similar to: write text file as output without quotes

Hello R users, i'm wondering how to trigger variable creation. Whenever a variable is created i want my own function myFun(...) to be started. if (exists("x")) {rm(x)} # after removal of x # any of these calls x<-10 # should call myFun x=10 # should call myFun assign(x,10) # should call myFun etc.

Dear all, I am attempting to perform a calculation which counts the number of positive (or negative) values based on the sample mean (on a per-row basis). If the mean is>0 then only positive values should be counted, and if the mean is <0 then only negative values should be counted. In cases where the mean is equal to zero, the value -99999 should be returned. The following is an example

Hi R users, I have a dataframe in the below format xyz 01/03/2007 15.25 USD xyz 01/04/2007 15.32 USD xyz 01/02/2008 23.22 USD abc 01/03/2007 45.2 EUR abc 01/04/2007 45.00 EUR

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 =

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 =

Hi R, I have a list > class(pp2) [1] "list" > length(pp2) [1] 1244 It is in the below format RIC Trade.Date Close.Price Currency.Code Convertion.Rate New.Price ABCD.SZ 2008/02/29 15.30 CNY 0.1408 2.154240 ABCD.SZ 2008/01/31 15.27 CNY 0.1392 2.125584 ABCD.SZ 2007/12/31 14.88 CNY 0.1371 2.040048

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 #

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]

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'),

Works fine for me. What do you object to in the following? Calling the above df "d", > dm <- as.matrix(d) > dm Sub_Pathways BMI_beta SAT_beta VAT_beta 1 "Alanine_and_Aspartate" " 0.23820" "-0.02409" " 0.94180" 2 "Alanine_and_Aspartate" "-0.31300" "-1.97510" "-2.22040" 3

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]'"

Hi, I apologize if the solution is right in front of me, but I can't find anything on how to convert a class of 'noquote' to 'matrix'. I've tried as.matrix and I've tried coercing it by 'class(x)<-matrix', but of course that didn't work. I've been using the function 'symnum' which returns an object of class 'noquote'. Here is an

Dear mailing list members, My question is maybe very basic, but I could not find the solution. I would like to do the following things 1) colnames(V1)[2] <- par$V2[1] colnames(V2)[2] <- par$V2[2] colnames(V3)[2] <- par$V2[3] ... colnames(V37)[2] <- par$V2[37] 2) V1 <- V1[,-1] V2 <- V2[,-1] V3 <- V3[,-1] ... V37 <- V37[,-1] 3) ms <- merge(V1,V2) ms <- merge(ms,V3)

Hi everybody, I would like to count how many times names in list L, nombreL, apear in list C, nombreC. Can I improve the next program? cuenta <- 0 topL <- length(nombreL) topC <- length(nombreC) for (i in 1:topL) { for (j in 1:topC) { k <- grep(noquote(nombreL[i]),nombreC[j])

Dear all, I am trying to read in and assign data from 50 tables in an automated fashion. I have the following code, which I created with the help of textbooks and the internet, but it only seems to read in the final data file over and over again. For example, when I type:> table_1951 I get the same values in the table as when I type> table_2000 despite the values in the source tables

Dear XpeRts, I prepared a no qoute Character string by the following command s<-noquote(paste (b1, collapse=",")) where, b1 is the vector of 24 intergers. > dput(b1) c(1L, 2L, 6L, 7L, 12L, 16L, 17L, 20L, 21L, 23L, 25L, 34L, 46L, 48L, 58L, 64L, 65L, 68L, 82L, 97L, 98L, 101L, 113L, 115L) > dput(s)

Hi, I have a string that I want to use in a variable call. How can I remove the quotes and/or the string properties of the string to use it in a variable call? Here's an example: library(lme) fm2 <- lme(distance ~ age, data = Orthodont, random = ~ 1) summary(fm2) I want to update the above regression to include new predictors according to what is in a string: predictors <-

Dear list, I'm using the brew package to generate a report containing various plots. I wrote a function that creates a plot in png and pdf formats, and outputs a suitable text string to insert the file in the final document using the asciidoc syntax, <% tmp <- 1 makePlot = function(p, name=paste("tmp",tmp,sep=""), width=300) {

I'm new to R (previously used SAS primarily) and I have a genetics data frame consisting of genotypes for each of 300+ subjects (ID1, ID2, ID3, ...) at 3000+ genetic locations (SNP1, SNP2, SNP3...). A small subset of the data is shown below: SNP_ID SNP1 SNP2 SNP3 SNP4 Maj_Allele C G C A Min_Allele T A T G ID1 CC GG CT AA ID2 CC GG CC AA ID3 CC GG nc AA

Is it possible to use "paste" to write out an expression and evaluate it? Suppose I want to add two vectors X1 and X2, defined as follows: X1 <- 1:6 X2 <- 6:1 If I write the following it looks like what I want but is a character: noquote(paste(paste("X", 1, sep = ""), paste("X", 2, sep = ""), sep = "+")) Is there a way to tell R