[R] exp / log scaled version of cut?
Dear list, For tasks involving people its often good to work with ages in exp or log scale, since either not much or very much happens in the first 10 years of life, and much less (or much more) happens the following years. For tables with age range summary statistics, I sometimes would like to have a function that would for instance cut a range of values into breaks exp() scale equidistance bins, but with unconverted values in the labels. Similar to the way log / exp scaling of figure axis tick marks work, but with a factor returned. i have not been able to find such a function. Is there one? Thanks! Fredrik Karlsson -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Variable labels
Hi,
I am sorry but I don't understand your question, Generally, "clicking" is
not something you can assume to be implemented for anything in R.
However, if you read the manual for the package
https://gdemin.github.io/expss/
you get an example at the bottom where an illustration of how the package
can be used to create Excel tables which would then be easy to interact
with through clicking.
Is that what you wanted?
Fredrik
On Thu, May 13, 2021 at 4:49 AM Steven Yen wrote:
> I insert variable with the expss function as shown below. No error
> message. My question is, how to save the variable labels in the data
> frame so that I can click to read the labels. Thank you.
>
> mydata<-read_excel("data/Excel/hseinv.xlsx",na=".")
> library(expss)
> mydata=apply_labels(mydata,
> year ="1947-1988",
> inv="real housing inv, millions $",
> pop="population, 1000s",
> price ="housing price index; 1982 = 1",
> linv ="log(inv)",
> lpop ="log(pop)",
> lprice ="log(price)",
> t ="time trend: t=1,...,42",
> invpc ="per capita inv: inv/pop",
> linvpc ="log(invpc)",
> lprice_1="lprice[_n-1]",
> linvpc_1="linvpc[_n-1]",
> gprice ="lprice - lprice_1",
> ginvpc ="linvpc - linvpc_1")
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
"Life is like a trumpet - if you don't put anything into it, you don't get
anything out of it."
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Re: [R] Problem with loop in folders
Hi,
I am sorry if I am misunderstanding what you are trying to do here, but can
you simplify it this way?
(unfortualtely, this is untested since I dont have a suitable set of files
and a directory structure to test against)
dbifiles <- list.files(pattern="*.dbi",recursive=TRUE)
csvfiles <- gsub("dbi$","csv",dbifiles)
for(i in seq_along(csvfiles)){
df <- read.dbf(dbfiles[i])
write.csv( df, file =csvfiles[i])
}
or something along these lines?
Fredrik
On Fri, Apr 24, 2020 at 4:08 PM Shubhasmita Sahani <
shubhasmita.sah...@gmail.com> wrote:
> Hi Everyone,
> I am trying to loop through the folders in the major working directory.
> Read the dbf file into the data frame then save the data frame as CSV file
> in another folder.
> For this, I have written this code, But not able to figure out where it is
> going wrong. Any ideas will be of great support.
>
>
> setwd(choose.dir())
> csvpath= "C:/plan/Learning/dummydata/csv/"
> a<-list.dirs()
> inpath<-"C:/workplan/Q2/Project1"
>
> for (folder in list.dirs()[-1]) {
>
>path<-setwd(paste0("inpath",folder))
>dbf<-list.files(path, pattern = "*ward.dbf")
>df <- read.dbf(dbf)
>dbfname<-basename(dbf)
>name<-file_path_sans_ext(dbfname) # get the name of the file like
> agra_ward
>write.csv( df, file = paste0("csvpath",name,"csv"))
>print(path)
>
> }
>
>
>
>
>
> --
> Thanks & Regards,
> Shubhasmita Sahani
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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.
>
--
"Life is like a trumpet - if you don't put anything into it, you don't get
anything out of it."
[[alternative HTML version deleted]]
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Re: [R] Creating Dummy Var in R for regression?
Hi,
please also have a look at the 'cut' function.Very handa function for these
types of situations.
Best,
Fredrik
On Sun, Aug 7, 2016 at 8:10 PM, Shivi Bhatia wrote:
> Thank you Jeremiah and all others for the assistance. This really helped.
>
> On Sat, Aug 6, 2016 at 5:01 AM, jeremiah rounds
> wrote:
>
> > Something like:
> >
> > d = data.frame(score = sample(1:10, 100, replace=TRUE))
> > d$score_t = "low"
> > d$score_t[d$score > 3] = "medium"
> > d$score_t[d$score >7 ] = "high"
> > d$score_t = factor(d$score_t, levels = c("low", "medium", "high"),
> > ordered=TRUE) #set ordered = FALSE for dummy variables
> > X = model.matrix(~score_t, data=d)
> > X
> >
> >
> >
> > On Fri, Aug 5, 2016 at 3:23 PM, Shivi Bhatia
> wrote:
> >
> >> Thanks you all for the assistance. This really helps.
> >>
> >> Hi Bert: While searching nabble i got to know R with factors variables
> >> there is no need to create dummy variable. However please consider this
> >> situation:
> >> I am in the process of building a logistic regression model on NPS data.
> >> The outcome variable is CE i.e. customer experience which has 3 rating
> so
> >> ordinal logistic regression will be used. However most of my variables
> are
> >> categorical. For instance one of the variable is agent knowledge which
> is
> >> a
> >> 10 point scale.
> >>
> >> This agent knowledge is again a 3 rated scale: high medium low hence i
> >> need
> >> to group these 10 values into 3 groups & then as you suggested i can
> >> directly enter them in the model without creating n-1 categories.
> >>
> >> I have worked on SAS extensively hence found this a bit confusing.
> >>
> >> Thanks for the help.
> >>
> >> On Sat, Aug 6, 2016 at 2:30 AM, Bert Gunter
> >> wrote:
> >>
> >> > Just commenting on the email subject, not the content (which you have
> >> > already been helped with): there is no need to *ever* create a dummy
> >> > variable for regression in R if what you mean by this is what is
> >> > conventionally meant. R will create the model matrix with appropriate
> >> > "dummy variables" for factors as needed. See ?contrasts and ?C for
> >> > relevant details and/or consult an appropriate R tutorial.
> >> >
> >> > Of course, if this is not what you meant, than ignore.
> >> >
> >> > Cheers,
> >> > Bert
> >> >
> >> >
> >> > Bert Gunter
> >> >
> >> > "The trouble with having an open mind is that people keep coming along
> >> > and sticking things into it."
> >> > -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
> >> >
> >> >
> >> > On Fri, Aug 5, 2016 at 1:49 PM, wrote:
> >> > > Hello,
> >> > >
> >> > > Your ifelse will never work because
> >> > > reasons$salutation== "Mr" & reasons$salutation=="Father" is always
> >> FALSE
> >> > > and so is reasons$salutation=="Mrs" & reasons$salutation=="Miss".
> >> > > Try instead | (or), not & (and).
> >> > >
> >> > > Hope this helps,
> >> > >
> >> > > Rui Barradas
> >> > >
> >> > >
> >> > >
> >> > > Citando Shivi Bhatia :
> >> > >
> >> > >> Dear Team,
> >> > >>
> >> > >> I need help with the below code in R:
> >> > >>
> >> > >> gender_rec<- c('Dr','Father','Mr'=1, 'Miss','MS','Mrs'=2, 3)
> >> > >>
> >> > >> reasons$salutation<- gender_rec[reasons$salutation].
> >> > >>
> >> > >> This code gives me the correct output but it overwrites the
> >> > >> reason$salutation variable. I need to create a new variable gender
> to
> >> > >> capture gender details and leave salutation as it is.
> >> > >>
> >> > >> i tried the below syntax but it is converting all to 1.
> >> > >>
> >> > >> reasons$gender<- ifelse(reasons$salutation== "Mr" &
> >> reasons$salutation==
> >> > >> "Father","Male", ifelse(reasons$salutation=="Mrs" &
> >> > reasons$salutation==
> >> > >> "Miss","Female",1))
> >> > >>
> >> > >> Please suggest.
> >> > >>
> >> > >> [[alternative HTML version deleted]]
> >> > >>
> >> > >> __
> >> > >> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >> > >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> > >> PLEASE do read the posting guide
> >> > >> http://www.R-project.org/posting-guide.htmland provide commented,
> >> > >> minimal, self-contained, reproducible code.
> >> > >
> >> > >
> >> > >
> >> > > [[alternative HTML version deleted]]
> >> > >
> >> > > __
> >> > > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >> > > 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@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> PLEASE do read the posting guide http://www.R-project.org/posti
> >> ng-guide.html
> >> and pro
[R] L1 penalized regression fails to predict from model
Dear list, Sorry for this cross-post from StackOverflow, but I see that SO was maybe the wrong forum for this question. Too package specific and Ok, what I am trying to do is to predict from an L1 penalized regression. This falls due to a data set dimension problem that I cannot figure out. The procedure I'm using is the following: require(penalized)# neg contains negative data# pos contains positive data Now, the procedure below aims to construct comparable (balanced in terms os positive and negative cases) training and validation data sets. # 50% negative training set negSamp <- neg %>% sample_frac(0.5) %>% as.data.frame()# Negative validation set negCompl <- neg[setdiff(row.names(neg),row.names(negSamp)),]# 50% positive training set posSamp <- pos %>% sample_frac(0.5) %>% as.data.frame()# Positive validation set posCompl <- pos[setdiff(row.names(pos),row.names(posSamp)),]# Combine sets validat <- rbind(negSamp,posSamp) training <- rbind(negCompl,posCompl) Ok, so here we now have two comparable sets. [1] FALSE TRUE> dim(training)[1] 1061 381> dim(validat)[1] 1060 381> identical(names(training),names(validat))[1] TRUE I fit the model to the training set without a problem (and I've tried using a range of Lambda1 values here). But, fitting the model to the validation data set fails, with a just odd error description. > fit <- > penalized(VoiceTremor,training[-1],data=training,lambda1=40,standardize=TRUE)# > nonzero coefficients: 13> fit2 <- predict(fit, penalized=validat[-1], > data=validat) Error in .local(object, ...) : row counts of "penalized", "unpenalized" and/or "data" do not match Just to make sure that this is not due to some NA's in the data set: > identical(validat,na.omit(validat))[1] TRUE Oddly enough, I may generate some new data that is comparable to the proper data set: > data.frame(VoiceTremor="NVT",matrix(rnorm(38),nrow=1000,ncol=380) ) -> neg > data.frame(VoiceTremor="VT",matrix(rnorm(38),nrow=1000,ncol=380) ) -> > pos> dim(pos)[1] 1000 381> dim(neg)[1] 1000 381 and run the procedure above, and then the prediction step works! How come? What could be wrong with my second (not training) data set? Fredrik -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] ggplot: geom_line with only pairs of points connected
Dear list,
I've been using plotmeans {gplots} a lot before, and found the
"connect" argument to be quite useful.
I've moved to ggplot for several reasons, but would still like to
connect lines conditionally, somehow.
Is it possible to do?
Small example:
df <- data.frame(cat=LETTERS[1:4],num=rnorm(4))
ggplot(df, aes(x=cat, y=num)) + geom_point() +
geom_line(aes(x=as.numeric(cat), y=num))
In this plot, I would like to be able to make the second line (the one
between B and C) not be there.
What I've come up with is this (instead using geom_segment):
df <- data.frame(cat=LETTERS[1:4],num=rnorm(4))
dfdf
<-cbind(df[seq(from=1,by=2,along.with=df),],df[seq(from=2,by=2,along.with=df),])
names(dfdf) <- c("cat","num","cat2","num2")
ggplot(df, aes(x=cat, y=num)) + geom_point() +
geom_segment(aes(x=cat,xend=cat2,y=num,yend=num2),data=dfdf)
Hoewever, in a more extended example (where df is more complex and
also contains factors used for grouping and faceting) this becomes a
bit cumbersome.
Is there a better way to do this?
/Fredrik
--
"Life is like a trumpet - if you don't put anything into it, you don't
get anything out of it."
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Math labels in .eps files
Dear Professor Ripley, Thank you for your prompt answer. Indeed, cairo_ps solved the problem for me. Thank you! Fredrik 5 jun 2012 kl. 11:51 skrev Prof Brian Ripley : > On 04/06/2012 17:25, Fredrik Karlsson wrote: >> Dear list, >> >> I have non-ascii labels that I want to include somethow in a .eps file for >> inclusion into a LateX document. Is this possible? >> >> What I see is that the (attached) png file looks ok but the .eps renders >> all non-ascii labels at the bottom as [..]. >> >> >> I've searched around enough to learn that including math labels in a .eps >> is a problem. Hower, is there a solution too somewhere? > > I don't think it is a problem: see ?plotmath. But the problem here is that > you seem to want glyphs not in the plotmath set. > > You have neither told us your platform nor the R code you used (as requested > in the posting guide), so we cannot reproduce this. But see ?postscript and > its section 'Encodings'. You need to specify a PostScript encoding which > contains your glyphs ... if there is one. Note too > > See Also: > > ‘cairo_ps’ for another device that can produce PostScript. > > -- > Brian D. Ripley, rip...@stats.ox.ac.uk > Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ > University of Oxford, Tel: +44 1865 272861 (self) > 1 South Parks Road, +44 1865 272866 (PA) > Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using ddply within a function by argument transfer
Dear list,
A follow up on this thread - the solution I ended up with does indeed
involve "parse()".
However, it does the job, and it is more efficient than what I came up with
in my other attempts.
suspicious.vowels <- function(data,factors,vowelcolumn,f1,f2) {
for(currfac in c(factors,vowelcolumn,f1,f2)){
if(!currfac %in% names(data)){
stop("Factor '",currfac,"' does not exist in the data frame. Please
provide only valid column names.")
}
}
# The dirty hack!
facs <- paste(factors,collapse=",")
mycall <-
paste("ddply(data,.(",facs,"),summarize,norms=vector.space(",f1,",",f2,")[[3]],angles=vector.space(",f2,",",f2,")[[4]])",sep="")
nADatap <- parse(text=mycall)
nAData <- eval(nADatap))
.
}
However, if anyone would like to weigh in with a more structured solution,
I would very much appreciate the opportunity to learn how this _should_
really be done.
/Fredrik
On Mon, Mar 5, 2012 at 2:11 AM, David Winsemius wrote:
>
> On Mar 4, 2012, at 6:28 PM, Fredrik Karlsson wrote:
>
> Hi Joshua,
>>
>> Yes, sorry - I attached an .rda file - maybe it was squashed.
>>
>
> The old version of the Posting Guide warned readers that only 'txt' files
> and 'pdf's would be acceptable attachments. I think the new modified Guide
> is less specific. You could ahve posted the output of dput on the object
> and given it an extension of .txt so that your email client would have
> labeled it properly at the right mime-type.
>
> --
>
> David Winsemius, MD
> West Hartford, CT
>
>
--
"Life is like a trumpet - if you don't put anything into it, you don't get
anything out of it."
[[alternative HTML version deleted]]
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Re: [R] Using ddply within a function by argument transfer
Hi Joshua,
Yes, sorry - I attached an .rda file - maybe it was squashed.
Anyway, yes, I agree with you that the function in the present state would
not be of very much help to an end user (and your right in thinking that
this is not the end implementation of the function, just an example). I too
cannot find a combination of candidate functions that work.
What I need is the ability for the user to specify the columns in data to
be processed by ddply and the data frame itself. So far I have not been
able to do this using ddply - I guess I'll have to roll my own using for
loops and data frame subsetting or something similar.
Thank you for trying!
/Fredrik
On Sun, Mar 4, 2012 at 12:50 PM, Joshua Wiley wrote:
> Hi,
>
> Still no data (either not attached or a format the mail server
> scrubs), but I made some up. ddply seems to be a bit tricky in that
> it holds off evaluating its arguments. I honestly am not sure how to
> get around that---no amount of fancy footwork with eval() or quote()
> is going to change the internals of ddply. The easiest approach seems
> to be to change the call to it in the first place. You cannot do this
> by passing arguments, but you can by using subsitute() to have your
> test function construct and then evaluate the ddply() call. Here is a
> working example:
>
> require(plyr)
>
> pb <- data.frame(Type = sample(6, 30, TRUE),
> Sex = sample(0:1, 30, TRUE),
> Speaker = sample(3, 30, TRUE),
> F1 = rnorm(30))
>
> insidefun <- function(x){
> return(x+1)
> }
>
> testfun <- function(data, factors, x, fun = quote(summarize)) {
> arglist <- list(data = data, factors = as.quoted(factors), fun = fun, x =
> x)
> myd <- substitute(ddply(.data=data,
>.variables=factors, .fun = fun, norm = x), arglist)
> out <- eval(myd)
> return(out)
> }
>
> testfun(pb, c("Type","Sex","Speaker"), x = quote(insidefun(F1)))
>
> By the way, in its present state, this is essentially a painfully
> complex wrapper to ddply and your code would probably be less buggy
> and easier to understand just using ddply, but presumably you are
> planning to have some other stuff go on in the real testfun().
>
> Cheers,
>
> Josh
>
> On Sun, Mar 4, 2012 at 9:06 AM, Fredrik Karlsson
> wrote:
> > Hi,
> >
> > Sorry all - I will provide a reproducable version of this. I am still
> > seeing the same problem - maybe it is due to me having to use summarize?
> > Anyway, here is an example (using the data set attached):
> >
> > Two test functions:
> >
> > insidefun <- function(x){
> >
> > return(x+1)
> > }
> >
> > testfun <- function(data,factors,x){
> >
> > datOrig <- ddply(.data=data,
> > .variables=as.quoted(factors),
> > .fun=summarize,
> > norm=insidefun(x)
> > )
> >
> > }
> >
> >> testfun(pb,c("Type","Sex","Speaker"),F1)
> > Error in insidefun(x) : object 'x' not found
> >
> > Now, if I call ddply directly, it works:
> >
> >> head(ddply(pb,.(Type,Sex,Speaker),summarize,norm=insidefun(F1)),4)
> > Type Sex Speaker norm
> > 1c f 62 461
> > 2c f 62 401
> > 3c f 62 601
> > 4c f 62 611
> >
> > I'm sure I'm doing something stupid - and of course I see that the ddply
> call
> > inside the function will have to do some processing in order to get the
> > arguments in the right mode/class - but I don't know how.
> >
> > /Fredrik
> >
> >
> >
> >
> > On Sun, Mar 4, 2012 at 6:44 AM, David Winsemius >wrote:
> >
> >>
> >> On Mar 4, 2012, at 12:20 AM, Fredrik Karlsson wrote:
> >>
> >> Hi Michael,
> >>>
> >>> No, sorry - that is neither the problem or the solution.
> >>>
> >>> suspicious.vowels(pb,c("Type",**"Sex","Vowel"),F1,F2)
> >>>>
> >>> Error in mean(y, na.rm = na.rm) : object 'f1' not found
> >>>
> >>>
> >> Obviously you have still failed to offer reproducible code to go along
> >> wiht your questions, but I would observe that in R that f1 is NOT going
> to
> >> equal F1.
> >>
> >> --
> >>
> >>>
> >>> /Fredrik
> >>>
> >>> On Sat, Mar 3, 2012 at 7:04 PM, R. Michael Weylandt <
> >>> michael.weyla...@gmail.com> wrote:
>
Re: [R] Using ddply within a function by argument transfer
Hi,
Sorry all - I will provide a reproducable version of this. I am still
seeing the same problem - maybe it is due to me having to use summarize?
Anyway, here is an example (using the data set attached):
Two test functions:
insidefun <- function(x){
return(x+1)
}
testfun <- function(data,factors,x){
datOrig <- ddply(.data=data,
.variables=as.quoted(factors),
.fun=summarize,
norm=insidefun(x)
)
}
> testfun(pb,c("Type","Sex","Speaker"),F1)
Error in insidefun(x) : object 'x' not found
Now, if I call ddply directly, it works:
> head(ddply(pb,.(Type,Sex,Speaker),summarize,norm=insidefun(F1)),4)
Type Sex Speaker norm
1c f 62 461
2c f 62 401
3c f 62 601
4c f 62 611
I'm sure I'm doing something stupid - and of course I see that the ddply call
inside the function will have to do some processing in order to get the
arguments in the right mode/class - but I don't know how.
/Fredrik
On Sun, Mar 4, 2012 at 6:44 AM, David Winsemius wrote:
>
> On Mar 4, 2012, at 12:20 AM, Fredrik Karlsson wrote:
>
> Hi Michael,
>>
>> No, sorry - that is neither the problem or the solution.
>>
>> suspicious.vowels(pb,c("Type",**"Sex","Vowel"),F1,F2)
>>>
>> Error in mean(y, na.rm = na.rm) : object 'f1' not found
>>
>>
> Obviously you have still failed to offer reproducible code to go along
> wiht your questions, but I would observe that in R that f1 is NOT going to
> equal F1.
>
> --
>
>>
>> /Fredrik
>>
>> On Sat, Mar 3, 2012 at 7:04 PM, R. Michael Weylandt <
>> michael.weyla...@gmail.com> wrote:
>>
>> Untested, but it might be simpler than that:
>>>
>>> suspicious.vowels(pb,c("Type",**"Sex","Vowel"),"F1",F2)
>>>
>>> Note that "F1" is in quotes but F2 isn't.
>>>
>>> Michael
>>>
>>> On Sat, Mar 3, 2012 at 5:46 PM, Fredrik Karlsson
>>> wrote:
>>>
>>>> Dear list,
>>>>
>>>> Sorry, but I cannot get my head around how and I could pass arguments
>>>>
>>> along
>>>
>>>> to high-level functions. What I have is a function that would benefit
>>>>
>>> from
>>>
>>>> me using ddply from the plyr package.
>>>> However, I cannot get the arguments passing part right.
>>>>
>>>> So, this is my function:
>>>>
>>>> suspicious.vowels <-
>>>>>
>>>> function(data,factors,f1,f2,**evaluate.original.params=TRUE) {
>>>>
>>>> datOrig <- ddply(.data=data,
>>>> .variables=as.quoted(factors),
>>>> .fun=summarize,
>>>> norm=vector.space(f1,f2)[["**Vector norms"]]
>>>> )
>>>>
>>>>
>>>> print(datOrig)
>>>>
>>>> }
>>>>
>>>> Of course, if I try to call this function, I get an error message
>>>> telling
>>>> me that the "f1" argument does not exist:
>>>>
>>>> suspicious.vowels(pb,c("Type",**"Sex","Vowel"),"F1",F2)
>>>>>
>>>> Error in mean(y, na.rm = na.rm) : object 'f1' not found
>>>>
>>>> However, the corresponding ddply call, when called from the console,
>>>> does
>>>> work:
>>>>
>>>>
>>>>>
>>>> head(ddply(pb,as.quoted(c("**Type","Sex","Vowel")),**
>>> summarize,norm=vector.space(**F1,F2)[["Vector
>>>
>>>> norms"]]),4)
>>>> Type Sex Vowel norm
>>>> 1c faa 250.1570
>>>> 2c faa 497.2711
>>>> 3c faa 172.3108
>>>> 4c faa 109.4464
>>>> ...
>>>>
>>>> So, how do I modify the function to pass the arguments that I supply
>>>> correctly? I cannot get my head around this enough to find the correct
>>>> combination of deparse /substitute /... to get this right.
>>>>
>>>> I would be thankful for all the help I could get on this.
>>>>
>>>> /Fredrik
>>>>
>>>> --
>>>> "Life is like a trumpet - if you don't put anything into it, you don't
>>>>
>>> get
>&g
Re: [R] Using ddply within a function by argument transfer
Hi Michael,
No, sorry - that is neither the problem or the solution.
> suspicious.vowels(pb,c("Type","Sex","Vowel"),F1,F2)
Error in mean(y, na.rm = na.rm) : object 'f1' not found
/Fredrik
On Sat, Mar 3, 2012 at 7:04 PM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:
> Untested, but it might be simpler than that:
>
> suspicious.vowels(pb,c("Type","Sex","Vowel"),"F1",F2)
>
> Note that "F1" is in quotes but F2 isn't.
>
> Michael
>
> On Sat, Mar 3, 2012 at 5:46 PM, Fredrik Karlsson
> wrote:
> > Dear list,
> >
> > Sorry, but I cannot get my head around how and I could pass arguments
> along
> > to high-level functions. What I have is a function that would benefit
> from
> > me using ddply from the plyr package.
> > However, I cannot get the arguments passing part right.
> >
> > So, this is my function:
> >
> >> suspicious.vowels <-
> > function(data,factors,f1,f2,evaluate.original.params=TRUE) {
> >
> > datOrig <- ddply(.data=data,
> >.variables=as.quoted(factors),
> >.fun=summarize,
> >norm=vector.space(f1,f2)[["Vector norms"]]
> >)
> >
> >
> > print(datOrig)
> >
> > }
> >
> > Of course, if I try to call this function, I get an error message telling
> > me that the "f1" argument does not exist:
> >
> >> suspicious.vowels(pb,c("Type","Sex","Vowel"),"F1",F2)
> > Error in mean(y, na.rm = na.rm) : object 'f1' not found
> >
> > However, the corresponding ddply call, when called from the console, does
> > work:
> >
> >>
> >
> head(ddply(pb,as.quoted(c("Type","Sex","Vowel")),summarize,norm=vector.space(F1,F2)[["Vector
> > norms"]]),4)
> > Type Sex Vowel norm
> > 1c faa 250.1570
> > 2c faa 497.2711
> > 3c faa 172.3108
> > 4c faa 109.4464
> > ...
> >
> > So, how do I modify the function to pass the arguments that I supply
> > correctly? I cannot get my head around this enough to find the correct
> > combination of deparse /substitute /... to get this right.
> >
> > I would be thankful for all the help I could get on this.
> >
> > /Fredrik
> >
> > --
> > "Life is like a trumpet - if you don't put anything into it, you don't
> get
> > anything out of it."
> >
> >[[alternative HTML version deleted]]
> >
> > __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
--
"Life is like a trumpet - if you don't put anything into it, you don't get
anything out of it."
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Using ddply within a function by argument transfer
Dear list,
Sorry, but I cannot get my head around how and I could pass arguments along
to high-level functions. What I have is a function that would benefit from
me using ddply from the plyr package.
However, I cannot get the arguments passing part right.
So, this is my function:
> suspicious.vowels <-
function(data,factors,f1,f2,evaluate.original.params=TRUE) {
datOrig <- ddply(.data=data,
.variables=as.quoted(factors),
.fun=summarize,
norm=vector.space(f1,f2)[["Vector norms"]]
)
print(datOrig)
}
Of course, if I try to call this function, I get an error message telling
me that the "f1" argument does not exist:
> suspicious.vowels(pb,c("Type","Sex","Vowel"),"F1",F2)
Error in mean(y, na.rm = na.rm) : object 'f1' not found
However, the corresponding ddply call, when called from the console, does
work:
>
head(ddply(pb,as.quoted(c("Type","Sex","Vowel")),summarize,norm=vector.space(F1,F2)[["Vector
norms"]]),4)
Type Sex Vowel norm
1c faa 250.1570
2c faa 497.2711
3c faa 172.3108
4c faa 109.4464
...
So, how do I modify the function to pass the arguments that I supply
correctly? I cannot get my head around this enough to find the correct
combination of deparse /substitute /... to get this right.
I would be thankful for all the help I could get on this.
/Fredrik
--
"Life is like a trumpet - if you don't put anything into it, you don't get
anything out of it."
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Multiple comparisons of lme model - interactions?
Dear list, Please excuse my ignorance, but I'm trying to model some data using the lme package. vot is a numeric response, and condition, location and obs are all categories. This works: > anova(vot.lme <- lme(vot ~ condition * location * obs,data=mergedCodesL,random= ~1 |patient)) numDF denDF F-value p-value (Intercept)1 1898 462.7519 <.0001 condition 2 1898 8.4126 0.0002 location 112 0.0272 0.8718 obs2 1898 472.5526 <.0001 condition:location 2 1898 1.0467 0.3513 condition:obs 4 1898 1.0683 0.3706 location:obs 2 1898 9.7067 0.0001 condition:location:obs 4 1898 4.6143 0.0010 If I then would like to do post-hoc testing, I found in the email archives that I could use the glht function of multcomp - something like summary(glht(vot.lme, linfct=mcp(obs = "Tukey"))) However, if I would like to investigate the condition:location:obs - interaction. What do I do then? Best! /Fredrik -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Finding suspicious data points?
Dear list, I would like to find data points that at least should be checked one more time before I process them further. I've had a look at the outliers package for this, and the outliers function in that package, but it appears to only return one value. An example: > outlier(c(1:3,rnorm(1000,mean=10,sd=300))) [1] 1 I think at least 1,2 and 3 should be checked in this case. Any ideas on how to achieve this in R? Actually, the real data I will be investigating consist of vector norms and angles (in an attempt to identify either very short, very long vectors, or vectors pointing in an odd angle for the category to which it has been assigned) so a 2D method would be even better. I would much appreciate any help I can get on this, /Fredrik -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Constructing an additional key inside of a lattice panel
Dear Deepayan,
Thank you for your quick and enlightening reply. I'll try making a modifies
version of the two functions and see what I end up with..
However, out of pure curiosity, what is wrong with the "legend" I
constructed below:
legend=list(corner=list(
fun=draw.key,
args=list(
key=list(text=
list(myFactorLevels),
lines=list(
lty=1:length(myFactorLevels))
),
draw=TRUE
)
),x=0,y=1
It gives me nothing (not even an error message :-)), so I have no idea about
what I did wrong here.
I would very much appreciate any insight on this.
/Fredrik
On Wed, Aug 17, 2011 at 2:57 PM, Deepayan Sarkar
wrote:
> On Tue, Aug 16, 2011 at 4:35 PM, Fredrik Karlsson
> wrote:
> > Hi,
> >
> > I would like to add an additional key inside of a panel based on a factor
> > that is not the "groups" argument.
> > I've tried using the panel.key function in latticeExtras, but I cannot
> get
> > the line types the way I want it.
> >
> > Using my factor myGroups, I've tried this:
> >
> > panel.key(text=levels(myGroups),lines=TRUE,points=FALSE,corner =
> > c(0,.98),key=list(lines=list(lty=1:length(levels(myGroups)
> >
> > I then get the key where I want it, the text is right, but line types are
> > not correct (always lty=1, I think).
> >
> > Any ideas on how I could solve this?
>
> Well, trying to add undocumented arguments and hoping they will
> magically work usually doesn't help.
>
> panel.key() works using simpleKey(), which by design is simple but not
> flexible. In particular, it will not allow you to set 'lty' directly,
> and instead use the values from the theme currently in use.
>
> You _can_ change the theme also, using a different argument; e.g.,
>
> library(lattice)
> library(latticeExtra)
>
> xyplot(1 ~ 1,
> panel = function(...) {
> panel.xyplot(...)
> panel.key(text = month.name[1:2],lines=TRUE,points=FALSE,
> corner = c(0,.98))
> },
> par.settings = simpleTheme(lty = 1:2))
>
> But I don't know if that interferes with the rest of your call.
>
> If all else fails, panel.key() is not a very complicated function, so
> you can take inspiration from it and write your own version that
> replaces the line
>
>key <- simpleKey(text, ...)
>
> with
>
>key <-
>
> where describes the legend that you want.
>
> -Deepayan
>
--
"Life is like a trumpet - if you don't put anything into it, you don't get
anything out of it."
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Constructing an additional key inside of a lattice panel
Dear list, following up on my own post, I have now started trying constructing a "legend" argument to xyplot that would work, based on the examples in Sarkar's book. I'm now at a state where I have a legend that does not throw an error, but no legend is displayed: legend=list(corner=list( fun=draw.key, args=list( key=list(text= list(myFactorLevels), lines=list( lty=1:length(myFactorLevels)) ), draw=TRUE ) ),x=0,y=1 Where myFactorLevels contains the levels I want displayed (e.i. the levels of myGroups in my previous attempt). Please, why is this legend not displayed? /Fredrik On Tue, Aug 16, 2011 at 1:05 PM, Fredrik Karlsson wrote: > Hi, > > I would like to add an additional key inside of a panel based on a factor > that is not the "groups" argument. > I've tried using the panel.key function in latticeExtras, but I cannot get > the line types the way I want it. > > Using my factor myGroups, I've tried this: > > panel.key(text=levels(myGroups),lines=TRUE,points=FALSE,corner = > c(0,.98),key=list(lines=list(lty=1:length(levels(myGroups) > > I then get the key where I want it, the text is right, but line types are > not correct (always lty=1, I think). > > Any ideas on how I could solve this? > > > > /Fredrik > > > > -- > "Life is like a trumpet - if you don't put anything into it, you don't get > anything out of it." > -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Constructing an additional key inside of a lattice panel
Hi, I would like to add an additional key inside of a panel based on a factor that is not the "groups" argument. I've tried using the panel.key function in latticeExtras, but I cannot get the line types the way I want it. Using my factor myGroups, I've tried this: panel.key(text=levels(myGroups),lines=TRUE,points=FALSE,corner = c(0,.98),key=list(lines=list(lty=1:length(levels(myGroups) I then get the key where I want it, the text is right, but line types are not correct (always lty=1, I think). Any ideas on how I could solve this? /Fredrik -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] current.panel.limits() of lattice returning NaN limits - why?
Hi,
I need a custom axis function for a plot, but it seems
that current.panel.limits() sometimes returns NaN limits for the plot, which
it much harder to calculate anything sensible.
An illustration:
Given this axis function:
vs.axis <- function(...){
xlim <- current.panel.limits()$xlim
ylim <- current.panel.limits()$ylim
# Debug code
print(list(ylim=ylim,xlim=xlim))
xat <- pretty(seq(xlim[1],xlim[2],100),n=5)
yat <- pretty(seq(ylim[1],ylim[2],100),n=4)
xlab <- sub("-","",as.character(xat))
ylab <- sub("-","",as.character(yat))
panel.axis(side="top",at=xat,labels=xlab)
panel.axis(side="right",at=yat,labels=ylab)
}
and the attached data set, I get this output:
> xyplot(F1 ~F2,data=pb,axis=vs.axis)
$ylim
[1] NaN NaN
$xlim
[1] 346.5 3823.5
Error in if (del == 0 && to == 0) return(to) :
missing value where TRUE/FALSE needed
What's wrong? Is there a more robust way of getting the x- and y- limits?
/Fredrik
--
"Life is like a trumpet - if you don't put anything into it, you don't get
anything out of it."
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Passing on "groups" argument to xyplot within a plotting function
Hi,
I am constructing a plotting function that I would like to behave like
plotting functions within the lattice package. It takes a "groups" argument,
which I process, and then I would like to pass that argument on to the
xyplot function for the actual plotting. However, what ever I do, get an
error that the variable is missing.
A short illustration:
Given the data set
> names(pb)
[1] "Type""Sex" "Speaker" "Vowel" "IPA" "F0" "F1"
[8] "F2" "F3"
and these test functions:
TESTFUN <- function(x,data,groups){
xyplot(x,data=data,groups=groups)
}
TESTFUN2 <- function(x,data,groups){
xyplot(x,data=data,groups=substitute(groups))
}
TESTFUN3 <- function(x,data,groups){
groups <- eval(substitute(groups), data, environment(x))
xyplot(x,data=data,groups=groups)
}
I fail to get "groups" to be passed on to xyplot correctly:
> TESTFUN(F1 ~ F2,data=pb,groups=Type)
Error in eval(expr, envir, enclos) : object 'groups' not found
> TESTFUN2(F1 ~ F2,data=pb,groups=Type)
Error in prepanel.default.function(groups = groups, x = c(2280L, 2400L, :
object 'groups' not found
> TESTFUN3(F1 ~ F2,data=pb,groups=Type)
Error in eval(expr, envir, enclos) : object 'groups' not found
Please help me understand what I am doing wrong.
/Fredrik
--
"Life is like a trumpet - if you don't put anything into it, you don't get
anything out of it."
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
[R] How do I make proper use of the by() function?
Dear list, I have a function that uses values from two vectors and spits out one new vector based on all the values of the two original vectors, and with the same length as them. Ok, I would now like to be able to apply that function simply on two columns in a data frame, divided by the levels of factors in the data frame. I'm trying to use by() for this, but the output is too hard to use. What I am doing is this: by(df, list(df$Factor1,df$Factor2),function(x) my_function(x$col1,x$col2),simplify=TRUE) and the output is too complex to be used in a simple way. Actually, I just want something like a data frame, where the results vectors are placed in one column and the conditions under which they were produced (i.e. the values of the factors according to which the data set were divided) in other columns. This does not seem to be provided by by(), and aggregate only provides me with the ability to use values from one column, right? So, are there other functions I could use? Thanks! /Fredrik -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Display data.frame as tree?
Dear list, I have a data.frame with aggregated frequencies of occurrence of two-way (and in one case three-way) features in my observations. Like this: location condition transient fricationmrallvoiced correctstop vpresent frequency 1 STNPOSTON TRUE FALSE FALSE FALSETRUE FALSE11 2 STN POSTOFF FALSE TRUE FALSE FALSE FALSE TRUE23 3 cZi POSTOFF FALSE TRUE FALSE FALSE FALSE TRUE34 4 STNPOSTON FALSE TRUE FALSE FALSE FALSE TRUE23 5 cZiPOSTON FALSE TRUE FALSE FALSE FALSE TRUE22 For visualisation, I would like to show this data as a (classification) tree (if possible). Like this: /\ location STN cZi / \ \ / \ \ condition PRE OFF ON PRE OFF ON /\ /\ /\ /\ /\ /\ .. and so on (I'm sure you get the idea) with frequencies in the leaves. Is there any way to do this already? Just thought I'd ask before I dive into one of the many tree drawing packages out there or roll my own in graphviz. Thank you! /Fredrik -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to reshape wide format data.frame to long format?
Dear list,
I need to convert this data.frame
> names(codesM)
[1] "key""AMR.pa1.M" "AMR.pa2.M" "AMR.pa3.M" "AMR.pa4.M"
[6] "AMR.pa5.M" "AMR.pa6.M" "AMR.pa7.M" "AMR.pa8.M" "AMR.pa9.M"
[11] "AMR.pa10.M" "AMR.ta1.M" "AMR.ta2.M" "AMR.ta3.M" "AMR.ta4.M"
[16] "AMR.ta5.M" "AMR.ta6.M" "AMR.ta7.M" "AMR.ta8.M" "AMR.ta9.M"
[21] "AMR.ta10.M" "AMR.ka1.M" "AMR.ka2.M" "AMR.ka3.M" "AMR.ka4.M"
[26] "AMR.ka5.M" "AMR.ka6.M" "AMR.ka7.M" "AMR.ka8.M" "AMR.ka9.M"
[31] "AMR.ka10.M" "SMR.pa1.M" "SMR.pa2.M" "SMR.pa3.M" "SMR.pa4.M"
[36] "SMR.pa5.M" "SMR.pa6.M" "SMR.pa7.M" "SMR.pa8.M" "SMR.pa9.M"
[41] "SMR.pa10.M" "SMR.ta1.M" "SMR.ta2.M" "SMR.ta3.M" "SMR.ta4.M"
[46] "SMR.ta5.M" "SMR.ta6.M" "SMR.ta7.M" "SMR.ta8.M" "SMR.ta9.M"
[51] "SMR.ta10.M" "SMR.ka1.M" "SMR.ka2.M" "SMR.ka3.M" "SMR.ka4.M"
[56] "SMR.ka5.M" "SMR.ka6.M" "SMR.ka7.M" "SMR.ka8.M" "SMR.ka9.M"
[61] "SMR.ka10.M"
> dim(codesM)
[1] 42 61
into a 3 x 2501 data.frame where the "key" variable is kept, the
values in columns 2-61 above is inserted into a "values" column and
the name of the column is inserted in a third column ("variable"
perhaps).
Like
key variable value
POSTOFF_1_1AMR.pa1.M 5
POSTOFF_1_1AMR.pa2.M 3
I think I should be able to do this using the "reshape" function, but
I cannot get it to work. I think I need some help to understand
this...
(If I could split the "variable" into three separate columns splitting
by ".", that would be even better.)
I appreciate all the help I could get.
/Fredrik
--
"Life is like a trumpet - if you don't put anything into it, you don't
get anything out of it."
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Any way to apply TWO functions with tapply()?
Hi,
What you can do is define your own function which takes a vector of
values, computes the statistics you want and then returns a string
which displays the output the way you want it. Then use this function
in your tapply call.
like (untested)
mySummary <- function(x) {
paste(mean(x),sd(x),sep=",")
}
tapply(data$response, list(data$targets, data$conditions) ,mySummary)
Of course, if you need a different output format, then you'll have to
adapt the paste call.
/Fredrik
On Fri, May 7, 2010 at 11:39 AM, Phil Wieland wrote:
>
> I need to compute the mean and the standard deviation of a data set and would
> like to have the results in one table/data frame. I call tapply() two times
> and do then merge the resulting tables to have them all in one table. Is
> there any way to tell tapply() to use the functions mean and sd within one
> function call? Something like tapply(data$response, list(data$targets,
> data$conditions), c(mean, sd)).
>
> Thanks in advance.
> --
> View this message in context:
> http://r.789695.n4.nabble.com/Any-way-to-apply-TWO-functions-with-tapply-tp2133924p2133924.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
"Life is like a trumpet - if you don't put anything into it, you don't
get anything out of it."
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Converting dollar value (factors) to numeric
Hi,
Something similar to this maybe?
> test <- as.factor("$111.11")
> test
[1] $111.11
Levels: $111.11
> as.numeric(substring(as.character(test),2))
[1] 111.11
To be applied to your data.frame columns.
/Fredrik
On Wed, May 5, 2010 at 9:16 AM, Wang, Kevin (SYD) wrote:
> Hi,
>
> I'm trying to read in a bunch of CSV files into R where many columns are
> coded like $111.11. When reading them in they are treated as factors.
>
> I'm wondering if there is an easy way to convert them into numeric in R
> (as I don't want to modify the source data)? I've done some searches
> and can't seem to find an easy way to do this.
>
> I apologise if this is a trivial question, I haven't been using R for a
> while.
>
> Many thanks in advance!
>
> Cheers
>
> Kev
>
> Kevin Wang
>> Senior Advisor, Health and Human Services Practice
>> Government Advisory Services
>>
>> KPMG
>> 10 Shelley Street
>> Sydney NSW 2000 Australia
>>
>> Tel +61 2 9335 8282
>> Fax +61 2 9335 7001
>>
> kevinw...@kpmg.com.au
>
>> Protect the environment: think before you print
>>
>>
>
>
> [[alternative HTML version deleted]]
>
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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[R] Best option for exporting data frame to SPSS?
Dear list,
I am leaving my old position and now need to convert my R data frames
into a format that can be used by an SPSS user replacing me, without
running into conversion problems.
The data set consists of strings in UTF8 encoding and values in double
precision floats. The data set is not terribly large, but I had bit
problems getting it into R due to the large number of unfortunate
characters in the strings (', #, " and so on) so I was just wondering
if there is any way to get the data into a SPSS friendly format (other
than tab-separated files) so that a minimum of conversion is done in
between the two systems.
A data base file (SQLite) would be ideal, but unfortunatelly, I don't
think the recieving end would be able to handle it, i.e. get the data
into SPSS.
Sorry for asking this on the list, but I have found lots of
information about getting data safelly _into_ R in the archive, but
far les about exporting data out of R.
Please give me your best tip.
/Fredrik
--
"Life is like a trumpet - if you don't put anything into it, you don't
get anything out of it."
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[R] Surpress one panel in lattice plot..?
Dear list, I have a two character vector with two different values in them (two each, that is). Naturally, when I use these vectors as grouping factors in a lattice plot, I get four panels. Now, one of the possible four combinations will never have data, so one panel will always be empty. So now my question is: can I force R / Lattice not to print it, and give me only the 3 filled panels? I have tried adding "layout=c(3,1)" to the list of arguments, but that gives me one plot that looks like I want it to, and then a new plot with just the empty panel. I use this in an odfWeave document, so just getting the last (empty) plot in the output is not really what I want. I do of course appreciate all the help I can get on this. /Fredrik -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Selecting groups with R
Hi John,
I would guess that your Color column is a factor, with three levels
("RED","BLUE","WHITE"), which means that they will all be included in
the output of a table() call, even if they are empty. Try
dataset <- transform(dataset, Color=as.character(Color))
or something similar and then create the table.
/Fredrik
On Fri, Aug 21, 2009 at 11:08 PM, jlwoodard wrote:
>
> I have a data set similar to the following:
>
> Color Score
> RED 10
> RED 13
> RED 12
> WHITE 22
> WHITE 27
> WHITE 25
> BLUE 18
> BLUE 17
> BLUE 16
>
> and I am trying to to select just the values of Color that are equal to RED
> or WHITE, excluding the BLUE.
>
> I've tried the following:
> myComp1<-subset(dataset, Color =="RED" | Color == "WHITE")
> myComp1<-subset(dataset, Color != "BLUE")
> myComp1<-dataset[which(dataset$Color != "BLUE"),]
>
> Each of the above lines successfully excludes the BLUE subjects, but the
> "BLUE" category is still present in my data set; that is, if I try
> table(Color) I get
>
> RED WHITE BLUE
> 82 151 0
>
> If I try to do a t-test (since I've presumably gone from three groups to two
> groups), I get:
> Error in if (stderr < 10 * .Machine$double.eps * max(abs(mx), abs(my)))
> stop("data are essentially constant") :
> missing value where TRUE/FALSE needed
> In addition: Warning message:
> In mean.default(y) : argument is not numeric or logical: returning NA
>
> and describe.by(score,Color) gives me descriptives for RED and WHITE, and
> BLUE also shows up as NULL.
>
> How can I eliminate the BLUE category completely so I can do a t-test using
> Color (with just the RED and WHITE subjects)?
>
> Many thanks in advance!!
>
> John
>
>
>
> --
> View this message in context:
> http://www.nabble.com/Selecting-groups-with-R-tp25088073p25088073.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
"Life is like a trumpet - if you don't put anything into it, you don't
get anything out of it."
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Stata file Import and Analysis in R
Hi, You should install the foreign package and try the read.dta function. I don't know whether the size of your fiel will be a problem, but you would try it at least. /Fredrik On Mon, Jun 29, 2009 at 1:58 PM, saurav pathak wrote: > Hi > I have a stata data set (.dta file) of size 600 MB, I need to import it in > R and do a 2SLS multilevel analysis on the data set. I would eb grateful if > help is provided for the first part of how to import this big file from > Stata to R and then how to open the imported Stata file in R? > Kindly help > Thanks in advance > > -- > Dr.Saurav Pathak > PhD, Univ.of.Florida > Mechanical Engineering > Doctoral Student > Innovation and Entrepreneurship > Imperial College Business School > s.patha...@imperial.ac.uk > 0044-7795321121 > > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How do I get just the two last tokens of each string in a vector?
HI Jim,
Thank you! Works perfectly.
/Fredrik
On Fri, Jun 26, 2009 at 1:00 PM, jim holtman wrote:
> This should do it
>
>> a <- c(" %L H*L L*H H%", "%L H* H%", "%L L*H %", "%L L*H %" )
>> # split the strings
>> x <- strsplit(a, ' ')
>> # get last two tokens
>> sapply(x, function(.tokens) paste(tail(.tokens, 2), collapse=' '))
> [1] "L*H H%" "H* H%" "L*H %" "L*H %"
>>
>>
>
>
> On Fri, Jun 26, 2009 at 7:21 AM, Fredrik Karlsson
> wrote:
>>
>> Dear list,
>>
>> Sorry for asking this very silly question on the list, but I seem to
>> have made my life complicated by going into string manipulation in
>> vectors.
>> What I need is to get the last part of a sting (the two last tokens,
>> separated by a space), and of course, this should be done for all
>> strings in a vector, creating a new vector of exual size.
>>
>> So,
>>
>> a <- c(" %L H*L L*H H%", "%L H* H%", "%L L*H %", "%L L*H %" )
>>
>> should be made into a vector
>>
>> c(" L*H H%", "H* H%", "L*H %", "L*H %" )
>>
>> I have tried strsplit, but it seems to produce a structure I cannot
>> get to work in this context. Any ideas on how to solve this?
>>
>> Thankful for all the help I can get.
>>
>> /Fredrik
>>
>>
>> --
>> "Life is like a trumpet - if you don't put anything into it, you don't
>> get anything out of it."
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem that you are trying to solve?
>
--
"Life is like a trumpet - if you don't put anything into it, you don't
get anything out of it."
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[R] How do I get just the two last tokens of each string in a vector?
Dear list,
Sorry for asking this very silly question on the list, but I seem to
have made my life complicated by going into string manipulation in
vectors.
What I need is to get the last part of a sting (the two last tokens,
separated by a space), and of course, this should be done for all
strings in a vector, creating a new vector of exual size.
So,
a <- c(" %L H*L L*H H%", "%L H* H%", "%L L*H %", "%L L*H %" )
should be made into a vector
c(" L*H H%", "H* H%", "L*H %", "L*H %" )
I have tried strsplit, but it seems to produce a structure I cannot
get to work in this context. Any ideas on how to solve this?
Thankful for all the help I can get.
/Fredrik
--
"Life is like a trumpet - if you don't put anything into it, you don't
get anything out of it."
__
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[R] Lattice group colors?
Dear list, I have been struggling to find how I would go about changing the bakground colors of groups in a lattice barchart in a way so that the auto.key generated also does the right thing and pick it up for the key. I have used the "col" argument (which I guess is sent to par()) in a way so that the colors are like I would want them, but now I guess I need to know which part part of the theme I need to change in order to make this change permanent for all the plots, and all the keys. I am of course thankful for all the help I can get. (Also, how does one know these things about the theme variables? Is there some documentation somewhere?) /Fredrik -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Fwd: Urgent - odfWeave produces graphs /images with Read-Error
-- Forwarded message -- From: Fredrik Karlsson Date: Wed, Jun 17, 2009 at 11:10 AM Subject: Re: [R] Urgent - odfWeave produces graphs /images with Read-Error To: Ronggui Huang Hi Ronggui, Thank you for this! In some of the graphs, the code that was copy-and-pasted did indeed not include the print command. Thank you for pointing this out! /Fredrik On Wed, Jun 17, 2009 at 10:41 AM, Ronggui Huang wrote: > > If you use lattice to produce the figures, you need to wrap them in print, > say, > <<>>= > print(xyplot(y~x,data=dat)) > @ > > Ronggui > > 2009/6/17 Fredrik Karlsson : > > Dear list, > > > > I have been working on a report with around 60 images in it, and > > everything has been fine, until now. I find that the image output that > > is produced by odfWeave produces images that are NOT readable by > > OpenOffice or NeoOffice. I get empty boxes with Read-Error written in > > them. While generating the file using odfWeave, I can see them being > > generated on the screen, and they all seem fine. > > > > Have anybody seen this before? Is there a way around it? > > > > /Fredrik > > > > -- > > "Life is like a trumpet - if you don't put anything into it, you don't > > get anything out of it." > > > > __ > > R-help@r-project.org mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > > and provide commented, minimal, self-contained, reproducible code. > > > > > > -- > HUANG Ronggui, Wincent > PhD Candidate > Dept of Public and Social Administration > City University of Hong Kong > Home page: http://asrr.r-forge.r-project.org/rghuang.html -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Urgent - odfWeave produces graphs /images with Read-Error
Dear list, I have been working on a report with around 60 images in it, and everything has been fine, until now. I find that the image output that is produced by odfWeave produces images that are NOT readable by OpenOffice or NeoOffice. I get empty boxes with Read-Error written in them. While generating the file using odfWeave, I can see them being generated on the screen, and they all seem fine. Have anybody seen this before? Is there a way around it? /Fredrik -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Creating a variable which is the sum of equal rows in a dataframe
Hi Cecilia,
You can use table for this:
#Generate a dataframe
> dat.df <- data.frame(Firm =
> sample(c(500600700,500600800,500600800),6,replace=TRUE),Banks=sample(c("Citybank","CGD","DES","Bank1"),6,replace=TRUE)
> )
#Get the counts in a table format
> with(dat.df, table(Firm,Banks))
Banks
FirmBank1 CGD Citybank DES
500600700 0 01 0
500600800 2 11 1
#Or, if you prefer the data.frame format, you can reshape it.
> as.data.frame(with(dat.df, table(Firm,Banks)))
FirmBanks Freq
1 500600700Bank10
2 500600800Bank12
3 500600700 CGD0
4 500600800 CGD1
5 500600700 Citybank1
6 500600800 Citybank1
7 500600700 DES0
8 500600800 DES1
Hope this helps.
/Fredrik
On Mon, May 4, 2009 at 1:19 PM, Cecilia Carmo wrote:
> Hi everyone:
>
> I need to count the number of banks of each firm in my data. The firm is
> identified by the fiscal number. The banks of each firm appears like this:
>
> Firm Banks
> 500600700 Citybank
> 500600700 CGD
> 500600700 BES
> 500600800 Citybank
> 500600800 Bank1
> 500600900 CGD
> …
>
> I want to obtain the following dataframe:
> Firm numberofbanks
> 500600700 3
> 500600800 2
> 500600900 1
> …
>
> This is a question of counting the times each firm appears, but I don’t know
> which function do this. If anyone could help me I appreciate.
>
> Thank you in advance for the help you could give me,
>
> Cecilia Carmo (Portugal)
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
"Life is like a trumpet - if you don't put anything into it, you don't
get anything out of it."
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Retrieving the context
Dear list,
I have a general problem that I really don't know how to solve efficiently
in R. Lets say we have a sequence of things, like for instance a string of
words, that is stored in a file. We need all the words in a table format, so
therefore we create an id for the word, that links the word to a file and
the position of the word within the file, like:
#In this case a very short file
> strsplit("This is a text string, wich is stored in the file myfile","
")[[1]] -> mystring
#Now, store in a data.frame
> mydf <- data.frame(strings=mystring,
word_id=paste("myfile",1:length(mystring),sep="_"))
> mydf
strings word_id
1 This myfile_1
2 is myfile_2
3a myfile_3
4 text myfile_4
5 string, myfile_5
6 wich myfile_6
7 is myfile_7
8 stored myfile_8
9 in myfile_9
10 the myfile_10
11file myfile_11
12 myfile myfile_12
Now, I would like to see all the words 'is' in a user defined window: so
see_context("is",mydf,1) would give
This is a
wich is stored
and see_context("is",mydf,2) would show two words before and after.. and so
on.
Any ideas on how to solve this kind of problem in R?
/Fredrik
--
"Life is like a trumpet - if you don't put anything into it, you don't get
anything out of it."
[[alternative HTML version deleted]]
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Re: [R] Analogy for %in% for the whole columns (rather than individual values)
Hi Dimitri,
What about this:
> with(a, x == y) -> a$equal
> a
xy equal
1 john mary FALSE
2 mary mary TRUE
3 mary john FALSE
4 john john TRUE
/Fredrik
On Thu, Mar 26, 2009 at 7:48 PM, Dimitri Liakhovitski wrote:
> Hello!
>
> I have a matrix a with 2 variables (see below) that contain character
> strings.
> I need to create a 3rd variable that contains True if the value in
> column x is equal to the value in column y. The code below does it.
>
> a<-data.frame(x=c("john", "mary", "mary",
> "john"),y=c("mary","mary","john","john"))
> a$x<-as.character(a$x)
> a$y<-as.character(a$y)
> a$equal<-NA
> for(i in 1:nrow(a)) {a[i,3]<-a[i,1] %in% a[i,2]}
>
> I was wondering if one can do it faster, without doing it row-by-row
> as I did it. Maybe there is a faster way to compare 2 columns?
>
> Thank you very much!
>
> --
> Dimitri Liakhovitski
> MarketTools, Inc.
> dimitri.liakhovit...@markettools.com
>
> __
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
"Life is like a trumpet - if you don't put anything into it, you don't get
anything out of it."
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Re: [R] odfWeave : change of text style of output is ignored
Dear Max, I was including the settings in my document, but I also tried setting it from the command line. This resulted in more of a change, but only in that the ttBlue now being defined as black rather than blue. So, some effect, but not all the way. Here is the session info: > sessionInfo() R version 2.8.1 (2008-12-22) i386-apple-darwin8.11.1 locale: sv_SE.UTF-8/sv_SE.UTF-8/C/C/sv_SE.UTF-8/sv_SE.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] car_1.2-12 nlme_3.1-90 odfWeave_0.7.9 XML_2.3-0 lattice_0.17-20 loaded via a namespace (and not attached): [1] MASS_7.2-46grid_2.8.1 latticeExtra_0.5-4 /Fredrik On Fri, Mar 20, 2009 at 4:06 PM, Max Kuhn wrote: > Did you do this within the document, or prior to running odfWeave? > > Also, please provide the output of sessionInfo() after starting odfWeave. > > Max > > On Fri, Mar 20, 2009 at 11:03 AM, Fredrik Karlsson > wrote: > > Dear list, > > > > The output of my embedded code sections come out in the style "ttBlue" > wich > > is Times New Roman and Blue when I open document. I would like to change > > this to a fixed witdth font, smaller and less blue by using style > > definitions in the document, but it really does not affect the output. > > > > I tried including > > > > getStyleDefs() -> styl > > styl$ttBlue$fontname <- "Courier" > > styl$ttBlue$fontSize <- "9pt" > > styl$ttBlue$fontColor <- "#00" > > setStyleDefs(styl) > > > > at the beginning of the document, but the output is still blue, not fixed > > width and still 10pt. > > > > Any ideas on how to get this to work? > > > > /Fredrik > > > > -- > > "Life is like a trumpet - if you don't put anything into it, you don't > get > > anything out of it." > > > >[[alternative HTML version deleted]] > > > > __ > > R-help@r-project.org mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > > and provide commented, minimal, self-contained, reproducible code. > > > > > > -- > > Max > -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] odfWeave : change of text style of output is ignored
Dear list, The output of my embedded code sections come out in the style "ttBlue" wich is Times New Roman and Blue when I open document. I would like to change this to a fixed witdth font, smaller and less blue by using style definitions in the document, but it really does not affect the output. I tried including getStyleDefs() -> styl styl$ttBlue$fontname <- "Courier" styl$ttBlue$fontSize <- "9pt" styl$ttBlue$fontColor <- "#00" setStyleDefs(styl) at the beginning of the document, but the output is still blue, not fixed width and still 10pt. Any ideas on how to get this to work? /Fredrik -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] ANOVA and TukeyHSD disagrees?
Dear list,
Sorry for posting a borderline statistical question on the list, but hte
SPSS people around me just stares at me blankly when refering to tests with
any term other than ANOVA and post-hoc. I would appreciate any insight on
how this all is possible:
I have a model fitted by aov() stored in "ppdur", which gives this result
when using ANOVA:
> anova(ppdur)
Analysis of Variance Table
Response: PAPositionPercentOfVoweldur
Df Sum Sq Mean Sq F value Pr(>F)
UtteranceType 4 247316183 2.7642 0.02696 *
SyllLable 1 14584 14584 6.5202 0.01094 *
Cycle 1 798 798 0.3566 0.55067
Speaker 299754987 2.2297 0.10855
Label 120082008 0.8979 0.34377
UtteranceType:SyllLable 4 152103803 1.7001 0.14854
UtteranceType:Cycle 4 131923298 1.4745 0.20855
SyllLable:Cycle 1 11306 11306 5.0545 0.02497 *
UtteranceType:Speaker 7 137211960 0.8764 0.52488
SyllLable:Speaker 21291 645 0.2885 0.74951
Cycle:Speaker 2 107535377 2.4038 0.09135 .
UtteranceType:Label 43579 895 0.4000 0.80871
SyllLable:Label 144994499 2.0114 0.15670
Cycle:Label 1 229 229 0.1022 0.74929
Speaker:Label 21241 620 0.2774 0.75788
UtteranceType:SyllLable:Cycle 3 473 158 0.0705 0.97571
UtteranceType:SyllLable:Speaker 6 139192320 1.0372 0.40006
UtteranceType:Cycle:Speaker 31221 407 0.1820 0.90865
SyllLable:Cycle:Speaker 21457 729 0.3258 0.72210
UtteranceType:SyllLable:Label 238231911 0.8545 0.42607
UtteranceType:Cycle:Label 385662855 1.2766 0.28160
SyllLable:Cycle:Label 135753575 1.5983 0.20669
UtteranceType:Speaker:Label 42658 664 0.2970 0.87990
SyllLable:Speaker:Label 2 139 70 0.0311 0.96938
Cycle:Speaker:Label 2 135996800 3.0400 0.04866 *
UtteranceType:SyllLable:Cycle:Speaker 220151008 0.4505 0.63757
UtteranceType:SyllLable:Cycle:Label 1 11 11 0.0051 0.94328
UtteranceType:SyllLable:Speaker:Label 1 603 603 0.2695 0.60386
Residuals 539 12056052237
---
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1
Ok now, when I want to know where the differences are, I get this result
from TukeyHSD:
> TukeyHSD(ppdur,c("Cycle:Speaker:Label" ),ordered=TRUE)
Tukey multiple comparisons of means
95% family-wise confidence level
factor levels have been ordered
Fit: aov(formula = PAPositionPercentOfVoweldur ~ UtteranceType * SyllLable *
Cycle * Speaker * Label, data =PABTSub)
$`Cycle:Speaker:Label`
difflwr upr p adj
3:Andrea:!H*L-1:Lavinia:!H*L 2.6069140 -37.499300 42.71313 1.000
1:Vito:!H*L-1:Lavinia:!H*L 8.7764075 -85.090794 102.64361 1.000
3:Andrea:H*L-1:Lavinia:!H*L 12.3411960 -18.883688 43.56608 0.9792814
1:Vito:H*L-1:Lavinia:!H*L 15.0416018 -32.746962 62.83017 0.9968844
1:Andrea:H*L-1:Lavinia:!H*L 15.2934976 -17.987036 48.57403 0.9379977
1:Lavinia:H*L-1:Lavinia:!H*L 16.9297832 -14.670124 48.52969 0.8394874
3:Lavinia:H*L-1:Lavinia:!H*L 18.3218965 -13.445765 50.08956 0.7631935
3:Lavinia:!H*L-1:Lavinia:!H*L 20.9338365 -19.932636 61.80031 0.8761167
3:Vito:!H*L-1:Lavinia:!H*L24.3874104 -18.890036 67.66486 0.7894161
3:Vito:H*L-1:Lavinia:!H*L 27.8865684 -6.758302 62.53144 0.2589397
1:Andrea:!H*L-1:Lavinia:!H*L 28.8093072 -18.979256 76.59787 0.7077134
1:Vito:!H*L-3:Andrea:!H*L 6.1694934 -87.982875 100.32186 1.000
3:Andrea:H*L-3:Andrea:!H*L 9.7342820 -22.337674 41.80624 0.9977466
1:Vito:H*L-3:Andrea:!H*L 12.4346877 -35.911602 60.78098 0.9995138
1:Andrea:H*L-3:Andrea:!H*L12.6865836 -21.389960 46.76313 0.9870844
1:Lavinia:H*L-3:Andrea:!H*L 14.3228692 -18.114318 46.76006 0.9529437
3:Lavinia:H*L-3:Andrea:!H*L 15.7149825 -16.885651 48.31562 0.9149770
3:Lavinia:!H*L-3:Andrea:!H*L 18.3269225 -23.190369 59.84421 0.9530401
3:Vito:!H*L-3:Andrea:!H*L 21.7804964 -22.112036 65.67303 0.8978922
3:Vito:H*L-3:Andrea:!H*L 25.2796544 -10.130570 60.68988 0.4470040
1:Andrea:!H*L-3:Andrea:!H*L 26.2023932 -22.143897 74.54868 0.821
3:Andrea:H*L-1:Vito:!H*L 3.5647885 -87.160916 94.29049 1.000
1:Vito:H*L-1:Vito:!H*L 6.2651943 -91.407252 103.93764 1.000
1:Andrea:H*L-1:Vito:!H*L 6.5170902 -84.936471 97.97065 1.000
1:Lavinia:H*L-1:Vito:!H*L 8.1533757 -82.702082 99.00883 1.000
3:Lavinia:
Re: [R] Last char of a string in a list -> list of strings
Dear Gabor,
Thank you very much for this very helpful suggestion. Indeed, I was
making it too complicated by spliting the string into a vector.
Thank you!
/Fredrik
On Wed, Mar 18, 2009 at 3:37 PM, Gabor Grothendieck
wrote:
> Try this:
>
>> L <- c("!H*L", "L%", "%L", "H*L", "L%", "%L", "H*L", "H*L", "L%",
> + "%L", "H*L", "!H*L", "L%", "%L", "%L", "%L", "%L", "H*L", "!H*L",
> + "L%", "H*", "%L", "H*L", "L_%", "%L", "%H", "%H", "!H*L", "%H",
> + "H*", "%H", "%H", "H%", "H*", "!H*L", "H*L", "!H*L", "H*", "H*L",
> + "L_%", "%L", "%L", "H*L", "L*H", "H%", "%H", "L*H", "H%", "%H",
> + "H*L", "%H", "%L", "H*L", "%H", "%L", "H*L", "L%", "%L", "H*L",
> + "!H*L", "L%", "%L", "H*L", "%L", "H*L", "%H", "H*L", "%L")
>>
>> sub(".*([^*_!%0])[*_!%0]*$", "\\1", x)
> [1] "L" "L" "L" "L" "L" "L" "L" "L" "L" "L" "L" "L" "L" "L" "L" "L"
> "L" "L" "L" "L" "H" "L" "L" "L" "L" "H" "H" "L" "H" "H" "H" "H" "H"
> "H" "L" "L" "L" "H"
> [39] "L" "L" "L" "L" "L" "H" "H" "H" "H" "H" "H" "L" "H" "L" "L" "H"
> "L" "L" "L" "L" "L" "L" "L" "L" "L" "L" "L" "H" "L" "L"
>
>
> On Wed, Mar 18, 2009 at 11:10 AM, Fredrik Karlsson wrote:
>> Dear list,
>>
>> This seems like a very simple problem, but I am failing to loose a
>> dimension (I think).
>> I have list, like this:
>>
>> ...
>> [2072] "!H*L" "L%" "%L" "H*L" "L%" "%L" "H*L" "H*L" "L%"
>> "%L" "H*L" "!H*L" "L%" "%L" "%L" "%L" "%L" "H*L" "!H*L"
>> [2091] "L%" "H*" "%L" "H*L" "L_%" "%L" "%H" "%H" "!H*L"
>> "%H" "H*" "%H" "%H" "H%" "H*" "!H*L" "H*L" "!H*L" "H*"
>> [2110] "H*L" "L_%" "%L" "%L" "H*L" "L*H" "H%" "%H" "L*H"
>> "H%" "%H" "H*L" "%H" "%L" "H*L" "%H" "%L" "H*L" "L%"
>> [2129] "%L" "H*L" "!H*L" "L%" "%L" "H*L" "%L" "H*L" "%H"
>> "H*L" "%L"
>>
>> What I would like to do is
>>
>> 1) Remove all the %!*0 characters
>> 2) Get the last character of what remains for each string
>> 3) Get a list, comparable to the one above, but with only one
>> character (so, the last three data points would become list(H,L,L)
>>
>> I tried
>>
>>> strsplit(gsub("[*_!%0]*","",TADescSub$ExtendedLabel,""),"")
>>
>> but that gives me something like:
>>
>> ...
>> [[2136]]
>> [1] "H" "L"
>>
>> [[2137]]
>> [1] "H"
>>
>> [[2138]]
>> [1] "H" "L"
>>
>> [[2139]]
>> [1] "L"
>>
>> So, what am I doing wrong?
>>
>> /Fredrik
>>
>> --
>> "Life is like a trumpet - if you don't put anything into it, you don't
>> get anything out of it."
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
--
"Life is like a trumpet - if you don't put anything into it, you don't
get anything out of it."
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Last char of a string in a list -> list of strings
Dear list,
This seems like a very simple problem, but I am failing to loose a
dimension (I think).
I have list, like this:
...
[2072] "!H*L" "L%" "%L" "H*L" "L%" "%L" "H*L" "H*L" "L%"
"%L" "H*L" "!H*L" "L%" "%L" "%L" "%L" "%L" "H*L" "!H*L"
[2091] "L%" "H*" "%L" "H*L" "L_%" "%L" "%H" "%H" "!H*L"
"%H" "H*" "%H" "%H" "H%" "H*" "!H*L" "H*L" "!H*L" "H*"
[2110] "H*L" "L_%" "%L" "%L" "H*L" "L*H" "H%" "%H" "L*H"
"H%" "%H" "H*L" "%H" "%L" "H*L" "%H" "%L" "H*L" "L%"
[2129] "%L" "H*L" "!H*L" "L%" "%L" "H*L" "%L" "H*L" "%H"
"H*L" "%L"
What I would like to do is
1) Remove all the %!*0 characters
2) Get the last character of what remains for each string
3) Get a list, comparable to the one above, but with only one
character (so, the last three data points would become list(H,L,L)
I tried
> strsplit(gsub("[*_!%0]*","",TADescSub$ExtendedLabel,""),"")
but that gives me something like:
...
[[2136]]
[1] "H" "L"
[[2137]]
[1] "H"
[[2138]]
[1] "H" "L"
[[2139]]
[1] "L"
So, what am I doing wrong?
/Fredrik
--
"Life is like a trumpet - if you don't put anything into it, you don't
get anything out of it."
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] odfWeave problem "Error in xmlEventParse"
Dear list,
Sorry for bothering you with a pure odfSweave question, but I just ran
into a problem that I cannot find the cause of.
Anyonse seen this before? This file "used to work", but not anymore.
Would apreciate all the help I could get.
/Fredrik
---
> odfWeave("Descriptives/IPDesc.odt","IPDescOut.odt")
Copying Descriptives/IPDesc.odt
Setting wd to
/var/folders/IV/IVKgtoUIF3CjOMG8RfyjoTU/-Tmp-//RtmpPbfKbO/odfWeave24150415398
Unzipping ODF file using unzip -o IPDesc.odt
Archive: IPDesc.odt
Removing IPDesc.odt
Creating a Pictures directory
Pre-processing the contents
extracting: mimetype
creating: Configurations2/statusbar/
inflating: Configurations2/accelerator/current.xml
creating: Configurations2/floater/
creating: Configurations2/popupmenu/
creating: Configurations2/progressbar/
creating: Configurations2/menubar/
creating: Configurations2/toolbar/
creating: Configurations2/images/Bitmaps/
inflating: layout-cache
inflating: content.xml
inflating: styles.xml
extracting: meta.xml
inflating: Thumbnails/thumbnail.png
inflating: settings.xml
inflating: META-INF/manifest.xml
Sweaving content.Rnw
Writing to file content_1.xml
Processing code chunks ...
1 : term verbatim(label=Init)
2 : term xml(label=IPTransSummary_SL)
3 : term xml(label=IPTransSummary_SP)
4 : term xml(label=IPTransSummary_SL_CY)
5 : term xml(label=IPTransSummary_SP_CY)
6 : term xml(label=IPTransSummary_SP_CY_SE)
7 : term xml(label=IPTransSummary_SL_IPL)
8 : term xml(label=IPTransSummary_SP_IPL)
9 : term xml(label=IP_CY_SL_mean)
10 : term xml(label=IP_CY_SL_sd)
11 : term verbatim(label=IPLenWords_CY_SL_bwplot)
12 : term verbatim(label=IPLenWords_SL_CY_bwplot)
13 : term xml(label=IP_CY_SP_mean)
14 : term xml(label=IP_CY_SP_sd)
15 : term verbatim(label=IPLenWords_CY_SP_bwplot)
16 : term verbatim(label=IPLenWords_SP_CY_bwplot)
17 : term xml(label=IP_CY_SL_mean)
18 : term xml(label=IP_CY_SL_sd)
19 : term verbatim(label=IPLenSylls_CY_SL_bwplot)
20 : term verbatim(label=IPLenSylls_SL_CY_bwplot)
21 : term xml(label=IP_CY_SP_mean)
22 : term xml(label=IP_CY_SP_sd)
23 : term verbatim(label=IPLenSylls_CY_SP_bwplot)
24 : term verbatim(label=IPLenSylls_SP_CY_bwplot)
25 : term verbatim(label=IplenCalc)
26 : term xml(label=WordLenSyll_CY_SL_mean)
27 : term xml(label=WordLenSyll_CY_SL_sd)
28 : term verbatim(label=WordLenSyll_CY_SL_bwplot)
29 : term verbatim(label=WordLenSyll_SL_CY_bwplot)
30 : term xml(label=WordLenSyll_CY_SP_mean)
31 : term xml(label=WordLenSyll_CY_SP_sd)
32 : term verbatim(label=WordLenSyll_CY_SP_bwplot)
33 : term verbatim(label=WordLenSyll_SP_CY_bwplot)
34 : term verbatim(label=CleanTA)
35 : term xml(label=IP_PA_CY_SL_mean)
36 : term xml(label=IP_PA_CY_SL_sd)
37 : term verbatim(label=IPLenTones_CY_SL_bwplot)
38 : term verbatim(label=IPLenTones_SL_CY_bwplot)
39 : term xml(label=IP_PA_CY_SP_mean)
40 : term xml(label=IP_PA_CY_SP_sd)
41 : term verbatim(label=IPLenTones_CY_SP_bwplot)
42 : term verbatim(label=IPLenTones_SP_CY_bwplot)
43 : term verbatim(label=IPClassrm)
44 : term verbatim(label=IPClassDistRelRM_PA_SL_CY_Fig)
45 : term verbatim(label=IPClassDistRelRM_PA_CY_SL_Fig)
46 : term verbatim(label=IPClassDistRelRM_PA_CY_SP_Fig)
47 : term xml(label=IPTrans_TS_SL_count)
48 :(label=IPTrans_TS_SP_count)
49 : term verbatim(label=close)
'content_1.xml' has been Sweaved
Removing content.xml
Post-processing the contents
Error in xmlEventParse(infile, handlers = handlers, trim = FALSE,
state = state) :
attempt to apply non-function
In addition: Warning message:
closing unused connection 3 (content.xml)
>
---
--
"Life is like a trumpet - if you don't put anything into it, you don't
get anything out of it."
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Grouped bwplots?
Dear list,
I am sorry for asking you this, but I am trying to do again what I
thought I have done before, although this time it does not work.
So, given the data set:
> testdf <- data.frame(grfak=sample(c("One","Two"),size=100,replace=TRUE),
> panfak= sample(c("Yes","No"),size=100,replace=TRUE), xfak=
> sample(c("Yep","Nope"),size=100,replace=TRUE), d=rnorm(100))
I would like to do:
> bwplot(d ~ xfak | panfak,data=testdf, groups= grfak)
where the groups argument makes a difference (as in dividing each box
into two by group).
Is this possible to do? If so, what I am doing wrong?
/Fredrik
--
"Life is like a trumpet - if you don't put anything into it, you don't
get anything out of it."
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Grayscale figures in odfWeave?
Dear list, I am sorry if this is a FAQ, but I have just started using odfWeave for report construction and it seems not to behave exactly like Sweave. I want to get my figures in grayscale, but the tip given in the FAQ of Sweave http://www.stat.uni-muenchen.de/~leisch/Sweave/FAQ.html#x1-1A.8 only produces gray scale images in the screen output, not in the odf document. This is my init code: <>= library(lattice) ltheme <- canonical.theme(color = FALSE) ## in-built B&W theme ltheme$strip.background$col <- "transparent" ## change strip bg lattice.options(default.theme = ltheme) ## set as default @ What am I doing wrong? /Fredrik -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R2HTML and output from a function
Dear list,
I am trying to construct a report function that would go through all
the object in the current environment and print them in HTML form.
What I have got is:
dataReport <- function(){
#First tables
tabs <- ls(sys.frame(), pattern=".*table")
for(currT in tabs){
cat("\n\n")
print(currT)
if(length(dim(get(currT))) > 2){
#We have to use ftable
require(Hmisc)
print(ftable(get(currT)))
}
else{
print(get(currT))
print(chisq.test(get(currT)))
}
}
#Next, the lme models
mods <- ls(sys.frame(), pattern=".*lme")
require(nlme)
for(currM in mods){
if(class(get(currM)) %in% c("lme","lm")){
cat("\n\n")
print(currM)
print(anova(get(currM)))
}
}
}
This generally works in the console. However, it I try to get the
output processed by R2HTML, I get only the output of one table.
So, now I have 2 questions:
1) Is there a way to get the output of all the print commands of the
function to be processed by R2HTML?
2) Is there a way to include saved plots in the output too?
/Fredrik
--
"Life is like a trumpet - if you don't put anything into it, you don't
get anything out of it."
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Change in Lattice bwplot?
Dear list, Sorry for asking this question, but has something changed in the syntax for bwplot in Lattice? In an old publication, I used > bwplot( VOTMS ~gender |type * group, data=merge(vot,words,by="ord"), nint=30, horizontal=F, layout=c(3,3), box.ratio=0.8) which produced a lovelly 3x3 lattice plot with one box/gender in each panel. Now, I try > bwplot( SyllableNucleusDiff ~ SourceLanguage,data=Hstar,horizontal=F) to get just a simple 1x1 panel plot with groups (which I will then of course make into a panel plot by adding | factor1 +factor2...), but I get a "Syntax error". So, has anything changed, or am I just doing something very silly? /Fredrik -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] TukeyHSD and 0.0000000
Hi, Sorry about that. Sure, here is some further information: > summary(InekeReduc) segorg intorg l2group mani score Belfast:3782 Belfast:3782 1: 2604 1 :2480 Min. : 1.000 L2 :7502 L2 :7440 2:12462 2 :2542 1st Qu.: 1.000 German :3782 German :3844 3 :2480 Median : 2.000 4 :2480 Mean : 3.011 5 :2480 3rd Qu.: 5.000 NA's:2604 Max. : 6.000 NA's :200.000 > summary( fit <- aov( score ~ segorg * intorg ,data=InekeReduc) ) Df Sum Sq Mean Sq F valuePr(>F) segorg2 150257512 2219.063 < 2.2e-16 *** intorg2 1019 509 150.441 < 2.2e-16 *** segorg:intorg 4164 41 12.086 8.299e-10 *** Residuals 14857 50297 3 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 200 observations deleted due to missingness > TukeyHSD(fit,"segorg:intorg") Tukey multiple comparisons of means 95% family-wise confidence level Fit: aov(formula = score ~ segorg * intorg, data = InekeReduc) $`segorg:intorg` diff lwrupr p adj L2:Belfast-Belfast:Belfast 1.59590485 1.33909758 1.8527121 0.000 German:Belfast-Belfast:Belfast 3.04430961 2.71827718 3.3703420 0.000 Belfast:L2-Belfast:Belfast 0.24375305 -0.01369296 0.5011991 0.0801617 L2:L2-Belfast:Belfast 1.55958957 1.30211188 1.8170673 0.000 German:L2-Belfast:Belfast 2.96873225 2.71126512 3.2261994 0.000 Belfast:German-Belfast:Belfast 0.44195992 0.12355001 0.7603698 0.0005667 L2:German-Belfast:Belfast 2.36192845 2.10440839 2.6194485 0.000 German:German-Belfast:Belfast 3.42962398 3.0098 3.7481470 0.000 German:Belfast-L2:Belfast 1.44840476 1.19092315 1.7058864 0.000 Belfast:L2-L2:Belfast -1.35215180 -1.51425695 -1.1900466 0.000 L2:L2-L2:Belfast -0.03631527 -0.19847075 0.1258402 0.9988643 German:L2-L2:Belfast1.37282740 1.21068871 1.5349661 0.000 Belfast:German-L2:Belfast -1.15394493 -1.40170389 -0.9061860 0.000 L2:German-L2:Belfast0.76602360 0.60380087 0.9282463 0.000 German:German-L2:Belfast1.83371914 1.58581484 2.0816234 0.000 Belfast:L2-German:Belfast -2.80055656 -3.05867524 -2.5424379 0.000 L2:L2-German:Belfast -1.48472003 -1.74287033 -1.2265697 0.000 German:L2-German:Belfast -0.07557736 -0.33371711 0.1825624 0.9925827 Belfast:German-German:Belfast -2.60234969 -2.92130373 -2.2833957 0.000 L2:German-German:Belfast -0.68238116 -0.94057371 -0.4241886 0.000 German:German-German:Belfast0.38531438 0.06624743 0.7043813 0.0056382 L2:L2-Belfast:L21.31583652 1.15267135 1.4790017 0.000 German:L2-Belfast:L22.72497920 2.56183071 2.8881277 0.000 Belfast:German-Belfast:L2 0.19820687 -0.05021410 0.4466278 0.2441352 L2:German-Belfast:L22.11817540 1.95494338 2.2814074 0.000 German:German-Belfast:L23.18587093 2.93730501 3.4344369 0.000 German:L2-L2:L2 1.40914267 1.24594419 1.5723412 0.000 Belfast:German-L2:L2 -1.11762966 -1.36608347 -0.8691758 0.000 L2:German-L2:L2 0.80233887 0.63905689 0.9656209 0.000 German:German-L2:L2 1.87003441 1.62143567 2.1186331 0.000 Belfast:German-German:L2 -2.52677233 -2.77521518 -2.2783295 0.000 L2:German-German:L2-0.60680380 -0.77006911 -0.4435385 0.000 German:German-German:L2 0.46089174 0.21230395 0.7094795 0.003 L2:German-Belfast:German1.91996853 1.67147082 2.1684662 0.000 German:German-Belfast:German2.98766407 2.67639017 3.2989380 0.000 German:German-L2:German 1.06769554 0.81905292 1.3163382 0.000 /Fredrik On Wed, Nov 5, 2008 at 1:15 PM, hadley wickham <[EMAIL PROTECTED]> wrote: > On Wed, Nov 5, 2008 at 5:42 AM, Fredrik Karlsson <[EMAIL PROTECTED]> wrote: >> Dear list, >> >> Sorry to ask you this, but I just ran a TukeyHSD on an model with a >> two thee level factors as independent variables and a numeric score >> dependent variable. >> The aov gives a significant interaction effect, and using the >> TukeyHSD, I get almost every row to be exactly 0.000. Should I be >> worried?? > > You should be worried that you did not provide any information to > reproduce your problem. How can you expect anyone else to give you > advice? > > Hadley > > > -- > http://had.co.nz/ > -- "Life is like a trumpet - if yo
[R] TukeyHSD and 0.0000000
Dear list, Sorry to ask you this, but I just ran a TukeyHSD on an model with a two thee level factors as independent variables and a numeric score dependent variable. The aov gives a significant interaction effect, and using the TukeyHSD, I get almost every row to be exactly 0.000. Should I be worried?? /Fredrik -- "Life is like a trumpet - if you don't put anything into it, you don't get anything out of it." __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Reversed but positive axis in trellis plots?
Hi, In my discpipline, it is common to plot one acoustic property on a positive scale but from top to bottom on the ordinate and the same for another measurement on the abscissa. So, the origin of the plot is on the top right of the plot, with increasing values to the left /down. This is to highlight the correlation between the acoustic measurement and the position of the forming structure, for instance when teaching it to students. The grouping ability of the trellis plot is quite handy whan plotting many instances of the same thing, so I was wondering if it is possible to make a trellis xyplot behave this way? Converting all values to negative and changing labels to the negative of the negative seems one solution to the reverseness of the axes, but how do I change the position? Is it possible? /Fredrik -- "Give up learning, and put an end to your troubles." __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
