). On Thu, 23 Nov 2006, Khaled Radhouane wrote: > help > I'd like to develop an VB 6 application and use R as > the "backgroud calculator". > How do I call R with its functions and link it to the VB interface? > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- Brian D. Ripley, [EMAIL PROTECTED] 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 __ R-help@stat.math.ethz.ch 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] show unique combination of two factor
Aimin Yan wrote:
> p factor have 5 levels
> aa factor have 19 levels.
> totally it should have 95 combinations.
> but I just find there are 92 combinations.
> Does anyone know how to code to find what combinations are missed?
Here is an example with fewer factor levels of one way you might do this:
df <- data.frame(p = rep(c("A","B","C","D"), each=10),
aa = rep(c("Yes","No"), 20))
df$aa <- replace(df$aa, df$p == "D", "No")
table(df)
aa
p No Yes
A 5 5
B 5 5
C 5 5
D 10 0
names(which(with(df, table(interaction(p, aa))) == 0))
[1] "D.Yes"
> Thanks,
>
> Aimin
>
> __
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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.
--
Chuck Cleland, Ph.D.
NDRI, Inc.
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 512-0171 (M, W, F)
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Re: [R] show unique combination of two factor
At 22:02 23/11/2006, Aimin Yan wrote: >p factor have 5 levels >aa factor have 19 levels. >totally it should have 95 combinations. >but I just find there are 92 combinations. >Does anyone know how to code to find what combinations are missed? Does which(table(p,aa) == 0, arr.ind=TRUE) do what you wanted. >Thanks, > >Aimin > > Michael Dewey http://www.aghmed.fsnet.co.uk __ R-help@stat.math.ethz.ch 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] random effect question and glm
At 21:52 23/11/2006, Aimin Yan wrote: >consider p as random effect with 5 levels, what is difference between these >two models? > > > p5.random.p <- lmer(Y >~p+(1|p),data=p5,family=binomial,control=list(usePQL=FALSE,msV=1)) > > p5.random.p1 <- lmer(Y >~1+(1|p),data=p5,family=binomial,control=list(usePQL=FALSE,msV=1)) Well, try them and see. Then if you cannot understand the output tell us a) what you found b) how it differed from what you expected >in addtion, I try these two models, it seems they are same. >what is the difference between these two model. Is this a cell means model? >m00 <- glm(sc ~aa-1,data = p5) >m000 <- glm(sc ~1+aa-1,data = p5) See above >thanks, > >Aimin Yan > > Michael Dewey http://www.aghmed.fsnet.co.uk __ R-help@stat.math.ethz.ch 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] Advice on Grid Data
On Fri, 24 Nov 2006, Lorenzo Isella wrote: > Dear All, > I would like to automate the analysis and plotting of data taken from a grid. > Typically I deal with 2 spatial coordinates and a scalar f(x,y), but > the spatial grid is not evenly spaced at all and usually given in this > form: > x y f(x,y) > 0.0 0.048979383 2.7659438106975056 > 0.0 0.044986665 2.603891585041688 > 0.0023807306 0.04787451 2.715949356768243 > 0.0 0.040993948 2.469223979694342 > 0.0023807306 0.043881793 2.5625191444824265 > 0.004761461 0.046769638 2.6629703119429022 > 0.0 0.03700123 2.361940994655468 > 0.0023807306 0.039889075 2.436480700580665 > 0.004761461 0.04277692 2.517958884562618 > 0.0071421918 0.045664765 2.606844303834078 > 0.0 0.060880877 3.470808435449538 > 0.0 0.05691371 3.1907723461238686 > 0.0020467786 0.059650626 3.3672237912200016 > 0.0 0.05294655 2.9558174712065237 > 0.0020467786 0.055683464 3.1075221272152054 > 0.004093557 0.05842038 3.268965886866726 > 0.0020467786 0.051716298 2.8929170062920653 > 0.004093557 0.054453213 3.029154848759894 > 0.006140336 0.057190128 3.1754081073235088 > 0.0 0.02473231 2.138648866573983 > 0.0 0.020556964 2.092324627395541 > > I tried the image plot and lattice but unsuccessfully. Now I am > reading about the sp package ( > http://cran.r-project.org/src/contrib/Descriptions/sp.html ), but I > mainly would like a piece of advice about what tools to use and how to > read and plot these data (I suppose it must be common e.g. in > geography to deal with this kind of problems). Your data are not on a 2D grid, there are (here) 7 unique x values, but 21 unique y values of 21. You can treat the data as a SpatialPointsDataFrame (see note in R News in 2005), but if you want to display them on an actual grid, you will have to interpolate. For more ideas, perhaps try the R-sig-geo mailing list. > Kind Regards > > Lorenzo > > __ > R-help@stat.math.ethz.ch 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. > -- Roger Bivand Economic Geography Section, Department of Economics, Norwegian School of Economics and Business Administration, Helleveien 30, N-5045 Bergen, Norway. voice: +47 55 95 93 55; fax +47 55 95 95 43 e-mail: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Diebold Mariano Test
Dear List Has anyone used R to distnguish between alternative forecasting models? In particular is the Diebold Mariano test available for use within R. Any assistance would be greatly appreciated. Graham Leask Lecturer in Strategy Economics & Strategy Group Aston University Aston Triangle Birmingham B4 7ET Tel: 0121 204 3150 E Mail: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] axis in an image() plot
Hi all, I've found quite usefull colored-grid created by image() but I'm facing a doubt I am not able to solve. Given the following data rectangle... 1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 12 22 0 7 2 1 0 2 0 2 6 -3 0 3 2 0 -1 0 9 3 -4 0 0 0 0 3 0 0 0 3 29 45 6 12 16 85 -2 0 -3 -4 89 -1 -1 1 4 2 9 3 6 17 3 -2 -9 -2 8 -1 0 0 0 5 44 16 -3 21 23 3 2 1 0 -2 13 18 -5 2 I am not able to draw x and y axis labeled 1 to 14 and 1 to 5 by 1. I've tried a number of options by using axis() to no avail. It will be also very helpfull to be able to draw the value of each x,y combination within its position in the grid. Text() seems to be the answer, but I am not able yet to get the correct position for each label. Please, could you point me in the right direction or offer some example? Thank you in advance, -- Ricardo Rodríguez Your XEN ICT Team __ R-help@stat.math.ethz.ch 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] soil.texture() function with bug?
Hi Patrick, Not sure what the problem is from your email. Can you send a code example which reproduces the error? Make sure you mention the version of R you are using! Also send your question/reply to/cc R-help as well. Then the rest of the world is there to help you too. Greetings, Sander. Patrick Kuss wrote: > Dear Sander Oom and Jim Lemon, > > thanks for putting the soils.texture() function into R. However, for > whatever reason I am not able to display the triangle correctly. Each of > the 27 tick labels shows as "c(10,20,30,40,50,60,70,80,90)" and thus > basically cover the whole triangle. Plotting points and adding graphical > paramters like 'col.symbols' works fine. I am also able to plot my soil > data using triax.plot() without any problems, but of course, the nice > feature of soil type areas is not available. > > Do you have any insights? > > Thanks a lot and cheers from Alaska > > Patrick > __ R-help@stat.math.ethz.ch 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] barplot help needed
Thank you very much for your help. I just don't understand the following line (which also gives me a dimension error later in the arrows command) etype <- rep(c(sd1, sd2, sd3, sd4), length(data1)) Antje (I don't see my emails to the mailinglist anymore... just the answers from other people... I don't understand???) Jacques VESLOT schrieb: > tab <- do.call(rbind, list(data1, data2, data3, data4)) > etype <- rep(c(sd1, sd2, sd3, sd4), length(data1)) > b <- barplot(tab, beside=T) > arrows(unlist(b), unlist(tab) - etype, unlist(b), unlist(tab) + etype, > code=3) > --- > Jacques VESLOT > > CNRS UMR 8090 > I.B.L (2ème étage) > 1 rue du Professeur Calmette > B.P. 245 > 59019 Lille Cedex > > Tel : 33 (0)3.20.87.10.44 > Fax : 33 (0)3.20.87.10.31 > > http://www-good.ibl.fr > --- > > Antje a écrit : >> hello, >> >> I would like to create the following barplot: >> >> I have 4 different data sets (same length + stddev for each data point) >> >> data1 >> sd1 >> data2 >> sd2 >> data3 >> sd3 >> data4 >> sd4 >> >> now, I'd like to plot in the following way: >> >> data1[1],data2[1],data3[1],data4[1] with it's sd-values side-by-side >> at one x-axis label (named "position 1") and each bar in different >> colors. >> >> data1[2],data2[2],data3[2],data4[2] at the next x-axis label (named >> "position 2") with the same color scheme >> >> and so on over the whole length. >> >> I managed to plot one set in the following way: >> >> par(mai=c(1.5,1,1,0.6)) >> plotInfo <- barplot(data1, las=2, ylim = c(0,plotMax+1), ylab = >> "Percentage") >> arrows(plotInfo,data1,plotInfo, data1 + sd1, length=0.1, angle=90) >> arrows(plotInfo,data1,plotInfo, data1 - sd1, length=0.1, angle=90) >> >> could anybody give me a help on this? >> >> Antje >> >> __ >> R-help@stat.math.ethz.ch mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide >> http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> > __ R-help@stat.math.ethz.ch 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] Diebold Mariano Test
Not using R, but you can have a look at the RATS source code DMARIANO.SRCavailable on page http://www.estima.com/Forecasting.shtml and translate it into R. Hannu On 11/24/06, Graham Leask <[EMAIL PROTECTED]> wrote: > > Dear List > > Has anyone used R to distnguish between alternative forecasting models? In > particular > is the Diebold Mariano test available for use within R. > > Any assistance would be greatly appreciated. > > > > Graham Leask > Lecturer in Strategy > Economics & Strategy Group > Aston University > Aston Triangle > Birmingham > B4 7ET > > Tel: 0121 204 3150 > E Mail: [EMAIL PROTECTED] > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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] Using roman and italic fonts in an xlab expression for a plot
or even shorter:
plot(1, xlab = "Level is" ~ italic(M))
On 11/23/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> Try this:
>
> plot(1, xlab = quote("Level is" ~ italic(M)))
>
> or
>
> plot(1, xlab = quote(Level ~ is ~ italic(M))
>
>
> On 11/23/06, Philip Boland <[EMAIL PROTECTED]> wrote:
> > Just wondering if it is possible to put something like
> >
> > "The excess level is M"
> >
> > in the xlab position of a plot but where the M is in italic font and
> > the rest in ordinary (or Roman) font.
> >
> > Thanks - Phil Boland
> >
> > __
> > R-help@stat.math.ethz.ch 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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[R] Plot with two seperate y axis
Hello, I would like to plot the following matrix: Wk=x achsis. Para 1 = left y-axis as a barplot para 2 right y-axis as a normal scatter plat. I could not find such a solution in any of my documentation. Can someone help me? Thanks a lot Thorsten WkPara 1 Para 2 312000 99.8 322005 99.0 332002 98.0 341090 98.5 352001 99.1 362010 97.0 372010 98.8 __ R-help@stat.math.ethz.ch 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] barplot help needed
thought sd1, sd2... were scalars but if not just do: etype <- c(sd1, sd2, sd3, sd4) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Antje a écrit : > Thank you very much for your help. > I just don't understand the following line (which also gives me a > dimension error later in the arrows command) > > etype <- rep(c(sd1, sd2, sd3, sd4), length(data1)) > > Antje > > (I don't see my emails to the mailinglist anymore... just the answers > from other people... I don't understand???) > > > Jacques VESLOT schrieb: > >>tab <- do.call(rbind, list(data1, data2, data3, data4)) >>etype <- rep(c(sd1, sd2, sd3, sd4), length(data1)) >>b <- barplot(tab, beside=T) >>arrows(unlist(b), unlist(tab) - etype, unlist(b), unlist(tab) + etype, >>code=3) >>--- >>Jacques VESLOT >> >>CNRS UMR 8090 >>I.B.L (2ème étage) >>1 rue du Professeur Calmette >>B.P. 245 >>59019 Lille Cedex >> >>Tel : 33 (0)3.20.87.10.44 >>Fax : 33 (0)3.20.87.10.31 >> >>http://www-good.ibl.fr >>--- >> >>Antje a écrit : >> >>>hello, >>> >>>I would like to create the following barplot: >>> >>>I have 4 different data sets (same length + stddev for each data point) >>> >>>data1 >>>sd1 >>>data2 >>>sd2 >>>data3 >>>sd3 >>>data4 >>>sd4 >>> >>>now, I'd like to plot in the following way: >>> >>>data1[1],data2[1],data3[1],data4[1] with it's sd-values side-by-side >>>at one x-axis label (named "position 1") and each bar in different >>>colors. >>> >>>data1[2],data2[2],data3[2],data4[2] at the next x-axis label (named >>>"position 2") with the same color scheme >>> >>>and so on over the whole length. >>> >>>I managed to plot one set in the following way: >>> >>>par(mai=c(1.5,1,1,0.6)) >>>plotInfo <- barplot(data1, las=2, ylim = c(0,plotMax+1), ylab = >>>"Percentage") >>>arrows(plotInfo,data1,plotInfo, data1 + sd1, length=0.1, angle=90) >>>arrows(plotInfo,data1,plotInfo, data1 - sd1, length=0.1, angle=90) >>> >>>could anybody give me a help on this? >>> >>>Antje >>> >>>__ >>>R-help@stat.math.ethz.ch mailing list >>>https://stat.ethz.ch/mailman/listinfo/r-help >>>PLEASE do read the posting guide >>>http://www.R-project.org/posting-guide.html >>>and provide commented, minimal, self-contained, reproducible code. >>> >> > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > __ R-help@stat.math.ethz.ch 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] Advice on Grid Data
On 24/11/06, Roger Bivand <[EMAIL PROTECTED]> wrote: > On Fri, 24 Nov 2006, Lorenzo Isella wrote: > > > Dear All, > > I would like to automate the analysis and plotting of data taken from a > > grid. > > Typically I deal with 2 spatial coordinates and a scalar f(x,y), but > > the spatial grid is not evenly spaced at all and usually given in this > > form: > > x y > > f(x,y) > > 0.0 0.048979383 2.7659438106975056 > > 0.0 0.044986665 2.603891585041688 > > 0.0023807306 0.04787451 2.715949356768243 > > 0.0 0.040993948 2.469223979694342 > > 0.0023807306 0.043881793 2.5625191444824265 > > 0.004761461 0.046769638 2.6629703119429022 > > 0.0 0.03700123 2.361940994655468 > > 0.0023807306 0.039889075 2.436480700580665 > > 0.004761461 0.04277692 2.517958884562618 > > 0.0071421918 0.045664765 2.606844303834078 > > 0.0 0.060880877 3.470808435449538 > > 0.0 0.05691371 3.1907723461238686 > > 0.0020467786 0.059650626 3.3672237912200016 > > 0.0 0.05294655 2.9558174712065237 > > 0.0020467786 0.055683464 3.1075221272152054 > > 0.004093557 0.05842038 3.268965886866726 > > 0.0020467786 0.051716298 2.8929170062920653 > > 0.004093557 0.054453213 3.029154848759894 > > 0.006140336 0.057190128 3.1754081073235088 > > 0.0 0.02473231 2.138648866573983 > > 0.0 0.020556964 2.092324627395541 > > > > I tried the image plot and lattice but unsuccessfully. Now I am > > reading about the sp package ( > > http://cran.r-project.org/src/contrib/Descriptions/sp.html ), but I > > mainly would like a piece of advice about what tools to use and how to > > read and plot these data (I suppose it must be common e.g. in > > geography to deal with this kind of problems). > > Your data are not on a 2D grid, there are (here) 7 unique x values, but 21 > unique y values of 21. You can treat the data as a SpatialPointsDataFrame > (see note in R News in 2005), but if you want to display them on an actual > grid, you will have to interpolate. For more ideas, perhaps try the > R-sig-geo mailing list. > > > Kind Regards > > > > Lorenzo > > > > __ > > R-help@stat.math.ethz.ch 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. > > > > -- > Roger Bivand > Economic Geography Section, Department of Economics, Norwegian School of > Economics and Business Administration, Helleveien 30, N-5045 Bergen, > Norway. voice: +47 55 95 93 55; fax +47 55 95 95 43 > e-mail: [EMAIL PROTECTED] > > Thanks for the advice, but I hope I have not been misleading: only part of the grid coordinates are reported here (the whole list is very long). Does your conclusion hold anyway? Cheers Lorenzo __ R-help@stat.math.ethz.ch 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] barplot help needed
Is the length of all your datasets equal? If not try etype <- rep(c(sd1, sd2, sd3, sd4), c(length(data1), length(data2), length(data3), length(data4)) Cheers, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Reseach Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 [EMAIL PROTECTED] www.inbo.be Do not put your faith in what statistics say until you have carefully considered what they do not say. ~William W. Watt A statistical analysis, properly conducted, is a delicate dissection of uncertainties, a surgery of suppositions. ~M.J.Moroney -Oorspronkelijk bericht- Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Namens Antje Verzonden: vrijdag 24 november 2006 13:42 Aan: r-help@stat.math.ethz.ch Onderwerp: Re: [R] barplot help needed Thank you very much for your help. I just don't understand the following line (which also gives me a dimension error later in the arrows command) etype <- rep(c(sd1, sd2, sd3, sd4), length(data1)) Antje (I don't see my emails to the mailinglist anymore... just the answers from other people... I don't understand???) Jacques VESLOT schrieb: > tab <- do.call(rbind, list(data1, data2, data3, data4)) > etype <- rep(c(sd1, sd2, sd3, sd4), length(data1)) > b <- barplot(tab, beside=T) > arrows(unlist(b), unlist(tab) - etype, unlist(b), unlist(tab) + etype, > code=3) > --- > Jacques VESLOT > > CNRS UMR 8090 > I.B.L (2ème étage) > 1 rue du Professeur Calmette > B.P. 245 > 59019 Lille Cedex > > Tel : 33 (0)3.20.87.10.44 > Fax : 33 (0)3.20.87.10.31 > > http://www-good.ibl.fr > --- > > Antje a écrit : >> hello, >> >> I would like to create the following barplot: >> >> I have 4 different data sets (same length + stddev for each data point) >> >> data1 >> sd1 >> data2 >> sd2 >> data3 >> sd3 >> data4 >> sd4 >> >> now, I'd like to plot in the following way: >> >> data1[1],data2[1],data3[1],data4[1] with it's sd-values side-by-side >> at one x-axis label (named "position 1") and each bar in different >> colors. >> >> data1[2],data2[2],data3[2],data4[2] at the next x-axis label (named >> "position 2") with the same color scheme >> >> and so on over the whole length. >> >> I managed to plot one set in the following way: >> >> par(mai=c(1.5,1,1,0.6)) >> plotInfo <- barplot(data1, las=2, ylim = c(0,plotMax+1), ylab = >> "Percentage") >> arrows(plotInfo,data1,plotInfo, data1 + sd1, length=0.1, angle=90) >> arrows(plotInfo,data1,plotInfo, data1 - sd1, length=0.1, angle=90) >> >> could anybody give me a help on this? >> >> Antje >> >> __ >> R-help@stat.math.ethz.ch mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide >> http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch 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] Plot with two seperate y axis
Assuming the data are in a data frame called dt, this should work: > plot(dt$Wk,dt$Para1,type="h") > par(new=TRUE) > plot(dt$Wk,dt$Para2,yaxt="n") > axis(4,at=97:100) On 24/11/06, Thorsten Muehge <[EMAIL PROTECTED]> wrote: > > Hello, > I would like to plot the following matrix: > Wk=x achsis. > Para 1 = left y-axis as a barplot > para 2 right y-axis as a normal scatter plat. > > I could not find such a solution in any of my documentation. > > Can someone help me? > > Thanks a lot > Thorsten > > WkPara 1 Para 2 > 312000 99.8 > 322005 99.0 > 332002 98.0 > 341090 98.5 > 352001 99.1 > 362010 97.0 > 372010 98.8 > > __ > R-help@stat.math.ethz.ch 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. > -- = David Barron Said Business School University of Oxford Park End Street Oxford OX1 1HP __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to find AUC in SVM (kernlab package)
Dear all,
I was wondering if someone can help me. I am learning SVM for
classification in my research with kernlab package. I want to know about
classification performance using Area Under Curve (AUC). I know ROCR
package can do this job but I found all example in ROCR package have
include prediction, for example, ROCR.hiv {ROCR}. My problem is how to
produce prediction in SVM and to find AUC.
Here is a simple example:
library(MASS)
library(kernlab)
library(ROCR)
pimamodel <- ksvm(type ~ .,data=Pima.tr,type="C-svc",C=10,prob.model=TRUE)
pimamodel
fitted(pimamodel)
pima.pred <- predict(pimamodel, Pima.te[,-8], type="probabilities")
pima.pred
# try to find AUC
#predid.no <- prediction(pima.pred[,1], Pima.te[,8])
#predid.yes <- prediction(pima.pred[,2], Pima.te[,8])
predid <- prediction(pima.pred, Pima.te[,8])
perfid <- performance(predid,"tpr","fpr")
perfid.auc <- performance(predid,"auc")
perfid.auc
Thank you very much for your help.
Best wishes, Muhammad Subianto
__
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Re: [R] Advice on Grid Data
On Fri, 24 Nov 2006, Lorenzo Isella wrote:
> On 24/11/06, Roger Bivand <[EMAIL PROTECTED]> wrote:
> > On Fri, 24 Nov 2006, Lorenzo Isella wrote:
> >
> > > Dear All,
> > > I would like to automate the analysis and plotting of data taken from a
> > > grid.
> > > Typically I deal with 2 spatial coordinates and a scalar f(x,y), but
> > > the spatial grid is not evenly spaced at all and usually given in this
> > > form:
> > > x y
> > > f(x,y)
> > > 0.0 0.048979383 2.7659438106975056
> > > 0.0 0.044986665 2.603891585041688
> > > 0.0023807306 0.04787451 2.715949356768243
> > > 0.0 0.040993948 2.469223979694342
> > > 0.0023807306 0.043881793 2.5625191444824265
> > > 0.004761461 0.046769638 2.6629703119429022
> > > 0.0 0.03700123 2.361940994655468
> > > 0.0023807306 0.039889075 2.436480700580665
> > > 0.004761461 0.04277692 2.517958884562618
> > > 0.0071421918 0.045664765 2.606844303834078
> > > 0.0 0.060880877 3.470808435449538
> > > 0.0 0.05691371 3.1907723461238686
> > > 0.0020467786 0.059650626 3.3672237912200016
> > > 0.0 0.05294655 2.9558174712065237
> > > 0.0020467786 0.055683464 3.1075221272152054
> > > 0.004093557 0.05842038 3.268965886866726
> > > 0.0020467786 0.051716298 2.8929170062920653
> > > 0.004093557 0.054453213 3.029154848759894
> > > 0.006140336 0.057190128 3.1754081073235088
> > > 0.0 0.02473231 2.138648866573983
> > > 0.0 0.020556964 2.092324627395541
> > >
> > > I tried the image plot and lattice but unsuccessfully. Now I am
> > > reading about the sp package (
> > > http://cran.r-project.org/src/contrib/Descriptions/sp.html ), but I
> > > mainly would like a piece of advice about what tools to use and how to
> > > read and plot these data (I suppose it must be common e.g. in
> > > geography to deal with this kind of problems).
> >
> > Your data are not on a 2D grid, there are (here) 7 unique x values, but 21
> > unique y values of 21. You can treat the data as a SpatialPointsDataFrame
> > (see note in R News in 2005), but if you want to display them on an actual
> > grid, you will have to interpolate. For more ideas, perhaps try the
> > R-sig-geo mailing list.
> >
> > > Kind Regards
> > >
> > > Lorenzo
> > >
> > > __
> > > R-help@stat.math.ethz.ch 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.
> > >
> >
> > --
> > Roger Bivand
> > Economic Geography Section, Department of Economics, Norwegian School of
> > Economics and Business Administration, Helleveien 30, N-5045 Bergen,
> > Norway. voice: +47 55 95 93 55; fax +47 55 95 95 43
> > e-mail: [EMAIL PROTECTED]
> >
> >
>
> Thanks for the advice, but I hope I have not been misleading: only
> part of the grid coordinates are reported here (the whole list is very
> long).
> Does your conclusion hold anyway?
Well, if you convert your data to a SpatialPointsDataFrame, and then try
to go to a SpatialPixelsDataFrame, you'll see if you succeed. A
SpatialGridDataFrame requires data at all the cells in the grid, which you
do not seem to have, so if you can get to a SpatialPixelsDataFrame, you
will have display methods. If you look at the example on:
?"SpatialPixels-class"
and try to get to:
data(meuse.grid)
pts = meuse.grid[c("x", "y")]
y = SpatialPixels(SpatialPoints(pts))
class(y)
y
then you should be able to do the same with your data present, see
?"SpatialPixelsDataFrame-class"
Roger
> Cheers
>
> Lorenzo
>
--
Roger Bivand
Economic Geography Section, Department of Economics, Norwegian School of
Economics and Business Administration, Helleveien 30, N-5045 Bergen,
Norway. voice: +47 55 95 93 55; fax +47 55 95 95 43
e-mail: [EMAIL PROTECTED]
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[R] Fwd: Dates Conversion/write.foreign
-- Forwarded message -- From: Shubha Vishwanath Karanth <[EMAIL PROTECTED]> Date: Nov 24, 2006 7:54 PM Subject: Dates Conversion/write.foreign To: Shubha Karanth <[EMAIL PROTECTED]>, Shubha Vishwanath Karanth <[EMAIL PROTECTED]> Hi R experts, I need an urgent help... I have an a dataframe caled idat. Below i give a snapshot of it. Datetime Volume_a Volume_b (11/21/06 12:50:00) (11/21/06 12:50:00) 00 (11/21/06 13:00:00) (11/21/06 13:00:00) 00 (11/21/06 13:10:00) (11/21/06 13:10:00) 00 (11/21/06 13:20:00) (11/21/06 13:20:00) 00 (11/21/06 13:30:00) (11/21/06 13:30:00) 342113 0 (11/21/06 13:40:00) (11/21/06 13:40:00) 695071 0 (11/21/06 13:50:00) (11/21/06 13:50:00) 470943 4690 (11/21/06 14:00:00) (11/21/06 14:00:00) 870072 0 (11/21/06 14:10:00) (11/21/06 14:10:00) 1010101 2000 (11/21/06 14:20:00) (11/21/06 14:20:00) 71428750 (11/21/06 14:30:00) (11/21/06 14:30:00) 388716 1780 (11/21/06 14:40:00) (11/21/06 14:40:00) 380038 1245 The type of each variable is given below: idat : 'data.frame':144 obs. of 3 variables: $ Datetime :Classes 'chron', 'dates', 'times' atomic [1:144] 13473 13473 13473 13473 13473 ... $ Volume_a : num 0 0 0 0 0 0 0 0 0 0 ... $ Volume_b : num 0 0 0 0 0 0 0 0 0 0 ... The variable i am interested is "Datetime". I exported this dataframe into SAS by the write.foreign command, write.foreign(idat,"Z:\\New\\idat_d.sas7bdat","Z:New\\idat_c.sas",package="SAS",dataname="f.idat") When I export, i see only the date value (Note: no time) in my SAS data set. Why is this? The write.foreign package says: "Numeric variables and factors are supported for all packages, dates and times (Date, dates, date, and POSIXt classes) are also supported for SAS and characters are supported for SPSS." So, what should i do to get my time value also apart from the datevalue? Do i need to convert the chron object, "Datetime" into POSIXt...or anything alse...so that SAS can read it as datetime. format? Thank you, Shubha. __ R-help@stat.math.ethz.ch 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] Fwd: Dates Conversion/write.foreign
Correct me if I'm wrong, but a few weeks ago my professor from the statistical computing class told us that SAS sometimes stores dates including the time but only displays the date. So it looks like the time isn't stored. This was the case with data imported from Excel. In our exercise we had to explicitly define the date as a day, month and year. So maybe you don't have a problem at all ;-) Cheers, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Reseach Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 [EMAIL PROTECTED] www.inbo.be Do not put your faith in what statistics say until you have carefully considered what they do not say. ~William W. Watt A statistical analysis, properly conducted, is a delicate dissection of uncertainties, a surgery of suppositions. ~M.J.Moroney -Oorspronkelijk bericht- Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Namens Shubha Karanth Verzonden: vrijdag 24 november 2006 15:32 Aan: r-help@stat.math.ethz.ch; [EMAIL PROTECTED]; [EMAIL PROTECTED] Onderwerp: [R] Fwd: Dates Conversion/write.foreign -- Forwarded message -- From: Shubha Vishwanath Karanth <[EMAIL PROTECTED]> Date: Nov 24, 2006 7:54 PM Subject: Dates Conversion/write.foreign To: Shubha Karanth <[EMAIL PROTECTED]>, Shubha Vishwanath Karanth <[EMAIL PROTECTED]> Hi R experts, I need an urgent help... I have an a dataframe caled idat. Below i give a snapshot of it. Datetime Volume_a Volume_b (11/21/06 12:50:00) (11/21/06 12:50:00) 00 (11/21/06 13:00:00) (11/21/06 13:00:00) 00 (11/21/06 13:10:00) (11/21/06 13:10:00) 00 (11/21/06 13:20:00) (11/21/06 13:20:00) 00 (11/21/06 13:30:00) (11/21/06 13:30:00) 342113 0 (11/21/06 13:40:00) (11/21/06 13:40:00) 695071 0 (11/21/06 13:50:00) (11/21/06 13:50:00) 470943 4690 (11/21/06 14:00:00) (11/21/06 14:00:00) 870072 0 (11/21/06 14:10:00) (11/21/06 14:10:00) 1010101 2000 (11/21/06 14:20:00) (11/21/06 14:20:00) 71428750 (11/21/06 14:30:00) (11/21/06 14:30:00) 388716 1780 (11/21/06 14:40:00) (11/21/06 14:40:00) 380038 1245 The type of each variable is given below: idat : 'data.frame':144 obs. of 3 variables: $ Datetime :Classes 'chron', 'dates', 'times' atomic [1:144] 13473 13473 13473 13473 13473 ... $ Volume_a : num 0 0 0 0 0 0 0 0 0 0 ... $ Volume_b : num 0 0 0 0 0 0 0 0 0 0 ... The variable i am interested is "Datetime". I exported this dataframe into SAS by the write.foreign command, write.foreign(idat,"Z:\\New\\idat_d.sas7bdat","Z:New\\idat_c.sas",packag e="SAS",dataname="f.idat") When I export, i see only the date value (Note: no time) in my SAS data set. Why is this? The write.foreign package says: "Numeric variables and factors are supported for all packages, dates and times (Date, dates, date, and POSIXt classes) are also supported for SAS and characters are supported for SPSS." So, what should i do to get my time value also apart from the datevalue? Do i need to convert the chron object, "Datetime" into POSIXt...or anything alse...so that SAS can read it as datetime. format? Thank you, Shubha. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch 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] Q. about the "solve" function in _package_ "Matrix"
> "yyan" == yyan liu <[EMAIL PROTECTED]>
> on Wed, 22 Nov 2006 15:04:33 -0800 (PST) writes:
yyan> Hi:
yyan> I have some problems when I use the function "solve" function in a
loop. In the following code, I have a diagonal martix "ttt" whose elements
change in every iteration in a loop. I defined a "dpoMatrix"class before the
loop so I do not need to define this class every time in the loop. The reason
is to save some computing time. The code is below. The inverse of the matrix
"ttt"
yyan> should change according to the change of "ttt" in the loop. However,
the values in "sigma.dpo.solve", which is the inverse of "ttt" does not change.
Can anybody figure out what is wrong with my code?
Well, you should not assign to the 'x' slot in this case
since if you look at
str(sigma.dpo)
you see that the matrix also has "cached" its Cholesky factor,
and the direct slot assignment does not ``see the need'' for
recomputing the Cholesky factor.
Of course, one could argue this to be a bit unfortunate,
but then, you really should only directly assign to the slots of
an S4 object if you know what you are doing ... which you did
not :-) ;-)
A more extreme view would say this to be a design flaw in the "Matrix"
*package* (not "library") that the authors have to consider.
Ideally, almost any slot reassignment of such a "Matrix" should
automatically annihilate the 'factors' slot.
yyan> Thanks a lot in advance!
you're welcome.
Martin Maechler, ETH Zurich
BTW, I hope that your real application does not work with
diagonal matrices, because these are much more efficiently
handled by Diagonal() and the "diagonalMatrix" class objects it
produces.
Here's your code amended to do what you want.
## ---
library(Matrix)
ttt <- diag(1,2)
str( sigma.dpo <- as(ttt, "dpoMatrix") )
## use one of these:
see.more <- TRUE
see.more <- FALSE
for(i in 1:5) {
cat("\n-",i,"---\n")
ttt <- diag(i,2)
## assigning to 'x' slot is ``dangerous''
## If you really want to do this, you also need to
## eliminate the cashed Cholesky factor:
[EMAIL PROTECTED] <- as.vector(ttt)
[EMAIL PROTECTED] <- list()
##
Isigma.dpo <- solve(sigma.dpo)
print(sigma.dpo)
if(see.more) ## see what's going on:
str(sigma.dpo)
print(Isigma.dpo)
}
1
## ---
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[R] odfTable: how to escape
Dear All, I'm appreciating odfWeave as a nice reporting tool, but I had some pain in producing tables with odfTable command where the first column began with < or > such as in age class heading, for example: <35 35-39 40-49 50-50 >60 In this case, to avoid a content.xml error, I had to change <35 in less than 35 and > 60 in over 60. Anyone knows how to escape those characters while producing a table in R or in a document chunk? Thanks in advance, Francesco Cernuto __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] don't put line in plot
Hi R-user, I have a problem when try to run the next code: plot(prueba$IC, type="l") but plot with type="p", there not problem, I don't know what is the problem?? Anybody can help me Regards, Fernando __ R-help@stat.math.ethz.ch 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] barplot help needed
Still, there is one problem. The SD-Values don't fit to the bar they belong to. I made the following experiment: > data1 <- c(2,4,6,2,5) > data2 <- data1 > sd1 <- c(0.5,1,1.5,1,2) > sd2 <- sd1 > tab <- do.call(rbind, list(data1, data2)) > etype <- c(sd1,sd2) > b <- barplot(tab, beside=T) > arrows(unlist(b), unlist(tab) - etype, unlist(b), unlist(tab) + etype, code=3) I expect the bars with the same height and the same stddev. The height is okay, but the stddev is messed up... if I do it like this: etype <- matrix(c(sd1,sd2), nrow=2, byrow=TRUE) it works (but maybe there is an easier way...) Antje Jacques VESLOT schrieb: > thought sd1, sd2... were scalars but if not just do: > etype <- c(sd1, sd2, sd3, sd4) > --- > Jacques VESLOT > > CNRS UMR 8090 > I.B.L (2ème étage) > 1 rue du Professeur Calmette > B.P. 245 > 59019 Lille Cedex > > Tel : 33 (0)3.20.87.10.44 > Fax : 33 (0)3.20.87.10.31 > > http://www-good.ibl.fr > --- > > > Antje a écrit : >> Thank you very much for your help. >> I just don't understand the following line (which also gives me a >> dimension error later in the arrows command) >> >> etype <- rep(c(sd1, sd2, sd3, sd4), length(data1)) >> >> Antje >> >> (I don't see my emails to the mailinglist anymore... just the answers >> from other people... I don't understand???) >> >> >> Jacques VESLOT schrieb: >> >>> tab <- do.call(rbind, list(data1, data2, data3, data4)) >>> etype <- rep(c(sd1, sd2, sd3, sd4), length(data1)) >>> b <- barplot(tab, beside=T) >>> arrows(unlist(b), unlist(tab) - etype, unlist(b), unlist(tab) + >>> etype, code=3) >>> --- >>> Jacques VESLOT >>> >>> CNRS UMR 8090 >>> I.B.L (2ème étage) >>> 1 rue du Professeur Calmette >>> B.P. 245 >>> 59019 Lille Cedex >>> >>> Tel : 33 (0)3.20.87.10.44 >>> Fax : 33 (0)3.20.87.10.31 >>> >>> http://www-good.ibl.fr >>> --- >>> >>> Antje a écrit : >>> hello, I would like to create the following barplot: I have 4 different data sets (same length + stddev for each data point) data1 sd1 data2 sd2 data3 sd3 data4 sd4 now, I'd like to plot in the following way: data1[1],data2[1],data3[1],data4[1] with it's sd-values side-by-side at one x-axis label (named "position 1") and each bar in different colors. data1[2],data2[2],data3[2],data4[2] at the next x-axis label (named "position 2") with the same color scheme and so on over the whole length. I managed to plot one set in the following way: par(mai=c(1.5,1,1,0.6)) plotInfo <- barplot(data1, las=2, ylim = c(0,plotMax+1), ylab = "Percentage") arrows(plotInfo,data1,plotInfo, data1 + sd1, length=0.1, angle=90) arrows(plotInfo,data1,plotInfo, data1 - sd1, length=0.1, angle=90) could anybody give me a help on this? Antje __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. >>> >> >> __ >> R-help@stat.math.ethz.ch mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide >> http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Error Message saying .Call("R_lazyLoadDBfetch", etc.
Hi,
I got the following error message when running a function of mine doing
intensive computations:
Erreur dans .Call("R_lazyLoadDBfetch", key, file, compressed, hook, PACKAGE =
"base") :
référence d'argument par défaut récursive
I haven't found neither where the problem lies nor what could be recursive.
Besides, I wonder whether it might or not be a problem of memory size.
Could you please give me any suggestion on how to interpret this message?
Thanks in advance,
jacques
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
http://www-good.ibl.fr
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Re: [R] barplot help needed
The arrows are messed up because they are partially outside the borders of the barplot. Try adding a good ylim to the barplot. Something like: b <- barplot(tab, beside=T, ylim = c(0, 8)) Cheers, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Reseach Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 [EMAIL PROTECTED] www.inbo.be Do not put your faith in what statistics say until you have carefully considered what they do not say. ~William W. Watt A statistical analysis, properly conducted, is a delicate dissection of uncertainties, a surgery of suppositions. ~M.J.Moroney -Oorspronkelijk bericht- Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Namens Antje Verzonden: vrijdag 24 november 2006 16:17 Aan: r-help@stat.math.ethz.ch Onderwerp: Re: [R] barplot help needed Still, there is one problem. The SD-Values don't fit to the bar they belong to. I made the following experiment: > data1 <- c(2,4,6,2,5) > data2 <- data1 > sd1 <- c(0.5,1,1.5,1,2) > sd2 <- sd1 > tab <- do.call(rbind, list(data1, data2)) > etype <- c(sd1,sd2) > b <- barplot(tab, beside=T) > arrows(unlist(b), unlist(tab) - etype, unlist(b), unlist(tab) + etype, code=3) I expect the bars with the same height and the same stddev. The height is okay, but the stddev is messed up... if I do it like this: etype <- matrix(c(sd1,sd2), nrow=2, byrow=TRUE) it works (but maybe there is an easier way...) Antje Jacques VESLOT schrieb: > thought sd1, sd2... were scalars but if not just do: > etype <- c(sd1, sd2, sd3, sd4) > --- > Jacques VESLOT > > CNRS UMR 8090 > I.B.L (2ème étage) > 1 rue du Professeur Calmette > B.P. 245 > 59019 Lille Cedex > > Tel : 33 (0)3.20.87.10.44 > Fax : 33 (0)3.20.87.10.31 > > http://www-good.ibl.fr > --- > > > Antje a écrit : >> Thank you very much for your help. >> I just don't understand the following line (which also gives me a >> dimension error later in the arrows command) >> >> etype <- rep(c(sd1, sd2, sd3, sd4), length(data1)) >> >> Antje >> >> (I don't see my emails to the mailinglist anymore... just the answers >> from other people... I don't understand???) >> >> >> Jacques VESLOT schrieb: >> >>> tab <- do.call(rbind, list(data1, data2, data3, data4)) >>> etype <- rep(c(sd1, sd2, sd3, sd4), length(data1)) >>> b <- barplot(tab, beside=T) >>> arrows(unlist(b), unlist(tab) - etype, unlist(b), unlist(tab) + >>> etype, code=3) >>> --- >>> Jacques VESLOT >>> >>> CNRS UMR 8090 >>> I.B.L (2ème étage) >>> 1 rue du Professeur Calmette >>> B.P. 245 >>> 59019 Lille Cedex >>> >>> Tel : 33 (0)3.20.87.10.44 >>> Fax : 33 (0)3.20.87.10.31 >>> >>> http://www-good.ibl.fr >>> --- >>> >>> Antje a écrit : >>> hello, I would like to create the following barplot: I have 4 different data sets (same length + stddev for each data point) data1 sd1 data2 sd2 data3 sd3 data4 sd4 now, I'd like to plot in the following way: data1[1],data2[1],data3[1],data4[1] with it's sd-values side-by-side at one x-axis label (named "position 1") and each bar in different colors. data1[2],data2[2],data3[2],data4[2] at the next x-axis label (named "position 2") with the same color scheme and so on over the whole length. I managed to plot one set in the following way: par(mai=c(1.5,1,1,0.6)) plotInfo <- barplot(data1, las=2, ylim = c(0,plotMax+1), ylab = "Percentage") arrows(plotInfo,data1,plotInfo, data1 + sd1, length=0.1, angle=90) arrows(plotInfo,data1,plotInfo, data1 - sd1, length=0.1, angle=90) could anybody give me a help on this? Antje __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. >>> >> >> __ >> R-help@stat.math.ethz.ch mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide >> http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-proj
[R] Using multiple objects in a for loop
I have a situation where I want to perform the same manipulations to multiple R
objects in series. I have constructed a vector () to serve as a list of the
objects. However, my assign() assigns the object name (on[j]) as a character
string to "temp" and not the object itself.
===
fn<-c("FT1.1.01.RData","FT1.1.02.RData")
on<-c("FT1101","FT1102")
b<-1
e<-2
for (i in 1:2){
for (j in b:e){
load(fn[j])
assign("temp",on[j])
.
.
.
rm(paste(stderr[i],".t",sep=""))
rm(temp)
rm(fn[i])
}
}
===
Daniel
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Re: [R] Rgraphviz -404 Page not found
Again i have problem in locating the package for clique-graphs
I tried with BioConductor under Browse for packages, it doesn't work atall.
Kindly guid me
Thanks
JJ
On 8/23/06, Seth Falcon <[EMAIL PROTECTED]> wrote:
>
> "j.joshua thomas" <[EMAIL PROTECTED]> writes:
>
> > Dear Robert,
> >
> > Thanks for your time.
> > I have downloaded Rgraphviz (windows binary) from www.bioconductor.org
> > and put inside R2.3.0 library then i installed from the local zip
> > its says package 'graph' couldnot be loaded.
> >
> > Am i doing the installation correctly? Still the new user.
> >
> > Can you guide me sir?
>
> Questions about BioC packages are best directed to the bioconductor
> mailing list.
>
> I would recommend trying:
>
> source("http://bioconductor.org/biocLite.R";)
> biocLite("Rgraphviz")
>
>
> Best,
>
> + seth
>
> __
> R-help@stat.math.ethz.ch 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.
>
--
Lecturer J. Joshua Thomas
KDU College Penang Campus
Research Student,
University Sains Malaysia
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Re: [R] barplot help needed
tab <- do.call(rbind, list(data1, data2, data3, data4)) etype <- do.call(rbind, list(sd1, sd2, sd3, sd4)) b <- barplot(tab, beside=T, ylim=c(0,max(tab+etype))) arrows(as.vector(b), as.vector(tab) - as.vector(etype), as.vector(b), as.vector(tab) + as.vector(etype), code=3) unlist() is not correct - sorry - since all are matrices - not data frames ! so use as.vector() --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Antje a écrit : > Still, there is one problem. The SD-Values don't fit to the bar they > belong to. I made the following experiment: > > > data1 <- c(2,4,6,2,5) > > data2 <- data1 > > sd1 <- c(0.5,1,1.5,1,2) > > sd2 <- sd1 > > tab <- do.call(rbind, list(data1, data2)) > > etype <- c(sd1,sd2) > > b <- barplot(tab, beside=T) > > arrows(unlist(b), unlist(tab) - etype, unlist(b), unlist(tab) + > etype, code=3) > > I expect the bars with the same height and the same stddev. The height > is okay, but the stddev is messed up... > > if I do it like this: > > etype <- matrix(c(sd1,sd2), nrow=2, byrow=TRUE) > > it works (but maybe there is an easier way...) > > Antje > > > > > Jacques VESLOT schrieb: > >>thought sd1, sd2... were scalars but if not just do: >>etype <- c(sd1, sd2, sd3, sd4) >>--- >>Jacques VESLOT >> >>CNRS UMR 8090 >>I.B.L (2ème étage) >>1 rue du Professeur Calmette >>B.P. 245 >>59019 Lille Cedex >> >>Tel : 33 (0)3.20.87.10.44 >>Fax : 33 (0)3.20.87.10.31 >> >>http://www-good.ibl.fr >>--- >> >> >>Antje a écrit : >> >>>Thank you very much for your help. >>>I just don't understand the following line (which also gives me a >>>dimension error later in the arrows command) >>> >>>etype <- rep(c(sd1, sd2, sd3, sd4), length(data1)) >>> >>>Antje >>> >>>(I don't see my emails to the mailinglist anymore... just the answers >>>from other people... I don't understand???) >>> >>> >>>Jacques VESLOT schrieb: >>> >>> tab <- do.call(rbind, list(data1, data2, data3, data4)) etype <- rep(c(sd1, sd2, sd3, sd4), length(data1)) b <- barplot(tab, beside=T) arrows(unlist(b), unlist(tab) - etype, unlist(b), unlist(tab) + etype, code=3) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Antje a écrit : >hello, > >I would like to create the following barplot: > >I have 4 different data sets (same length + stddev for each data point) > >data1 >sd1 >data2 >sd2 >data3 >sd3 >data4 >sd4 > >now, I'd like to plot in the following way: > >data1[1],data2[1],data3[1],data4[1] with it's sd-values side-by-side >at one x-axis label (named "position 1") and each bar in different >colors. > >data1[2],data2[2],data3[2],data4[2] at the next x-axis label (named >"position 2") with the same color scheme > >and so on over the whole length. > >I managed to plot one set in the following way: > >par(mai=c(1.5,1,1,0.6)) >plotInfo <- barplot(data1, las=2, ylim = c(0,plotMax+1), ylab = >"Percentage") >arrows(plotInfo,data1,plotInfo, data1 + sd1, length=0.1, angle=90) >arrows(plotInfo,data1,plotInfo, data1 - sd1, length=0.1, angle=90) > >could anybody give me a help on this? > >Antje > >__ >R-help@stat.math.ethz.ch mailing list >https://stat.ethz.ch/mailman/listinfo/r-help >PLEASE do read the posting guide >http://www.R-project.org/posting-guide.html >and provide commented, minimal, self-contained, reproducible code. > >>>__ >>>R-help@stat.math.ethz.ch mailing list >>>https://stat.ethz.ch/mailman/listinfo/r-help >>>PLEASE do read the posting guide >>>http://www.R-project.org/posting-guide.html >>>and provide commented, minimal, self-contained, reproducible code. >>> >> > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do rea
Re: [R] How to find AUC in SVM (kernlab package)
Hi
you need predict.ksvm() function.
for more information see The kernlab package here:
http://lib.stat.cmu.edu/R/CRAN/doc/packages/kernlab.pdf
cheers,
Amir
Muhammad Subianto <[EMAIL PROTECTED]> wrote: Dear all,
I was wondering if someone can help me. I am learning SVM for
classification in my research with kernlab package. I want to know about
classification performance using Area Under Curve (AUC). I know ROCR
package can do this job but I found all example in ROCR package have
include prediction, for example, ROCR.hiv {ROCR}. My problem is how to
produce prediction in SVM and to find AUC.
Here is a simple example:
library(MASS)
library(kernlab)
library(ROCR)
pimamodel <- ksvm(type ~ .,data=Pima.tr,type="C-svc",C=10,prob.model=TRUE)
pimamodel
fitted(pimamodel)
pima.pred <- predict(pimamodel, Pima.te[,-8], type="probabilities")
pima.pred
# try to find AUC
#predid.no <- prediction(pima.pred[,1], Pima.te[,8])
#predid.yes <- prediction(pima.pred[,2], Pima.te[,8])
predid <- prediction(pima.pred, Pima.te[,8])
perfid <- performance(predid,"tpr","fpr")
perfid.auc <- performance(predid,"auc")
perfid.auc
Thank you very much for your help.
Best wishes, Muhammad Subianto
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Re: [R] How to find AUC in SVM (kernlab package)
On this day 11/24/2006 05:03 PM, Amir Safari wrote:
> Hi
> you need predict.ksvm() function.
>
Yes, like this is below I do to predict (kernlab package,
http://www.jstatsoft.org/v11/i09/v11i09.pdf):
pima.pred <- predict(pimamodel, Pima.te[,-8], type="probabilities")
pima.pred
My problems is how to find AUC (with ROCR package, or other ROC
functions) from predict above.
Regards, Muhammad Subianto
> for more information see The kernlab package here:
> http://lib.stat.cmu.edu/R/CRAN/doc/packages/kernlab.pdf
>
> cheers,
> Amir
>
>
> */Muhammad Subianto <[EMAIL PROTECTED]>/* wrote:
>
> Dear all,
> I was wondering if someone can help me. I am learning SVM for
> classification in my research with kernlab package. I want to know
> about
> classification performance using Area Under Curve (AUC). I know ROCR
> package can do this job but I found all example in ROCR package have
> include prediction, for example, ROCR.hiv {ROCR}. My problem is
> how to
> produce prediction in SVM and to find AUC.
>
> Here is a simple example:
>
> library(MASS)
> library(kernlab)
> library(ROCR)
>
> pimamodel <- ksvm(type ~
> .,data=Pima.tr,type="C-svc",C=10,prob.model=TRUE)
> pimamodel
> fitted(pimamodel)
>
> pima.pred <- predict(pimamodel, Pima.te[,-8], type="probabilities")
> pima.pred
>
> # try to find AUC
> #predid.no <- prediction(pima.pred[,1], Pima.te[,8])
> #predid.yes <- prediction(pima.pred[,2], Pima.te[,8])
> predid <- prediction(pima.pred, Pima.te[,8])
> perfid <- performance(predid,"tpr","fpr")
> perfid.auc <- performance(predid,"auc")
> perfid.auc
>
> Thank you very much for your help.
>
> Best wishes, Muhammad Subianto
>
> __
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> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
>
> Cheap Talk? Check out
>
>
> Yahoo! Messenger's low PC-to-Phone call rates.
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Re: [R] Error Message saying .Call("R_lazyLoadDBfetch", etc.
Jacques VESLOT <[EMAIL PROTECTED]> writes:
> Hi,
>
> I got the following error message when running a function of mine doing
> intensive computations:
> Erreur dans .Call("R_lazyLoadDBfetch", key, file, compressed, hook, PACKAGE =
> "base") :
> référence d'argument par défaut récursive
>
> I haven't found neither where the problem lies nor what could be recursive.
>
> Besides, I wonder whether it might or not be a problem of memory size.
>
> Could you please give me any suggestion on how to interpret this message?
The prototypical situation is this
> f<-function(a=a)a
> f(2)
[1] 2
> f()
Error in f() : recursive default argument reference
The default for argument a ends up needing the value of a. This can
also happens more indirectly when multiple arguments refer to
eachother.
BTW: Is that error message proper French? Should it not be
référence récursive d'argument par défaut
or so.
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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[R] Splitting criterion in tree and rpart
Dear all, I'm interested in fitting a classification tree by using a user-specified splitting criterion. If I am right, Tree package allows to use only "deviance" or "gini" while rpart package offers "anova", "poisson", "class" or "exp". and tries to make an intelligent guess if method is missing. I also found that, in rpart, method can be a list of functions named init, split and eval but I really don't know how can I do it. Does someone can suggest me some documentation I can have a look do define my splitting criterion and pass it to rpart ? Thanks Paolo Paolo Radaelli Dipartimento di Metodi Quantitativi per le Scienze Economiche ed Aziendali Facoltà di Economia Università degli Studi di Milano-Bicocca e-mail [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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] Rgraphviz -404 Page not found
"j.joshua thomas" <[EMAIL PROTECTED]> writes: > Again i have problem in locating the package for clique-graphs > I tried with BioConductor under Browse for packages, it doesn't work atall. > > Kindly guid me I think you already got an answer on the Bioconductor list: RBGL is likely the package you are looking for. __ R-help@stat.math.ethz.ch 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] SNA packages/network package
> Bagatti Davide wrote:
>> Hello everyone,
>>
>> I am an italian student who is working with packages SNA (social network
>> analysis) and network.
>> I ask you if there is a simple way to write a R-script using these packages
>> to extract from an adjacency matrix the following things:
>>
>> -number of cliques which are in the network;
>> -number of k-cores (e.g. 3-cores, 4-cores);
>> -cycles (e.g. 3-cycles, 4-cycles)
>> -hub & authorities.
You might take a look at the 'graph' and 'RBGL' packages (both are
Bioconductor packages).
There is a graphAM class that can be initialized with an adjacency
matrix. Most functions in RBGL take either any graph instance or a
graphNEL. So you might have (approx. and untested):
library(graph)
library(RBGL)
myAdjMat
g1 = new("graphAM", myAdjMat)
g2 = as(g1, "graphNEL")
## call some RBGL function on g2
+ seth
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[R] Error in Calling C++ function from R!!!
Hello,
I tried to call an external function of R from the following code in C++:
void prodgdot(double *x, double *y, int *n, double *output)
{
int i;
*output=0;
for (i=0;i<*n;i++)
{
*output+=x[i]*y[i];
}
}
I compiled it using from my working directory in linux terminal and I think
it's ok:
[EMAIL PROTECTED]:~/mysrc/meus_testes_iniciais$ R CMD SHLIB codigoprova.cc
# The output
g++ -I/usr/share/R/include -I/usr/share/R/include -fpic -g -O2 -c
codigoprova.cc -o codigoprova.o
g++ -shared -o codigoprova.so codigoprova.o -L/usr/lib/R/lib -lR
After, I tried to call R from my working directory an Error occurs:
> x<-c(1,4,6,2)
> y<-c(3,2.4,1,9)
> dyn.load("codigoprova.so")
> is.loaded("codigoprova.so")
[1] FALSE
> product<-.C("prodgdot",myx=x,muy=y,myn=NROW(x),myoutput=as.double(0))
Error in .C("prodgdot", myx = x, muy = y, myn = NROW(x), myoutput =
as.double(0)) :
C symbol name "prodgdot" not in load table
>
Does anyone know what is the problem?
Thank's in advance!
Gilberto.
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[R] McQuitty method
Is anyone able to explain the McQuitty method in hclust? Thanks in advance, Jeff Miller [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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] Error in Calling C++ function from R!!!
>>product<-.C("prodgdot",myx=x,muy=y,myn=NROW(x),myoutput=as.double(0))
>
> Error in .C("prodgdot", myx = x, muy = y, myn = NROW(x), myoutput =
> as.double(0)) :
> C symbol name "prodgdot" not in load table
>
>
> Does anyone know what is the problem?
C++ name mangling?
http://cran.r-project.org/doc/manuals/R-exts.html#Interfacing-C_002b_002b-code
Solution: wrap in extern "C" { ... } as discussed there.
Barry
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[R] Random effects ANOVA
Hi, I used the command lmer, in package lme4, to fit a random effects ANOVA. But i didn't get the p-values of significance tests of variance components. Does anyone know how to do it? Thanks, Luis Ernesto - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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] using nested ifelse and rowSums to create new variable?
Thanks Dimitris and Marc. The solution to my problem was found in a
a combination of your suggestions. I created a count vector
separately and referenced it as part of the nested ifelse
statements (see below).
Cheers,
Tony
> count <- rowSums(DF[, 3:6] < 4)
> count
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
3 1 2 2 3 2 2 0 1 2 3 2 0 3 3 2 3 2 2 4
> DF2<-data.frame(DF, count)
> DF2
x1 x2 x3 x4 x5 x6 count
1 5 1 1 1 2 5 3
2 5 5 5 5 4 1 1
3 5 1 1 5 4 1 2
4 5 5 1 4 5 1 2
5 1 1 1 2 4 1 3
6 1 1 1 4 1 5 2
7 5 5 1 4 1 5 2
8 1 5 5 5 5 5 0
9 5 1 5 4 2 5 1
10 5 1 1 1 4 5 2
11 1 1 1 1 4 1 3
12 5 5 5 1 2 5 2
13 5 5 5 4 4 5 0
14 5 1 1 2 1 5 3
15 1 5 1 1 2 5 3
16 1 1 5 5 2 1 2
17 1 1 5 2 2 1 3
18 1 5 5 2 4 1 2
19 5 1 1 2 5 5 2
20 1 1 1 2 2 1 4
> dep<-with(DF2,
+ ifelse((x1==5) & (x2==5), -2,
+ ifelse((x1==1 & x2==1), -1,
+ ifelse((x1==1 & x2==5) | (x1==5 & x2==1) &
+ (count==0), 0,
+ ifelse((x1==1 & x2==5) | (x1==5 & x2==1) &
+ (count==1), 1,
+ ifelse((x1==1 & x2==5) | (x1==5 & x2==1) &
+ (count==2), 2,
+ ifelse((x1==1 & x2==5) | (x1==5 & x2==1) &
+ (count==3), 3,
+ ifelse((x1==1 & x2==5) | (x1==5 & x2==1) &
+ (count==4), 4,
+ 99
> table(dep)
dep
-2 -1 0 1 2 3
5 6 3 1 3 2
--On Tuesday, November 21, 2006 3:42 PM -0600 Marc Schwartz
<[EMAIL PROTECTED]> wrote:
> On Tue, 2006-11-21 at 15:24 -0600, Marc Schwartz wrote:
>> On Tue, 2006-11-21 at 14:26 -0600, Tony N. Brown wrote:
>> > Dear R-help community,
>> >
>> > If I have a data.frame df as follows:
>> >
>> > > df
>> >x1 x2 x3 x4 x5 x6
>> > 1 5 5 1 1 2 1
>> > 2 5 5 5 5 1 5
>> > 3 1 5 5 5 5 5
>> > 4 5 5 1 4 5 5
>> > 5 5 1 5 2 4 1
>> > 6 5 1 5 4 5 1
>> > 7 5 1 5 4 4 5
>> > 8 5 1 1 1 1 5
>> > 9 1 5 1 1 2 5
>> > 10 5 1 5 4 5 5
>> > 11 1 5 5 2 1 1
>> > 12 5 5 5 4 4 1
>> > 13 1 5 1 4 4 1
>> > 14 1 1 5 4 5 5
>> > 15 1 5 5 4 5 1
>> > 16 1 1 5 5 5 1
>> > 17 5 5 5 2 2 5
>> > 18 1 5 1 5 5 5
>> > 19 5 5 5 2 4 5
>> > 20 1 1 5 2 4 5
>> >
>> > How can I create a variable that captures the pattern of
>> > responses and counts across rows?
>> >
>> > I used the ifelse function and that works fine for the first
>> > two conditions (see R code below). But I need help figuring
>> > out how to count the number of scores in each row for columns
>> > x3, x4, x5, and x6 that are less than 4, conditional upon an
>> > ifelse. I then want to assign a value to the new variable
>> > based upon the count.
>> >
>> > The new variable I want to create is called dep. Here's my R
>> > code:
>> >
>> > dep<-with(df,
>> > ifelse((x1==5) & (x2==5), 0,
>> > ifelse((x1==1 & x2==1), 1,
>> >
>> > ifelse((x1==1 & x2==5) | (x1==5 & x2==1) &
>> > (rowSums(df[ ,c(x3, x4, x5, x6)]<4) ==1), 2,
>> > ifelse((x1==1 & x2==5) | (x1==5 & x2==1) &
>> > (rowSums(df[ ,c(x3, x4, x5, x6)]<4) ==2), 3,
>> > ifelse((x1==1 & x2==5) | (x1==5 & x2==1) &
>> > (rowSums(df[ ,c(x3, x4, x5, x6)]<4) ==3), 4,
>> >
>> > 99))
>> >
>> > dep
>> > 0 1 2 99
>> > 6 3 6 5
>> >
>> > I expected dep to range from 0 to 4 and its length to be equal
>> > to 20.
>> >
>> > Thanks in advance for your help and suggestions.
>>
>> I may be a bit off in what your desired end result is, but
>> perhaps this will provide some insight. I renamed your data
>> frame to DF, since df is a function:
>>
>> > apply(DF[, -(1:2)], 1, function(x) sum(x < 4))
>> 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
>> 4 1 0 1 2 1 0 3 3 0 3 1 2 0 1 1 2 1 1 1
>
>
> Thanks to Dimitris' post tweaking my neurons, the above can of
> course be simplified to:
>
>> rowSums(DF[, -(1:2)] < 4)
> 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
> 4 1 0 1 2 1 0 3 3 0 3 1 2 0 1 1 2 1 1 1
>
> Note that rowSums() will accept an all numeric data frame as an
> argument, so the initial coercion is not required.
>
> Hence:
>
> TAB <- table(rowSums(DF[, -(1:2)] < 4),
> factor(rowSums(DF[, 1:2]),
> labels = c("(1,1)", "(5,1)|(1,5)", "(5,5)")))
>
> TAB <- addmargins(TAB)
>
>
>> TAB
>
> (1,1) (5,1)|(1,5) (5,5) Sum
> 0 1 3 0 4
> 1 2 3 4 9
> 2 0 2 1 3
> 3 0 3 0 3
> 4 0 0 1 1
> Sum 3 11 6 20
>
>
> HTH,
>
> Marc
>
>
>
---
Tony N. Brown, Ph.D.
Assistant Professor of Sociology
Vanderbilt University
Phone: (615) 322-7518
Fax:(615) 322-7505
Email: [EMAIL PROTECTED]
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Re: [R] don't put line in plot
fernando espindola schrieb: > Hi R-user, > > I have a problem when try to run the next code: > > plot(prueba$IC, type="l") > > but plot with type="p", there not problem, I don't know what is the > problem?? Anybody can help me > normally there is no problem with type="l" maybe you should send a reproducible code With regards Carmen [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] low-variance warning in lmer
For block effects with small variance, lmer will sometimes
estimate the variance as being very close to zero and issue
a warning. I don't have a problem with this -- I've explored
things a bit with some simulations (see below) and conclude that
this is probably inevitable when trying to incorporate
random effects with not very much data (the means and medians
of estimates are plausibly close to the nominal values:
the fraction of runs with warnings/near-zero estimates drops
from about 50% when the between-block variance is 1% of
the total (with 2 treatments, 12 blocks nested within treatment,
3 replicates per block), to 15% when between=30% of total,
to near zero when between=50% of total.
My question is: what should I suggest to students when this
situation comes up? Can anyone point me to appropriate references?
(I haven't found anything relevant in Pinheiro and Bates, but
I may not have looked in the right place ...)
thanks
Ben Bolker
self-contained but unnecessarily complicated simulation
code/demonstration:
---
library(lme4)
library(lattice)
simfun <- function(reefeff,ntreat=2,nreef=12,
nreefpertreat=3,
t.eff=10,
totvar=25,seed=NA) {
if (!is.na(seed)) set.seed(seed)
ntot = nreef*nreefpertreat
npertreat=ntot/ntreat
reef = gl(nreef,nreefpertreat)
treat = gl(ntreat,npertreat)
r.sd = sqrt(totvar*reefeff)
e.sd = sqrt(totvar*(1-reefeff))
y.det = ifelse(treat==1,0,t.eff)
r.vals = rnorm(nreef,sd=r.sd)
e.vals = rnorm(ntot,sd=e.sd)
y <- y.det+r.vals[as.numeric(reef)]+e.vals
data.frame(y,treat,reef)
}
getranvar <- function(x) {
vc <- VarCorr(x)
c(diag(vc[[1]]),attr(vc,"sc")^2)
}
estfun <- function(reefeff,...) {
x <- simfun(reefeff=reefeff,...)
ow <- options(warn=2)
f1 <- try(lmer(y~treat+(1|reef),data=x))
w <- (class(f1)=="try-error" && length(grep("effectively zero",f1))>0)
options(ow)
f2 <- lmer(y~treat+(1|reef),data=x)
c(getranvar(f2),as.numeric(w))
}
rvec <- rep(c(0.01,0.05,0.1,0.15,0.2,0.3,0.5),each=100)
X <- t(sapply(rvec,estfun))
colnames(X) <- c("reefvar","resvar","warn")
rfrac <- X[,"reefvar"]/(X[,"reefvar"]+X[,"resvar"])
fracwarn <- tapply(X[,"warn"],rvec,mean)
est.mean <- tapply(rfrac,rvec,mean)
op <- par(mfrow=c(1,2))
plot(rvec,rfrac,type="n",xlim=c(-0.02,0.55),axes=FALSE)
axis(side=2)
box()
boxplot(rfrac~rvec,at=unique(rvec),add=TRUE,pars=list(boxwex=0.03),
col="gray")
points(jitter(rvec),rfrac,col=X[,"warn"]+1)
lines(unique(rvec),fracwarn,col="blue",type="b",lwd=2)
text(0.4,0.1,"fraction\nzero",col="blue")
abline(a=0,b=1,lwd=2)
points(unique(rvec),est.mean,cex=1.5,col=5)
##
plot(rvec,rfrac,type="n",xlim=c(-0.02,0.55),axes=FALSE,log="y")
axis(side=2)
axis(side=1,at=unique(rvec))
box()
points(jitter(rvec),rfrac,col=X[,"warn"]+1)
curve(1*x,add=TRUE,lwd=2)
print(densityplot(~rfrac,groups=rvec,from=0,to=0.9,
panel=function(...) {
panel.densityplot(...)
cols <- trellis.par.get("superpose.line")$col
lpoints(unique(rvec),rep(8,length(rvec)),type="h",
col=cols[1:length(rvec)])
}))
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[R] [R-pkgs] New package `np' - nonparametric kernel smoothing methods for mixed datatypes
Dear R users, A new package titled `np' is now available from CRAN. The package implements recently developed kernel methods that seamlessly handle the mix of continuous, unordered, and ordered factor datatypes often found in applied settings. The package also allows users to create their own nonparametric/semiparametric routines using high-level function calls (via the function npksum()) rather than writing their own C or Fortran code. Much of the code underlying the package is written in C including the function npksum(). Currently, a range of methods can be found in the package including - multivariate nonparametric unconditional and conditional density estimation - multivariate nonparametric conditional mean and gradient estimation (local constant and local linear) - multivariate nonparametric conditional distribution, quantile and gradient estimation - nonparametric model specification tests for testing correctness of parametric regression models - nonparametric significance tests for nonparametric regression - semiparametric multivariate regression methods (partially linear, index models, average derivative estimation, varying/smooth coefficient models) A function npplot() is implemented that allows users to visualize the resulting objects. A variety of methods for computing standard errors and error bounds are implemented including asymptotic and resampling-based methods (the latter employing the `boot' package which is required). A range of examples can be found in the examples section of the accompanying help files. We caution potential users that multivariate data-driven bandwidth selection methods can be numerically intensive. For this reason we are now using some of the functionality contained in the Rmpi package to develop an MPI-aware version of the np package that we have tentatively titled the `npRmpi' package. The np package implements a number of methods found in the newly released publication by Li and Racine (2007) titled Nonparametric Econometrics: Theory and Practice, Princeton University Press. See press.princeton.edu/titles/8355.html for further details. Best regards, Jeff. -- Professor J. S. Racine Phone: (905) 525 9140 x 23825 Department of EconomicsFAX:(905) 521-8232 McMaster Universitye-mail: [EMAIL PROTECTED] 1280 Main St. W.,Hamilton, URL: http://www.economics.mcmaster.ca/racine/ Ontario, Canada. L8S 4M4 `The generation of random numbers is too important to be left to chance' ___ R-packages mailing list R-packages@stat.math.ethz.ch https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Nonlinear statistical modeling -- a comparison of R and AD Model Builder
There has recently been some discussion on the list about AD Model builder and the suitability of R for constructing the types of models used in fisheries management. https://stat.ethz.ch/pipermail/r-help/2006-January/086841.html https://stat.ethz.ch/pipermail/r-help/2006-January/086858.html I think that many R users understimate the numerical challenges that some of the typical nonlinear statistical model used in different fields present. R may not be a suitable platform for development for such models. Around 10 years ago John Schnute, Laura Richards, and Norm Olsen with Canadian federal fisheries undertook an investigation comparing various statistical modeling packages for a simple age-structured statistical model of the type commonly used in fisheries. They compared AD Mdel Builder, Gauss, Matlab, and Splus. Unfortunately a working model could not be produced with Splus so its times could not be included in the comparison. It is possible to produce a working model with the present day version of R so that R can now be directly compared with AD Model Builder for this type of model. I have put the results of the test together with the original Schnute and Richards paper and the working R and AD Model Builder codes on Otter's web site http://otter-rsch.ca/tresults.htm The results are that AD Model builder is roughly 1000 times faster than R for this problem. ADMB takes about 2 seconds to converge while R takes over 90 minutes. This is a simple toy example. Real fisheries models are often hundred of times more computationally intensive as this one. Cheers, Dave ~ -- David A. Fournier P.O. Box 2040, Sidney, B.C. V8l 3S3 Canada Phone/FAX 250-655-3364 http://otter-rsch.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Sunflower plot error; how to deal with NA
I suspect the problem stems from the fact that there are a couple of NA values. > sunflowerplot(lastoto,maxear) Error in rep.int(i.multi, number[number > 1]) : invalid number of copies in rep.int() So I used the subset command to get rid of the cases with NA hell<-subset(ChinOtoMayB,is.na(lastoto)==FALSE) Then it worked perfectly sunflowerplot(hell$lastoto,hell$maxear) Is there a method with greater finesse to deal with NA values? In correlations one can state use="complete.obs" But that did not work in sunflowerplot -- Farrel Buchinsky [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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] Nonlinear statistical modeling -- a comparison of R and AD Model Builder
dave fournier <[EMAIL PROTECTED]> wrote: > I think that many R users understimate the numerical challenges > that some of the typical nonlinear statistical model used in different > fields present. R may not be a suitable platform for development for > such models. > > Around 10 years ago John Schnute, Laura Richards, and Norm Olsen > with Canadian federal fisheries undertook an investigation > comparing various statistical modeling packages for a simple > age-structured statistical model of the type commonly used in > fisheries. [...] It is possible > to produce a working model with the present day version of R so that > R can now be directly compared with AD Model Builder for this type of model. > > The results are that AD Model builder is roughly 1000 times faster than > R for this problem. ADMB takes about 2 seconds to converge while > R takes over 90 minutes. Our group's experiences reflect, at least qualitatively, what Dave says above. We use R for analyzing results from models written in his AD Model Builder, and a couple of years ago, we started programming one of our models directly in R. We quickly abandoned that idea because of lengthy execution time under R. That is not a judgement of either piece of software. R and ADMB are designed for different types of task, and it seems to me that they complement each other well. That experience was in part the genesis of our X2R software (now at CRAN -- pardon the plug), which saves results from ADMB models into a format that R can read as a list. We feel that now we have the best of both worlds -- fast execution with ADMB, followed by the programming ease and excellent graphics of R for analysis of results and projections under numerous scenarios. -- Mike Prager, NOAA, Beaufort, NC * Opinions expressed are personal and not represented otherwise. * Any use of tradenames does not constitute a NOAA endorsement. __ R-help@stat.math.ethz.ch 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] Nonlinear statistical modeling -- a comparison of R and AD Model Builder
Did you try supplying gradient information to nlminb? (I note that nlminb is used for the optimization, but I don't see any gradient information supplied to it.) I would suspect that supplying gradient information would greatly speed up the computation (as you note in comments at http://otter-rsch.ca/tresults.htm.) I'm curious -- when you say "R may not be a suitable platform for development for such models", what aspect of R do you feel is lacking? Is it the specific optimization routines available, or is it some other more general aspect? Also, another optimization algorithm available in R is the "L-BFGS-B" method for optim() in the MASS package. I've had extremely good experiences with using this code in S-PLUS. It can take box constraints, and can use gradient information. It is my first choice for most optimization problems, and I believe it is very widely used. Did you try using that optimization routine with this problem? -- Tony Plate dave fournier wrote: > There has recently been some discussion on the list about > AD Model builder and the suitability of R for constructing the > types of models used in fisheries management. > >https://stat.ethz.ch/pipermail/r-help/2006-January/086841.html > >https://stat.ethz.ch/pipermail/r-help/2006-January/086858.html > > I think that many R users understimate the numerical challenges > that some of the typical nonlinear statistical model used in different > fields present. R may not be a suitable platform for development for > such models. > > Around 10 years ago John Schnute, Laura Richards, and Norm Olsen > with Canadian federal fisheries undertook an investigation > comparing various statistical modeling packages for a simple > age-structured statistical model of the type commonly used in > fisheries. They compared AD Mdel Builder, Gauss, Matlab, and > Splus. Unfortunately a working model could not be produced with Splus > so its times could not be included in the comparison. It is possible > to produce a working model with the present day version of R so that > R can now be directly compared with AD Model Builder for this type of model. > > I have put the results of the test together with the original > Schnute and Richards paper and the working R and AD Model Builder > codes on Otter's web site > > http://otter-rsch.ca/tresults.htm > > The results are that AD Model builder is roughly 1000 times faster than > R for this problem. ADMB takes about 2 seconds to converge while > R takes over 90 minutes. > > This is a simple toy example. Real fisheries models are often hundred of > times more computationally intensive as this one. > > Cheers, > > Dave > ~ __ R-help@stat.math.ethz.ch 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] axis in an image() plot
If you data is a matrix, then try: image(1:5, 1:14, data.rect) text(row(data.rect), col(data.rect), data.rect) On 11/24/06, Ricardo Rodríguez - Your XEN ICT Team <[EMAIL PROTECTED]> wrote: > Hi all, > > I've found quite usefull colored-grid created by image() but I'm facing a > doubt I am not able to solve. > > Given the following data rectangle... > > 1 2 3 4 5 6 7 8 9 10 11 12 13 14 > 1 12 22 0 7 2 1 0 2 0 2 6 -3 0 3 > 2 0 -1 0 9 3 -4 0 0 0 0 3 0 0 0 > 3 29 45 6 12 16 85 -2 0 -3 -4 89 -1 -1 1 > 4 2 9 3 6 17 3 -2 -9 -2 8 -1 0 0 0 > 5 44 16 -3 21 23 3 2 1 0 -2 13 18 -5 2 > > I am not able to draw x and y axis labeled 1 to 14 and 1 to 5 by 1. I've > tried a number of options by using axis() to no avail. > > It will be also very helpfull to be able to draw the value of each x,y > combination within its position in the grid. Text() seems to be the answer, > but I am not able yet to get the correct position for each label. > > Please, could you point me in the right direction or offer some example? > > Thank you in advance, > > -- > Ricardo Rodríguez > Your XEN ICT Team > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] vector problem
Hello out there,
i am not yet that experienced and trying to my best on a real survey.
but i am stuck with a little matrix / vector problem.
my vector of answers could have a length of 3 or only one. i want to
rbind all the answers into one matrix. (one vector for each participant)
answers vectors for one participant could look like:
p1: 100
p2: 20 80
p3: 40 10 50
i have the following loop which should rbind them but i get no proper
matrix.
i´d like to have something like this
100 0 0
20 80 0
40 10 50
Here´s my loop:
for(s in 1:length(fr))
{
### ansmini is a complete survey of one participant
ansmini3=answers[relevant[s,],]
for(x in 1:length(qidsb3))
{
# outputs all answer ids out of all data which belong to the
qids
out of the question id vector
ansmin3=ansmini3[,1][ansmini3[,3]==qidsb3[x]]
newb3=rbind(newb3,ansmin3)
}
}
this is doing the job for question with only one answer, so i have
only one answer id.
in this new case i have 1-3 possible answers.. that´s why i´am not
getting a nice newb3 matrix...
i´d really be happy about some advice!
thanks in advance
matthias
[[alternative HTML version deleted]]
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Re: [R] Random effects ANOVA
Dear Luis, I believe that lmer() doesn't print out standard errors or Wald tests for variance-covariance components because these tests aren't reliable. You can omit each random effect from the model in turn, and use anova() to perform a likelihood-ratio test for the corresponding variance and covariance components, comparing each null model with the full model. I hope this helps, John John Fox Department of Sociology McMaster University Hamilton, Ontario Canada L8S 4M4 905-525-9140x23604 http://socserv.mcmaster.ca/jfox > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Luis Salasar > Sent: Friday, November 24, 2006 1:57 PM > To: r-help@stat.math.ethz.ch > Subject: [R] Random effects ANOVA > > Hi, > > I used the command lmer, in package lme4, to fit a random > effects ANOVA. But i didn't get the p-values of significance > tests of variance components. Does anyone know how to do it? Thanks, > > Luis Ernesto > > > - > > [[alternative HTML version deleted]] > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch 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] vector problem
merge.zoo can do that:
> p1 <- 100
> p2 <- c(20, 80)
> p3 <- c(40, 10, 50)
> library(zoo)
> t(merge(p1 = zoo(p1), p2 = zoo(p2), p3 = zoo(p3), fill = 0))
1 2 3
p1 100 0 0
p2 20 80 0
p3 40 10 50
On 11/24/06, bunny , lautloscrew.com <[EMAIL PROTECTED]> wrote:
> my vector of answers could have a length of 3 or only one. i want to
> rbind all the answers into one matrix. (one vector for each participant)
> answers vectors for one participant could look like:
>
> p1: 100
> p2: 20 80
> p3: 40 10 50
>
> i have the following loop which should rbind them but i get no proper
> matrix.
> i´d like to have something like this
>
> 100 0 0
> 20 80 0
> 40 10 50
>
> Here´s my loop:
>
>
> for(s in 1:length(fr))
>{
>### ansmini is a complete survey of one participant
>ansmini3=answers[relevant[s,],]
>
>for(x in 1:length(qidsb3))
>{
># outputs all answer ids out of all data which belong to the
> qids
> out of the question id vector
>ansmin3=ansmini3[,1][ansmini3[,3]==qidsb3[x]]
>newb3=rbind(newb3,ansmin3)
>}
>}
>
> this is doing the job for question with only one answer, so i have
> only one answer id.
> in this new case i have 1-3 possible answers.. that´s why i´am not
> getting a nice newb3 matrix...
>
> i´d really be happy about some advice!
> thanks in advance
>
> matthias
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Re: [R] Sunflower plot error; how to deal with NA
I only specified one of the variables and it worked fine since I was using the subset command. On 11/24/06, Farrel Buchinsky <[EMAIL PROTECTED]> wrote: > > I suspect the problem stems from the fact that there are a couple of NA > values. > > > sunflowerplot(lastoto,maxear) > > Error in rep.int(i.multi, number[number > 1]) : > invalid number of copies in rep.int() > So I used the subset command to get rid of the cases with NA > > hell<-subset(ChinOtoMayB,is.na(lastoto)==FALSE) > > Then it worked perfectly > > sunflowerplot(hell$lastoto,hell$maxear) > > Is there a method with greater finesse to deal with NA values? In > correlations one can state use="complete.obs" > But that did not work in sunflowerplot > > -- > Farrel Buchinsky > -- Farrel Buchinsky Mobile: (412) 779-1073 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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] Nonlinear statistical modeling -- a comparison of R and AD Model Builder
Dave > Did you try supplying gradient information to nlminb? (I note that nlminb is used for the optimization, but I don't see any gradient information supplied to it.) I would suspect that supplying gradient information would greatly speed up the computation (as you note in comments at http://otter-rsch.ca/tresults.htm.) Actually you should probably ask Norm Olsen these questions. I am not proficient in R and am just using his code. However I can say that providing derivatives for such a model is a highly nontrivial exercise. As I said in my posting, the R script and data are available to anyone who feels that the exercise was not carried out properly and would like to improve on it. Also one does not need to provide derivatives to the AD Model Builder program. Finally suppose that you are very good at calculating derivatives and manage to get them right. Then someone else comes along who wants to modify the model. Unless they are also very good at calculating derivatives there will be trouble. > > I'm curious -- when you say "R may not be a suitable platform for development for such models", what aspect of R do you feel is lacking? Is it the specific optimization routines available, or is it some other more general aspect? 2 seconds vs 90 minutes. For a real problem of tihs type the timings would probably be something like 10 minutes vs more than 2,700 minutes. > > Also, another optimization algorithm available in R is the "L-BFGS-B" method for optim() in the MASS package. I've had extremely good experiences with using this code in S-PLUS. It can take box constraints, and can use gradient information. It is my first choice for most optimization problems, and I believe it is very widely used. Did you try using that optimization routine with this problem? > > -- Tony Plate > > dave fournier wrote: >> There has recently been some discussion on the list about >> AD Model builder and the suitability of R for constructing the >> types of models used in fisheries management. >> >>https://stat.ethz.ch/pipermail/r-help/2006-January/086841.html >> >>https://stat.ethz.ch/pipermail/r-help/2006-January/086858.html >> >> I think that many R users understimate the numerical challenges >> that some of the typical nonlinear statistical model used in different >> fields present. R may not be a suitable platform for development for >> such models. >> >> Around 10 years ago John Schnute, Laura Richards, and Norm Olsen >> with Canadian federal fisheries undertook an investigation >> comparing various statistical modeling packages for a simple >> age-structured statistical model of the type commonly used in >> fisheries. They compared AD Mdel Builder, Gauss, Matlab, and >> Splus. Unfortunately a working model could not be produced with Splus >> so its times could not be included in the comparison. It is possible >> to produce a working model with the present day version of R so that >> R can now be directly compared with AD Model Builder for this type of model. >> >> I have put the results of the test together with the original >> Schnute and Richards paper and the working R and AD Model Builder >> codes on Otter's web site >> >> http://otter-rsch.ca/tresults.htm >> >> The results are that AD Model builder is roughly 1000 times faster than >> R for this problem. ADMB takes about 2 seconds to converge while >> R takes over 90 minutes. >> >> This is a simple toy example. Real fisheries models are often hundred of >> times more computationally intensive as this one. >> >> Cheers, >> >> Dave >> ~ > -- David A. Fournier P.O. Box 2040, Sidney, B.C. V8l 3S3 Canada Phone/FAX 250-655-3364 http://otter-rsch.com -- David A. Fournier P.O. Box 2040, Sidney, B.C. V8l 3S3 Canada Phone/FAX 250-655-3364 http://otter-rsch.com __ R-help@stat.math.ethz.ch 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] Nonlinear statistical modeling -- a comparison of R and AD Model Builder
Hi, Mike & Dave: Have you considered nonlinear mixed effects models for the types of problems considered in the comparison paper you cite? Those "benchmark trials" consider "T years of data ... for A age classes and the total number of parameters is m = T+A+5". Without knowing more about the problem, I suspect that the T year parameters and the A age class parameters might be better modeled as random effects. If this were done, the optimization problem would then involve 7 parameters, the 5 fixed-effect parameters suggested by the computation of "m" plus two variance parameters, one for the random "year" effects and another for the random "age class" effect. This would replace the problem of maximizing, e.g., a likelihood over T+A+5 parameters with one of maximizing a marginal likelihood over 2+5 parameters after integrating out the T and A random effects. These integrations may not be easy, and I might stick with the fixed-effects solution if I couldn't get answers in the available time using a model I thought would be theoretically more appropriate. Also, I might use the fixed-effects solution to get starting values for an attempt to maximize a more appropriate marginal likelihood. For the latter, I might first try 'nlmle'. If that failed, I might explore Markov Chain Monte Carlo (MCMC). I have not done MCMC myself, but the "MCMCpack" R package looks like it might make it feasible for the types of problems considered in this comparison. The CRAN summary of that package led me to an Adobe Acrobat version of a PPT slide presentation that seemed to consider just this type of problem (e.g., http://mcmcpack.wustl.edu/files/MartinQuinnMCMCpackslides.pdf). Have you considered that? Hope this helps. Spencer Graves Mike Prager wrote: > dave fournier <[EMAIL PROTECTED]> wrote: > > >> I think that many R users understimate the numerical challenges >> that some of the typical nonlinear statistical model used in different >> fields present. R may not be a suitable platform for development for >> such models. >> >> Around 10 years ago John Schnute, Laura Richards, and Norm Olsen >> with Canadian federal fisheries undertook an investigation >> comparing various statistical modeling packages for a simple >> age-structured statistical model of the type commonly used in >> fisheries. [...] It is possible >> to produce a working model with the present day version of R so that >> R can now be directly compared with AD Model Builder for this type of model. >> >> The results are that AD Model builder is roughly 1000 times faster than >> R for this problem. ADMB takes about 2 seconds to converge while >> R takes over 90 minutes. >> > > Our group's experiences reflect, at least qualitatively, what > Dave says above. We use R for analyzing results from models > written in his AD Model Builder, and a couple of years ago, we > started programming one of our models directly in R. We quickly > abandoned that idea because of lengthy execution time under R. > That is not a judgement of either piece of software. R and ADMB > are designed for different types of task, and it seems to me > that they complement each other well. > > That experience was in part the genesis of our X2R software (now > at CRAN -- pardon the plug), which saves results from ADMB > models into a format that R can read as a list. We feel that > now we have the best of both worlds -- fast execution with ADMB, > followed by the programming ease and excellent graphics of R for > analysis of results and projections under numerous scenarios. > > __ R-help@stat.math.ethz.ch 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] low-variance warning in lmer
Ben Bolker zoo.ufl.edu> writes: > For block effects with small variance, lmer will sometimes ... > My question is: what should I suggest to students when this > situation comes up? Can anyone point me to appropriate references? > (I haven't found anything relevant in Pinheiro and Bates, but > I may not have looked in the right place ...) Gelman has quite some similar examples. Take a look at his schools example in the Bayesian book. Gregor __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] nlme
-- David A. Fournier P.O. Box 2040, Sidney, B.C. V8l 3S3 Canada Phone/FAX 250-655-3364 http://otter-rsch.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] nlme sorry about that
Sorry list I twitched and sent the wrong stuff. Maybe I had enough fun for today. -- David A. Fournier P.O. Box 2040, Sidney, B.C. V8l 3S3 Canada Phone/FAX 250-655-3364 http://otter-rsch.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] trimming plotting area in a boxplot
Hello all, I am trying to trim the plotting area in a boxplot, such that the space between the plot margins (left and right) are of identical size as between the boxes. In the example below I want to get rid of the space outside of the abline(). I appreciate any suggestions. Factor = LETTERS[1:5] A = c(1:5); B = c(2:6); C = c(1:5); D = c(3:7); E = c(1:5); F = c(4:8) df = data.frame(A,B,C,D,E,F) boxplot(df) abline(v=c(0.5,6.5)) Cheers, Patrick __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Multiple Conditional Tranformations
Greetings,
I'm learning R and I'm stuck on a basic concept: how to specify a
logical condition once and then perform multiple transformations under
that condition. The program below is simplified to demonstrate the goal.
Its results are exactly what I want, but I would like to check the
logical state of gender only once and create both (or any number of)
scores at once.
mystring<-
("id,group,gender,q1,q2,q3,q4
01,1,f,2,2,5,4
02,2,f,2,1,4,5
03,1,f,2,2,4,4
04,2,f,1,1,5,5
05,1,m,4,5,4,
06,2,m,5,4,5,5
07,1,m,3,3,4,5
08,2,m,5,5,5,4")
mydata<-read.table(textConnection(mystring),header=TRUE,sep=",",row.name
s="id")
mydata
#Create score1 so that it differs for males and females:
mydata$score1 <- ifelse( mydata$gender=="f" ,
(mydata$score1 <- (2*mydata$q1)+mydata$q2),
ifelse( mydata$gender=="m",
(mydata$score1 <- (3*mydata$q1)+mydata$q2), NA )
)
mydata
#Create score2 so that it too differs for males and females:
mydata$score2 <- ifelse( mydata$gender=="f" ,
(mydata$score2 <- (2.5*mydata$q1)+mydata$q2),
ifelse( mydata$gender=="m",
(mydata$score2 <- (3.5*mydata$q1)+mydata$q2), NA )
)
mydata
Thanks!
Bob
=
Bob Muenchen (pronounced Min'-chen), Manager
Statistical Consulting Center
U of TN Office of Information Technology
200 Stokely Management Center, Knoxville, TN 37996-0520
Voice: (865) 974-5230
FAX: (865) 974-4810
Email: [EMAIL PROTECTED]
Web: http://oit.utk.edu/scc ,
News: http://listserv.utk.edu/archives/statnews.html
=
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch 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] Rgraphviz -404 Page not found
THX Group. On 11/25/06, Seth Falcon <[EMAIL PROTECTED]> wrote: > > "j.joshua thomas" <[EMAIL PROTECTED]> writes: > > > Again i have problem in locating the package for clique-graphs > > I tried with BioConductor under Browse for packages, it doesn't work > atall. > > > > Kindly guid me > > I think you already got an answer on the Bioconductor list: RBGL is > likely the package you are looking for. > -- Lecturer J. Joshua Thomas KDU College Penang Campus Research Student, University Sains Malaysia [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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] Multiple Conditional Tranformations
Mark,
Here's what I get when I try that approach.
Thanks,
Bob
> mydata$score1<-numeric(mydata$q1) #just initializing.
> mydata$score2<-numeric(mydata$q1)
> mydata$score1<-NA
> mydata$score2<-NA
> mydata
group gender q1 q2 q3 q4 score1 score2
1 1 f 2 2 5 4 NA NA
2 2 f 2 1 4 5 NA NA
3 1 f 2 2 4 4 NA NA
4 2 f 1 1 5 5 NA NA
5 1 m 4 5 4 NA NA NA
6 2 m 5 4 5 5 NA NA
7 1 m 3 3 4 5 NA NA
8 2 m 5 5 5 4 NA NA
> mydata$score1[mydata$gender == "f"]<-2*mydata$q1 + mydata$q2
Warning message:
number of items to replace is not a multiple of replacement length
> mydata$score2[mydata$gender == "f"]<-2.5*mydata$q1 + mydata$q2
Warning message:
number of items to replace is not a multiple of replacement length
> mydata$score1[mydata$gender == "m"]<-3*mydata$q1 + mydata$q2
Warning message:
number of items to replace is not a multiple of replacement length
> mydata$score2[mydata$gender == "m"]<-3.5*mydata$q1 + mydata$q2
Warning message:
number of items to replace is not a multiple of replacement length
>
-Original Message-
From: Leeds, Mark (IED) [mailto:[EMAIL PROTECTED]
Sent: Friday, November 24, 2006 8:45 PM
To: Muenchen, Robert A (Bob)
Subject: RE: [R] Multiple Conditional Tranformations
I'm not sure if I understand your question but I don't think you need
iflelse statements.
myscore<-numeric(q1) ( because I'm not sure how to initialize a list so
initialize a vector with q1 elements )
myscore<-NA ( I think this should set all the values in myscore to NA )
myscore[mydata$gender == f]<-2*mydata$q1 + mydata$q2
myscore[mydata$gender == m]<-3*mydata$q1 + mydata$q2
the above should do what you do in the first part of your code but I
don't know if that was your question ?
also, it does it making myscore a vector because I didn't know how to
initialize a list.
Someone else may goive a better solution. I'm no expert.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Muenchen, Robert
A (Bob)
Sent: Friday, November 24, 2006 8:27 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Multiple Conditional Tranformations
Greetings,
I'm learning R and I'm stuck on a basic concept: how to specify a
logical condition once and then perform multiple transformations under
that condition. The program below is simplified to demonstrate the goal.
Its results are exactly what I want, but I would like to check the
logical state of gender only once and create both (or any number of)
scores at once.
mystring<-
("id,group,gender,q1,q2,q3,q4
01,1,f,2,2,5,4
02,2,f,2,1,4,5
03,1,f,2,2,4,4
04,2,f,1,1,5,5
05,1,m,4,5,4,
06,2,m,5,4,5,5
07,1,m,3,3,4,5
08,2,m,5,5,5,4")
mydata<-read.table(textConnection(mystring),header=TRUE,sep=",",row.name
s="id")
mydata
#Create score1 so that it differs for males and females:
mydata$score1 <- ifelse( mydata$gender=="f" ,
(mydata$score1 <- (2*mydata$q1)+mydata$q2),
ifelse( mydata$gender=="m",
(mydata$score1 <- (3*mydata$q1)+mydata$q2), NA )
)
mydata
#Create score2 so that it too differs for males and females:
mydata$score2 <- ifelse( mydata$gender=="f" ,
(mydata$score2 <- (2.5*mydata$q1)+mydata$q2),
ifelse( mydata$gender=="m",
(mydata$score2 <- (3.5*mydata$q1)+mydata$q2), NA )
)
mydata
Thanks!
Bob
=
Bob Muenchen (pronounced Min'-chen), Manager Statistical Consulting
Center U of TN Office of Information Technology 200 Stokely Management
Center, Knoxville, TN 37996-0520
Voice: (865) 974-5230
FAX: (865) 974-4810
Email: [EMAIL PROTECTED]
Web: http://oit.utk.edu/scc ,
News: http://listserv.utk.edu/archives/statnews.html
=
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch 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.
This is not an offer (or solicitation of an offer) to buy/sell the
securities/instruments mentioned or an official confirmation. Morgan
Stanley may deal as principal in or own or act as market maker for
securities/instruments mentioned or may advise the issuers. This is not
research and is not from MS Research but it may refer to a research
analyst/research report. Unless indicated, these views are the author's
and may differ from those of Morgan Stanley research or others in the
Firm. We do not represent this is accurate or complete and we may not
update this. Past performance is not indicative of future returns. For
additional information, research
Re: [R] Multiple Conditional Tranformations
Mark,
I finally got that approach to work by spreading the logical condition
everywhere. That gets the lengths to match. Still, I can't help but
think there must be a way to specify the logic once per condition.
Thanks,
Bob
mydata$score1<-numeric(mydata$q1) #just initializing.
mydata$score2<-numeric(mydata$q1)
mydata$score1<-NA
mydata$score2<-NA
mydata
mydata$score1[mydata$gender == "f"]<- 2*mydata$q1[mydata$gender=="f"] +
mydata$q2[mydata$gender=="f"]
mydata$score2[mydata$gender == "f"]<-2.5*mydata$q1[mydata$gender=="f"] +
mydata$q2[mydata$gender=="f"]
mydata$score1[mydata$gender == "m"]<-3*mydata$q1[mydata$gender=="m"] +
mydata$q2[mydata$gender=="m"]
mydata$score2[mydata$gender == "m"]<-3.5*mydata$q1[mydata$gender=="m"] +
mydata$q2[mydata$gender=="m"]
mydata
=
Bob Muenchen (pronounced Min'-chen), Manager
Statistical Consulting Center
U of TN Office of Information Technology
200 Stokely Management Center, Knoxville, TN 37996-0520
Voice: (865) 974-5230
FAX: (865) 974-4810
Email: [EMAIL PROTECTED]
Web: http://oit.utk.edu/scc,
News: http://listserv.utk.edu/archives/statnews.html
=
-Original Message-
From: Leeds, Mark (IED) [mailto:[EMAIL PROTECTED]
Sent: Friday, November 24, 2006 8:45 PM
To: Muenchen, Robert A (Bob)
Subject: RE: [R] Multiple Conditional Tranformations
I'm not sure if I understand your question but I don't think you need
iflelse statements.
myscore<-numeric(q1) ( because I'm not sure how to initialize a list so
initialize a vector with q1 elements )
myscore<-NA ( I think this should set all the values in myscore to NA )
myscore[mydata$gender == f]<-2*mydata$q1 + mydata$q2
myscore[mydata$gender == m]<-3*mydata$q1 + mydata$q2
the above should do what you do in the first part of your code but I
don't know if that was your question ?
also, it does it making myscore a vector because I didn't know how to
initialize a list.
Someone else may goive a better solution. I'm no expert.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Muenchen, Robert
A (Bob)
Sent: Friday, November 24, 2006 8:27 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Multiple Conditional Tranformations
Greetings,
I'm learning R and I'm stuck on a basic concept: how to specify a
logical condition once and then perform multiple transformations under
that condition. The program below is simplified to demonstrate the goal.
Its results are exactly what I want, but I would like to check the
logical state of gender only once and create both (or any number of)
scores at once.
mystring<-
("id,group,gender,q1,q2,q3,q4
01,1,f,2,2,5,4
02,2,f,2,1,4,5
03,1,f,2,2,4,4
04,2,f,1,1,5,5
05,1,m,4,5,4,
06,2,m,5,4,5,5
07,1,m,3,3,4,5
08,2,m,5,5,5,4")
mydata<-read.table(textConnection(mystring),header=TRUE,sep=",",row.name
s="id")
mydata
#Create score1 so that it differs for males and females:
mydata$score1 <- ifelse( mydata$gender=="f" ,
(mydata$score1 <- (2*mydata$q1)+mydata$q2),
ifelse( mydata$gender=="m",
(mydata$score1 <- (3*mydata$q1)+mydata$q2), NA )
)
mydata
#Create score2 so that it too differs for males and females:
mydata$score2 <- ifelse( mydata$gender=="f" ,
(mydata$score2 <- (2.5*mydata$q1)+mydata$q2),
ifelse( mydata$gender=="m",
(mydata$score2 <- (3.5*mydata$q1)+mydata$q2), NA )
)
mydata
Thanks!
Bob
=
Bob Muenchen (pronounced Min'-chen), Manager Statistical Consulting
Center U of TN Office of Information Technology 200 Stokely Management
Center, Knoxville, TN 37996-0520
Voice: (865) 974-5230
FAX: (865) 974-4810
Email: [EMAIL PROTECTED]
Web: http://oit.utk.edu/scc ,
News: http://listserv.utk.edu/archives/statnews.html
=
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch 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.
This is not an offer (or solicitation of an offer) to buy/sell the
securities/instruments mentioned or an official confirmation. Morgan
Stanley may deal as principal in or own or act as market maker for
securities/instruments mentioned or may advise the issuers. This is not
research and is not from MS Research but it may refer to a research
analyst/research report. Unless indicated, these views are the author's
and may differ from those of Morgan Stanley research or others in the
Firm. We do not represent this is accurate or complete and we may not
update this. Past performance is not indicative
Re: [R] Multiple Conditional Tranformations
Good idea. I'm still getting used to how flexible R is on substitutions
like that! -Bob
-Original Message-
From: Leeds, Mark (IED) [mailto:[EMAIL PROTECTED]
Sent: Friday, November 24, 2006 10:20 PM
To: Muenchen, Robert A (Bob)
Subject: RE: [R] Multiple Conditional Tranformations
You could set temp<-which(my$gender[my$gender == "f"]) and then temp
will have the female indices and
Then you could just put temp everywhere instead of the statement but I
think that's the best you can do.
Definitely, someone will reply and there may be a shorter way that I am
unaware of.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Muenchen, Robert
A (Bob)
Sent: Friday, November 24, 2006 10:09 PM
To: r-help@stat.math.ethz.ch
Subject: Re: [R] Multiple Conditional Tranformations
Mark,
I finally got that approach to work by spreading the logical condition
everywhere. That gets the lengths to match. Still, I can't help but
think there must be a way to specify the logic once per condition.
Thanks,
Bob
mydata$score1<-numeric(mydata$q1) #just initializing.
mydata$score2<-numeric(mydata$q1)
mydata$score1<-NA
mydata$score2<-NA
mydata
mydata$score1[mydata$gender == "f"]<- 2*mydata$q1[mydata$gender=="f"] +
mydata$q2[mydata$gender=="f"]
mydata$score2[mydata$gender == "f"]<-2.5*mydata$q1[mydata$gender=="f"] +
mydata$q2[mydata$gender=="f"]
mydata$score1[mydata$gender == "m"]<-3*mydata$q1[mydata$gender=="m"] +
mydata$q2[mydata$gender=="m"]
mydata$score2[mydata$gender == "m"]<-3.5*mydata$q1[mydata$gender=="m"] +
mydata$q2[mydata$gender=="m"]
mydata
=
Bob Muenchen (pronounced Min'-chen), Manager Statistical Consulting
Center U of TN Office of Information Technology 200 Stokely Management
Center, Knoxville, TN 37996-0520
Voice: (865) 974-5230
FAX: (865) 974-4810
Email: [EMAIL PROTECTED]
Web: http://oit.utk.edu/scc,
News: http://listserv.utk.edu/archives/statnews.html
=
-Original Message-
From: Leeds, Mark (IED) [mailto:[EMAIL PROTECTED]
Sent: Friday, November 24, 2006 8:45 PM
To: Muenchen, Robert A (Bob)
Subject: RE: [R] Multiple Conditional Tranformations
I'm not sure if I understand your question but I don't think you need
iflelse statements.
myscore<-numeric(q1) ( because I'm not sure how to initialize a list so
initialize a vector with q1 elements )
myscore<-NA ( I think this should set all the values in myscore to NA )
myscore[mydata$gender == f]<-2*mydata$q1 + mydata$q2
myscore[mydata$gender == m]<-3*mydata$q1 + mydata$q2
the above should do what you do in the first part of your code but I
don't know if that was your question ?
also, it does it making myscore a vector because I didn't know how to
initialize a list.
Someone else may goive a better solution. I'm no expert.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Muenchen, Robert
A (Bob)
Sent: Friday, November 24, 2006 8:27 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Multiple Conditional Tranformations
Greetings,
I'm learning R and I'm stuck on a basic concept: how to specify a
logical condition once and then perform multiple transformations under
that condition. The program below is simplified to demonstrate the goal.
Its results are exactly what I want, but I would like to check the
logical state of gender only once and create both (or any number of)
scores at once.
mystring<-
("id,group,gender,q1,q2,q3,q4
01,1,f,2,2,5,4
02,2,f,2,1,4,5
03,1,f,2,2,4,4
04,2,f,1,1,5,5
05,1,m,4,5,4,
06,2,m,5,4,5,5
07,1,m,3,3,4,5
08,2,m,5,5,5,4")
mydata<-read.table(textConnection(mystring),header=TRUE,sep=",",row.name
s="id")
mydata
#Create score1 so that it differs for males and females:
mydata$score1 <- ifelse( mydata$gender=="f" ,
(mydata$score1 <- (2*mydata$q1)+mydata$q2),
ifelse( mydata$gender=="m",
(mydata$score1 <- (3*mydata$q1)+mydata$q2), NA )
)
mydata
#Create score2 so that it too differs for males and females:
mydata$score2 <- ifelse( mydata$gender=="f" ,
(mydata$score2 <- (2.5*mydata$q1)+mydata$q2),
ifelse( mydata$gender=="m",
(mydata$score2 <- (3.5*mydata$q1)+mydata$q2), NA )
)
mydata
Thanks!
Bob
=
Bob Muenchen (pronounced Min'-chen), Manager Statistical Consulting
Center U of TN Office of Information Technology 200 Stokely Management
Center, Knoxville, TN 37996-0520
Voice: (865) 974-5230
FAX: (865) 974-4810
Email: [EMAIL PROTECTED]
Web: http://oit.utk.edu/scc ,
News: http://listserv.utk.edu/archives/statnews.html
=
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listi
[R] predict and arima
Hi all,
Forecasting from an arima model is easy with predict.
But I can't manage to backcast : invent data from the model before the
begining of the sample.
The theory is easy : take your parameters, reverse your data, forecast, and
then reverse the forecast
I've tried to adapt the predict function to do that (i'm not sure that the
statistical procedure is fine (with the residuals), but that's not my point
right now) :
mav.backcast.arima<-function(model,n.backcast,...)
{
if (class(model)[1]!="Arima") stop("argument must be an object
of class 'Arima' (see ?arima)")
model2<-model
model$residuals<-rev(model$residuals)
if (is.ts(model2$residuals))
model$residuals<-ts(model$residuals,start=start(model2$residuals),
frequency=frequency(model2$residuals))
pred.before<-predict(model,n.ahead=n.backcast,...)
freq<-frequency(model$residuals)
startingdate<-per.sub(start(model2$residuals),n.backcast,freq=freq)
pred<-ts(rev(pred.before$pred),start=startingdate,freq=freq)
se<-ts(rev(pred.before$se),start=startingdate,freq=freq)
return((list(pred = pred, se =se)))
}
This function does not work : it gives always the same result, it does not
depend on the residuals (i've tried to insert
a model$residuals<-rep(1,100) after the definition, to check that).
Then i look at the code, with getS3method("predict","Arima"). And i get even
more confused (!) :
where does data play a role in the function ? residuals are loaded into rsd,
but this variable is not used after...
I looked at KalmanForecast and at the C code of KalmanFore, but it did not
help me understand what was going on.
thanks
Franck A.
btw, it has nothing to do with it, but i've done some stuff on time series
(filtering with Hodrick prescott or Baxter King, for instance) that you can
find on http://arnaud.ensae.net
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch 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] Multiple Conditional Tranformations
Try this:
transform(mydata,
score1 = (2 + (gender == "m")) * q1 + q2,
score2 = (2.5 + (gender == "m")) * q1 + q2
)
On 11/24/06, Muenchen, Robert A (Bob) <[EMAIL PROTECTED]> wrote:
> Mark,
>
> I finally got that approach to work by spreading the logical condition
> everywhere. That gets the lengths to match. Still, I can't help but
> think there must be a way to specify the logic once per condition.
>
> Thanks,
> Bob
>
> mydata$score1<-numeric(mydata$q1) #just initializing.
> mydata$score2<-numeric(mydata$q1)
> mydata$score1<-NA
> mydata$score2<-NA
> mydata
>
> mydata$score1[mydata$gender == "f"]<- 2*mydata$q1[mydata$gender=="f"] +
>
> mydata$q2[mydata$gender=="f"]
> mydata$score2[mydata$gender == "f"]<-2.5*mydata$q1[mydata$gender=="f"] +
>
> mydata$q2[mydata$gender=="f"]
> mydata$score1[mydata$gender == "m"]<-3*mydata$q1[mydata$gender=="m"] +
> mydata$q2[mydata$gender=="m"]
> mydata$score2[mydata$gender == "m"]<-3.5*mydata$q1[mydata$gender=="m"] +
>
> mydata$q2[mydata$gender=="m"]
> mydata
>
> =
> Bob Muenchen (pronounced Min'-chen), Manager
> Statistical Consulting Center
> U of TN Office of Information Technology
> 200 Stokely Management Center, Knoxville, TN 37996-0520
> Voice: (865) 974-5230
> FAX: (865) 974-4810
> Email: [EMAIL PROTECTED]
> Web: http://oit.utk.edu/scc,
> News: http://listserv.utk.edu/archives/statnews.html
> =
>
>
> -Original Message-
> From: Leeds, Mark (IED) [mailto:[EMAIL PROTECTED]
> Sent: Friday, November 24, 2006 8:45 PM
> To: Muenchen, Robert A (Bob)
> Subject: RE: [R] Multiple Conditional Tranformations
>
> I'm not sure if I understand your question but I don't think you need
> iflelse statements.
>
> myscore<-numeric(q1) ( because I'm not sure how to initialize a list so
> initialize a vector with q1 elements )
>
> myscore<-NA ( I think this should set all the values in myscore to NA )
> myscore[mydata$gender == f]<-2*mydata$q1 + mydata$q2
> myscore[mydata$gender == m]<-3*mydata$q1 + mydata$q2
>
> the above should do what you do in the first part of your code but I
> don't know if that was your question ?
> also, it does it making myscore a vector because I didn't know how to
> initialize a list.
> Someone else may goive a better solution. I'm no expert.
>
>
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Muenchen, Robert
> A (Bob)
> Sent: Friday, November 24, 2006 8:27 PM
> To: r-help@stat.math.ethz.ch
> Subject: [R] Multiple Conditional Tranformations
>
> Greetings,
>
>
>
> I'm learning R and I'm stuck on a basic concept: how to specify a
> logical condition once and then perform multiple transformations under
> that condition. The program below is simplified to demonstrate the goal.
> Its results are exactly what I want, but I would like to check the
> logical state of gender only once and create both (or any number of)
> scores at once.
>
>
>
> mystring<-
>
> ("id,group,gender,q1,q2,q3,q4
>
> 01,1,f,2,2,5,4
>
> 02,2,f,2,1,4,5
>
> 03,1,f,2,2,4,4
>
> 04,2,f,1,1,5,5
>
> 05,1,m,4,5,4,
>
> 06,2,m,5,4,5,5
>
> 07,1,m,3,3,4,5
>
> 08,2,m,5,5,5,4")
>
>
>
> mydata<-read.table(textConnection(mystring),header=TRUE,sep=",",row.name
> s="id")
>
> mydata
>
>
>
> #Create score1 so that it differs for males and females:
>
> mydata$score1 <- ifelse( mydata$gender=="f" ,
>
> (mydata$score1 <- (2*mydata$q1)+mydata$q2),
>
> ifelse( mydata$gender=="m",
>
> (mydata$score1 <- (3*mydata$q1)+mydata$q2), NA )
>
> )
>
> mydata
>
>
>
> #Create score2 so that it too differs for males and females:
>
> mydata$score2 <- ifelse( mydata$gender=="f" ,
>
> (mydata$score2 <- (2.5*mydata$q1)+mydata$q2),
>
> ifelse( mydata$gender=="m",
>
> (mydata$score2 <- (3.5*mydata$q1)+mydata$q2), NA )
>
> )
>
> mydata
>
>
>
>
>
> Thanks!
>
> Bob
>
> =
> Bob Muenchen (pronounced Min'-chen), Manager Statistical Consulting
> Center U of TN Office of Information Technology 200 Stokely Management
> Center, Knoxville, TN 37996-0520
> Voice: (865) 974-5230
> FAX: (865) 974-4810
> Email: [EMAIL PROTECTED]
> Web: http://oit.utk.edu/scc ,
> News: http://listserv.utk.edu/archives/statnews.html
>
> =
>
>
>
>
>[[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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> Stanley may de
Re: [R] Multiple Conditional Tranformations
And here is a variation:
transform(mydata,
score1 = (2 + (gender == "m")) * q1 + q2,
score2 = score1 + 0.5 * q1
)
or
transform(
transform(mydata, score1 = (2 + (gender == "m")) * q1 + q2),
score2 = score1 + 0.5 * q1
)
On 11/25/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> Try this:
>
>
> transform(mydata,
> score1 = (2 + (gender == "m")) * q1 + q2,
> score2 = (2.5 + (gender == "m")) * q1 + q2
> )
>
>
> On 11/24/06, Muenchen, Robert A (Bob) <[EMAIL PROTECTED]> wrote:
> > Mark,
> >
> > I finally got that approach to work by spreading the logical condition
> > everywhere. That gets the lengths to match. Still, I can't help but
> > think there must be a way to specify the logic once per condition.
> >
> > Thanks,
> > Bob
> >
> > mydata$score1<-numeric(mydata$q1) #just initializing.
> > mydata$score2<-numeric(mydata$q1)
> > mydata$score1<-NA
> > mydata$score2<-NA
> > mydata
> >
> > mydata$score1[mydata$gender == "f"]<- 2*mydata$q1[mydata$gender=="f"] +
> >
> > mydata$q2[mydata$gender=="f"]
> > mydata$score2[mydata$gender == "f"]<-2.5*mydata$q1[mydata$gender=="f"] +
> >
> > mydata$q2[mydata$gender=="f"]
> > mydata$score1[mydata$gender == "m"]<-3*mydata$q1[mydata$gender=="m"] +
> > mydata$q2[mydata$gender=="m"]
> > mydata$score2[mydata$gender == "m"]<-3.5*mydata$q1[mydata$gender=="m"] +
> >
> > mydata$q2[mydata$gender=="m"]
> > mydata
> >
> > =
> > Bob Muenchen (pronounced Min'-chen), Manager
> > Statistical Consulting Center
> > U of TN Office of Information Technology
> > 200 Stokely Management Center, Knoxville, TN 37996-0520
> > Voice: (865) 974-5230
> > FAX: (865) 974-4810
> > Email: [EMAIL PROTECTED]
> > Web: http://oit.utk.edu/scc,
> > News: http://listserv.utk.edu/archives/statnews.html
> > =
> >
> >
> > -Original Message-
> > From: Leeds, Mark (IED) [mailto:[EMAIL PROTECTED]
> > Sent: Friday, November 24, 2006 8:45 PM
> > To: Muenchen, Robert A (Bob)
> > Subject: RE: [R] Multiple Conditional Tranformations
> >
> > I'm not sure if I understand your question but I don't think you need
> > iflelse statements.
> >
> > myscore<-numeric(q1) ( because I'm not sure how to initialize a list so
> > initialize a vector with q1 elements )
> >
> > myscore<-NA ( I think this should set all the values in myscore to NA )
> > myscore[mydata$gender == f]<-2*mydata$q1 + mydata$q2
> > myscore[mydata$gender == m]<-3*mydata$q1 + mydata$q2
> >
> > the above should do what you do in the first part of your code but I
> > don't know if that was your question ?
> > also, it does it making myscore a vector because I didn't know how to
> > initialize a list.
> > Someone else may goive a better solution. I'm no expert.
> >
> >
> > -Original Message-
> > From: [EMAIL PROTECTED]
> > [mailto:[EMAIL PROTECTED] On Behalf Of Muenchen, Robert
> > A (Bob)
> > Sent: Friday, November 24, 2006 8:27 PM
> > To: r-help@stat.math.ethz.ch
> > Subject: [R] Multiple Conditional Tranformations
> >
> > Greetings,
> >
> >
> >
> > I'm learning R and I'm stuck on a basic concept: how to specify a
> > logical condition once and then perform multiple transformations under
> > that condition. The program below is simplified to demonstrate the goal.
> > Its results are exactly what I want, but I would like to check the
> > logical state of gender only once and create both (or any number of)
> > scores at once.
> >
> >
> >
> > mystring<-
> >
> > ("id,group,gender,q1,q2,q3,q4
> >
> > 01,1,f,2,2,5,4
> >
> > 02,2,f,2,1,4,5
> >
> > 03,1,f,2,2,4,4
> >
> > 04,2,f,1,1,5,5
> >
> > 05,1,m,4,5,4,
> >
> > 06,2,m,5,4,5,5
> >
> > 07,1,m,3,3,4,5
> >
> > 08,2,m,5,5,5,4")
> >
> >
> >
> > mydata<-read.table(textConnection(mystring),header=TRUE,sep=",",row.name
> > s="id")
> >
> > mydata
> >
> >
> >
> > #Create score1 so that it differs for males and females:
> >
> > mydata$score1 <- ifelse( mydata$gender=="f" ,
> >
> > (mydata$score1 <- (2*mydata$q1)+mydata$q2),
> >
> > ifelse( mydata$gender=="m",
> >
> > (mydata$score1 <- (3*mydata$q1)+mydata$q2), NA )
> >
> > )
> >
> > mydata
> >
> >
> >
> > #Create score2 so that it too differs for males and females:
> >
> > mydata$score2 <- ifelse( mydata$gender=="f" ,
> >
> > (mydata$score2 <- (2.5*mydata$q1)+mydata$q2),
> >
> > ifelse( mydata$gender=="m",
> >
> > (mydata$score2 <- (3.5*mydata$q1)+mydata$q2), NA )
> >
> > )
> >
> > mydata
> >
> >
> >
> >
> >
> > Thanks!
> >
> > Bob
> >
> > =
> > Bob Muenchen (pronounced Min'-chen), Manager Statistical Consulting
> > Center U of TN Office of Information Technology 200 Stokely Management
> > Center, Knoxville, TN 37996-0520
> > Voice: (865) 974-5230
> > FAX: (865) 974-4810
> > Email: [EMAIL PROTECTED]
> > Web: http://oit.utk.edu/scc ,
> > News: http://listserv.utk.edu/archives/statnews.html
> >
