similar to: correlation help

Hello, everybody Say I have nm1 <- c(rep(1,10), rep(0,10)) then I can do: diff(nm1) to see where I have shift in value but what if I have nm2 <- c(rep("SPZ8", 10), rep("SPX9", 10)) how can I produce the same ouput as diff(nm1) does, that is zeros everywhere except for one place where SPZ8 changes to SPX9 (there should be 1 there)? What if I have a matrix of characters

Hi Listers, This has a simple answer but it has been eluding me nonetheless. I have been building a PCA plot from scratch with the ability to plot predefined groups in different colors. This has worked fine but when I try to get a polygon drawn around each of the groups it is not recognising my colour file correctly and is only printing the first colour in the file....code is below

Hi. I am trying to use the linear optimizer from package lpSolve in R 2.4.1 on Windows XP (Version 5.1). The problem I am trying to solve has 2843 variables (2841 integer, 2 continuous) and 8524 constraints, and I have 2 Gb of memory. After I load the input data into R, I have at most 1.5 Gb of memory available. If I start the lp with significantly less memory available (say 1 Gb), I get

Hi. I have defined a function, f1, that calls another function, f2. Inside f1 an intermediate variable called nm1 is created; it is a matrix. f2 takes a matrix argument, and I defined f2 (schematically) as follows: f2<-function(nmArg1=nm1,...){nC<-ncol(nmArg1); ... } so that it expects nm1 as the default value of its argument. f1 is defined (schematically) as:

Dear R-users, I am trying to reshape the DF "dat2" in the "long" format, but can't figure out how to use the "split"-option: > dat2 a.1995.z b.1995.z a.1996.z var 1 100.00000 100.00000 100.00000 Neue Anlagen insgesamt 2 40.09904 23.60890 40.88960 Neue Ausr??stungen 3 59.90096 76.39110 59.11040 Neue Bauten This

Hello, colnames seems to be not optimized well for data.frame. It escapes processing for data.frame in if (is.data.frame(x) && do.NULL) return(names(x)) but only when do.NULL true. This makes huge difference when do.NULL false. Minimal edit to `colnames`: if (is.data.frame(x)) { nm <- names(x) if (do.NULL || !is.null(nm)) return(nm) else

Hi people, If there is a fn "colMeans" why isn't there a "colSDs" etc Thanks, Phil. -- Philip Rhoades Pricom Pty Limited (ACN 003 252 275) GPO Box 3411 Sydney NSW 2001 Australia Mobile: +61:0411-185-652 Fax: +61:2:8923-5363 E-mail: pri at chu.com.au -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

Dear Bert, thank you for suggestion. I am aware of R-help function. I must apologize, my earlier question could lead to assumptions otherwise. As you can see below, I only get Std. error but not lower and Upper CIs. I was wondering if there is another argument or method, could give CIs which I cannot find anywhere. Thanks Adrian > Nsurv <-

Dear Group: I have 72 files (.txt). Each file has 2 columns and column 1 is always identical for all 70 files. Each file has 90,799 rows and is standard across all files. I want to create a matrix 40(rows) x 70 columns. I tried : temp = list.files(pattern="*.txt") named.list <- lapply(temp, read.delim) library(data.table) files.matrix <-rbindlist(named.list) >

Dear all, Myself Vijaykumar Muley working as senior research fellow. By training I am a computational biologist with not a strong knowledge of statistics. I have done some analysis which is explained as follows, I have 10340 (X) profiles of binary vectors with same length(N=845), I will call then "gene profiles" for example... v1 v2 v3 v4.....vN a 1 0 1 0 1 b 0

I need to calculate a large number of t statistics, and would like to do so via matrix operations. So far I have figured out a way to calculate the mean of each row of the matrix: d <- matrix(runif(100000,1,10), 1000, 10) # some test data s <- rep(1,ncol(d)) # a sum vector to use for matrix multiplication means <- (d%*%s)/ncol(d) This is at least 1 order of magnitude faster than

Did you not notice the conf.type = "none" argument to your survfit call and the associated documentation in the survfit help? -- Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Sat, Aug 26, 2017 at 5:18 PM, Adrian Johnson

I am using the R 2.2.1 in a Windows XP environment. I have a dataframe with 12 columns and 1,000 rows. (Some of the rows have 1 or fewer values.) I am trying to use rowVars to calculate the variance of each row. I am getting the following message: ?Error in na.remove.default(x) : length of 'dimnames' [1] not equal to array extent? Is there a good work-around?

Hi I am not sure hiw to get thr CI from summary function on the survfit object. I can get the percent survival for 2 years from survfit object but I dont get Confidence intervals Could anyone suggest a hint Thanks Adrian [[alternative HTML version deleted]]

Is it possible to compute the variance of every column in a matrix without a loop? -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch

??? Both ?survit.object (linked in the see also section of ?survfit) and ?summary.survfit give you this information. Do you not know how to use R's help faciities -- in which case you should learn them now; see ?help -- or have I misunderstood your query? Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it."

Hello group, my question is deciding what test would be appropriate for following question. An experiment 'A' yielded 3200 observations of which 431 are significant. Similarly, using same method, another experiment 'B' on a different population yielded 2541 observations of which 260 are significant. There are 180 observations that are common between significant observations of A

Hi: I am looking for some help in making two boxplots next to each other. I have a data like this: N1 T1 N2 T2 N3 T3 N4 T4 ... Nn Tn 7 8.2 4 5 8 10 4 5 ..... 10 11 I want to have box plot for all Normal samples (N1,N2,N3,N4,,,,Nn) and another box plot for all tumors (T1,T2,T3,T4,...Tn). I have data in a numeric class. If data is represented as N1

Hi everyone. Probably this is statistical question rather than an R, but it involves packages from R I am asking here since I am unable to find an answer. In the parametric modeling packages like glmnet, lasso etc......., we are able to obtain the coeffcients that have entered the model. for eg in glmnet if we are working on a dataset containing 15 variables the coeffecient parameters output

Full_Name: Franc Brglez Version: R 2.9.1 GUI 1.28 Tiger build 32-bit (5444) OS: MacOSX -- 10.6.2 Submission from: (NULL) (24.148.163.114) I am demonstrating what may be a bug or my lack of experience. Please review as it would help to hear from someone. MANY THANKS -- Franc Brglez The function "binS2binV" returns what I consider a wrong value -- see the terminal output binS2binV =