similar to: referencing headers to extract components of a data frame

Dear all, I have the following data frame: > dat V1 V2 V3 V4 V5 V6 V7 V8 V9 1 1 AAAACACCCACCCCCCCCCCCCCCCCCCCCCCCC 9.0 18 12.00 18.0 15.0 12.0 6.0 2 1 ACGATACGGCGACCACCGAGATCTACACTCTTCC 18.0 8 12.00 18.0 15.0 12.0 18.0 3 1 ACTACTGCTCCCCCCCCACTCCCCCCCCCCCCCC 15.0 8 12.00 12.0 18.0 12.0 12.0 4 1 ACTTATACGGCGACCACCGAGATCTACACTCTTT 15.0

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 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 =

In S, NULL and list() are not the same. In R they are (I think). --------------------------------------------------- At least, is.list(NULL) #-> 'F' in S; 'TRUE' in R Yes: I had an instance where this broke correct S code: match(c("xlab","ylab"), names(list(...))) when '...' is empty, gives an error in R, but gives c(NA,NA) in S.

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 #

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, 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]

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

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

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, I would like to check which rows of 'types.prev' matrix pop up in 'types', following R in-built procedure. I tried 'match' function but it works in case of the one dimensional vectors. Will appreciate any suggestions. best, robert > types edate K [1,] 20060819 12.5 [2,] 20060819 17.5 [3,] 20060819 22.5 [4,] 20070217 12.5 [5,] 20060617 10.0

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)

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 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) {

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])

Hello All: I have a vector of data, z > z [1] 0.1 0.1 0.1 0.1 0.2 0.2 0.3 0.3 0.3 0.4 0.5 0.5 0.5 0.7 0.7 0.7 0.9 0.9 1.1 [20] 1.1 1.2 1.3 1.4 The first 4 elements have values of 0.1 followed 2 elements with values 0.2. When I invoke rank(z), I expected to get (1+2+3+4)/4 = 2.5 for the first 4 elements in the ranking and (5+6)/2 = 5.5 for elements 5 and 6. But what I do

Hi, I plan to upgrade/replace my somewhat crippled and outdated samba 4.7.4 domain controller. The OS is an openSUSE-Leap-42.3 which had no packages for a samba-ad-dc. These packages have been introduced in successor openSUSE releases starting with Leap-15.0. Leap-15.0 comes with samba 4.7.11. So I set up a new Leap-15.0 host and joint it as a dc controller. I set up the sysvol replication

My code below makes a data frame with columns for date, time, and price. Time on each date runs from 1 to 4. I'd like to add a new column "ts$closingprice", which would have the closing price for that date. To find the closing price, I'd like to take the price in the row having the greatest time value for each date. Then I'd like to fill that closing price into the

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