similar to: coercing factors to matrix() --> num/char ? --

Does R, or will R, be integrating the changes to the Chambers/Lucent S language under their version 5.0? If not already, then when? John Thaden Little Rock, Arkansas, USA -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe"

The old factor() wars...... {{maybe don't just report that factor()s are broken by design..}} If ff is a factor, e.g., ff <- as.factor(rep(1:2,3)) f2 <- ff; levels(f2) <- c("Lo","Hi") 1) we don't allow {in R as in S-plus 3.x} arithmetic on factors, i.e. 1 + ff gives an error, and I think most of us believe this is as desired. {{if one

I use the step() function occasionally, and I think I understand its objective, proper use, and limitations. Now I see stepwise model selection being used in what seems to be an unusual way, and I wonder if it is right or wrong. May I describe? Genetic mapping tries to find where in an animal's genome are genetic elements that influence a particular physical trait. Say there are 100

Hi there, I'm running R 2.0.0 on Windows 95. I'm trying to coerce a column of factors within a data frame to numeric. This is not a problem with a vector, but I can't find a way to index a column within a data frame to achieve this. All the examples from 'An introduction to R', 'S-plus 6 programmers guide', etc, use simple vectors. I'm sure I'm missing

The documentation for assignment says: In all the assignment operator expressions, 'x' can be a name or an expression defining a part of an object to be replaced (e.g., 'z[[1]]'). A syntactic name does not need to be quoted, though it can be (preferably by backticks). But the implementation allows assignment to a character string (i.e. not a name), which it

Dear R-helpers, I have found a slightly annoying problem when trying to plot lines on graphs. I first created my data using tapply, thus:- y1=as.vector(fit1$coef$random$id) x1=tapply(o1,id,median,na.rm=T) x2=tapply(o2,id,median,na.rm=T) #then I plot the data, thus:- plot(x1[x2==0],y[x2==0]) #if I now try to fit the linear regression, R 'hangs up'

My dear R-helpers: I am a novice in R and have the following text string manipulation question. Is there a function that performs the job described below? Say, wanted_output <- c("ab", "cd", "ef") #the function_wanted can generate c("ab", "cd", "ef") using ab cd ef as the single input argument wanted_output <- function_wanted(ab

Hello Im struggling on something... I have one continuous variable (A), and I need to explain it with 4 factors, and maybe one continuous covariate. And of course, my variable A is not normal at all (it's a duration in seconds, whole numbers). What can I do? I would know how do deal with it if I had one factor, maybe two, but in that case Im really not sure what I am supposed to do to

Hi all, I read in a column which looks like "chr1:000889594-000889638", and need to break them into three columns like "chr1:", "000889594" and "000889638". How shall I do in R. Thanks a lot for your suggestions! Bill

Dear mailing list I have a data that have 20,000 rows and 20 columns. Io wonted to extract the 10th row only. Example the 10th, 20th, 30th 40th…..20000 th. can you please help me how do I do that.Than kyou. Example is below. Inpute: AG GG GG AG CC CC CC CC CT CC CT CT GG GG GG GG CC CC CC CC GG GG GG GG CC CC CC CC GG CG CG GG GG GG GG GG *CC CC CC CC* AA AG AG AA AA AA AA AA GG AG AG GG GG AG AG

Dear All I have a data.frame like that: structure(list(name = c("Gga_rs10722041", "Gga_rs10722249", "Gga_rs10722565", "Gga_rs10723082", "Gga_rs10723993", "Gga_rs10724555", "Gga_rs10726238", "Gga_rs10726461", "Gga_rs10726774", "Gga_rs10726967", "Gga_rs10727581", "Gga_rs10728004",

Hello! I am writing some numbers and character vectors to an ascii file and would like to get rid of [1] ... as shown bellow (a dummy example) R> runif(20) [1] 0.653574 0.164053 0.036031 0.127208 0.134274 0.103252 0.506480 0.547759 [9] 0.912421 0.584382 0.987208 0.996846 0.666760 0.053637 0.327590 0.370737 [17] 0.505706 0.412316 0.887421 0.812151 I have managed to work up to remove quotes

Hi, I am trying to split multiple columns. One column works just fine, but I want to do it for multiple columns??? Example > a ID V2 V3 V4 V5 V6 V7 V8 V9 V10 1 PBBA0644 -- GG AA -- AA -- AA GG GG 2 PBBA1010 -- GG AA -- AA -- AA GG GG 3 0127ATPR -- GG AA -- AA -- AA GG GG 4 0128EHAB -- GG AA -- AG -- AA AG GG 5 PBBA0829 -- GG AA -- AA -- AA GG AG

Something has changed in the S4 default for coercing. Am I now suppose to use setAs, or is there something else I should do to make this work: R version 2.7.0 (2008-04-22) .... > require("methods") > setClassUnion("OptionalPOSIXct", c("POSIXct", "NULL")) [1] "OptionalPOSIXct" > setClass("TSmetax",

Dear list, I am trying to evaluate how many elements in a vector equal a certain value. The vectors are the columns of a data.frame, read in using read.table(): > dim(data) [1] 2600 742 > data[1:5,1:5] SNP001 SNP002 SNP003 SNP004 SNP005 1 GG AA TT TT GG 2 GG AA TC TT GG 3 GG AC CC TT GG 4 AG AA TT TT GG 5

I HAVE A VERY BIG DATA OF 67 COLMS AND 25000 ROWS AND WOULD LIKE TO TRANSPOSE IT THE R HELP WAS NOT ENOUGH INFORMATION BECOUSE I AM NOT A PROGRAMMER AND FIRST TIME R USER. SO CAN YOU GIVE SOME HINTS OF CODING, AA TT GG GG CC AA TT GG GG CC AA TT GG GG CC AA TT GG GG CC AA TT GG GG CC TO AA AA AA AA AA TT TT TT TT TT GG GG GG GG GG GG GG GG GG GG CC CC CC CC CC [[alternative HTML

Dear all, Coercing a logical vector to a numeric one is easy. The as.numeric function is used. However what do we use when we have a matrix or an array? Sumona

# Your mailer is set to "none" (default on Windows), # hence we cannot send the bug report directly from R. # Please copy the bug report (after finishing it) to # your favorite email program and send it to # # r-bugs@biostat.ku.dk # ###################################################### In version 0.99.0 for Windows NT I get the following: > .Random.seed [1] 0 21012

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

Hello, I have the following code using rbinom, but I don't understand what *"+1"* means in the code. Could someone help? Thanks so much, > X1<-c("A","B")[rbinom(n,1,0.6)+1] > X2<-c("C","D")[rbinom(n,1,0.1)+1] -- View this message in context: http://r.789695.n4.nabble.com/a-quick-question-about-rbinom-tp4262977p4262977.html Sent