similar to: Using paste to create and evaluate a variable expression

Hi, I am using R a lot to make plots relating to radioactivity, I am often using expression() to label the plots with nuclide names written with superscripts, e.g. expression(paste("Releases of ", { }^{99},Tc," (TBq/year)"))->ywtext But, is there any simple way to change the number and name of the nuclide through a variable? I tried nuccode=expression({ }^{99},Tc)

It is very wierd... Can some of you confirm the following behavior ? It is a new bug (feature ?) which was not yet in 0.49 ... noquote <- function(obj) { ## constructor for a useful "minor" class if(!inherits(obj,"noquote")) class(obj) <- c(class(obj),"noquote") obj } "[.noquote" <- function (x, subs) structure(unclass(x)[subs], class =

The following recursion is about 120 times faster in C#. I know R is not known for its speed with recursions but I'm wondering if anyone has a tip about how to speed things up in R. #"T" is a vector and "m" is a number between 1 and sum(T) A <- function(T,m) { lt <- length(T) if (lt == 1) { if (0 <= m & m <= T[1]) { return(1) } else { return(0) }

Dear All I am trying to convert from the type "expression" to the type "numeric". The following works: > x <- expression(6.2) > as.numeric(as.character(x)) [1] 6.2 However, the following does not work: > x <- expression(62/100) > as.numeric(as.character(x)) [1] NA Warning message: NAs introduced by coercion Any idea about how to deal with the second case?

#I have a dataframe called "tests" that contain "character expressions". These characters are rules that use data from within another dataframe. Is there any way within R I can access the rules in the dataframe called tests, and then "evaluate" these "rules"? #An example may better explain what I am trying to accomplish: tests <-

Anyone know if coin can run a permutation test based on a (user-defined) statistic other than the mean difference? The function independence_test does the permutation t-test via difference in means. I'm wondering if it's possible to use independence_test to run a permutation test for some other statistic than the difference in means. For example, I'd like to run a permutation test

Dear list-subscriber, in the process of writing a general code snippet to extract coefficients in an expression (in the example below: 0.5 and -0.7), I stumbled over the following peculiar (at least peculiar to me:-) ) sorting behaviour of the function all.names(): > expr1 <- expression(x3 = 0.5 * x1 - 0.7 * x2) > all.names(expr1) [1] "-" "*" "x1"

The source to the noquote() function looks like this: noquote <- function(obj, right = FALSE) { ## constructor for a useful "minor" class if(!inherits(obj,"noquote")) class(obj) <- c(attr(obj, "class"), if(right) c(right = "noquote") else "noquote") obj } Notice what happens with right =

Dear list, I have a spacialPolygonDataFrame where variables were unnecessarily imported as factors. So I am trying to unfactor variables from spatialPolygonDataFrame at data with a loop for (i in (1:length(names( spatialPolygonDataFrame)))){ command<-paste("spatialPolygonDataFrame$names(spatialPolygonDataFrame at data[",i,"])<-as.character(

Hi to all I use a loop to plot 9 different histograms of 9 different transformations of one dataset (x). I want to label the histograms with the mathematical expression of each transformation (e.g. x^3). For that I've prepared a vector with the labeling names for "expression". > trans.expr <- c("x^3", "x^2", "x", "frac(1,x)",

I currently have a program that automates 2-way ANOVA on a series of endpoints, but before the ANOVA is carried out I want the code to test the assumptions of normality and equal variance and report along with each anova result in the output file.  How can I do this? I have pasted below the code that I currently use.   library(car) numFiles = x #

Hey, everybody--- Ignorance CAN be bliss, at least for a while, but, .... Just so you know... A week or two ago, some upgrades to the expression parser (you know, the expressions you put in $[ ... ] in your extensions.conf file) that I submitted, have been merged into the CVS HEAD of the source. Hopefully, for around 99.9% of you, it won't make any difference to you. The Makefile has also

Dear R fellows, Assume I define a <- expression(fn+tp) sen <- expression(tp/a) Now I'd like to know, which variables are necessary for calculating sen all.vars(sen) This results in a vector c(tp,a). But I'd like all.vars to evaluate the sen-object down to the ground level, which would result in a vector c(tp,fn) (because a was defined as fn+tp). In other words, I'd like

Hello, I'm having difficulty passing an object name to a lapply function. Can somebody tell me the trick to make this work? #Works T13702 <- TRACKDATA[["13702.xls"]][["data"]] min(unlist(lapply(list(T13702), function(x) mdy.date(x[1, 2], x[1, 1], x[1, 3])))) 16553 #Works d<-2 assign(paste("T",substr(names(TRACKDATA)[d],1,(nchar(names(TRACKDATA)[d]

This may have been asked before, but is there an elegant way to check whether an variable/argument passed to a function is a "parse tree" for an (unevaluated) expression or not, *without* evaluating it if not? Currently, I do various rather ad hoc eval()+substitute() tricks for this that most likely only work under certain circumstances. Ideally, I'm looking for a isParseTree()

Hi R, When I use the below to write the text file try=data.frame(rep("a",5), rep("b",5)) write.table(try,"z:\\try.txt",row.names=F,col.names=F,sep="\t") the output contains two columns with quotes! Is there a way to write without quotes? I tried try[,1]=noquote(try[,1]) try[,2]=noquote(try[,2]) Thank you, Regards, Ravi Shankar

I haven't been able to get anywhere tracking this down. It seems that c.noquote() does something strange with its third (and subsequent) parameters: R-2.7.0 under NetBSD, R-2.6.0 under Solaris, and R-2.8.0 (unstable) (2008-06-10 r45893) under WinXP, I get: > c(noquote('z'), 'y', 'x', '*') [1] z y x * x * > or: > c(noquote('z'),

Hello all, I've run into what I bet is a silly problem; however, I've been trying to get around it now for a couple weeks and every time I think I have the answer it still doesn't work. So I apologize in advance if this is painfully obvious, but I've run out of ideas and would really appreciate any input. My situation is this, I'm importing a number of tab

Hi, I'm generating the name of the variable with paste function and then using that variable name further to get the specific position value from the data.frame, here is the snippet from my code: modelResults <- extractModelParameters("C:/PilotStudy/Mplus_Input/Test", recursive=TRUE) #extractModelParameters reads all the output files from the Test folder and create the

G'day all, I am trying to use a string as an argument in a selection but things are not working as I expect, seems the selection is not seeing the expanded string and I do not know how to make it. Perhaps the noquote class value that is returned is the problem. Here is an example. > selection #this is my string [1] "attackprogress$Se=='Toona ciliata [19825: JMM35]'"