Re: [R] "R CMD check" does not find a mistake
> "CG" == Christophe Genolini > on Tue, 09 Jun 2009 16:17:15 +0200 writes: CG> Hi the list, I build a package. They was a mistake in CG> it, but R CMD check did not find it. Is that normal ? CG> Here is what Kurt gets (which is right, I did this CG> mistake): CG> --- 8< CG> * checking for code/documentation mismatches ... WARNING CG> S4 class codoc mismatches from documentation object 'LongData-class': CG> Slots for class 'LongData' CG> Code: id other time traj varName CG> Docs: id time traj varName CG> --- 8< Others have already hinted at the solution: The problem only shows in "R-devel" (2.10.x). Note that I had sent an explicit message to this list *exactly* in order to advise all package authors : Subject: [Rd] R-devel:codocClasses() now finds more --> R CMD check warnings Date: Wed, 3 Jun 2009 11:45:45 +0200 i.e. only 6 days before you saw the problem yes, I am sometimes too optimistic, assuming that people actually read what I write Regards anyway! Martin -- Martin Maechler, ETH Zurich __ 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 to get output from a nested loop
How about like this:
> t1 <- data.frame(row.names=c('c1','c2','c3','c4'), mk1=c(1,1,0,0),
> mk2=c(0,0,0,1), mk3=c(1,1,1,1), mk4=c(0,0,0,0), mk5=c(0,0,0,1), S=c(4,5,3,2))
> t1
mk1 mk2 mk3 mk4 mk5 S
c1 1 0 1 0 0 4
c2 1 0 1 0 0 5
c3 0 0 1 0 0 3
c4 0 1 1 0 1 2
> apply(combn(1:5, 2), 2, function(x) t1[,c(x[1], 6, x[2])])
[[1]]
mk1 S mk2
c1 1 4 0
c2 1 5 0
c3 0 3 0
c4 0 2 1
... ...
On Wed, Jun 10, 2009 at 1:10 PM, Scott Hermann wrote:
> Dear all,
>
> I imagine that this is a trival question, but it has perplexed for most of
> the day. Any help would be greatly appreciated.
>
> Below is an example of what I'm trying to do.
>
> Essentially I want to produce all unique 1 x 1 combinations of certain
> columns from a dataframe, and join these to other columns from the same
> dataframe. I'm having problems with the nested loop as I can only output
> data from the last "cycle" of the loop. I realise that the problem is with
> the st1[[i]] but I'm not sure how to define it differently.
>
> ##I want to make a list file of all 1x1 combinations of "mk" columns, and add
> "clone" and "S" to these combinations.
>
> clone<-c("c1","c2","c3","c4")
> mk1<-c(1,1,0,0)
> mk2<-c(0,0,0,1)
> mk3<-c(1,1,1,1)
> mk4<-c(0,0,0,0)
> mk5<-c(0,0,0,1)
> S<-c(4,5,3,2)
> t1<-as.data.frame(cbind(clone,mk1,mk2,mk3,mk4, mk5,S))
> row.names(t1)<-t1$clone
> t1<-t1[,-1]
> t1
>
> ###A nested loop. I'm trying to get all combinations of columns 1:5 and join
> each of them with column 7.
>
> st1 <- list()
> for(i in 1:4) {
> for(j in (i+1):5){
> st1[[i]] <- cbind(t1[,c(i,6)],t1[,j])
> }
> }
> st1
>
> Thanks for your help,
>
> Scott
> BSES Limited Disclaimer
>
>
>
> This email and any files transmitted with it are confide...{{dropped:13}}
>
> __
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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] Creating a specific skewed distribution
All, Can someone help me create a skewed distribution, mean = 30, with probability of selecting a random number from the distribution greater than or equal 60 equal to 10%? I need the probability density function to equal zero at zero, and have a maximum height at or near 30. Is this possible? And if it is possible, how can I adjust the distribution so that the probability of selecting a random number greater than or equal to 60 is p. Thanks. No idea how to start. David [[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] poly regression
hi, I want to do a polynomial regression of y on x of degree 2, as following > x<-1:10 > y<-x^2 > lm(y~poly(x,2)) Call: lm(formula = y ~ poly(x, 2)) Coefficients: (Intercept) poly(x, 2)1 poly(x, 2)2 38.5099.9122.98 Which is not what i had expected. If I wrote the expression in an explicit form, y~1+x+I(x^2), I could get the expected result: > lm(y~1+x+I(x^2)) Call: lm(formula = y ~ 1 + x + I(x^2)) Coefficients: (Intercept)x I(x^2) 001 What is the diff between them? __ 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 get output from a nested loop
Dear all,
I imagine that this is a trival question, but it has perplexed for most of the
day. Any help would be greatly appreciated.
Below is an example of what I'm trying to do.
Essentially I want to produce all unique 1 x 1 combinations of certain columns
from a dataframe, and join these to other columns from the same dataframe. I'm
having problems with the nested loop as I can only output data from the last
"cycle" of the loop. I realise that the problem is with the st1[[i]] but I'm
not sure how to define it differently.
##I want to make a list file of all 1x1 combinations of "mk" columns, and add
"clone" and "S" to these combinations.
clone<-c("c1","c2","c3","c4")
mk1<-c(1,1,0,0)
mk2<-c(0,0,0,1)
mk3<-c(1,1,1,1)
mk4<-c(0,0,0,0)
mk5<-c(0,0,0,1)
S<-c(4,5,3,2)
t1<-as.data.frame(cbind(clone,mk1,mk2,mk3,mk4, mk5,S))
row.names(t1)<-t1$clone
t1<-t1[,-1]
t1
###A nested loop. I'm trying to get all combinations of columns 1:5 and join
each of them with column 7.
st1 <- list()
for(i in 1:4) {
for(j in (i+1):5){
st1[[i]] <- cbind(t1[,c(i,6)],t1[,j])
}
}
st1
Thanks for your help,
Scott
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[R] Unable to load package:lme4 [ Ubuntu 9.04 ]
Hi folks, When I try to load package 'lme4' on my Linux box (64-bit Ubuntu 9.04), I get the following error: - > library(lme4) Error in dyn.load(file, DLLpath = DLLpath, ...) : function 'cholmod_start' not provided by package 'Matrix' Error: package/namespace load failed for 'lme4' I removed and re-installed both 'Matrix' and 'lme4', but that doesn't seem to solve the problem. If this is something for the [R-sig-ME] mailing list, please let me know. Thanks, -Girish > sessionInfo() R version 2.8.1 (2008-12-22) x86_64-pc-linux-gnu locale: LC_CTYPE=en_IN;LC_NUMERIC=C;LC_TIME=en_IN;LC_COLLATE=en_IN;LC_MONETARY=C;LC_MESSAGES=en_IN;LC_PAPER=en_IN;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_IN;LC_IDENTIFICATION=C attached base packages: [1] splines stats graphics grDevices utils datasets methods [8] base other attached packages: [1] Matrix_0.999375-17 lattice_0.17-25Design_2.2-0 survival_2.35-4 [5] Hmisc_3.6-0nlme_3.1-92car_1.2-14 MASS_7.2-47 loaded via a namespace (and not attached): [1] cluster_1.12.0 grid_2.8.1 tools_2.8.1 __ 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] searchpaths
No. This function is used to get "a list of 'attach'ed _packages_ " rather than search files in your computer. 2009/6/10 : > May I use "searchpaths()" with arguments partially matching file names that > are found in different directories ? > My question is whether this is th R function equivalent of Linux "find" or > Windows "search". > Both O.S. calls are given a starting point so that they search all diectories > from then downwards looking for files whose > names match the searching criteria. > Thank you. > Maura > > > > tutti i telefonini TIM! > > > [[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. > -- 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 __ 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] R on EC2 and R CMD javareconf
Hello, I am using EC2 to launch several instances. On each instance, I perform the following commands yum -y install R (installs R-2.6) R CMD javareconf sleep 10 wget rpackage.tgz R CMD INSTALL rpackage.tgz Now, rpackage.tgz needs to be built with jni libraries. When the instance is fully up, i log in and check if rpackage was successfully installed and find out it wasn't. Running R CMD INSTALL rpackage.tgz again, i get the response that I should run R CMD javareconf, which I do and then successfully install rpackage (using R CMD...) Q: Why doesn't it work the first time? I am using a small instance and Fedora 8 AMI Regards Saptarshi __ 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] searchpaths
May I use "searchpaths()" with arguments partially matching file names that are found in different directories ? My question is whether this is th R function equivalent of Linux "find" or Windows "search". Both O.S. calls are given a starting point so that they search all diectories from then downwards looking for files whose names match the searching criteria. Thank you. Maura tutti i telefonini TIM! [[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] Good Programming Practice Question - Functions in Different Files
library("R.utils");
sourceDirectory("C:/appropriatepath/R/", modifiedOnly=TRUE);
On Tue, Jun 9, 2009 at 8:09 PM, spencerg wrote:
> Sundar Dorai-Raj taught me to do the following:
>
> Rdir <- "c:\appropriatepath\R"
>
> Rfiles <- dir(Rdir, pattern='\\.R$', full.names=TRUE)
>
> invisible(lapply(aTR, source))
>
>
> The "invisible" suppresses the garbage while still displaying error
> messages.
>
> Hope this helps. Spencer
>
> Paul Hiemstra wrote:
>>
>> Hi,
>>
>> I always use source for this kinds of things. If you have a lot of code,
>> you could consider wrapping it into an R-package. This would allow you to
>> load all the code by using the library(myPackage) command. But this might be
>> unnecessary for your situation.
>>
>> cheers,
>> Paul
>>
>> Jason Rupert wrote:
>>>
>>> I've gotten to the point wih an R script where I would like to
>>> encapsulate several blocks of codes in R functions.
>>>
>>> In order to keep the top level script simple I would like to put them in
>>> a separate file. This should help the readability of the top level main
>>> script.
>>>
>>> Is source(...) the best way to load those functions in the top level
>>> script? For example, in the top level script
>>> Line#
>>> 01 source("FunctionsFile.R")
>>> 02
>>> 03 callfunctionfromfunctionfile(...)
>>>
>>> Thanks for any feedback and insights.
>>> Also, I found the following:
>>> http://www.stat.auckland.ac.nz/~paul/ItDT/HTML/node71.html
>>>
>>> But are there other guides out there that speak to good practices when
>>> developing in R?
>>>
>>> __
>>> 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-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Good Programming Practice Question - Functions in Different Files
Sundar Dorai-Raj taught me to do the following:
Rdir <- "c:\appropriatepath\R"
Rfiles <- dir(Rdir, pattern='\\.R$', full.names=TRUE)
invisible(lapply(aTR, source))
The "invisible" suppresses the garbage while still displaying
error messages.
Hope this helps.
Spencer
Paul Hiemstra wrote:
Hi,
I always use source for this kinds of things. If you have a lot of
code, you could consider wrapping it into an R-package. This would
allow you to load all the code by using the library(myPackage)
command. But this might be unnecessary for your situation.
cheers,
Paul
Jason Rupert wrote:
I've gotten to the point wih an R script where I would like to
encapsulate several blocks of codes in R functions.
In order to keep the top level script simple I would like to put them
in a separate file. This should help the readability of the top
level main script.
Is source(...) the best way to load those functions in the top level
script?
For example, in the top level script
Line#
01 source("FunctionsFile.R")
02
03 callfunctionfromfunctionfile(...)
Thanks for any feedback and insights.
Also, I found the following:
http://www.stat.auckland.ac.nz/~paul/ItDT/HTML/node71.html
But are there other guides out there that speak to good practices
when developing in R?
__
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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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem : solving a equation with R , fail with uniroot function
Peter Alspach wrote:
>
> Tena koe Yann
>
> It may not be the only problem, but you are missing an operator between
> (bb0+x+1) and (bb1-x-1).
>
>
And fixing that typo and viewing the function shows that it
is a bit problematic on this interval anyway ... (diverges
at x = 1, closer look shows it's pretty nasty -- I haven't stopped
to think about its behavior very carefully, but it will clearly
take some effort. If you're lucky there is a typo and the
real function is better behaved!)
test <- function(x,bb0=-3,bb1=5,c0=2,r0=0) {
((exp(c0-r0)*(bb0+x)*(bb1-x))/((bb0+x+1)*(bb1-x-1))-1)}
uniroot(test,c(-100,100))$root
curve(test(x),from=-100,to=100)
curve(test(x),from=1.1,to=10)
abline(h=0,col=2)
--
View this message in context:
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Re: [R] Problem : solving a equation with R , fail with uniroot function
Tena koe Yann
It may not be the only problem, but you are missing an operator between
(bb0+x+1) and (bb1-x-1).
HTH
Peter Alspach
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of yann lancien
> Sent: Wednesday, 10 June 2009 9:31 a.m.
> To: r-help@r-project.org
> Subject: [R] Problem : solving a equation with R , fail with
> uniroot function
>
> Hi ,
> I would like to know if a R function have the same behaviour
> than the matlab solve function.
> I tried something with uniroot but I have some problems:
> The equation I need to solve is :
> exp(c0-r0)*(bb0+x)*(bb1-x)=(bb0+x+1)(bb1-x-1)
>
> So I tried this:
>
> STEP 1: my function test
> test <- function(x,bb0=-3,bb1=5,c0=2,r0=0) {
> + ((exp(c0-r0)*(bb0+x)*(bb1-x))/((bb0+x+1)(bb1-x-1))-1)}
>
> STEP 2:
> > uniroot(test,c(-100,100))$root*
> Erreur dans f(lower, ...) :
> tentative d'appliquer un objet qui n'est pas une fonction*
>
> R tells me that test is not a function ...
>
> STEP 3:
> > is.function(test)
> [1] TRUE
>
>
> Thanks,
> Yann
>
> [[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.
>
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[R] LS graph
Hi all, I´m new using R. I wanto to print in a unique plot four results of four different ordinary least squares regressions. You see, I want to plot (on each box of the for parts of the graph, actual, fitted values and on the bottom the residuals). On the other side,if I only want to plot the for different residuals of each on the four estimations, how I should proceed?Thanks in advance. Imagine I have called e1,e2,e3,e4 And for titles I want to put (title) on both x axis and y axis. How should I put a main title on each graph. I hope I have explainned good. Anyway, thanks in advance for your help. -- View this message in context: http://www.nabble.com/LS-graph-tp23951942p23951942.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.
[R] Problem : solving a equation with R , fail with uniroot function
Hi ,
I would like to know if a R function have the same behaviour than the matlab
solve function.
I tried something with uniroot but I have some problems:
The equation I need to solve is :
exp(c0-r0)*(bb0+x)*(bb1-x)=(bb0+x+1)(bb1-x-1)
So I tried this:
STEP 1: my function test
test <- function(x,bb0=-3,bb1=5,c0=2,r0=0) {
+ ((exp(c0-r0)*(bb0+x)*(bb1-x))/((bb0+x+1)(bb1-x-1))-1)}
STEP 2:
> uniroot(test,c(-100,100))$root*
Erreur dans f(lower, ...) :
tentative d'appliquer un objet qui n'est pas une fonction*
R tells me that test is not a function ...
STEP 3:
> is.function(test)
[1] TRUE
Thanks,
Yann
[[alternative HTML version deleted]]
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[R] list to data frame; how to store short elements?
This is programming style, best practices type of question. When converting a list to a data frame (eg to use in ggplot), what is "best" way to store descriptive elements? For example, a survfit object mostly maps to a data frame, but I would also like to carry along the type, call, etc. I'm inclined to make these attributes of the data frame. Any better technique? TIA, Jim Porzak TGN.com San Francisco, CA www.linkedin.com/in/jimporzak use R! Group SF: www.meetup.com/R-Users/ [[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] generating new data with for loop
Bugzilla from gordon.holtslan...@usask.ca wrote:
>
> I'm new at R ...
> I've not done for loops in R - so this is very new to me.
>
> One of our students has a data frame that contains two columns data
>
> 1. unixtime time of an event (in unix time - #of seconds)
> 2. duration of event in seconds.
>
> We need to create new data - the unixtime (seconds) that these events
> occurred.
>
> We want to create a for loop (or nested for loops) that goes through the
> first column, and then creates a new unixtime value for each second of
> duration for that event
>
> In a single row of data -
> if the duration was three seconds one iteration of this loop should create
> these values
>
> unixtime
> unixtime = unixtime +1
> unixtime = unixtime +1
> unixtime = unixtime +1
>
> I need these four data values (and all the successive data values created
> by
> the loop(s)) saved into a new dataframe
>
> How do I get R to create a new data object and add data to it on each
> interation of the for loop.
>
> My test for loops only save the data from last iteration of the loop.
>
> Am I missing something really simple?
>
> Any recommended reference on loop and control structures in R?
>
> Thanks,
>
Hard to see what you tried to do, so hard to correct it.
## "data"
d = data.frame(unixtime=5:10,duration=3:8)
## method 1 (magic)
d2 = data.frame(basetime=unlist(apply(d,1,function(x)
{rep(x["unixtime"],1+x["duration"])})),
time=unlist(apply(d,1,function(x)
{x["unixtime"]+0:x["duration"]})))
## method 2 (straightforward)
time <- numeric(sum(d$duration)+nrow(d))
count <- 0
for (i in 1:nrow(d)) {
dur <- d$duration[i]
time[(count+1):(count+dur+1)] <- d$unixtime[i]+0:dur
count <- count+dur+1
}
## method 3 (slowest but simplest)
time <- numeric(0)
for (i in 1:nrow(d)) {
dur <- d$duration[i]
time <- c(time,d$unixtime[i]+0:dur)
}
--
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and provide commented, minimal, self-contained, reproducible code.
[R] RODBC sqlSave does not append the records
I am running R version 2.8.1 on Windows XP OS. recs is a one-row, eighteen-column data.frame I want to insert into a DB2 table. sqlSave(channel, recs, tablename = "testappend_slt", append = TRUE, + rownames = FALSE, colnames = FALSE, + verbose = TRUE, oldstyle = FALSE, + safer = TRUE, addPK = FALSE, + fast = FALSE, test = FALSE, nastring = NULL) Query: INSERT INTO "testappend_slt" ( "RECORDNO", "HEADERID", "SITE", "TESTER", "TESTCELL", "PRODUCT", "OPERATION", "OPNUM", "PRODMODE", "VERSION", "REVISION", "UUT", "MCTL", "OPERATOR", "STARTTIME", "ENDTIME", "ENDSTATUS", "CREATED" ) VALUES ( 99, 1125080002314, 13, 'nextra14 ', 'cell14 ', 'Nextra Gen2', 'IVT Gen2 ', 'T710', 'E', 'PNX10.0b ', '1104a ', 00023, 'FORGETIT', 'Elaine ', 1227627900, 1228685220, 'C', '12/9/2008 11:06:11' ) > I don't get any error message, but when I check, the record has not been added to the table. This observation troubles me: The third and fourth columns have class & mode= character but they are not enclosed in single quotes like the other character columns. This also applies to the UUT column. I notice the values in these columns contain including only numerical digits, Any suggestions for how to resolve are appreciated! Sincerely, Elaine McGovern Jones ISC Tape and DASD Storage Products Characterization and Failure Analysis Engineering Phone: 408 707 9588 jon...@us.ibm.com [[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] calculate effect size from p-value?
Dear R-List! Is there any function (apparently for BioStat there is!) to calculate effect sizes with the p-values and sample sizes? That would be really useful for meta-analysis! Thanks for the help! Jokel __ 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] generating new data with for loop
Try this: x <- data.frame(u=c(1:5), seconds=sample(5)) transform(x[rep(1:nrow(x),x$seconds),], seconds = unlist(lapply(split(x$seconds, x$u), seq))) On Tue, Jun 9, 2009 at 6:11 PM, Gordon J Holtslander < gordon.holtslan...@usask.ca> wrote: > I'm new at R ... > I've not done for loops in R - so this is very new to me. > > One of our students has a data frame that contains two columns data > > 1. unixtime time of an event (in unix time - #of seconds) > 2. duration of event in seconds. > > We need to create new data - the unixtime (seconds) that these events > occurred. > > We want to create a for loop (or nested for loops) that goes through the > first column, and then creates a new unixtime value for each second of > duration for that event > > In a single row of data - > if the duration was three seconds one iteration of this loop should create > these values > > unixtime > unixtime = unixtime +1 > unixtime = unixtime +1 > unixtime = unixtime +1 > > I need these four data values (and all the successive data values created > by > the loop(s)) saved into a new dataframe > > How do I get R to create a new data object and add data to it on each > interation of the for loop. > > My test for loops only save the data from last iteration of the loop. > > Am I missing something really simple? > > Any recommended reference on loop and control structures in R? > > Thanks, > > Gord > -- > Gordon J. Holtslander > gordon.holtslan...@usask.ca > Dept. of Biology > University of Saskatchewan > tel 306 966-4433 > fax 306 966-4461 > > __ > 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. > -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40" S 49° 16' 22" O [[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] Help with if statements
> Error in if (ALLRESULTS[i, 16] <= 0.05) significance_count = significance_count + : > missing value where TRUE/FALSE needed > > The script is included below > > it works if i convert the NA values to zero but this is not appropriate as it includes the zero as significant. > > ANY SUGGESTIONS significance.count <- rowSums(ALLRESULTS[,16:22] <= .05, na.rm=TRUE) ?rowSums Good solution. I would like to comment on the OP's naivete: since he knows that zero is a value of interest to his algorithm, why the heck replace "NA" with zero? In general, if there is some reason not to use na.rm or na.omit, start out by thinking about what NA values mean to your analysis. If they are the equivalent of "FALSE" conditions, then replace them with a numerical value which is "FALSE". In this example, replacing NA with 0.051 would do just fine. __ 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] generating new data with for loop
I'm new at R ... I've not done for loops in R - so this is very new to me. One of our students has a data frame that contains two columns data 1. unixtime time of an event (in unix time - #of seconds) 2. duration of event in seconds. We need to create new data - the unixtime (seconds) that these events occurred. We want to create a for loop (or nested for loops) that goes through the first column, and then creates a new unixtime value for each second of duration for that event In a single row of data - if the duration was three seconds one iteration of this loop should create these values unixtime unixtime = unixtime +1 unixtime = unixtime +1 unixtime = unixtime +1 I need these four data values (and all the successive data values created by the loop(s)) saved into a new dataframe How do I get R to create a new data object and add data to it on each interation of the for loop. My test for loops only save the data from last iteration of the loop. Am I missing something really simple? Any recommended reference on loop and control structures in R? Thanks, Gord -- Gordon J. Holtslander gordon.holtslan...@usask.ca Dept. of Biology University of Saskatchewan tel 306 966-4433 fax 306 966-4461 __ 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] IP-Address
Hi Peter,
I hope you could help. I am stuck with this. The last problem I have is:
I have table like:
id rank color status ip
138 29746 yellow yes 162.131.58.1
138 29746 yellow yes 162.131.58.2
138 29746 yellow yes 162.131.58.3
138 29746 yellow yes 162.131.58.4
138 29746 yellow yes 162.131.58.5
138 29746 yellow yes 162.131.58.6
138 29746 yellow yes 162.132.58.20
138 29746 yellow yes 162.252.20.21
138 29746 yellow yes 162.254.20.22
138 29746 yellow yes 163.253.7.23
138 29746 yellow yes 163.253.20.25
138 29746 yellow yes 161.138.45.226
How can i make this to a range:
138 29746 yellow yes 162.131.58.1-162.131.58.6
138 29746 yellow yes 162.132.58.20-163.253.7.23
138 29746 yellow yes 162.252.20.21-162.254.20.22
138 29746 yellow yes 163.253.20.25-161.138.45.226
Kind regards,
Eddie
> edw...@web.de wrote:
> > Hi,
> >
> >
> > Unfortunately, they can't handle NA. Any suggestion? Some row for Ip
> > don't have ip address. This cause an error/ wrong result.
>
> A quick fix could be to substitute "..." or "0.0.0.0" for the "NA"
> entries. (Use something like
>
> ipch <- as.character(df$ip)
> ipch[is.na(df$ip)] <- "..."
> connection <- textConnection(ipch)
>
> )
>
> > Eddie
> >
> >> library(gsubfn)
> >> library(gtools)
> >> library(rbenchmark)
> >>
> >> n <- 1
> >> df <- data.frame(
> >> a = rnorm(n),
> >> b = rnorm(n),
> >> c = rnorm(n),
> >> ip = replicate(n, paste(sample(255, 4), collapse='.'), simplify=TRUE)
> >> )
> >>
> >> res <- benchmark(columns=c('test', 'elapsed'), replications=10,
> >
> > order=NULL,
> >
> >> peda = {
> >> connection <- textConnection(as.character(df$ip))
> >> o <- do.call(order, read.table(connection, sep='.'))
> >> close(connection)
> >> df[o, ]
> >> },
> >>
> >> peda2 = {
> >> connection <- textConnection(as.character(df$ip))
> >> dfT <- read.table(connection, sep='.', colClasses=rep("integer",
> >> 4), quote="", na.strings=NULL, blank.lines.skip=FALSE)
> >> close(connection)
> >> o <- do.call(order, dfT)
> >> df[o, ]
> >> },
> >>
> >> hb = {
> >> ip <- strsplit(as.character(df$ip), split=".", fixed=TRUE)
> >> ip <- unlist(ip, use.names=FALSE)
> >> ip <- as.integer(ip)
> >> dim(ip) <- c(4, nrow(df))
> >> ip <- 256^3*ip[1,] + 256^2*ip[2,] + 256*ip[3,] + ip[4,]
> >> o <- order(ip)
> >> df[o, ]
> >> },
> >>
> >> hb2 = {
> >> ip <- strsplit(as.character(df$ip), split=".", fixed=TRUE)
> >> ip <- unlist(ip, use.names=FALSE)
> >> ip <- as.integer(ip);
> >> dim(ip) <- c(4, nrow(df))
> >> o <- sort.list(ip[4,], method="radix", na.last=TRUE)
> >> for (kk in 3:1) {
> >> o <- o[sort.list(ip[kk,o], method="radix", na.last=TRUE)]
> >> }
> >> df[o, ]
> >> }
> >> )
> >>
> >> print(res)
> >>
> >> test elapsed
> >> 1 peda 4.12
> >> 2 peda2 4.08
> >> 3 hb 0.28
> >> 4 hb2 0.25
> >>
> >>
> >> On Sun, May 31, 2009 at 12:42 AM, Wacek Kusnierczyk
> >>
> >> wrote:
> >> > edwin Sendjaja wrote:
> >> >> Hi VQ,
> >> >>
> >> >> Thank you. It works like charm. But I think Peter's code is faster.
> >
> > What
> >
> >> >> is the difference?
> >> >
> >> > i think peter's code is more r-elegant, though less generic. here's a
> >> > quick test, with not so surprising results. gsubfn is implemented in
> >> > r, not c, and it is painfully slow in this test. i also added gabor's
> >> > suggestion.
> >> >
> >> >library(gsubfn)
> >> >library(gtools)
> >> >library(rbenchmark)
> >> >
> >> >n = 1000
> >> >df = data.frame(
> >> > a=rnorm(n),
> >> > b = rnorm(n),
> >> > c = rnorm(n),
> >> > ip = replicate(n, paste(sample(255, 4), collapse='.'),
> >> > simplify=TRUE))
> >> >benchmark(columns=c('test', 'elapsed'), replications=10,
> >> > order=NULL, peda={
> >> > connection = textConnection(as.character(df$ip))
> >> > o = do.call(order, read.table(connection, sep='.'))
> >> > close(connection)
> >> > df[o, ] },
> >> > waku=df[order(gsubfn(perl=TRUE,
> >> > '[0-9]+',
> >> > ~ sprintf('%03d', as.integer(x)),
> >> > as.character(df$ip))), ],
> >> > gagr=df[mixedorder(df$ip), ] )
> >> >
> >> ># peda 0.070
> >> ># waku 7.070
> >> ># gagr 4.710
> >> >
> >> >
> >> > vQ
> >> >
> >> > __
> >> > 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-help@r-project.org mailing list
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> PLEASE do read the posting guide
> >> http://www.R-project.org/posting-guide.html and provide commented,
> >
> > minimal,
> >
> >> self-contained, reproducible code.
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Re: [R] Black-Litterman model
JannaB schrieb: > Anyone know where I can obtain an R implementation of the Black- > Litterman portfolio model? > > There is the BLCOP package. hth Stefan __ 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] surveillance package useful for daily surveillance?
The surveillance package (see below) seems to be mainly designed for weekly or monthly surveillance (although I am still learning about it.) Can it be used for daily surveillance? I had hoped to try using it with respect to daily visits to walk-in clinics in our area. Package ‘surveillance’ April 19, 2009 Title Outbreak detection algorithms for surveillance data Version 1.0-2 Date 2009-05-03 Author M. Höhle with contributions from T. Correa, M. Hofmann, C. Lang, M. Paul, A. Riebler and V. Wimmer Thanks. -- Christopher W. Ryan, MD SUNY Upstate Medical University Clinical Campus at Binghamton 40 Arch Street, Johnson City, NY 13790 cryanatbinghamtondotedu "If you want to build a ship, don't drum up the men to gather wood, divide the work and give orders. Instead, teach them to yearn for the vast and endless sea." [Antoine de St. Exupery] __ 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] Cross correlations on many imported files.
Hello everyone,
I am trying to import all of the csv files from a particular folder and then
run cross correlations on each of them. i.e In the end have a matrix like
structure with cross correlations.
So far I have been able to import all of the data and run the ccf's but I
need a way to store the data in an array. I know that this could be done in
matlab by using something like both a for and nested for loop with the
results being place in results[i,j].
Below I only have tried to create an array and not the entire matrix
structure.
files <- list.files(pattern=".csv$")
for(i in seq(along=files)) { # start for loop with numeric or character
vector; numeric vector is often more flexible
x <- read.csv(files[i], skip=4,header=TRUE)
assign(paste('in',i,sep=""),x) # generates data frame object
and names it
after content in variable 'i'
print(paste(files[i],paste('in',i,sep="")), quote=F)
}
library(tseries)
x<-get(paste('in',1,sep=""))
for(i in 1:length(files)-1){
y<-get(paste('in',i+1,sep=""))
xy<-ccf(x[[5]],y[[5]],lag.max=NULL,na.action=na.exclude)
*** result[i] = xy
}
Any help or advice would be greatly apperciated. Thanks in advance,
Sean
--
View this message in context:
http://www.nabble.com/Cross-correlations-on-many-imported-files.-tp23950345p23950345.html
Sent from the R help mailing list archive at Nabble.com.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] last.warning and Sweave?
Not sure either where the warning gets trapped, but you could use
something like that:
\documentclass{article}
\begin{document}
Test {\tt last.warning}
<<>>=
withCallingHandlers( {
x <- log(-1)
}, warning = function(w){
assign( "last.warning", w, env = baseenv() )
})
@
<<>>=
last.warning
@
\end{document}
Romain
Dieter Menne wrote:
Ben Bolker ufl.edu> writes:
Sweave does something clever with warnings, which I have so far been
...
I thought I could get around this with last.warning , but apparently
that doesn't work -- this file doesn't produce the desired output of
reproducing the last warning ...
\documentclass{article}
Missing \begin{document}
Test {\tt last.warning}
<<>>=
x <- log(-1)
L <- last.warning
@
<<>>=
L
@
\end{document}
==
No real answer, but I inserted a
print(exits("last.warning"))
which gives FALSE when sweaved, but TRUE when the tangled R file is run.
Dieter
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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.
--
Romain Francois
Independent R Consultant
+33(0) 6 28 91 30 30
http://romainfrancois.blog.free.fr
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and provide commented, minimal, self-contained, reproducible code.
[R] Black-Litterman model
Anyone know where I can obtain an R implementation of the Black- Litterman portfolio model? __ 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] last.warning and Sweave?
Ben Bolker ufl.edu> writes:
> Sweave does something clever with warnings, which I have so far been
...
> I thought I could get around this with last.warning , but apparently
> that doesn't work -- this file doesn't produce the desired output of
> reproducing the last warning ...
>
>
> \documentclass{article}
Missing \begin{document}
> Test {\tt last.warning}
>
> <<>>=
> x <- log(-1)
> L <- last.warning
> @
>
> <<>>=
> L
> @
>
> \end{document}
> ==
No real answer, but I inserted a
print(exits("last.warning"))
which gives FALSE when sweaved, but TRUE when the tangled R file is run.
Dieter
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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] scatterplot (car) legend modification
Dear Sam, It's not exactly clear to me what you're doing here, since you appear to be specifying some of the arguments to scatterplot() (and to other functions) in the command used to invoke R. I'm not sure why you would want to do that. That said, you could suppress the legend that scatterplot() produces and use the legend() command to make your own. Given what you're doing, however, it would probably be simpler to use plot() rather than scatterplot(). You could add the ellipses and the points with the data.ellipse() function from the car package, and use legend() for the legend. scatterplot() and the other plotting functions in the car package aren't really meant for making customized graphs, but rather for easily producing informative plots for exploring data in the context of regression analysis. Regards, John > -Original Message- > From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On > Behalf Of sam stein > Sent: June-08-09 11:49 PM > To: r-help@r-project.org > Subject: [R] scatterplot (car) legend modification > > hi, > > new to R and using the car package to do some scatterplots with ellipses > hoping to add the area and center points of each ellipse to the legend? > > looking for some direction / ideas here is the script, the data is > where > golf shots end up by club. > > x (dispersion), y (distance), group (Club) > > thanks, sam > > ## usage: Rscript shotScatter.R infile outfile level1 level2 minX maxX > minY maxY > ## load the library > library(car) > > ## get the args > args = commandArgs(TRUE); > > ## get the data format is: x, y, group > Shot_Data <- read.table(args[1], > sep = ",", > header=T) > > ## setup the pdf file name > pdf(file = args[2]) > > ## do the plot > scatterplot(y ~ x | Club, data=Shot_Data, > ellipse = TRUE, > levels = c(as.numeric(args[3]), as.numeric(args[4])), > robust = TRUE, > boxplots = "", > xlab = "Dispersion", > ylab = "Distance", > xlim = c(as.numeric(args[5]), as.numeric(args[6])), > ylim = c(as.numeric(args[7]), as.numeric(args[8])), > smooth = FALSE, > reg.line = FALSE) > > __ > 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-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] Convex bounds in the function "arms" of package "HI"
Dear Sir/Madam,
i am learning to use a statistical sampling-modle(arms in HI)
But i have a problem with the "indFunc"(bounds of the convex support).
I have found a good example of c code using "arms" function and try to
program the example in R.
If you are interested, you can find the c code from arms usage(P.155-162)
But i always get the error message:
error at convex.bounds(y.start,dir=1,indFunc,...): x not in the support of
indFunc
the R code :
simulate intercept mu
#starting values of mu: mu<-c(0,-2,-4,.)
for (k in 1:K){
for (t in 1:25){
if(k==1){
n_MU[t]<-firmsCCC[t] -used in the density function(Mudens)
N_MU[t]<-defCCC[t] -used in the density function(Mudens)
MUl=mu[2]
MUr=50
}
else if(k==2){
n_MU[t]<-firmsB[t]
N_MU[t]<-defB[t]
MUl=mu[3]
MUr=mu[1]
}
else if...
...
}
Muprev<-mu[k]
mu[k]<- arms(Muprev,Mudens,function(xm,...)(xm>MUl)*(xmhttps://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] OT: a weighted rank-based, non-paired test statistic ?
On Tuesday 09 June 2009, Torsten Hothorn wrote:
> > Date: Fri, 5 Jun 2009 16:09:42 -0700 (PDT)
> > From: Thomas Lumley
> > To: dylan.beaude...@gmail.com
> > Cc: "'r-h...@stat.math.ethz.ch'"
> > Subject: Re: [R] OT: a weighted rank-based, non-paired test statistic ?
> >
> > On Fri, 5 Jun 2009, Dylan Beaudette wrote:
> >> Is anyone aware of a rank-based, non-paired test such as the
> >> Krustal-Wallis,
> >> that can accommodate weights?
> >
> > You don't say what sort of weights, but basically, no.
> >
> > Whether you have precision weights or sampling weights, the test will no
> > longer be distribution-free.
> >
> >> Alternatively, would it make sense to simulate a dataset by duplicating
> >> observations in proportion to their weight, and then using the
> >> Krustal-Wallis
> >> test?
> >
> > No.
>
> well, if you have case weights, i.e., w[i] == 5 means: there are five
> observations which look exactly like observation i, then there are several
>
> ways to do it:
> > library("coin")
> >
> > set.seed(29)
> > x <- gl(3, 10)
> > y <- rnorm(length(x), mean = c(0, 0, 1)[x])
> > d <- data.frame(y = y, x = x)
> > w <- rep(2, nrow(d)) ### double each obs
> >
> > ### all the same
> > kruskal_test(y ~ x, data = rbind(d, d))
>
> Asymptotic Kruskal-Wallis Test
>
> data: y by x (1, 2, 3)
> chi-squared = 12.1176, df = 2, p-value = 0.002337
>
> > kruskal_test(y ~ x, data = d[rep(1:nrow(d), w),])
>
> Asymptotic Kruskal-Wallis Test
>
> data: y by x (1, 2, 3)
> chi-squared = 12.1176, df = 2, p-value = 0.002337
>
> > kruskal_test(y ~ x, data = d, weights = ~ w)
>
> Asymptotic Kruskal-Wallis Test
>
> data: y by x (1, 2, 3)
> chi-squared = 12.1176, df = 2, p-value = 0.002337
>
> the first two work by duplicating data, the latter one is more memory
> efficient since it computes weighted statistics (and their distribution).
>
> However, as Thomas pointed out, other forms of weights are more difficult
> to deal with.
>
> Best wishes,
>
> Torsten
Thanks Torsten. This looks like the solution I was after.
Cheers,
Dylan
> > -thomas
> >
> > Thomas Lumley Assoc. Professor, Biostatistics
> > tlum...@u.washington.eduUniversity of Washington, Seattle
> >
> > __
> > 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.
--
Dylan Beaudette
Soil Resource Laboratory
http://casoilresource.lawr.ucdavis.edu/
University of California at Davis
530.754.7341
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and provide commented, minimal, self-contained, reproducible code.
[R] reading in multiple files
Dear R People: I have about 6000 files to be read in that I'd like to go to one matrix. There are two columns, 1 line in each file. Is there a way to bring them in to produce one matrix or data frame, please? Thanks, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodg...@gmail.com __ R-help@r-project.org mailing list 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] reading in multiple files
You could try:
do.call("rbind", lapply(list.files("path/to/files", full = TRUE), read.csv))
And add more arguments to lapply if the files are not csv, have no header, etc.
--sundar
On Tue, Jun 9, 2009 at 11:18 AM, Erin Hodgess wrote:
> Dear R People:
>
> I have about 6000 files to be read in that I'd like to go to one
> matrix. There are two columns, 1 line in each file.
>
> Is there a way to bring them in to produce one matrix or data frame, please?
>
> Thanks,
> Erin
>
>
> --
> Erin Hodgess
> Associate Professor
> Department of Computer and Mathematical Sciences
> University of Houston - Downtown
> mailto: erinm.hodg...@gmail.com
>
> __
> R-help@r-project.org mailing list
> 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
and provide commented, minimal, self-contained, reproducible code.
Re: [R] using regular expressions to retrieve a digit-digit-dot structure from a string
This is just to you.
You might want to read r-help through gmane.org or one of the other
online readers or get a gmail or other online email account just
for reading newsgroups to circumvent your thread-challenged
email client.
Regards.
On Tue, Jun 9, 2009 at 1:42 PM, Greg Snow wrote:
> Yes, I already apologized to Wacek for missing that and pointing out what he
> had already said.
>
> Given everything in this thread (though it is hard to keep track of all of
> it, my e-mail client does not keep all the parts of the thread together),
> this is probably one of those few tasks that R is not the best tool for.
> There is a Perl module called Lingua::DE::Sentence with the description:
> "Perl extension for tokenizing german texts into their sentences" which seems
> to be exactly what the original poster was looking for. So the best option
> may be to use Perl and the above module to preprocess his texts, then use R
> for later steps.
>
> --
> Gregory (Greg) L. Snow Ph.D.
> Statistical Data Center
> Intermountain Healthcare
> greg.s...@imail.org
> 801.408.8111
>
>
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
>> project.org] On Behalf Of Gabor Grothendieck
>> Sent: Tuesday, June 09, 2009 11:27 AM
>> To: Greg Snow
>> Cc: Wacek Kusnierczyk; r-help@r-project.org; Mark Heckmann
>> Subject: Re: [R] using regular expressions to retrieve a digit-digit-
>> dot structure from a string
>>
>> Wacek already mentioned that; however, its still
>> arguably more complex to specify delimiters
>> than to specify content. Aside from having
>> to specify perl = TRUE and ungreedy matching
>> the content-based regexp is entirely straight forward
>> but for lookbehind (including \K) one has the added
>> complexity of distinguishing between matching and returned
>> values.
>>
>> On Tue, Jun 9, 2009 at 12:36 PM, Greg Snow wrote:
>> > You can sometimes fake variable width look behinds with Perl regexs
>> using '\K':
>> >
>> >> gregexpr('\\b[0-9]+\\K[.]', 'a. 1. a1. 11.', perl=TRUE)
>> > [[1]]
>> > [1] 5 13
>> > attr(,"match.length")
>> > [1] 1 1
>> >
>> >
>> > --
>> > Gregory (Greg) L. Snow Ph.D.
>> > Statistical Data Center
>> > Intermountain Healthcare
>> > greg.s...@imail.org
>> > 801.408.8111
>> >
>> >
>> >> -Original Message-
>> >> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
>> >> project.org] On Behalf Of Wacek Kusnierczyk
>> >> Sent: Tuesday, June 09, 2009 1:05 AM
>> >> To: Gabor Grothendieck
>> >> Cc: r-help@r-project.org; Mark Heckmann
>> >> Subject: Re: [R] using regular expressions to retrieve a digit-
>> digit-
>> >> dot structure from a string
>> >>
>> >> Gabor Grothendieck wrote:
>> >> > On Mon, Jun 8, 2009 at 7:18 PM, Wacek
>> >> > Kusnierczyk wrote:
>> >> >
>> >> >> Gabor Grothendieck wrote:
>> >> >>
>> >> >>> Try this. See ?regex for more.
>> >> >>>
>> >> >>>
>> >> >>>
>> >> x <- 'This happened in the 21. century." (the dot behind 21 is'
>> >> regexpr("(?![0-9]+)[.]", x, perl = TRUE)
>> >>
>> >>
>> >> >>> [1] 24
>> >> >>> attr(,"match.length")
>> >> >>> [1] 1
>> >> >>>
>> >> >>>
>> >> >> yes, but
>> >> >>
>> >> >> gregexpr('(?![0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
>> >> >> # 2 5 9
>> >> >>
>> >> >
>> >> > Yes, it should be:
>> >> >
>> >> >
>> >> >> gregexpr('(?<=[0-9])[.]', 'a. 1. a1.', perl=TRUE)
>> >> >>
>> >> > [[1]]
>> >> > [1] 5 9
>> >> > attr(,"match.length")
>> >> > [1] 1 1
>> >> >
>> >> > which displays the position of every dot that is preceded
>> >> > immediately by a digit. Or just replace gregexpr with regexpr
>> >> > if its intended that it match only one.
>> >> >
>> >>
>> >> i guess what was needed was something like
>> >>
>> >> gregexpr('(?<=\\b[0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
>> >> # 5
>> >>
>> >> which won't work, however, because pcre does not support variable-
>> width
>> >> lookbehinds.
>> >>
>> >> >
>> >> >> which, i guess, is not what you want. if what you want is to
>> match
>> >> all
>> >> >> and only dots that follow at least one digit preceded by a word
>> >> >> boundary, then the following should do, as far as i can see:
>> >> >>
>> >> >> gregexpr('\\b[0-9]+\\K[.]', 'a. 1. a1.', perl=TRUE)
>> >> >> # 5
>> >> >>
>> >> >> vQ
>> >> >>
>> >>
>> >> __
>> >> 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-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-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEA
Re: [R] using regular expressions to retrieve a digit-digit-dot structure from a string
If there were significant advantage to that perl module
I would recommend interfacing R to it rather than
suffer with perl.
For example, see xls2csv (and read.xls) in the gdata package
for an example of interfacing to a perl program.
I don't want to turn this into an R vs. perl thread but there are
certainly many people using R for linguistics and, like perl
which has books on perl and linguistics, there are books
specifically on R and linguistics. One is mentioned on the
gsubfn site. Also there are many linguistics packages in
R that could be explored:
http://cran.r-project.org/web/views/NaturalLanguageProcessing.html
and there would be a big advantage of being able to leverage all
of R's other capabilities.
On Tue, Jun 9, 2009 at 1:42 PM, Greg Snow wrote:
> Yes, I already apologized to Wacek for missing that and pointing out what he
> had already said.
>
> Given everything in this thread (though it is hard to keep track of all of
> it, my e-mail client does not keep all the parts of the thread together),
> this is probably one of those few tasks that R is not the best tool for.
> There is a Perl module called Lingua::DE::Sentence with the description:
> "Perl extension for tokenizing german texts into their sentences" which seems
> to be exactly what the original poster was looking for. So the best option
> may be to use Perl and the above module to preprocess his texts, then use R
> for later steps.
>
> --
> Gregory (Greg) L. Snow Ph.D.
> Statistical Data Center
> Intermountain Healthcare
> greg.s...@imail.org
> 801.408.8111
>
>
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
>> project.org] On Behalf Of Gabor Grothendieck
>> Sent: Tuesday, June 09, 2009 11:27 AM
>> To: Greg Snow
>> Cc: Wacek Kusnierczyk; r-help@r-project.org; Mark Heckmann
>> Subject: Re: [R] using regular expressions to retrieve a digit-digit-
>> dot structure from a string
>>
>> Wacek already mentioned that; however, its still
>> arguably more complex to specify delimiters
>> than to specify content. Aside from having
>> to specify perl = TRUE and ungreedy matching
>> the content-based regexp is entirely straight forward
>> but for lookbehind (including \K) one has the added
>> complexity of distinguishing between matching and returned
>> values.
>>
>> On Tue, Jun 9, 2009 at 12:36 PM, Greg Snow wrote:
>> > You can sometimes fake variable width look behinds with Perl regexs
>> using '\K':
>> >
>> >> gregexpr('\\b[0-9]+\\K[.]', 'a. 1. a1. 11.', perl=TRUE)
>> > [[1]]
>> > [1] 5 13
>> > attr(,"match.length")
>> > [1] 1 1
>> >
>> >
>> > --
>> > Gregory (Greg) L. Snow Ph.D.
>> > Statistical Data Center
>> > Intermountain Healthcare
>> > greg.s...@imail.org
>> > 801.408.8111
>> >
>> >
>> >> -Original Message-
>> >> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
>> >> project.org] On Behalf Of Wacek Kusnierczyk
>> >> Sent: Tuesday, June 09, 2009 1:05 AM
>> >> To: Gabor Grothendieck
>> >> Cc: r-help@r-project.org; Mark Heckmann
>> >> Subject: Re: [R] using regular expressions to retrieve a digit-
>> digit-
>> >> dot structure from a string
>> >>
>> >> Gabor Grothendieck wrote:
>> >> > On Mon, Jun 8, 2009 at 7:18 PM, Wacek
>> >> > Kusnierczyk wrote:
>> >> >
>> >> >> Gabor Grothendieck wrote:
>> >> >>
>> >> >>> Try this. See ?regex for more.
>> >> >>>
>> >> >>>
>> >> >>>
>> >> x <- 'This happened in the 21. century." (the dot behind 21 is'
>> >> regexpr("(?![0-9]+)[.]", x, perl = TRUE)
>> >>
>> >>
>> >> >>> [1] 24
>> >> >>> attr(,"match.length")
>> >> >>> [1] 1
>> >> >>>
>> >> >>>
>> >> >> yes, but
>> >> >>
>> >> >> gregexpr('(?![0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
>> >> >> # 2 5 9
>> >> >>
>> >> >
>> >> > Yes, it should be:
>> >> >
>> >> >
>> >> >> gregexpr('(?<=[0-9])[.]', 'a. 1. a1.', perl=TRUE)
>> >> >>
>> >> > [[1]]
>> >> > [1] 5 9
>> >> > attr(,"match.length")
>> >> > [1] 1 1
>> >> >
>> >> > which displays the position of every dot that is preceded
>> >> > immediately by a digit. Or just replace gregexpr with regexpr
>> >> > if its intended that it match only one.
>> >> >
>> >>
>> >> i guess what was needed was something like
>> >>
>> >> gregexpr('(?<=\\b[0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
>> >> # 5
>> >>
>> >> which won't work, however, because pcre does not support variable-
>> width
>> >> lookbehinds.
>> >>
>> >> >
>> >> >> which, i guess, is not what you want. if what you want is to
>> match
>> >> all
>> >> >> and only dots that follow at least one digit preceded by a word
>> >> >> boundary, then the following should do, as far as i can see:
>> >> >>
>> >> >> gregexpr('\\b[0-9]+\\K[.]', 'a. 1. a1.', perl=TRUE)
>> >> >> # 5
>> >> >>
>> >> >> vQ
>> >> >>
>> >>
>> >> __
>> >> 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 provi
Re: [R] Regex question to find a string that contains 5-9 alpha-numeric characters, at least one of which is a number
Tan, Richard wrote:
> Sorry I did not give some examples in my previous posting to make my
> question clear. It's not exactly 1 digit, but at least one digit. Here
> are some examples:
>
>
>> input = c(none='0foo f0oo foo0 foofoofoo0 0foofoofoo TOOL9NGG
>>
> NONUMBER',all='foob0 fo0o0b 0foob 0foobardo foob4rdoo foobardo0')
>
>> gsub(x=input, replacement='x', perl=TRUE,pattern=something)
>>
>
> none
> all
> "0foo f0oo foo0 foo00 f0o0o foofoofoo0 0foofoofoo TOOL9NGG NONUMBER"
> "x x x x x x"
>
ok, then to my simple mind the following should do:
input = c(
none='0foo f0oo foo0 foofoofoo0 0foofoofoo TOOL9NGG NONUMBER',
all='foob0 fo0o0b 0foob 0foobardo foob4rdoo foobardo0 123456789')
gsub('(?=[[:alpha:]]{0,8}[[:digit:]])\\b[[:alnum:]]{5,9}\\b', 'x',
input, perl=TRUE)
# none -> '0foo f0oo foo0 foofoofoo0 0foofoofoo TOOL9NGG NONUMBER',
# all -> 'x x x x x x x')
where the regex reads 'if there is ahead of you a digit following at
most 8 letters, match 5 to 9 alphanumerics (digits and/or letters).
vQ
__
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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] Sweave and accents
1) Please send reproducible code. In this case possibly with a link to the
Snw file.
2) Try setting the charset in LaTeX: \usepackage[utf8]{inputenc}
On Tue, Jun 9, 2009 at 12:07 PM, Arnau Mir Torres wrote:
> Hello.
>
> I want to write my notes in Sweave in my own language (spanish). But my
> language has accents and when I run Sweave in R to translate my Snw file
> into the tex file the accents are translated into unrecognizable characters.
> For example, the word "camión" (truck) is translated into "camión"
>
> Somebody knows how can I do it?
>
> One solution I don't like is using a perl program to substitute my accented
> letters into "normal" accents in LaTeX in my Snw file before running Sweave.
> That is:
>
> á -> \'a
> é -> \'e
> ...
>
> The problem is when I run the spelling program, it doesn't work because the
> spelling doesn't recognize the words with "normal" accents.
>
> There is a solution but it is tedious:
>
> 1) substitute the accented letters letter into "normal" accents in LaTeX.
> 2) run Sweave and create the tex file.
> 3) subtitute the "normal" accents into the accented letters in the tex
> file.
>
>
> My question is:
> Is there another way to do it?
>
>
> Arnau.
>
>
> Arnau Mir Torres
> Edifici A. Turmeda
> Campus UIB
> Ctra. Valldemossa, km. 7,5
> 07122 Palma de Mca.
> tel: (+34) 971172987
> fax: (+34) 971173003
> email: arnau@uib.es
> URL: http://dmi.uib.es/~arnau
>
>
>
[[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] calibration curve options
David A.G wrote: Hi R-users, can anyone explain me how to play around with the options of the bootstrap calibration curve obtained using the calibrate() function in Design package? I am trying to colour the diagonal, i.e. the ideal curve, in red, and also hide the bias-corrected curve. Thanks, Dave The source code is there so you can create a local copy with any customizations you want, e.g., mycal <- calibrate.default (for every type except psm and cph) modify mycal Frank _ Show them the way! Add maps and directions to your party invites. [[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. -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ 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 regular expressions to retrieve a digit-digit-dot structure from a string
Yes, I already apologized to Wacek for missing that and pointing out what he
had already said.
Given everything in this thread (though it is hard to keep track of all of it,
my e-mail client does not keep all the parts of the thread together), this is
probably one of those few tasks that R is not the best tool for. There is a
Perl module called Lingua::DE::Sentence with the description: "Perl extension
for tokenizing german texts into their sentences" which seems to be exactly
what the original poster was looking for. So the best option may be to use
Perl and the above module to preprocess his texts, then use R for later steps.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Gabor Grothendieck
> Sent: Tuesday, June 09, 2009 11:27 AM
> To: Greg Snow
> Cc: Wacek Kusnierczyk; r-help@r-project.org; Mark Heckmann
> Subject: Re: [R] using regular expressions to retrieve a digit-digit-
> dot structure from a string
>
> Wacek already mentioned that; however, its still
> arguably more complex to specify delimiters
> than to specify content. Aside from having
> to specify perl = TRUE and ungreedy matching
> the content-based regexp is entirely straight forward
> but for lookbehind (including \K) one has the added
> complexity of distinguishing between matching and returned
> values.
>
> On Tue, Jun 9, 2009 at 12:36 PM, Greg Snow wrote:
> > You can sometimes fake variable width look behinds with Perl regexs
> using '\K':
> >
> >> gregexpr('\\b[0-9]+\\K[.]', 'a. 1. a1. 11.', perl=TRUE)
> > [[1]]
> > [1] 5 13
> > attr(,"match.length")
> > [1] 1 1
> >
> >
> > --
> > Gregory (Greg) L. Snow Ph.D.
> > Statistical Data Center
> > Intermountain Healthcare
> > greg.s...@imail.org
> > 801.408.8111
> >
> >
> >> -Original Message-
> >> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> >> project.org] On Behalf Of Wacek Kusnierczyk
> >> Sent: Tuesday, June 09, 2009 1:05 AM
> >> To: Gabor Grothendieck
> >> Cc: r-help@r-project.org; Mark Heckmann
> >> Subject: Re: [R] using regular expressions to retrieve a digit-
> digit-
> >> dot structure from a string
> >>
> >> Gabor Grothendieck wrote:
> >> > On Mon, Jun 8, 2009 at 7:18 PM, Wacek
> >> > Kusnierczyk wrote:
> >> >
> >> >> Gabor Grothendieck wrote:
> >> >>
> >> >>> Try this. See ?regex for more.
> >> >>>
> >> >>>
> >> >>>
> >> x <- 'This happened in the 21. century." (the dot behind 21 is'
> >> regexpr("(?![0-9]+)[.]", x, perl = TRUE)
> >>
> >>
> >> >>> [1] 24
> >> >>> attr(,"match.length")
> >> >>> [1] 1
> >> >>>
> >> >>>
> >> >> yes, but
> >> >>
> >> >> gregexpr('(?![0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
> >> >> # 2 5 9
> >> >>
> >> >
> >> > Yes, it should be:
> >> >
> >> >
> >> >> gregexpr('(?<=[0-9])[.]', 'a. 1. a1.', perl=TRUE)
> >> >>
> >> > [[1]]
> >> > [1] 5 9
> >> > attr(,"match.length")
> >> > [1] 1 1
> >> >
> >> > which displays the position of every dot that is preceded
> >> > immediately by a digit. Or just replace gregexpr with regexpr
> >> > if its intended that it match only one.
> >> >
> >>
> >> i guess what was needed was something like
> >>
> >> gregexpr('(?<=\\b[0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
> >> # 5
> >>
> >> which won't work, however, because pcre does not support variable-
> width
> >> lookbehinds.
> >>
> >> >
> >> >> which, i guess, is not what you want. if what you want is to
> match
> >> all
> >> >> and only dots that follow at least one digit preceded by a word
> >> >> boundary, then the following should do, as far as i can see:
> >> >>
> >> >> gregexpr('\\b[0-9]+\\K[.]', 'a. 1. a1.', perl=TRUE)
> >> >> # 5
> >> >>
> >> >> vQ
> >> >>
> >>
> >> __
> >> 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-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-
> guide.html
> and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Non-linear regression/Quantile regression
The coefficients are different but the predictions are the same. On Tue, Jun 9, 2009 at 12:41 PM, Greg Snow wrote: > poly by default uses orthogonal polynomials which work better mathematically > but are harder to interpret. See ?poly > > -- > Gregory (Greg) L. Snow Ph.D. > Statistical Data Center > Intermountain Healthcare > greg.s...@imail.org > 801.408.8111 > > >> -Original Message- >> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- >> project.org] On Behalf Of despaired >> Sent: Tuesday, June 09, 2009 9:59 AM >> To: r-help@r-project.org >> Subject: Re: [R] Non-linear regression/Quantile regression >> >> >> Hi, >> >> thanks, it works :-) >> But where is the difference between demand ~ Time + I(Time^2) and >> demand ~ >> poly(Time, 2) ? >> Or: How do I have to interpret the results? (I get different results >> for the >> two methods) >> >> Thank you again! >> >> >> Gabor Grothendieck wrote: >> > >> > Those are linear in the coefficients so try these: >> > >> > library(quantreg) >> > >> > rq1 <- rq(demand ~ Time + I(Time^2), data = BOD, tau= 1:3/4); rq1 >> > >> > # or >> > rq2 <- rq(demand ~ poly(Time, 2), data = BOD, tau = 1:3/4); rq2 >> > >> > >> > On Tue, Jun 9, 2009 at 10:55 AM, despaired >> wrote: >> >> >> >> Hi, >> >> >> >> I'm relatively new to R and need to do a quantile regression. Linear >> >> quantile regression works, but for my data I need some quadratic >> >> function. >> >> So I guess, I have to use a nonlinear quantile regression. I tried >> the >> >> example on the help page for nlrq with my data and it worked. But >> the >> >> example there was with a SSlogis model. Trying to write >> >> >> >> dat.nlrq <- nlrq(BM ~ I(Regen100^2), data=dat, tau=0.25, trace=TRUE) >> >> >> >> or >> >> >> >> dat.nlrq <- nlrq(BM ~ poly(Regen100^2), data=dat, tau=0.25, >> trace=TRUE) >> >> >> >> (I don't know the difference) both gave me the following error >> message: >> >> >> >> error in getInitial.default(func, data, mCall = >> as.list(match.call(func, >> >> : >> >> no 'getInitial' method found for "function" objects >> >> >> >> Looking in getInitial, it must have to do something with the >> starting >> >> parameters or selfStart model. But I have no idea, what this is and >> how I >> >> handle this problem. Can anyone please help? >> >> >> >> Thanks a lot in advance! >> >> -- >> >> View this message in context: >> >> http://www.nabble.com/Non-linear-regression-Quantile-regression- >> tp23944530p23944530.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. >> >> >> > >> > __ >> > 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. >> > >> > >> >> -- >> View this message in context: http://www.nabble.com/Non-linear- >> regression-Quantile-regression-tp23944530p23945900.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. > __ > 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-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 regular expressions to retrieve a digit-digit-dot structure from a string
Wacek already mentioned that; however, its still
arguably more complex to specify delimiters
than to specify content. Aside from having
to specify perl = TRUE and ungreedy matching
the content-based regexp is entirely straight forward
but for lookbehind (including \K) one has the added
complexity of distinguishing between matching and returned
values.
On Tue, Jun 9, 2009 at 12:36 PM, Greg Snow wrote:
> You can sometimes fake variable width look behinds with Perl regexs using
> '\K':
>
>> gregexpr('\\b[0-9]+\\K[.]', 'a. 1. a1. 11.', perl=TRUE)
> [[1]]
> [1] 5 13
> attr(,"match.length")
> [1] 1 1
>
>
> --
> Gregory (Greg) L. Snow Ph.D.
> Statistical Data Center
> Intermountain Healthcare
> greg.s...@imail.org
> 801.408.8111
>
>
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
>> project.org] On Behalf Of Wacek Kusnierczyk
>> Sent: Tuesday, June 09, 2009 1:05 AM
>> To: Gabor Grothendieck
>> Cc: r-help@r-project.org; Mark Heckmann
>> Subject: Re: [R] using regular expressions to retrieve a digit-digit-
>> dot structure from a string
>>
>> Gabor Grothendieck wrote:
>> > On Mon, Jun 8, 2009 at 7:18 PM, Wacek
>> > Kusnierczyk wrote:
>> >
>> >> Gabor Grothendieck wrote:
>> >>
>> >>> Try this. See ?regex for more.
>> >>>
>> >>>
>> >>>
>> x <- 'This happened in the 21. century." (the dot behind 21 is'
>> regexpr("(?![0-9]+)[.]", x, perl = TRUE)
>>
>>
>> >>> [1] 24
>> >>> attr(,"match.length")
>> >>> [1] 1
>> >>>
>> >>>
>> >> yes, but
>> >>
>> >> gregexpr('(?![0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
>> >> # 2 5 9
>> >>
>> >
>> > Yes, it should be:
>> >
>> >
>> >> gregexpr('(?<=[0-9])[.]', 'a. 1. a1.', perl=TRUE)
>> >>
>> > [[1]]
>> > [1] 5 9
>> > attr(,"match.length")
>> > [1] 1 1
>> >
>> > which displays the position of every dot that is preceded
>> > immediately by a digit. Or just replace gregexpr with regexpr
>> > if its intended that it match only one.
>> >
>>
>> i guess what was needed was something like
>>
>> gregexpr('(?<=\\b[0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
>> # 5
>>
>> which won't work, however, because pcre does not support variable-width
>> lookbehinds.
>>
>> >
>> >> which, i guess, is not what you want. if what you want is to match
>> all
>> >> and only dots that follow at least one digit preceded by a word
>> >> boundary, then the following should do, as far as i can see:
>> >>
>> >> gregexpr('\\b[0-9]+\\K[.]', 'a. 1. a1.', perl=TRUE)
>> >> # 5
>> >>
>> >> vQ
>> >>
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-
>> guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] calibration curve options
It's rather simple to get the code and add modifications. You wanted the dashed line to be red? (To find out what the underlying function name is, you use methods(plot): The dashed ideal line plotting function is not set up to accept parameters. so you need to alter the code. Use this as a guide: >plot.calibrate <- function( Hi R-users, can anyone explain me how to play around with the options of the bootstrap calibration curve obtained using the calibrate() function in Design package? I am trying to colour the diagonal, i.e. the ideal curve, in red, and also hide the bias-corrected curve. Thanks, Dave _ Show them the way! Add maps and directions to your party invites. -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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] Regex question to find a string that contains 5-9 alpha-numeric characters, at least one of which is a number
Sorry I did not give some examples in my previous posting to make my
question clear. It's not exactly 1 digit, but at least one digit. Here
are some examples:
> input = c(none='0foo f0oo foo0 foofoofoo0 0foofoofoo TOOL9NGG
NONUMBER',all='foob0 fo0o0b 0foob 0foobardo foob4rdoo foobardo0')
> gsub(x=input, replacement='x', perl=TRUE,pattern=something)
none
all
"0foo f0oo foo0 foo00 f0o0o foofoofoo0 0foofoofoo TOOL9NGG NONUMBER"
"x x x x x x"
-Original Message-
From: Wacek Kusnierczyk [mailto:waclaw.marcin.kusnierc...@idi.ntnu.no]
Sent: Tuesday, June 09, 2009 1:06 PM
To: Greg Snow
Cc: Marc Schwartz; Barry Rowlingson; r-help@r-project.org; Tan, Richard
Subject: Re: [R] Regex question to find a string that contains 5-9
alpha-numeric characters, at least one of which is a number
Greg Snow wrote:
> Here is one way using a single pattern (so can be used in a
substitution), it uses Perl's positive look ahead patters:
>
>
>> test <-
>> c("SHRT","5HRT","M1TCH","M1TCH5","LONG3RS","NONUMBER","TOOLNGG","
>> ooops.3")
>>
>> sub( '(?=[a-zA-Z]{0,8}[0-9])[a-zA-Z0-9]{5,9}', 'xxx', test,
>> perl=TRUE)
>>
yes, but:
sub( '(?=[a-zA-Z]{0,8}[0-9])[a-zA-Z0-9]{5,9}', 'x', '12345',
perl=TRUE)
# "x"
which is not what was expected -- as far as i understand, the point was
to match 5-9 character strings with exactly 1 digit.
vQ
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] using regular expressions to retrieve a digit-digit-dot structure from a string
On Tue, Jun 9, 2009 at 7:44 AM, Mark Heckmann wrote: > Thanks for your help. Your answers solved the problem I posted and that is > just when I noticed that I misspecified the problem ;) > My problem is to separate a German texts by sentences. Unfortunately I > haven't found an R package doing this kind of text separation in German, so > I try it "manually". > > Just using the dot as separator fails in occasions like: > txt <- "One January 1. I saw Rick. He was born in the 19. century." > Sentence boundary disambiguation is a non-trivial problem, as you can see in your above example (cf. "I arrived on January 1. I saw Rick."). You can get ~95% accuracy fairly straightforwardly, but the last 5% are hard. Take a look at http://en.wikipedia.org/wiki/Sentence_boundary_disambiguation, which points to other good resources. -s [[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] Regex question to find a string that contains 5-9 alpha-numeric characters, at least one of which is a number
Greg Snow wrote:
> Here is one way using a single pattern (so can be used in a substitution), it
> uses Perl's positive look ahead patters:
>
>
>> test <-
>> c("SHRT","5HRT","M1TCH","M1TCH5","LONG3RS","NONUMBER","TOOLNGG","ooops.3")
>>
>> sub( '(?=[a-zA-Z]{0,8}[0-9])[a-zA-Z0-9]{5,9}', 'xxx', test, perl=TRUE)
>>
yes, but:
sub( '(?=[a-zA-Z]{0,8}[0-9])[a-zA-Z0-9]{5,9}', 'x', '12345',
perl=TRUE)
# "x"
which is not what was expected -- as far as i understand, the point was
to match 5-9 character strings with exactly 1 digit.
vQ
__
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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] Non-linear regression/Quantile regression
poly by default uses orthogonal polynomials which work better mathematically but are harder to interpret. See ?poly -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 > -Original Message- > From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- > project.org] On Behalf Of despaired > Sent: Tuesday, June 09, 2009 9:59 AM > To: r-help@r-project.org > Subject: Re: [R] Non-linear regression/Quantile regression > > > Hi, > > thanks, it works :-) > But where is the difference between demand ~ Time + I(Time^2) and > demand ~ > poly(Time, 2) ? > Or: How do I have to interpret the results? (I get different results > for the > two methods) > > Thank you again! > > > Gabor Grothendieck wrote: > > > > Those are linear in the coefficients so try these: > > > > library(quantreg) > > > > rq1 <- rq(demand ~ Time + I(Time^2), data = BOD, tau= 1:3/4); rq1 > > > > # or > > rq2 <- rq(demand ~ poly(Time, 2), data = BOD, tau = 1:3/4); rq2 > > > > > > On Tue, Jun 9, 2009 at 10:55 AM, despaired > wrote: > >> > >> Hi, > >> > >> I'm relatively new to R and need to do a quantile regression. Linear > >> quantile regression works, but for my data I need some quadratic > >> function. > >> So I guess, I have to use a nonlinear quantile regression. I tried > the > >> example on the help page for nlrq with my data and it worked. But > the > >> example there was with a SSlogis model. Trying to write > >> > >> dat.nlrq <- nlrq(BM ~ I(Regen100^2), data=dat, tau=0.25, trace=TRUE) > >> > >> or > >> > >> dat.nlrq <- nlrq(BM ~ poly(Regen100^2), data=dat, tau=0.25, > trace=TRUE) > >> > >> (I don't know the difference) both gave me the following error > message: > >> > >> error in getInitial.default(func, data, mCall = > as.list(match.call(func, > >> : > >> no 'getInitial' method found for "function" objects > >> > >> Looking in getInitial, it must have to do something with the > starting > >> parameters or selfStart model. But I have no idea, what this is and > how I > >> handle this problem. Can anyone please help? > >> > >> Thanks a lot in advance! > >> -- > >> View this message in context: > >> http://www.nabble.com/Non-linear-regression-Quantile-regression- > tp23944530p23944530.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. > >> > > > > __ > > 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. > > > > > > -- > View this message in context: http://www.nabble.com/Non-linear- > regression-Quantile-regression-tp23944530p23945900.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. __ 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] Sweave and accents
On 6/9/2009 12:07 PM, Arnau Mir Torres wrote: Hello. I want to write my notes in Sweave in my own language (spanish). But my language has accents and when I run Sweave in R to translate my Snw file into the tex file the accents are translated into unrecognizable characters. For example, the word "camión" (truck) is translated into "camión" Somebody knows how can I do it? That looks as though you are using Latin1 encoding, and Sweave is giving you UTF-8. (If not those two, then a similar sort of problem.) The handling of encodings is tricky, because it depends on your system settings (what sort of locale you are running in). The l10n_info() and sessionInfo() functions may be helpful in diagnosing the problem. Duncan Murdoch One solution I don't like is using a perl program to substitute my accented letters into "normal" accents in LaTeX in my Snw file before running Sweave. That is: á -> \'a é -> \'e ... The problem is when I run the spelling program, it doesn't work because the spelling doesn't recognize the words with "normal" accents. There is a solution but it is tedious: 1) substitute the accented letters letter into "normal" accents in LaTeX. 2) run Sweave and create the tex file. 3) subtitute the "normal" accents into the accented letters in the tex file. My question is: Is there another way to do it? Arnau. Arnau Mir Torres Edifici A. Turmeda Campus UIB Ctra. Valldemossa, km. 7,5 07122 Palma de Mca. tel: (+34) 971172987 fax: (+34) 971173003 email: arnau@uib.es URL: http://dmi.uib.es/~arnau __ 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-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] Dependency between packages for Windows-binaries
Hi,
I have already asked a similar question without response
(https://stat.ethz.ch/pipermail/r-help/2009-June/200300.html) so I am
here reformulating in the hope that someone is able to help. If
something is unclear, please ask.
I am working on the development of two packages, pkg1 and pkg2 (based on
work in two different projects). pkg1 is quite generic, pkg2 tries to
solve a particular problem within same field (geostatistics). Therefore,
there might be users who want to use pkg2 as an add-on package to
increase the functionality of pkg1. In other words, functions in pkg1
are based on the S3 class system, and I want pkg2 to offer methods for
pkg2-objects to functions defined in pkg1, for users having both
packages installed. Merging the packages or making pkg2 directly
dependent pkg1 would be the easiest solution, but it is a bad solution
as most users will only be interested in one of the packages.
I thought this could be solved by including the following in NAMESPACE
of pkg2:
if ("pkg1" %in% rownames(utils:::installed.packages()) {
importFrom(pkg1, fun1, fun2, fun3)
S3method(fun1, class2)
S3method(fun2, class2)
S3method(fun3, class2)
}
Unfortunately, this doesn't seem to work as I expected, particularly
when I try to build a binary for Windows-users (R CMD build -binary
pkg2). It seems for me that the dependency between pkg2 and pkg1 is
defined at the time of building a binary, not when the package is loaded
with library(pkg2):
If I have pkg1 installed when I build pkg2, the package will not load
for users who do not have pkg1 installed:
> install.packages("d:/pkg2_0.1-1.zip", repos=NULL)
> library(pkg2)
(...)
Error : object 'fun1' not found whilst loading namespace 'pkg2'
Error: package/namespace load failed for 'pkg2'
If I remove pkg1 from the library before building pkg2, pkg2 can be
installed and loaded by everyone, but the new method is not visible for
those with pkg1 installed:
> install.packages("d:/pkg2_0.1-2.zip", repos=NULL)
> library(pkg1)
> library(pkg2)
> methods(fun1)
[1] fun1.class1*
Non-visible functions are asterisked
I have tried with and without one of the following:
Suggests: pkg1
Enhances: pkg1
in the DESCRIPTION file of pkg2, but I cannot see that either of them
make a difference.
I can of course solve this problem temporarily, building two different
versions of pkg2, with and without having pkg1 installed on my computer.
But first of all, that seems unnecessary complicated, and more
important, the plan is also to upload these packages to CRAN where I
doubt that such a two-version solution is possible or accepted.
Is there a way to build one binary that works both with and without
having pkg1 installed? I have tried to search in mail archives and
"Writing R extensions", but either it is not there, or I have missed it.
Thanks,
Jon
BTW, I am using R version 2.9.0 and Rtools version 29 under Windows XP-SP3
__
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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] using regular expressions to retrieve a digit-digit-dot structure from a string
You can sometimes fake variable width look behinds with Perl regexs using '\K':
> gregexpr('\\b[0-9]+\\K[.]', 'a. 1. a1. 11.', perl=TRUE)
[[1]]
[1] 5 13
attr(,"match.length")
[1] 1 1
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Wacek Kusnierczyk
> Sent: Tuesday, June 09, 2009 1:05 AM
> To: Gabor Grothendieck
> Cc: r-help@r-project.org; Mark Heckmann
> Subject: Re: [R] using regular expressions to retrieve a digit-digit-
> dot structure from a string
>
> Gabor Grothendieck wrote:
> > On Mon, Jun 8, 2009 at 7:18 PM, Wacek
> > Kusnierczyk wrote:
> >
> >> Gabor Grothendieck wrote:
> >>
> >>> Try this. See ?regex for more.
> >>>
> >>>
> >>>
> x <- 'This happened in the 21. century." (the dot behind 21 is'
> regexpr("(?![0-9]+)[.]", x, perl = TRUE)
>
>
> >>> [1] 24
> >>> attr(,"match.length")
> >>> [1] 1
> >>>
> >>>
> >> yes, but
> >>
> >>gregexpr('(?![0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
> >># 2 5 9
> >>
> >
> > Yes, it should be:
> >
> >
> >> gregexpr('(?<=[0-9])[.]', 'a. 1. a1.', perl=TRUE)
> >>
> > [[1]]
> > [1] 5 9
> > attr(,"match.length")
> > [1] 1 1
> >
> > which displays the position of every dot that is preceded
> > immediately by a digit. Or just replace gregexpr with regexpr
> > if its intended that it match only one.
> >
>
> i guess what was needed was something like
>
> gregexpr('(?<=\\b[0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
> # 5
>
> which won't work, however, because pcre does not support variable-width
> lookbehinds.
>
> >
> >> which, i guess, is not what you want. if what you want is to match
> all
> >> and only dots that follow at least one digit preceded by a word
> >> boundary, then the following should do, as far as i can see:
> >>
> >>gregexpr('\\b[0-9]+\\K[.]', 'a. 1. a1.', perl=TRUE)
> >># 5
> >>
> >> vQ
> >>
>
> __
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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.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with if statements
On 6/9/2009 12:17 PM, Amit Patel wrote:
> Hi
> I am trying to create a column in a data frame which gives a sigificane score
> from 0-7. It should read values from 7 different colums and add 1 to the
> counter if the value is <=0.05. I get an error message saying
>
> Error in if (ALLRESULTS[i, 16] <= 0.05) significance_count =
> significance_count + :
> missing value where TRUE/FALSE needed
>
> The script is included below
>
> it works if i convert the NA values to zero but this is not appropriate as it
> includes the zero as significant.
>
> ANY SUGGESTIONS
significance.count <- rowSums(ALLRESULTS[,16:22] <= .05, na.rm=TRUE)
?rowSums
> #SCRIPT STARTS
> for (i in 1:length(ALLRESULTS[,1])) {
> significance_count = 0
>
> if (ALLRESULTS[i,16] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
> if (ALLRESULTS[i,17] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
> if (ALLRESULTS[i,18] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
> if (ALLRESULTS[i,19] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
> if (ALLRESULTS[i,20] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
> if (ALLRESULTS[i,21] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
> if (ALLRESULTS[i,22] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
>
> ALLRESULTS[i,23] <- significance_count}
>
>
>
>
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> 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.
--
Chuck Cleland, Ph.D.
NDRI, Inc. (www.ndri.org)
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 512-0171 (M, W, F)
fax: (917) 438-0894
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Re: [R] Help with if statements
Dear Amit,
Try this:
significance_count <- apply( ALLRESULTS[,16:22], 1, function(x) sum( x <=
0.05 ) ) )
HTH,
Jorge
On Tue, Jun 9, 2009 at 12:17 PM, Amit Patel wrote:
>
> Hi
> I am trying to create a column in a data frame which gives a sigificane
> score from 0-7. It should read values from 7 different colums and add 1 to
> the counter if the value is <=0.05. I get an error message saying
>
> Error in if (ALLRESULTS[i, 16] <= 0.05) significance_count =
> significance_count + :
> missing value where TRUE/FALSE needed
>
> The script is included below
>
> it works if i convert the NA values to zero but this is not appropriate as
> it includes the zero as significant.
>
> ANY SUGGESTIONS
>
>
>
> #SCRIPT STARTS
> for (i in 1:length(ALLRESULTS[,1])) {
> significance_count = 0
>
> if (ALLRESULTS[i,16] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
> if (ALLRESULTS[i,17] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
> if (ALLRESULTS[i,18] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
> if (ALLRESULTS[i,19] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
> if (ALLRESULTS[i,20] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
> if (ALLRESULTS[i,21] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
> if (ALLRESULTS[i,22] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
>
> ALLRESULTS[i,23] <- significance_count}
>
>
>
>
> __
> 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.
>
[[alternative HTML version deleted]]
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Re: [R] Help with if statements
Try this:
for (i in 1:dim(ALLRESULTS)[1]) {
ALLRESULTS[i,23] <- length(ALLRESULTS[i,][ALLRESULTS[i,16:22] <= 0.05])
}
On Wed, Jun 10, 2009 at 12:17 AM, Amit Patel wrote:
>
> Hi
> I am trying to create a column in a data frame which gives a sigificane score
> from 0-7. It should read values from 7 different colums and add 1 to the
> counter if the value is <=0.05. I get an error message saying
>
> Error in if (ALLRESULTS[i, 16] <= 0.05) significance_count =
> significance_count + :
> missing value where TRUE/FALSE needed
>
> The script is included below
>
> it works if i convert the NA values to zero but this is not appropriate as it
> includes the zero as significant.
>
> ANY SUGGESTIONS
>
>
>
> #SCRIPT STARTS
> for (i in 1:length(ALLRESULTS[,1])) {
> significance_count = 0
>
> if (ALLRESULTS[i,16] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
> if (ALLRESULTS[i,17] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
> if (ALLRESULTS[i,18] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
> if (ALLRESULTS[i,19] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
> if (ALLRESULTS[i,20] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
> if (ALLRESULTS[i,21] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
> if (ALLRESULTS[i,22] <= 0.05 ) significance_count = significance_count +1
> else significance_count = significance_count
>
> ALLRESULTS[i,23] <- significance_count}
>
>
>
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regex question to find a string that contains 5-9 alpha-numeric characters, at least one of which is a number
Here is one way using a single pattern (so can be used in a substitution), it
uses Perl's positive look ahead patters:
> test <-
> c("SHRT","5HRT","M1TCH","M1TCH5","LONG3RS","NONUMBER","TOOLNGG","ooops.3")
>
> sub( '(?=[a-zA-Z]{0,8}[0-9])[a-zA-Z0-9]{5,9}', 'xxx', test, perl=TRUE)
[1] "SHRT""5HRT""xxx" "xxx" "xxx"
[6] "NONUMBER""TOOLNGG" "ooops.3"
>
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Marc Schwartz
> Sent: Monday, June 08, 2009 6:33 PM
> To: Barry Rowlingson
> Cc: r-help@r-project.org; Tan, Richard
> Subject: Re: [R] Regex question to find a string that contains 5-9
> alpha-numeric characters, at least one of which is a number
>
>
> On Jun 8, 2009, at 5:27 PM, Barry Rowlingson wrote:
>
> > On Mon, Jun 8, 2009 at 10:40 PM, Tan, Richard
> > wrote:
> >> Hi,
> >>
> >> This is not exactly an R question but I am trying to use gsub to
> >> replace
> >> a string that contains 5-9 alpha-numeric characters, at least one of
> >> which is a number. Is there a good way to write it in a one line
> >> regex?
> >
> > The only way I can think of is to spell out all the possible
> > expressions, somethinglike:
> >
> > [0-9][a-z0-9]{4} | [a-z0-9][0-9][a-z0-9]{3} |
> > [a-z0-9]{2}[0-9][a-z0-9]{2} and so on. That is, have a regex
> > component for every possible 5, 6, 7, 8, and 9 character expression
> > with [0-9] in each place. I'm not sure this qualifies as 'good',
> > though..
> >
> > Better to do it in two stages, one to check for 5-9 alphanumerics,
> > and then another to check for a number.
> >
> > Here's something on a test vector 's':
> >
> >> cbind(s,grepl("^[A-Z0-9]{5,9}$",s),grepl("[0-9]",s))
> > s
> > [1,] "SHRT""FALSE" "FALSE"
> > [2,] "5HRT""FALSE" "TRUE"
> > [3,] "M1TCH" "TRUE" "TRUE"
> > [4,] "M1TCH5" "TRUE" "TRUE"
> > [5,] "LONG3RS" "TRUE" "TRUE"
> > [6,] "NONUMBER""TRUE" "FALSE"
> > [7,] "TOOLNGG" "FALSE" "FALSE"
> >
> > The ones you want give two TRUE values. Extending to lower-case is
> > left as an exercise...
> >
> > Barry
>
>
> I was trying to think of a way to do this with only a single grep(),
> but it has been too long of a day.
>
> So here is a bit of a simplification on the two stage approach:
>
> > vec
> [1] "SHRT""5HRT""M1TCH" "M1TCH5"
> "LONG3RS" "NONUMBER""TOOLNGG"
>
>
> > grep("[0-9]", vec[grep("^[[:alnum:]]{5,9}$", vec)], value = TRUE)
> [1] "M1TCH" "M1TCH5" "LONG3RS"
>
>
> HTH,
>
> Marc Schwartz
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-
> guide.html
> and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Non-linear regression/Quantile regression
Try `poly(Time, 2, raw=TRUE)' Here is an example: Time <- runif(100) demand <- 1 + 0.5 * Time - 1.2 * Time^2 + rt(100, df=4) rq1 <- rq(demand ~ Time + I(Time^2), tau = 1:3/4) rq2 <- rq(demand ~ poly(Time, 2, raw=TRUE), tau = 1:3/4) all.equal(c(rq1$coef), c(rq2$coef)) Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: rvarad...@jhmi.edu Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty_personal_pages/Varadhan.h tml -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of despaired Sent: Tuesday, June 09, 2009 11:59 AM To: r-help@r-project.org Subject: Re: [R] Non-linear regression/Quantile regression Hi, thanks, it works :-) But where is the difference between demand ~ Time + I(Time^2) and demand ~ poly(Time, 2) ? Or: How do I have to interpret the results? (I get different results for the two methods) Thank you again! Gabor Grothendieck wrote: > > Those are linear in the coefficients so try these: > > library(quantreg) > > rq1 <- rq(demand ~ Time + I(Time^2), data = BOD, tau= 1:3/4); rq1 > > # or > rq2 <- rq(demand ~ poly(Time, 2), data = BOD, tau = 1:3/4); rq2 > > > On Tue, Jun 9, 2009 at 10:55 AM, despaired wrote: >> >> Hi, >> >> I'm relatively new to R and need to do a quantile regression. Linear >> quantile regression works, but for my data I need some quadratic >> function. >> So I guess, I have to use a nonlinear quantile regression. I tried >> the example on the help page for nlrq with my data and it worked. But >> the example there was with a SSlogis model. Trying to write >> >> dat.nlrq <- nlrq(BM ~ I(Regen100^2), data=dat, tau=0.25, trace=TRUE) >> >> or >> >> dat.nlrq <- nlrq(BM ~ poly(Regen100^2), data=dat, tau=0.25, >> trace=TRUE) >> >> (I don't know the difference) both gave me the following error message: >> >> error in getInitial.default(func, data, mCall = >> as.list(match.call(func, >> : >> no 'getInitial' method found for "function" objects >> >> Looking in getInitial, it must have to do something with the starting >> parameters or selfStart model. But I have no idea, what this is and >> how I handle this problem. Can anyone please help? >> >> Thanks a lot in advance! >> -- >> View this message in context: >> http://www.nabble.com/Non-linear-regression-Quantile-regression-tp239 >> 44530p23944530.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. >> > > __ > 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. > > -- View this message in context: http://www.nabble.com/Non-linear-regression-Quantile-regression-tp23944530p2 3945900.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. __ 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] Isolating a single plot from plots produced simultaneously
On Tue, 9 Jun 2009, Laura Bonnett wrote: I'd like to isolate the second plot produced (the estimated functional form of the influence of age on the log relative hazard) so that I can use the 'points' function to add the linear predictors for the untransformed and the log-transformed models. In the usual situation one would produce a plot and then type: I think predict(, type="terms") does what you want. -thomas Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle __ 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] Help with if statements
Hi
I am trying to create a column in a data frame which gives a sigificane score
from 0-7. It should read values from 7 different colums and add 1 to the
counter if the value is <=0.05. I get an error message saying
Error in if (ALLRESULTS[i, 16] <= 0.05) significance_count = significance_count
+ :
missing value where TRUE/FALSE needed
The script is included below
it works if i convert the NA values to zero but this is not appropriate as it
includes the zero as significant.
ANY SUGGESTIONS
#SCRIPT STARTS
for (i in 1:length(ALLRESULTS[,1])) {
significance_count = 0
if (ALLRESULTS[i,16] <= 0.05 ) significance_count = significance_count +1 else
significance_count = significance_count
if (ALLRESULTS[i,17] <= 0.05 ) significance_count = significance_count +1 else
significance_count = significance_count
if (ALLRESULTS[i,18] <= 0.05 ) significance_count = significance_count +1 else
significance_count = significance_count
if (ALLRESULTS[i,19] <= 0.05 ) significance_count = significance_count +1 else
significance_count = significance_count
if (ALLRESULTS[i,20] <= 0.05 ) significance_count = significance_count +1 else
significance_count = significance_count
if (ALLRESULTS[i,21] <= 0.05 ) significance_count = significance_count +1 else
significance_count = significance_count
if (ALLRESULTS[i,22] <= 0.05 ) significance_count = significance_count +1 else
significance_count = significance_count
ALLRESULTS[i,23] <- significance_count}
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[R] quantile of a mixture of bivriate normal distributions
Hi, Does anyone know how to compute the quantile of a mixture of four bivariate normal distriutions? Many thanks! Hannah [[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] Comparing R and SAs
Hi: For those of you who are adept at both SAS and R, I have the following questions: a) What are some reasons / tasks for which you would use R over SAS and vice versa? b) What are some things for which R is a must have that SAS cannot fulfill the requirements? I am on the ramp up on both of them. The general feeling that I am getting by following this group is that R updates to the product are at a much faster pace and therefore, this would be better for someone who wants the bleeding edge (correct me if I am wrong). But I am also interested in what is inherently better in R that SAS cannot offer perhaps because of the design. Thanks. Satish __ 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] calibration curve options
Hi R-users, can anyone explain me how to play around with the options of the bootstrap calibration curve obtained using the calibrate() function in Design package? I am trying to colour the diagonal, i.e. the ideal curve, in red, and also hide the bias-corrected curve. Thanks, Dave _ Show them the way! Add maps and directions to your party invites. [[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] Sweave and accents
Hello. I want to write my notes in Sweave in my own language (spanish). But my language has accents and when I run Sweave in R to translate my Snw file into the tex file the accents are translated into unrecognizable characters. For example, the word "camión" (truck) is translated into "camión" Somebody knows how can I do it? One solution I don't like is using a perl program to substitute my accented letters into "normal" accents in LaTeX in my Snw file before running Sweave. That is: á -> \'a é -> \'e ... The problem is when I run the spelling program, it doesn't work because the spelling doesn't recognize the words with "normal" accents. There is a solution but it is tedious: 1) substitute the accented letters letter into "normal" accents in LaTeX. 2) run Sweave and create the tex file. 3) subtitute the "normal" accents into the accented letters in the tex file. My question is: Is there another way to do it? Arnau. Arnau Mir Torres Edifici A. Turmeda Campus UIB Ctra. Valldemossa, km. 7,5 07122 Palma de Mca. tel: (+34) 971172987 fax: (+34) 971173003 email: arnau@uib.es URL: http://dmi.uib.es/~arnau __ 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 regular expressions to retrieve a digit-digit-dot structure from a string
Another way to handle is to match the contents rather than the
delimiters using strapply in gsubfn (http://gsubfn.googlecode.com).
Below a sentence is defined as starting with a non-space followed by
anything followed by an alpha followed by dot, question mark or
exclamation mark.
The (?U) means that we use ungreedy matching so that the first
sentence terminator matches rather than the last.
This is an arguably simpler regexp as it does not involve lookbehind.
> txt <- "One January 1. I saw Rick. He was born in the 19. century."
> library(gsubfn)
> strapply(txt, "(?U)([^ ].*[[:alpha:]][.?!])", c, perl = TRUE)
[[1]]
[1] "One January 1. I saw Rick." "He was born in the 19. century."
This won't be as fast as the other solutions but typically speed is
not a real consideration unless you have huge amounts of text. Also
note that the development version of strapply runs 5x faster than
the production version on certain sample problems.
On Tue, Jun 9, 2009 at 10:40 AM, Mark Heckmann wrote:
>
> Thanks,
>
> Now it works great. I modified it a bit so the sentences will be split by
> questionmarks (.?!:), etc. as well.
>
> strsplit(gsub("([[:alpha:]][\\.\\?\\!\\:])", "\\1*", txt), "\\* *") [[1]]
>
> e.g.
>
>> strsplit(gsub("([[:alpha:]][\\.\\?\\!\\:])", "\\1*", txt), "\\* *") [[1]]
> [1] "One January 1. I saw Rick?" "He was born in the 19. century."
>
> ---
>
> Mark Heckmann
> + 49 (0) 421 - 1614618
> www.markheckmann.de
> R-Blog: http://ryouready.wordpress.com
>
>
>
>
> -Ursprüngliche Nachricht-
> Von: Marc Schwartz [mailto:marc_schwa...@me.com]
> Gesendet: Dienstag, 9. Juni 2009 14:17
> An: Mark Heckmann
> Cc: r-help@r-project.org; 'Gabor Grothendieck';
> waclaw.marcin.kusnierc...@idi.ntnu.no
> Betreff: Re: AW: [R] using regular expressions to retrieve a digit-digit-dot
> structure from a string
>
> On Jun 9, 2009, at 6:44 AM, Mark Heckmann wrote:
>
>> Hey all,
>>
>> Thanks for your help. Your answers solved the problem I posted and
>> that is
>> just when I noticed that I misspecified the problem ;)
>> My problem is to separate a German texts by sentences. Unfortunately I
>> haven't found an R package doing this kind of text separation in
>> German, so
>> I try it "manually".
>>
>> Just using the dot as separator fails in occasions like:
>> txt <- "One January 1. I saw Rick. He was born in the 19. century."
>>
>> Here I want the algorithm to separate the string only at the
>> positions where
>> the dot is not preceded by a digit. The R-snippets posted pick out
>> "1." and
>> "19."
>>
>> txt <- "One January 1. I saw Rick. He was born in the 19. century."
>>> gregexpr('(?<=[0-9])[.]',txt, perl=T)
>> [[1]]
>> [1] 14 49
>> attr(,"match.length")
>> [1] 1 1
>>
>> But I just need it the other way round. So I tried:
>>
>>> strsplit(txt, "[[:alpha:]]\\." , perl=T)
>> [[1]]
>> [1] "One January 1. I saw Ric" " He was born in the 19. centur"
>>
>> But this erases the last letter from each sentence. Does someone
>> know a
>> solution?
>>
>> TIA
>> Mark
>
>
>
> This is one of those rare? times where it might be nice for strsplit()
> to have an option to retain the split regex at the end of each parsed
> segment, rather than removing it.
>
> There may be a better way, but trying to both avoid a loop over vector
> indices and trying to stay with R functions that use .Internal() for
> speed, you may be able to use something like this:
>
> > strsplit(gsub("([[:alpha:]]\\.)", "\\1*", txt), "\\* *")
> [[1]]
> [1] "One January 1. I saw Rick." "He was born in the 19. century."
>
> What I am essentially doing is to add an "*" to the ending of each
> sentence (you can use other characters) such that strsplit() can split
> on that character without affecting the rest of the sentence. So as
> an intermediate result, you get:
>
> > gsub("([[:alpha:]]\\.)", "\\1*", txt)
> [1] "One January 1. I saw Rick.* He was born in the 19. century.*"
>
> which then makes the strsplit() parsing a bit easier. Since both
> strsplit() and grep() use .Internal()s, hopefully this would still be
> reasonably fast. Note that I have strsplit() split on the "*" possibly
> followed by one or more " ", which is required for mid-line splits.
>
> HTH,
>
> Marc Schwartz
>
> __
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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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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Non-linear regression/Quantile regression
Hi, thanks, it works :-) But where is the difference between demand ~ Time + I(Time^2) and demand ~ poly(Time, 2) ? Or: How do I have to interpret the results? (I get different results for the two methods) Thank you again! Gabor Grothendieck wrote: > > Those are linear in the coefficients so try these: > > library(quantreg) > > rq1 <- rq(demand ~ Time + I(Time^2), data = BOD, tau= 1:3/4); rq1 > > # or > rq2 <- rq(demand ~ poly(Time, 2), data = BOD, tau = 1:3/4); rq2 > > > On Tue, Jun 9, 2009 at 10:55 AM, despaired wrote: >> >> Hi, >> >> I'm relatively new to R and need to do a quantile regression. Linear >> quantile regression works, but for my data I need some quadratic >> function. >> So I guess, I have to use a nonlinear quantile regression. I tried the >> example on the help page for nlrq with my data and it worked. But the >> example there was with a SSlogis model. Trying to write >> >> dat.nlrq <- nlrq(BM ~ I(Regen100^2), data=dat, tau=0.25, trace=TRUE) >> >> or >> >> dat.nlrq <- nlrq(BM ~ poly(Regen100^2), data=dat, tau=0.25, trace=TRUE) >> >> (I don't know the difference) both gave me the following error message: >> >> error in getInitial.default(func, data, mCall = as.list(match.call(func, >> : >> no 'getInitial' method found for "function" objects >> >> Looking in getInitial, it must have to do something with the starting >> parameters or selfStart model. But I have no idea, what this is and how I >> handle this problem. Can anyone please help? >> >> Thanks a lot in advance! >> -- >> View this message in context: >> http://www.nabble.com/Non-linear-regression-Quantile-regression-tp23944530p23944530.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. >> > > __ > 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. > > -- View this message in context: http://www.nabble.com/Non-linear-regression-Quantile-regression-tp23944530p23945900.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.
Re: [R] Splicing factors without losing levels
On Tue, Jun 9, 2009 at 11:16 AM, Titus von der Malsburg wrote: > On Tue, Jun 09, 2009 at 11:04:03AM -0400, Stavros Macrakis wrote: > > This may seem like a minor point, but I think it is worthwhile using > > descriptive names for functions. > > Makes sense. I thought I've seen this use somewhere else (probably in > Lisp?). What better name do you suggest for this operation? > The two meanings I can think of in Lisp for splicing are 1) The backquote operator ,@X, which means to insert the value of X as part of the surrounding list rather than as an element of the list, e.g. `(a b ,@'(c d) e f) == (append '(a b) '(c d) '(e f)) => (a b c d e f), as opposed to `(a b ,'(c d) e f) == (append '(a b) (list '(c d)) '(e f)) => (a b (c d) e f). 2) The notion of inserting (typically destructively) one list in the middle of another. I would suggest a name like 'intersperse' or 'alternate'. -s [[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] using regular expressions to retrieve a digit-digit-dotstructure from a string
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Mark Heckmann
> Sent: Tuesday, June 09, 2009 4:45 AM
> To: r-help@r-project.org
> Cc: waclaw.marcin.kusnierc...@idi.ntnu.no; marc_schwa...@me.com
> Subject: Re: [R] using regular expressions to retrieve a
> digit-digit-dotstructure from a string
>
> Hey all,
>
> Thanks for your help. Your answers solved the problem I
> posted and that is
> just when I noticed that I misspecified the problem ;)
> My problem is to separate a German texts by sentences. Unfortunately I
> haven't found an R package doing this kind of text separation
> in German, so
> I try it "manually".
>
> Just using the dot as separator fails in occasions like:
> txt <- "One January 1. I saw Rick. He was born in the 19. century."
>
> Here I want the algorithm to separate the string only at the
> positions where
> the dot is not preceded by a digit. The R-snippets posted
> pick out "1." and
> "19."
>
> txt <- "One January 1. I saw Rick. He was born in the 19. century."
> > gregexpr('(?<=[0-9])[.]',txt, perl=T)
> [[1]]
> [1] 14 49
> attr(,"match.length")
> [1] 1 1
>
> But I just need it the other way round. So I tried:
>
> > strsplit(txt, "[[:alpha:]]\\." , perl=T)
> [[1]]
> [1] "One January 1. I saw Ric" " He was born in the 19. centur"
>
> But this erases the last letter from each sentence. Does
> someone know a
> solution?
In S+ strsplit() has an argument called subpattern that lets you
specify which parenthesized part of the regular expression
to use as the split point. It is the akin to the \\ used in the
replacement argument of sub and gsub. E.g., to split the string
at the sequence of spaces after a period, but not after period preceded
by a digit do:
> txt <- "One January 1. I saw Rick. He was born in the 19. century."
> strsplit(txt, "[^[:digit:]]\\.([[:space:]]+)", subpattern=1)
[[1]]:
[1] "One January 1. I saw Rick." "He was born in the 19. century."
subpattern=0, the default, means text matched by the entire regular
expression. regexpr has the same argument. Would such an argument
solve your problem?
Bill Dunlap
TIBCO Software Inc - Spotfire Division
wdunlap tibco.com
> TIA
> Mark
>
> ---
>
> Mark Heckmann
> + 49 (0) 421 - 1614618
> www.markheckmann.de
> R-Blog: http://ryouready.wordpress.com
>
>
>
>
> -Ursprüngliche Nachricht-
> Von: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
> Gesendet: Dienstag, 9. Juni 2009 12:48
> An: Wacek Kusnierczyk
> Cc: Mark Heckmann; r-help@r-project.org
> Betreff: Re: [R] using regular expressions to retrieve a
> digit-digit-dot
> structure from a string
>
> On Tue, Jun 9, 2009 at 3:04 AM, Wacek
> Kusnierczyk wrote:
> > Gabor Grothendieck wrote:
> >> On Mon, Jun 8, 2009 at 7:18 PM, Wacek
> >> Kusnierczyk wrote:
> >>
> >>> Gabor Grothendieck wrote:
> >>>
> Try this. See ?regex for more.
>
>
>
> > x <- 'This happened in the 21. century." (the dot behind 21 is'
> > regexpr("(?![0-9]+)[.]", x, perl = TRUE)
> >
> >
> [1] 24
> attr(,"match.length")
> [1] 1
>
>
> >>> yes, but
> >>>
> >>> gregexpr('(?![0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
> >>> # 2 5 9
> >>>
> >>
> >> Yes, it should be:
> >>
> >>
> >>> gregexpr('(?<=[0-9])[.]', 'a. 1. a1.', perl=TRU
> E)
> >>>
> >> [[1]]
> >> [1] 5 9
> >> attr(,"match.length")
> >> [1] 1 1
> >>
> >> which displays the position of every dot that is preceded
> >> immediately by a digit. Or just replace gregexpr with regexpr
> >> if its intended that it match only one.
> >>
> >
> > i guess what was needed was something like
> >
> > gregexpr('(?<=\\b[0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
> > # 5
> >
> > which won't work, however, because pcre does not support
> variable-width
> > lookbehinds.
>
> No, what I wrote was what I intended. I don't think we are
> discussing the answer
> at this point but just the interpretation of what was intended. You
> are including
> the word boundary in the question and I am not. I think its
> also possible
> that
> regexpr is what is wanted, not gregexpr, but at this point I think the
> poster has
> enough answers that he can complete it himself by considering
> what he wants
> and using one of ours or a suitable modification.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Isolating a single plot from plots produced simultaneously
This is almost certainly a device-dependent issue, which is the reason the Posting Guide requests that you include such details. On a Mac using the GUI interface, one can scroll backwards through the graphics windows with -. With the plot of interest as the focus one can then issue additonal plotting function calls such as lines() or points() and then save the plots as pdf if the default Qartz device has been used. I suspect there are equivalent facilities for Windows and Linux installations and encourage you to try reading the help pages: ?Devices #should bring up useful (installation-specific) information -- David On Jun 9, 2009, at 10:36 AM, Laura Bonnett wrote: Dear R-Help, I am using the 'mfp' package. It produces three plots (as I am using the Cox model) simultaneously which can be viewed together using the following code: fit <- mfp(Surv(rem.Remtime,rem.Rcens)~fp(age) +strata(rpa),family=cox,data=nearma,select=0.05,verbose=TRUE) par(mfrow=c(2,2)) plot(fit) They can be viewed separately but the return key must be pressed after each graph appears (Click or hit ENTER for next page). I'd like to isolate the second plot produced (the estimated functional form of the influence of age on the log relative hazard) so that I can use the 'points' function to add the linear predictors for the untransformed and the log-transformed models. In the usual situation one would produce a plot and then type: coxfitu <- coxph(Surv(rem.Remtime,rem.Rcens)~age +strata(rpa),data=nearma) points(coxfitu$linear.predictor,col=2) coxfitl <- coxph(Surv(rem.Remtime,rem.Rcens)~log(age) +strata(rpa),data=nearma) points(coxfitl$linear.predictor,col=3) Can anyone tell me how to isolate just the second plot produced? Thank you for your help, Laura __ 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. David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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] Non-linear regression/Quantile regression
Those are linear in the coefficients so try these: library(quantreg) rq1 <- rq(demand ~ Time + I(Time^2), data = BOD, tau= 1:3/4); rq1 # or rq2 <- rq(demand ~ poly(Time, 2), data = BOD, tau = 1:3/4); rq2 On Tue, Jun 9, 2009 at 10:55 AM, despaired wrote: > > Hi, > > I'm relatively new to R and need to do a quantile regression. Linear > quantile regression works, but for my data I need some quadratic function. > So I guess, I have to use a nonlinear quantile regression. I tried the > example on the help page for nlrq with my data and it worked. But the > example there was with a SSlogis model. Trying to write > > dat.nlrq <- nlrq(BM ~ I(Regen100^2), data=dat, tau=0.25, trace=TRUE) > > or > > dat.nlrq <- nlrq(BM ~ poly(Regen100^2), data=dat, tau=0.25, trace=TRUE) > > (I don't know the difference) both gave me the following error message: > > error in getInitial.default(func, data, mCall = as.list(match.call(func, : > no 'getInitial' method found for "function" objects > > Looking in getInitial, it must have to do something with the starting > parameters or selfStart model. But I have no idea, what this is and how I > handle this problem. Can anyone please help? > > Thanks a lot in advance! > -- > View this message in context: > http://www.nabble.com/Non-linear-regression-Quantile-regression-tp23944530p23944530.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. > __ 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] Splicing factors without losing levels
On Tue, Jun 09, 2009 at 11:04:03AM -0400, Stavros Macrakis wrote: > This may seem like a minor point, but I think it is worthwhile using > descriptive names for functions. Makes sense. I thought I've seen this use somewhere else (probably in Lisp?). What better name do you suggest for this operation? Titus __ 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] Splicing factors without losing levels
Various people have provided technical solutions to your problem. May I suggest, though, that 'splice' isn't quite the right word for this operation? Splicing two pieces of rope / movie film / audio tape / wires / etc. means connecting them at their ends, either at an extremity or in the middle, e.g. X: Y: Extremity splice: xx or yyxx Middle splice: xxxyyyx or yyyxxx The splice itself is the point of connection (xy or yx) between two things. In normal English, splicing never refers to interspersing alternate members of X and Y. This may seem like a minor point, but I think it is worthwhile using descriptive names for functions. -s On Tue, Jun 9, 2009 at 5:12 AM, Titus von der Malsburg wrote: > An operation that I often need is splicing two vectors: > > > splice(1:3, 4:6) > [1] 1 4 2 5 3 6 > [[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] using regular expressions to retrieve a digit-digit-dot structure from a string
Hey all,
Thanks for your help. Your answers solved the problem I posted and that is
just when I noticed that I misspecified the problem ;)
My problem is to separate a German texts by sentences. Unfortunately I
haven't found an R package doing this kind of text separation in German, so
I try it "manually".
Just using the dot as separator fails in occasions like:
txt <- "One January 1. I saw Rick. He was born in the 19. century."
Here I want the algorithm to separate the string only at the positions where
the dot is not preceded by a digit. The R-snippets posted pick out "1." and
"19."
txt <- "One January 1. I saw Rick. He was born in the 19. century."
> gregexpr('(?<=[0-9])[.]',txt, perl=T)
[[1]]
[1] 14 49
attr(,"match.length")
[1] 1 1
But I just need it the other way round. So I tried:
> strsplit(txt, "[[:alpha:]]\\." , perl=T)
[[1]]
[1] "One January 1. I saw Ric" " He was born in the 19. centur"
But this erases the last letter from each sentence. Does someone know a
solution?
TIA
Mark
---
Mark Heckmann
+ 49 (0) 421 - 1614618
www.markheckmann.de
R-Blog: http://ryouready.wordpress.com
-Ursprüngliche Nachricht-
Von: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Gesendet: Dienstag, 9. Juni 2009 12:48
An: Wacek Kusnierczyk
Cc: Mark Heckmann; r-help@r-project.org
Betreff: Re: [R] using regular expressions to retrieve a digit-digit-dot
structure from a string
On Tue, Jun 9, 2009 at 3:04 AM, Wacek
Kusnierczyk wrote:
> Gabor Grothendieck wrote:
>> On Mon, Jun 8, 2009 at 7:18 PM, Wacek
>> Kusnierczyk wrote:
>>
>>> Gabor Grothendieck wrote:
>>>
Try this. See ?regex for more.
> x <- 'This happened in the 21. century." (the dot behind 21 is'
> regexpr("(?![0-9]+)[.]", x, perl = TRUE)
>
>
[1] 24
attr(,"match.length")
[1] 1
>>> yes, but
>>>
>>> gregexpr('(?![0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
>>> # 2 5 9
>>>
>>
>> Yes, it should be:
>>
>>
>>> gregexpr('(?<=[0-9])[.]', 'a. 1. a1.', perl=TRU
E)
>>>
>> [[1]]
>> [1] 5 9
>> attr(,"match.length")
>> [1] 1 1
>>
>> which displays the position of every dot that is preceded
>> immediately by a digit. Or just replace gregexpr with regexpr
>> if its intended that it match only one.
>>
>
> i guess what was needed was something like
>
> gregexpr('(?<=\\b[0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
> # 5
>
> which won't work, however, because pcre does not support variable-width
> lookbehinds.
No, what I wrote was what I intended. I don't think we are
discussing the answer
at this point but just the interpretation of what was intended. You
are including
the word boundary in the question and I am not. I think its also possible
that
regexpr is what is wanted, not gregexpr, but at this point I think the
poster has
enough answers that he can complete it himself by considering what he wants
and using one of ours or a suitable modification.
__
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] Using ADF.Test
Hi,
I am quite new to R and would appreciate some guidance, if possible.
I have imported a csv file: spread <- read.csv("Spread.csv")
I get the following error when I try to run adf.test:
> adf.test(spread,alternative = c("stationary", "explosive"),0)
Error in embed(y, k) : 'x' is not a vector or matrix
Why is this?
--
View this message in context:
http://www.nabble.com/Using-ADF.Test-tp23940824p23940824.html
Sent from the R help mailing list archive at Nabble.com.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] using regular expressions to retrieve a digit-digit-dot structure from a string
Thanks,
Now it works great. I modified it a bit so the sentences will be split by
questionmarks (.?!:), etc. as well.
strsplit(gsub("([[:alpha:]][\\.\\?\\!\\:])", "\\1*", txt), "\\* *") [[1]]
e.g.
> strsplit(gsub("([[:alpha:]][\\.\\?\\!\\:])", "\\1*", txt), "\\* *") [[1]]
[1] "One January 1. I saw Rick?" "He was born in the 19. century."
---
Mark Heckmann
+ 49 (0) 421 - 1614618
www.markheckmann.de
R-Blog: http://ryouready.wordpress.com
-Ursprüngliche Nachricht-
Von: Marc Schwartz [mailto:marc_schwa...@me.com]
Gesendet: Dienstag, 9. Juni 2009 14:17
An: Mark Heckmann
Cc: r-help@r-project.org; 'Gabor Grothendieck';
waclaw.marcin.kusnierc...@idi.ntnu.no
Betreff: Re: AW: [R] using regular expressions to retrieve a digit-digit-dot
structure from a string
On Jun 9, 2009, at 6:44 AM, Mark Heckmann wrote:
> Hey all,
>
> Thanks for your help. Your answers solved the problem I posted and
> that is
> just when I noticed that I misspecified the problem ;)
> My problem is to separate a German texts by sentences. Unfortunately I
> haven't found an R package doing this kind of text separation in
> German, so
> I try it "manually".
>
> Just using the dot as separator fails in occasions like:
> txt <- "One January 1. I saw Rick. He was born in the 19. century."
>
> Here I want the algorithm to separate the string only at the
> positions where
> the dot is not preceded by a digit. The R-snippets posted pick out
> "1." and
> "19."
>
> txt <- "One January 1. I saw Rick. He was born in the 19. century."
>> gregexpr('(?<=[0-9])[.]',txt, perl=T)
> [[1]]
> [1] 14 49
> attr(,"match.length")
> [1] 1 1
>
> But I just need it the other way round. So I tried:
>
>> strsplit(txt, "[[:alpha:]]\\." , perl=T)
> [[1]]
> [1] "One January 1. I saw Ric" " He was born in the 19. centur"
>
> But this erases the last letter from each sentence. Does someone
> know a
> solution?
>
> TIA
> Mark
This is one of those rare? times where it might be nice for strsplit()
to have an option to retain the split regex at the end of each parsed
segment, rather than removing it.
There may be a better way, but trying to both avoid a loop over vector
indices and trying to stay with R functions that use .Internal() for
speed, you may be able to use something like this:
> strsplit(gsub("([[:alpha:]]\\.)", "\\1*", txt), "\\* *")
[[1]]
[1] "One January 1. I saw Rick." "He was born in the 19. century."
What I am essentially doing is to add an "*" to the ending of each
sentence (you can use other characters) such that strsplit() can split
on that character without affecting the rest of the sentence. So as
an intermediate result, you get:
> gsub("([[:alpha:]]\\.)", "\\1*", txt)
[1] "One January 1. I saw Rick.* He was born in the 19. century.*"
which then makes the strsplit() parsing a bit easier. Since both
strsplit() and grep() use .Internal()s, hopefully this would still be
reasonably fast. Note that I have strsplit() split on the "*" possibly
followed by one or more " ", which is required for mid-line splits.
HTH,
Marc Schwartz
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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] adonis formula
Hello, I have a pretty large data set of insect community data from a tree plantation. There are 14 subraces (fixed) of tree, each with 10 families fixed within subrace. There are then 10 reps (random) of each family, which are in specific part of the plantation. I am trying to determine the proper formula for adonis for testing for differences in communities within and between rep, subrace and family fixed in subrace. Please let me know id you can help. Also, does adonis support unbalanced designs? Thanks for your help. Ben __ 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] Writing Reports from R in Office Open XML format (ooxmlWeave?)
Wow, It sounds great. Looking forward to it. 2009/6/9 Duncan Temple Lang : > Yes. We will release a version in the next few weeks > when I have time to wrap it all up. > There is also a Docbook-based version that uses > R extensions to Docbook for authoring structured > documents. > > D. > > Tobias Sing wrote: >> >> Dear all, >> >> has someone implemented functionality for writing reports from R in >> Office Open XML format (*), similar to what odfWeave does for the ODF >> format of OpenOffice? It would be great to have a kind of >> "ooxmlWeave" at least for those of us who are forced to work in an >> MS ecosystem. >> >> (*) Office Open XML is the default, XML-based, file format for MS >> Word: http://en.wikipedia.org/wiki/Office_Open_XML >> >> Kind regards, >> Tobias >> >> __ >> 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-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 __ 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] Non-linear regression/Quantile regression
Hi, I'm relatively new to R and need to do a quantile regression. Linear quantile regression works, but for my data I need some quadratic function. So I guess, I have to use a nonlinear quantile regression. I tried the example on the help page for nlrq with my data and it worked. But the example there was with a SSlogis model. Trying to write dat.nlrq <- nlrq(BM ~ I(Regen100^2), data=dat, tau=0.25, trace=TRUE) or dat.nlrq <- nlrq(BM ~ poly(Regen100^2), data=dat, tau=0.25, trace=TRUE) (I don't know the difference) both gave me the following error message: error in getInitial.default(func, data, mCall = as.list(match.call(func, : no 'getInitial' method found for "function" objects Looking in getInitial, it must have to do something with the starting parameters or selfStart model. But I have no idea, what this is and how I handle this problem. Can anyone please help? Thanks a lot in advance! -- View this message in context: http://www.nabble.com/Non-linear-regression-Quantile-regression-tp23944530p23944530.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.
Re: [R] Cairo Package Installation
Install the cairo headers? Probably in some cairo or cairo-devel package
for your OS.
Uwe Ligges
Douglas M. Hultstrand wrote:
I am trying to install the Cairo package on a linux machine, the Cairo
package did not install correctly (could not find cairo.h), I am new to
R and linux any help on the installation would be great. Below are
output from trying to install the Cairo package, thought this might help.
> install.packages("Cairo")
Warning in install.packages("Cairo") :
argument 'lib' is missing: using
'/home/tyep/R/x86_64-unknown-linux-gnu-library/2.8'
trying URL
'http://streaming.stat.iastate.edu/CRAN/src/contrib/Cairo_1.4-4.tar.gz'
Content type 'application/x-gzip' length 75200 bytes (73 Kb)
opened URL
==
downloaded 73 Kb
* Installing *source* package 'Cairo' ...
checking for gcc... gcc -std=gnu99
checking for C compiler default output file name... a.out
checking whether the C compiler works... yes
checking whether we are cross compiling... no
checking for suffix of executables...
checking for suffix of object files... o
checking whether we are using the GNU C compiler... yes
checking whether gcc -std=gnu99 accepts -g... yes
checking for gcc -std=gnu99 option to accept ISO C89... none needed
checking how to run the C preprocessor... gcc -std=gnu99 -E
checking for grep that handles long lines and -e... /bin/grep
checking for egrep... /bin/grep -E
checking for ANSI C header files... yes
checking for sys/wait.h that is POSIX.1 compatible... yes
checking for sys/types.h... yes
checking for sys/stat.h... yes
checking for stdlib.h... yes
checking for string.h... yes
checking for memory.h... yes
checking for strings.h... yes
checking for inttypes.h... yes
checking for stdint.h... yes
checking for unistd.h... yes
checking for string.h... (cached) yes
checking sys/time.h usability... yes
checking sys/time.h presence... yes
checking for sys/time.h... yes
checking for unistd.h... (cached) yes
checking for an ANSI C-conforming const... yes
checking for pkg-config... /usr/bin/pkg-config
checking whether pkg-config knows about cairo... no
configure: CAIRO_LIBS is unset, attempting to guess it.
configure: CAIRO_CFLAGS=
checking if R was compiled with the RConn patch... no
checking cairo.h usability... no
checking cairo.h presence... no
checking for cairo.h... no
configure: error: Cannot find cairo.h! Please install cairo
(http://www.cairographics.org/) and/or set CAIRO_CFLAGS/LIBS
correspondingly.
ERROR: configuration failed for package 'Cairo'
** Removing '/home/tyep/R/x86_64-unknown-linux-gnu-library/2.8/Cairo'
The downloaded packages are in
/tmp/RtmpILmmXw/downloaded_packages
Warning message:
In install.packages("Cairo") :
installation of package 'Cairo' had non-zero exit status
>
Thanks,
Doug
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] "R CMD check" does not find a mistake
Henrik Bengtsson wrote: "using R version 2.9.0 beta (2009-04-04 r48290)" As a start, Kurt et al. are for sure using a much more recent version (probably also the very latest patched version). and if not only then at least also R-devel. Uwe /H On Tue, Jun 9, 2009 at 7:17 AM, Christophe Genolini wrote: Hi the list, I build a package. They was a mistake in it, but R CMD check did not find it. Is that normal ? Here is what Kurt gets (which is right, I did this mistake): --- 8< * checking for code/documentation mismatches ... WARNING S4 class codoc mismatches from documentation object 'LongData-class': Slots for class 'LongData' Code: id other time traj varName Docs: id time traj varName --- 8< But here is what I get : --- 8< R CMD check longitudinalData * checking for working pdflatex ... OK * using log directory 'C:/Documents and Settings/Christophe/Mes documents/Recher che/Trajectoires/kmeal/longitudinalData.Rcheck' * using R version 2.9.0 beta (2009-04-04 r48290) * using session charset: ISO8859-1 * checking for file 'longitudinalData/DESCRIPTION' ... OK * checking extension type ... Package * this is package 'longitudinalData' version '0.5' * package encoding: latin1 * checking package dependencies ... OK * checking if this is a source package ... OK * checking for .dll and .exe files ... OK * checking whether package 'longitudinalData' can be installed ... OK * checking package directory ... OK * checking for portable file names ... OK * checking DESCRIPTION meta-information ... OK * checking top-level files ... OK * checking index information ... OK * checking package subdirectories ... OK * checking R files for non-ASCII characters ... OK * checking R files for syntax errors ... OK * checking whether the package can be loaded ... OK * checking whether the package can be loaded with stated dependencies ... OK * checking for unstated dependencies in R code ... OK * checking S3 generic/method consistency ... OK * checking replacement functions ... OK * checking foreign function calls ... OK * checking R code for possible problems ... OK * checking Rd files ... OK * checking Rd files against version 2 parser ... OK * checking Rd cross-references ... OK * checking for missing documentation entries ... OK * checking for code/documentation mismatches ... OK * checking Rd \usage sections ... OK * checking examples ... OK * checking PDF version of manual ... OK R CMD build longitudinalData * checking for file 'longitudinalData/DESCRIPTION' ... OK * preparing 'longitudinalData': * checking DESCRIPTION meta-information ... OK * removing junk files * checking for LF line-endings in source and make files * checking for empty or unneeded directories * building 'longitudinalData_0.5.tar.gz' --- 8< Christophe __ 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-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-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] "R CMD check" does not find a mistake
try issueing this before calling R CMD check export _R_CHECK_CODOC_S4_METHODS_=true hth, Ingmar On 9 Jun 2009, at 16:17, Christophe Genolini wrote: Hi the list, I build a package. They was a mistake in it, but R CMD check did not find it. Is that normal ? Here is what Kurt gets (which is right, I did this mistake): --- 8< * checking for code/documentation mismatches ... WARNING S4 class codoc mismatches from documentation object 'LongData-class': Slots for class 'LongData' Code: id other time traj varName Docs: id time traj varName --- 8< But here is what I get : --- 8< R CMD check longitudinalData * checking for working pdflatex ... OK * using log directory 'C:/Documents and Settings/Christophe/Mes documents/Recher che/Trajectoires/kmeal/longitudinalData.Rcheck' * using R version 2.9.0 beta (2009-04-04 r48290) * using session charset: ISO8859-1 * checking for file 'longitudinalData/DESCRIPTION' ... OK * checking extension type ... Package * this is package 'longitudinalData' version '0.5' * package encoding: latin1 * checking package dependencies ... OK * checking if this is a source package ... OK * checking for .dll and .exe files ... OK * checking whether package 'longitudinalData' can be installed ... OK * checking package directory ... OK * checking for portable file names ... OK * checking DESCRIPTION meta-information ... OK * checking top-level files ... OK * checking index information ... OK * checking package subdirectories ... OK * checking R files for non-ASCII characters ... OK * checking R files for syntax errors ... OK * checking whether the package can be loaded ... OK * checking whether the package can be loaded with stated dependencies ... OK * checking for unstated dependencies in R code ... OK * checking S3 generic/method consistency ... OK * checking replacement functions ... OK * checking foreign function calls ... OK * checking R code for possible problems ... OK * checking Rd files ... OK * checking Rd files against version 2 parser ... OK * checking Rd cross-references ... OK * checking for missing documentation entries ... OK * checking for code/documentation mismatches ... OK * checking Rd \usage sections ... OK * checking examples ... OK * checking PDF version of manual ... OK R CMD build longitudinalData * checking for file 'longitudinalData/DESCRIPTION' ... OK * preparing 'longitudinalData': * checking DESCRIPTION meta-information ... OK * removing junk files * checking for LF line-endings in source and make files * checking for empty or unneeded directories * building 'longitudinalData_0.5.tar.gz' --- 8< Christophe __ 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-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] Cairo Package Installation
I am trying to install the Cairo package on a linux machine, the Cairo
package did not install correctly (could not find cairo.h), I am new to
R and linux any help on the installation would be great. Below are
output from trying to install the Cairo package, thought this might help.
> install.packages("Cairo")
Warning in install.packages("Cairo") :
argument 'lib' is missing: using
'/home/tyep/R/x86_64-unknown-linux-gnu-library/2.8'
trying URL
'http://streaming.stat.iastate.edu/CRAN/src/contrib/Cairo_1.4-4.tar.gz'
Content type 'application/x-gzip' length 75200 bytes (73 Kb)
opened URL
==
downloaded 73 Kb
* Installing *source* package 'Cairo' ...
checking for gcc... gcc -std=gnu99
checking for C compiler default output file name... a.out
checking whether the C compiler works... yes
checking whether we are cross compiling... no
checking for suffix of executables...
checking for suffix of object files... o
checking whether we are using the GNU C compiler... yes
checking whether gcc -std=gnu99 accepts -g... yes
checking for gcc -std=gnu99 option to accept ISO C89... none needed
checking how to run the C preprocessor... gcc -std=gnu99 -E
checking for grep that handles long lines and -e... /bin/grep
checking for egrep... /bin/grep -E
checking for ANSI C header files... yes
checking for sys/wait.h that is POSIX.1 compatible... yes
checking for sys/types.h... yes
checking for sys/stat.h... yes
checking for stdlib.h... yes
checking for string.h... yes
checking for memory.h... yes
checking for strings.h... yes
checking for inttypes.h... yes
checking for stdint.h... yes
checking for unistd.h... yes
checking for string.h... (cached) yes
checking sys/time.h usability... yes
checking sys/time.h presence... yes
checking for sys/time.h... yes
checking for unistd.h... (cached) yes
checking for an ANSI C-conforming const... yes
checking for pkg-config... /usr/bin/pkg-config
checking whether pkg-config knows about cairo... no
configure: CAIRO_LIBS is unset, attempting to guess it.
configure: CAIRO_CFLAGS=
checking if R was compiled with the RConn patch... no
checking cairo.h usability... no
checking cairo.h presence... no
checking for cairo.h... no
configure: error: Cannot find cairo.h! Please install cairo
(http://www.cairographics.org/) and/or set CAIRO_CFLAGS/LIBS
correspondingly.
ERROR: configuration failed for package 'Cairo'
** Removing '/home/tyep/R/x86_64-unknown-linux-gnu-library/2.8/Cairo'
The downloaded packages are in
/tmp/RtmpILmmXw/downloaded_packages
Warning message:
In install.packages("Cairo") :
installation of package 'Cairo' had non-zero exit status
>
Thanks,
Doug
--
-
Douglas M. Hultstrand, MS
Senior Hydrometeorologist
Metstat, Inc. Windsor, Colorado
voice: 970.686.1253
email: dmhul...@metstat.com
web: http://www.metstat.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] "R CMD check" does not find a mistake
"using R version 2.9.0 beta (2009-04-04 r48290)" As a start, Kurt et al. are for sure using a much more recent version (probably also the very latest patched version). /H On Tue, Jun 9, 2009 at 7:17 AM, Christophe Genolini wrote: > Hi the list, > > I build a package. They was a mistake in it, but R CMD check did not find > it. Is that normal ? > > Here is what Kurt gets (which is right, I did this mistake): > > --- 8< > * checking for code/documentation mismatches ... WARNING > S4 class codoc mismatches from documentation object 'LongData-class': > Slots for class 'LongData' > Code: id other time traj varName > Docs: id time traj varName > --- 8< > > But here is what I get : > > --- 8< > R CMD check longitudinalData > * checking for working pdflatex ... OK > * using log directory 'C:/Documents and Settings/Christophe/Mes > documents/Recher > che/Trajectoires/kmeal/longitudinalData.Rcheck' > * using R version 2.9.0 beta (2009-04-04 r48290) > * using session charset: ISO8859-1 > * checking for file 'longitudinalData/DESCRIPTION' ... OK > * checking extension type ... Package > * this is package 'longitudinalData' version '0.5' > * package encoding: latin1 > * checking package dependencies ... OK > * checking if this is a source package ... OK > * checking for .dll and .exe files ... OK > * checking whether package 'longitudinalData' can be installed ... OK > * checking package directory ... OK > * checking for portable file names ... OK > * checking DESCRIPTION meta-information ... OK > * checking top-level files ... OK > * checking index information ... OK > * checking package subdirectories ... OK > * checking R files for non-ASCII characters ... OK > * checking R files for syntax errors ... OK > * checking whether the package can be loaded ... OK > * checking whether the package can be loaded with stated dependencies ... OK > * checking for unstated dependencies in R code ... OK > * checking S3 generic/method consistency ... OK > * checking replacement functions ... OK > * checking foreign function calls ... OK > * checking R code for possible problems ... OK > * checking Rd files ... OK > * checking Rd files against version 2 parser ... OK > * checking Rd cross-references ... OK > * checking for missing documentation entries ... OK > * checking for code/documentation mismatches ... OK > * checking Rd \usage sections ... OK > * checking examples ... OK > * checking PDF version of manual ... OK > > > R CMD build longitudinalData > * checking for file 'longitudinalData/DESCRIPTION' ... OK > * preparing 'longitudinalData': > * checking DESCRIPTION meta-information ... OK > * removing junk files > * checking for LF line-endings in source and make files > * checking for empty or unneeded directories > * building 'longitudinalData_0.5.tar.gz' > --- 8< > > Christophe > > __ > 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-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] is it possible to combine multiple barplots?
On Tue, Jun 09, 2009 at 04:39:29PM +0200, Titus von der Malsburg wrote: > > is there a way to plot both of them in one plot, so that the bars for > > value "1" (dataA: 3, dataB: 5) would appear side by side, followed by > > the bars for value 2 etc.? Oh, I see you want something different. I should've read your message more closely. I found this example in the gallery: http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=54 Maybe it's close enough to what you want to do. Titus __ 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] is it possible to combine multiple barplots?
You can use barchart in package lattice. Here's a rough sketch: library(lattice) dataA <- rep(1:4, c(3,2,2,4)) dataB <- rep(1:4, c(5,4,3,2)) da <- data.frame(table(dataA)) db <- data.frame(table(dataB)) da$cond <- "a" db$cond <- "b" colnames(da)[1] <- "data" colnames(db)[1] <- "data" d <- rbind(da, db) barchart(Freq~data|cond, d) Titus On Tue, Jun 09, 2009 at 04:23:32PM +0200, Philipp Schmidt wrote: > i am working with two sets of likert scale type (4 distinct values) data: > > dataA <- rep(1:4, c(3,2,2,4)) > dataB <- rep(1:4, c(5,4,3,2)) > > i can now (bar)plot both of these separately and compare the distributions. > > plot(table(dataA), type='h') > plot(table(dataB), type='h') > > is there a way to plot both of them in one plot, so that the bars for > value "1" (dataA: 3, dataB: 5) would appear side by side, followed by > the bars for value 2 etc.? > > thanks! > > best - P > > __ > 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-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] Isolating a single plot from plots produced simultaneously
Dear R-Help, I am using the 'mfp' package. It produces three plots (as I am using the Cox model) simultaneously which can be viewed together using the following code: fit <- mfp(Surv(rem.Remtime,rem.Rcens)~fp(age)+strata(rpa),family=cox,data=nearma,select=0.05,verbose=TRUE) par(mfrow=c(2,2)) plot(fit) They can be viewed separately but the return key must be pressed after each graph appears (Click or hit ENTER for next page). I'd like to isolate the second plot produced (the estimated functional form of the influence of age on the log relative hazard) so that I can use the 'points' function to add the linear predictors for the untransformed and the log-transformed models. In the usual situation one would produce a plot and then type: coxfitu <- coxph(Surv(rem.Remtime,rem.Rcens)~age+strata(rpa),data=nearma) points(coxfitu$linear.predictor,col=2) coxfitl <- coxph(Surv(rem.Remtime,rem.Rcens)~log(age)+strata(rpa),data=nearma) points(coxfitl$linear.predictor,col=3) Can anyone tell me how to isolate just the second plot produced? Thank you for your help, Laura __ 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] is it possible to combine multiple barplots?
i am working with two sets of likert scale type (4 distinct values) data: dataA <- rep(1:4, c(3,2,2,4)) dataB <- rep(1:4, c(5,4,3,2)) i can now (bar)plot both of these separately and compare the distributions. plot(table(dataA), type='h') plot(table(dataB), type='h') is there a way to plot both of them in one plot, so that the bars for value "1" (dataA: 3, dataB: 5) would appear side by side, followed by the bars for value 2 etc.? thanks! best - P __ 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] Writing Reports from R in Office Open XML format (ooxmlWeave?)
Yes. We will release a version in the next few weeks when I have time to wrap it all up. There is also a Docbook-based version that uses R extensions to Docbook for authoring structured documents. D. Tobias Sing wrote: Dear all, has someone implemented functionality for writing reports from R in Office Open XML format (*), similar to what odfWeave does for the ODF format of OpenOffice? It would be great to have a kind of "ooxmlWeave" at least for those of us who are forced to work in an MS ecosystem. (*) Office Open XML is the default, XML-based, file format for MS Word: http://en.wikipedia.org/wiki/Office_Open_XML Kind regards, Tobias __ 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-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] "R CMD check" does not find a mistake
Hi the list, I build a package. They was a mistake in it, but R CMD check did not find it. Is that normal ? Here is what Kurt gets (which is right, I did this mistake): --- 8< * checking for code/documentation mismatches ... WARNING S4 class codoc mismatches from documentation object 'LongData-class': Slots for class 'LongData' Code: id other time traj varName Docs: id time traj varName --- 8< But here is what I get : --- 8< R CMD check longitudinalData * checking for working pdflatex ... OK * using log directory 'C:/Documents and Settings/Christophe/Mes documents/Recher che/Trajectoires/kmeal/longitudinalData.Rcheck' * using R version 2.9.0 beta (2009-04-04 r48290) * using session charset: ISO8859-1 * checking for file 'longitudinalData/DESCRIPTION' ... OK * checking extension type ... Package * this is package 'longitudinalData' version '0.5' * package encoding: latin1 * checking package dependencies ... OK * checking if this is a source package ... OK * checking for .dll and .exe files ... OK * checking whether package 'longitudinalData' can be installed ... OK * checking package directory ... OK * checking for portable file names ... OK * checking DESCRIPTION meta-information ... OK * checking top-level files ... OK * checking index information ... OK * checking package subdirectories ... OK * checking R files for non-ASCII characters ... OK * checking R files for syntax errors ... OK * checking whether the package can be loaded ... OK * checking whether the package can be loaded with stated dependencies ... OK * checking for unstated dependencies in R code ... OK * checking S3 generic/method consistency ... OK * checking replacement functions ... OK * checking foreign function calls ... OK * checking R code for possible problems ... OK * checking Rd files ... OK * checking Rd files against version 2 parser ... OK * checking Rd cross-references ... OK * checking for missing documentation entries ... OK * checking for code/documentation mismatches ... OK * checking Rd \usage sections ... OK * checking examples ... OK * checking PDF version of manual ... OK R CMD build longitudinalData * checking for file 'longitudinalData/DESCRIPTION' ... OK * preparing 'longitudinalData': * checking DESCRIPTION meta-information ... OK * removing junk files * checking for LF line-endings in source and make files * checking for empty or unneeded directories * building 'longitudinalData_0.5.tar.gz' --- 8< Christophe __ 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] Writing Reports from R in Office Open XML format (ooxmlWeave?)
Dear all, has someone implemented functionality for writing reports from R in Office Open XML format (*), similar to what odfWeave does for the ODF format of OpenOffice? It would be great to have a kind of "ooxmlWeave" at least for those of us who are forced to work in an MS ecosystem. (*) Office Open XML is the default, XML-based, file format for MS Word: http://en.wikipedia.org/wiki/Office_Open_XML Kind regards, Tobias __ 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] rpart - the xval argument in rpart.control and in xpred.rpart
Usually 10-fold cross validation is performed more than once to get an estimate of the misclassification rate thus I thought "number of cross-validations" was different from the "number of cross-validation groups". So, if I want to perform 10-fold cross-validation more than once (say 5) in order to estimate the miscalssification rate I have to run xpred.rpart 5 times ? Thanks Paolo I have some problems in understanding the meaning of the xval argument in the two functions rpart.control and xpred.rpart. In the former it is defined as the number of cross-validations while in the latter it is defined as the number of cross-validation groups. It is the same thing. If xval=10 then the data is divided into 10 disjoint groups. A model is fit with group 1 left out and that model is used to predict the observations in group 1; then a model is fit with group 2 left out; then group 3, ... So 10 groups = 10 fits of the model. Actually I thought that in rpart.control Terry Therneau __ 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] rpart - the xval argument in rpart.control and in xpred.rpart
> I have some problems in understanding the meaning of the xval argument in > the two functions rpart.control and xpred.rpart. In the former it is defined > as the number of cross-validations while in the latter it is defined as the > number of cross-validation groups. It is the same thing. If xval=10 then the data is divided into 10 disjoint groups. A model is fit with group 1 left out and that model is used to predict the observations in group 1; then a model is fit with group 2 left out; then group 3, ... So 10 groups = 10 fits of the model. Terry Therneau __ 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] removing elements from a "unit" vector
On 6/9/2009 9:02 AM, baptiste auguie wrote: Dear list, I'm quite surprised by this, unit(1:5,"char")[-c(1:2)] #4char 3char # what's going on?? while I expected something like, c(1:5)[-c(1:2)] # 3 4 5 Note that, unit(1:5,"char")[c(1:2)] # 1char 2char # fine ?unit warns about unit.c for concatenating, but also says, It is possible to subset unit objects in the normal way (e.g., unit(1:5,"npc")[2:4]) Am I missing something obvious? No, I think the grid:::unit.[ method has a bug. It does the indexing of an object x by making sure the indices are integers, then class(x) <- NULL x <- x[(index - 1)%%this.length + 1] which assumes that index has positive entries. Duncan Murdoch Thanks, baptiste sessionInfo() R version 2.9.0 (2009-04-17) i386-apple-darwin8.11.1 locale: en_GB.UTF-8/en_US.UTF-8/C/C/en_GB.UTF-8/en_GB.UTF-8 attached base packages: [1] datasets stats grid graphics grDevices utils methods base __ 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-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] rpart - the xval argument in rpart.control and in xpred.rpart
Dear R users, I'm working with the rpart package and want to evaluate the performance of user defined split functions. I have some problems in understanding the meaning of the xval argument in the two functions rpart.control and xpred.rpart. In the former it is defined as the number of cross-validations while in the latter it is defined as the number of cross-validation groups. If I am correct this means that for xpred.rpart, xval denots the number V of one V-fold cross validation while for rpart.control it denotes the number of times that the cross-validation is performed. Am I right ? If so, how many subsets compose the learning sample when performing one of the xval cross-validations ? Is it a 10-fold cross-validation repeated xval=10 times ? Thank you for your help Paolo Paolo Radaelli Dipartimento di Metodi Quantitativi per le Scienze Economiche ed Aziendali Facoltà di Economia Università degli Studi di Milano-Bicocca Via Bicocca degli Arcimboldi, 8 20126 Milano Italy e-mail paolo.radae...@unimib.it Tel +39 02 6448 3163 Fax +39 02 6448 3105 __ 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] R command to join data.frames rows with identical keys?
An example: > schoolZone1_df <- data.frame(address=101, schoolZone="Sherman") > schoolZone2_df <- data.frame(address=108, schoolZone="Baker") > schoolZone_df <- rbind(schoolZone1_df, schoolZone2_df) > schoolZone_df address schoolZone 1 101Sherman 2 108 Baker > neighborhoodInfo1_df <- data.frame(address=101, squareFootage=2000, > lotsize=0.75) > neighborhoodInfo2_df <- data.frame(address=108, squareFootage=3000, > lotsize=1.25) > neighborhoodInfo_df <- rbind(neighborhoodInfo1_df, neighborhoodInfo2_df) > neighborhoodInfo_df address squareFootage lotsize 1 101 20000.75 2 108 30001.25 > merge(schoolZone_df, neighborhoodInfo_df, by="address") address schoolZone squareFootage lotsize 1 101Sherman 20000.75 2 108 Baker 30001.25 Titus __ 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] removing elements from a "unit" vector
Dear list, I'm quite surprised by this, unit(1:5,"char")[-c(1:2)] #4char 3char # what's going on?? while I expected something like, c(1:5)[-c(1:2)] # 3 4 5 Note that, unit(1:5,"char")[c(1:2)] # 1char 2char # fine ?unit warns about unit.c for concatenating, but also says, It is possible to subset unit objects in the normal way (e.g., unit(1:5,"npc")[2:4]) Am I missing something obvious? Thanks, baptiste sessionInfo() R version 2.9.0 (2009-04-17) i386-apple-darwin8.11.1 locale: en_GB.UTF-8/en_US.UTF-8/C/C/en_GB.UTF-8/en_GB.UTF-8 attached base packages: [1] datasets stats grid graphics grDevices utils methods base __ 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] R command to join data.frames rows with identical keys?
Have a look at the merge function.
"Merge two data frames by common columns or row names, or do other
versions of database _join_ operations."
Titus
On Tue, Jun 09, 2009 at 05:48:05AM -0700, Jason Rupert wrote:
>
> I've got two data.frames and, when certain "keys" match, I would like to add
> the column values from one data frame to the other data.frame.
>
> Below I list the two data.frames, i.e. neighborhoodInfo_df, and
> schoolZone_df. Based on the "address" key I would like to add the
> "schoolZone" key to the neighborhoodInfo_df data.frame.
>
> By any chance is there an R command to accomplish this in one or two steps.
> I think I could do this in a "for" loop or something, but think there is
> might be another way in R to accomplish it smarter.
>
> Thanks for any info.
>
>
> neighborhoodInfo1_df<-data.frame(address<-c(101),squareFootage<-c(2000),lotsize<-c(0.75))
>
> neighborhoodInfo2_df<-data.frame(address<-c(108),squareFootage<-c(3000),
> lotsize<-c(1.25))
>
> neighborhoodInfo_df<-rbind(neighborhoodInfo1_df, neighborhoodInfo2_df)
>
>
> schoolZone1_df<-data.frame(address<-c(101), schoolZone<-c("Sherman"))
>
> schoolZone2_df<-data.frame(address<-c(108), schoolZone<-c("Baker"))
>
>
> schoolZone_df<-rbind(schoolZone1_df, schoolZone2_df)
>
> __
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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] (no subject)
__ 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] R command to join data.frames rows with identical keys?
I've got two data.frames and, when certain "keys" match, I would like to add
the column values from one data frame to the other data.frame.
Below I list the two data.frames, i.e. neighborhoodInfo_df, and schoolZone_df.
Based on the "address" key I would like to add the "schoolZone" key to the
neighborhoodInfo_df data.frame.
By any chance is there an R command to accomplish this in one or two steps. I
think I could do this in a "for" loop or something, but think there is might be
another way in R to accomplish it smarter.
Thanks for any info.
neighborhoodInfo1_df<-data.frame(address<-c(101),squareFootage<-c(2000),lotsize<-c(0.75))
neighborhoodInfo2_df<-data.frame(address<-c(108),squareFootage<-c(3000),
lotsize<-c(1.25))
neighborhoodInfo_df<-rbind(neighborhoodInfo1_df, neighborhoodInfo2_df)
schoolZone1_df<-data.frame(address<-c(101), schoolZone<-c("Sherman"))
schoolZone2_df<-data.frame(address<-c(108), schoolZone<-c("Baker"))
schoolZone_df<-rbind(schoolZone1_df, schoolZone2_df)
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Re: [R] using regular expressions to retrieve a digit-digit-dot structure from a string
On Jun 9, 2009, at 6:44 AM, Mark Heckmann wrote:
Hey all,
Thanks for your help. Your answers solved the problem I posted and
that is
just when I noticed that I misspecified the problem ;)
My problem is to separate a German texts by sentences. Unfortunately I
haven't found an R package doing this kind of text separation in
German, so
I try it "manually".
Just using the dot as separator fails in occasions like:
txt <- "One January 1. I saw Rick. He was born in the 19. century."
Here I want the algorithm to separate the string only at the
positions where
the dot is not preceded by a digit. The R-snippets posted pick out
"1." and
"19."
txt <- "One January 1. I saw Rick. He was born in the 19. century."
gregexpr('(?<=[0-9])[.]',txt, perl=T)
[[1]]
[1] 14 49
attr(,"match.length")
[1] 1 1
But I just need it the other way round. So I tried:
strsplit(txt, "[[:alpha:]]\\." , perl=T)
[[1]]
[1] "One January 1. I saw Ric" " He was born in the 19. centur"
But this erases the last letter from each sentence. Does someone
know a
solution?
TIA
Mark
This is one of those rare? times where it might be nice for strsplit()
to have an option to retain the split regex at the end of each parsed
segment, rather than removing it.
There may be a better way, but trying to both avoid a loop over vector
indices and trying to stay with R functions that use .Internal() for
speed, you may be able to use something like this:
> strsplit(gsub("([[:alpha:]]\\.)", "\\1*", txt), "\\* *")
[[1]]
[1] "One January 1. I saw Rick." "He was born in the 19. century."
What I am essentially doing is to add an "*" to the ending of each
sentence (you can use other characters) such that strsplit() can split
on that character without affecting the rest of the sentence. So as
an intermediate result, you get:
> gsub("([[:alpha:]]\\.)", "\\1*", txt)
[1] "One January 1. I saw Rick.* He was born in the 19. century.*"
which then makes the strsplit() parsing a bit easier. Since both
strsplit() and grep() use .Internal()s, hopefully this would still be
reasonably fast. Note that I have strsplit() split on the "*" possibly
followed by one or more " ", which is required for mid-line splits.
HTH,
Marc Schwartz
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Re: [R] using regular expressions to retrieve a digit-digit-dot structure from a string
Gabor Grothendieck wrote:
> On Tue, Jun 9, 2009 at 3:04 AM, Wacek
> Kusnierczyk wrote:
>
>> Gabor Grothendieck wrote:
>>
>>> On Mon, Jun 8, 2009 at 7:18 PM, Wacek
>>> Kusnierczyk wrote:
>>>
>>>
Gabor Grothendieck wrote:
> Try this. See ?regex for more.
>
>
>
>
>> x <- 'This happened in the 21. century." (the dot behind 21 is'
>> regexpr("(?![0-9]+)[.]", x, perl = TRUE)
>>
>>
>>
> [1] 24
> attr(,"match.length")
> [1] 1
>
>
>
yes, but
gregexpr('(?![0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
# 2 5 9
>>> Yes, it should be:
>>>
>>>
>>>
gregexpr('(?<=[0-9])[.]', 'a. 1. a1.', perl=TRU
> E)
>
>>> [[1]]
>>> [1] 5 9
>>> attr(,"match.length")
>>> [1] 1 1
>>>
>>> which displays the position of every dot that is preceded
>>> immediately by a digit. Or just replace gregexpr with regexpr
>>> if its intended that it match only one.
>>>
>>>
>> i guess what was needed was something like
>>
>>gregexpr('(?<=\\b[0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
>># 5
>>
>> which won't work, however, because pcre does not support variable-width
>> lookbehinds.
>>
>
> No, what I wrote was what I intended. I don't think we are
> discussing the answer
> at this point but just the interpretation of what was intended.
which amounts to discussing whether the answer is appropriate ;)
> You
> are including
> the word boundary in the question and I am not.
indeed, and i think this was essential. but irrespectively of whether
it really was or not, this sort of problem shows the insufficiency of a
lookbehind, and illustrates the use of the \K operator, so it will
hopefully be easier for the op and others to design the right pattern in
similar future cases.
vQ
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Re: [R] using regular expressions to retrieve a digit-digit-dot structure from a string
On Tue, Jun 9, 2009 at 3:04 AM, Wacek
Kusnierczyk wrote:
> Gabor Grothendieck wrote:
>> On Mon, Jun 8, 2009 at 7:18 PM, Wacek
>> Kusnierczyk wrote:
>>
>>> Gabor Grothendieck wrote:
>>>
Try this. See ?regex for more.
> x <- 'This happened in the 21. century." (the dot behind 21 is'
> regexpr("(?![0-9]+)[.]", x, perl = TRUE)
>
>
[1] 24
attr(,"match.length")
[1] 1
>>> yes, but
>>>
>>> gregexpr('(?![0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
>>> # 2 5 9
>>>
>>
>> Yes, it should be:
>>
>>
>>> gregexpr('(?<=[0-9])[.]', 'a. 1. a1.', perl=TRU
E)
>>>
>> [[1]]
>> [1] 5 9
>> attr(,"match.length")
>> [1] 1 1
>>
>> which displays the position of every dot that is preceded
>> immediately by a digit. Or just replace gregexpr with regexpr
>> if its intended that it match only one.
>>
>
> i guess what was needed was something like
>
> gregexpr('(?<=\\b[0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
> # 5
>
> which won't work, however, because pcre does not support variable-width
> lookbehinds.
No, what I wrote was what I intended. I don't think we are
discussing the answer
at this point but just the interpretation of what was intended. You
are including
the word boundary in the question and I am not. I think its also possible that
regexpr is what is wanted, not gregexpr, but at this point I think the
poster has
enough answers that he can complete it himself by considering what he wants
and using one of ours or a suitable modification.
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Re: [R] Splicing factors without losing levels
On Tue, Jun 09, 2009 at 11:23:36AM +0200, ONKELINX, Thierry wrote:
> For factors, you better convert them first back to character strings.
>
> splice <- function(x, y) {
> x <- levels(x)[x]
> y <- levels(y)[y]
> factor(as.vector(rbind(x, y)))
> }
Thank you very much, Thierry!
I failed to mention something important in my last mail: x and y have
the same levels. (I assume that the integer to level name mapping of
a factor defines its class and that it only makes sense to combine
factors of the same class.)
Say
> x <- factor(c(2,2,4,4), levels=1:4, labels=c("a","b","c","d"))
then
> x
[1] b b d d
Levels: a b c d
> as.integer(x)
[1] 2 2 4 4
but
> splice(x,x)
[1] b b b b d d d d
Levels: b d
> as.integer(splice(x,x))
[1] 1 1 1 1 2 2 2 2
I'd like to have a splice function that retains the level to label
mapping. One candidate for a solution is:
splice <- function(x,y) {
xy <- as.vector(rbind(x, y))
if (is.factor(x) && is.factor(y))
xy <- factor(xy, levels=1:length(levels(x)), labels=levels(x))
xy
}
However, this relies on assumtions about the implementation of
factors that are neither mentioned nor guaranteed in the man page:
Levels are underlyingly integers starting from one and going to
length(levels). levels(x) gives me the labels of these integers in an
order corresponding to 1:length(levels(x)).
Without these assumptions I see no way to recover the integer to level
name mapping for levels that are defined in a factor but do not occur.
I'd be happy if somebody could clarify this issue!
Titus
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