similar to: Variable passed to function not used in function in select=... in subset

Hallo! I have a problem with fitting data with nls. The first example with y1 (data frame df1) shows an error, the second works fine. Is there a possibility to get a fit (e.g. JMP can fit also data I can not manage to fit with R). Sometimes I also got an error singularity with starting parameters. # x-values x<-c(-1,5,8,11,13,15,16,17,18,19,21,22) # y1-values (first data set)

Hi, How do I tell an if statement to generate two seperate outputs. E.g If X>5 I want to create df1 and df2: if (X>5) {df1<-c(4,5,6,7,8) AND df2<-c(9,10,11,12,13)} Thanks, James -- View this message in context: http://www.nabble.com/If-statement-generates-two-outputs-tp22650844p22650844.html Sent from the R help mailing list archive at Nabble.com.

Hi, I am interested in generating two different optimized versions of a function into a single binary, so that I can dynamically select the function to execute at runtime. Can anyone shed some insight into the difficulties in achieving this in the LLVM infrastructure? Thank you! vj. -------------- next part -------------- An HTML attachment was scrubbed... URL:

Hello Group, I'm trying to learn R and am having a problem getting output from a function I'm trying to write. The problem is clearly one of scope, but I can't find the documentation that tells me how to get around the issue. Here is an example of my problem. testfunc<-function(x) { y<-10 print(y) print(x) } testfunc(4) The variables x and y are accessible during execution

It sounds as thought you are interested in Hoyt's Anova which is a form of generalizability theory. This is usually estimated using by getting the variance components from ANOVA. > -----Original Message----- > From: r-help-bounces at stat.math.ethz.ch > [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Karl Knoblick > Sent: Tuesday, May 16, 2006 6:10 AM > To: r-help at

given what ?identical says, i find the following odd: x = 1:10 y = 1:10 all.equal(x,y) [1] TRUE identical(x,y) [1] TRUE y[11] = 11 y = y[1:10] all.equal(x,y) [1] TRUE identical(x,y) [1] FALSE y [1] 1 2 3 4 5 6 7 8 9 10 length(y) [1] 10 looks like a bug. platform i686-pc-linux-gnu arch i686 os linux-gnu system

Hallo! GENERAL QUESTION: I'm trying to change the tick marks of the x-axis in a grouped data plot (nlme). CONCRETE EXAMPLE: In the example (see below) I want the x-axis to have tick marks at 0, 6, 12, 18, 24. How can I do this? WHAT I TRIED I tried "normal" methods like axis(...) but this does not work with this plot. And I also tried xlim=c(0,24) but the ticks are unchanged and

the following is a trivialized version of some functional code i tried to use in r: (funcs = lapply(1:5, function(i) function() i)) # a list of no-parameter functions, each with its own closure environment, # each supposed to return the corresponding index when applied to no arguments sapply(funcs, function(func) func()) # supposed to return c(1,2,3,4,5) there is absolutely nothing unusual in

Hello! I want to calculate the intra- and interrater reliability of my study. The design is very simple, 5 raters rated a diagnostic score 3 times for 19 patients. Are there methods/funtions in R? I only found packages to calculate interrater variability and intraclass correlation coefficients for matrices of n*m (n subjects, m raters) - I have n subjects, m raters and r repetitions. Can

Following Duncan's suggestion, I forward the below to R-devel. vQ -------- Original Message -------- Subject: Re: [R] Randomly remove condition-selected rows from a matrix Date: Fri, 02 Jan 2009 10:34:52 -0500 From: Duncan Murdoch <murdoch at stats.uwo.ca> To: Wacek Kusnierczyk <Waclaw.Marcin.Kusnierczyk at idi.ntnu.no> CC: R help <R-help at stat.math.ethz.ch>

Full_Name: Wacek Kusnierczyk Version: 2.8.0 and 2.10.0 r48242 OS: Ubuntu 8.04 Linux 32 bit Submission from: (NULL) (129.241.110.161) In the following code: duplicated(data.frame(), incomparables=NA) # Error in if (!is.logical(incomparables) || incomparables) .NotYetUsed("incomparables != FALSE") : # missing value where TRUE/FALSE needed the raised error is clearly not the

Karl, If you use one of the specialized packages to calculate your ICC, make sure that you know what you're getting. (I haven't checked the packages out myself, so I don't know either.) You might want to read David Futrell's article in the May 1995 issue of Quality Progress where he describes six different ways to calculate ICCs from the same data set, all with different

m = matrix(1:4, 2) apply(m, 1, cat, '\n') # 1 2 # 3 4 # NULL why the null? vQ

Dear R Gurus: How do I find the functions which are primitives, please? Thanks, Edna Bell

Full_Name: Wacek Kusnierczyk Version: 2.10.0 r48365 OS: Ubuntu 8.04 Linux 32bit Submission from: (NULL) (129.241.110.141) sprintf has a documented limit on strings included in the output using the format '%s'. It appears that there is a limit on the length of strings included with, e.g., the format '%d' beyond which surprising things happen (output modified for conciseness):

Hallo! I want to understand / recalculate what is done to get the CI of the logistic regression evaluated with lrm. As far as I came back, my problem is the variance-covariance matrix fit$var of the fit (fit<-lrm(...), fit$var). Here what I found and where I stucked: ----------------- library(Design) # data D<-c(rep("a", 20), rep("b", 20)) V<-0.25*(1:40) V[1]<-25

Full_Name: Wacek Kusnierczyk Version: 2.10.0 r48269 OS: Ubuntu 8.04 Linux 32 bit Submission from: (NULL) (129.241.199.164) In the following example (and many other cases): quote(a=1) # 1 the argument matching is apparently incorrect wrt. the documentation (The R Language Definition, v 2.8.1, sec. 4.3.2, p. 23), which specifies the following algorithm for argument matching: 1. Attempt to

Hi all I have the following regular expression problem: I want to find complete elements of a vector that end in a repeated character but where the repetition doesn't make up the whole word. That is, for the vector vec: vec<-c("aaaa", "baaa", "bbaa", "bbba", "baamm", "aa") I would like to get "baaa" "bbaa"

Full_Name: Wacek Kusnierczyk Version: 2.8.0 and 2.10.0 r48242 OS: Ubuntu 8.04 Linux 32 bit Submission from: (NULL) (80.202.30.36) The following raises an error: if (as.raw(1)) 1 # error: unimplemented type 'raw' in 'asLogical' However, ?'if' says: " Arguments: cond: A length-one logical vector that is not 'NA'. Conditions of length

Dear R Gurus: I read somewhere that functions are considered vectors. Is this true, please? thanks Edna Bell