search for: lemmen

...- c(subject.bin.means, bin.means); } detach(data); if ( !length(subject.bin.means) == bins*length(l) ) { stop("Inappropriate number of means calculated."); } else { subject.bin.means } } ---------- Forwarded Message ---------- Date: dinsdag 27 mei 2003 14:53 +0200 From: Paul Lemmens <P.Lemmens at nici.kun.nl> To: r-help at stat.math.ethz.ch Subject: [R] Numbers that look equal, should be equal, but if() doesn't see as equal Hi! After a lot of testing and debugging I'm falling silent in figuring out what goes wrong in the following. I'm implementing the V...

...ata) # But now, I would like to do here what I'd do on the console as # wery$ok <- factor(wery$ok), so here data$facts[2] <- factor(data$facts[2]) # This won't work here. How do I continu? # Or perhaps also # data.tmp <- data$resp[data$facts[1] == 'i'] } thank you, Paul Lemmens P.S: str(wery) `data.frame': 150 obs. of 11 variables: $ V1 : int 1 1 1 1 1 1 1 1 1 1 ... $ V2 : int 1 2 3 4 5 6 7 8 9 10 ... $ V3 : int 960 1060 980 1060 1020 1010 1060 1010 1090 1090 ... $ V4 : int 520 450 470 440 440 530 580 530 560 540 ... $ V5 : int 1483 3753 5758...

...oesn't work like this, but having tried all kinds of combinations of eval(), parse(), substitute(), and friends I was unable to get the contents of clipname as the element name of the list that I return in f(). Is this possible to do so at all and if so, what am I missing? Kind regards, Paul Lemmens

...ualified to say that, be warned]. Interestingly, at some point both methods for `is.na<-` perform this operation: x[value] <- NA. Ahem. By the way, `is.na(x) <- FALSE` will leave x unchanged (including leaving it as NA ! how bad is that ?!) > -----Original Message----- > From: Paul Lemmens [mailto:P.Lemmens at nici.kun.nl] > Sent: 14 October 2003 16:10 > To: r-help at stat.math.ethz.ch > Subject: RE: is.na(v)<-b (was: Re: [R] Beginner's query - segmentation > fault) > > > Security Warning: > If you are not sure an attachment is safe to open please co...

...q)) r <- length %% parts if ( r > 0 ) { p[r+1:length(p)] <- NA z <- cbind(z,p) } z <- na.omit(as.vector(t(z))) } and then sections(32,5) -> a As I see it, rows in result are 5 and the vector length of p (which is 5) is a multiple of 5. kind regards, Paul -- Paul Lemmens NICI, University of Nijmegen ASCII Ribbon Campaign /"\ Montessorilaan 3 (B.01.03) Against HTML Mail \ / NL-6525 HR Nijmegen X The Netherlands / \ Phonenumber +31-24-...

...c==1, drop=TRUE) table(dd$c) 1 2 5 0 So to lose the second level of dd$c, in method B I still need to 'dd$c <- dd$c[, drop=TRUE]', while the manual seems to imply that with the drop argument to subset() this would not be necessary. Could you comment? kind regards, Paul -- Paul Lemmens NICI, University of Nijmegen ASCII Ribbon Campaign /"\ Montessorilaan 3 (B.01.05) Against HTML Mail \ / NL-6525 HR Nijmegen X The Netherlands / \ Phonenumber +31-24-...

...en I insert a op <- par(new=TRUE) either before or immediately after the first plot statement (the type='n' one) in the above code fragment, the resulting graph still only shows one column of data. Have I misinterpreted the instructions or the functionality of new=TRUE? Thank you, Paul Lemmens

...ays "Calculates type-II or type-III analysis-of-variance tables for model objects produced by 'lm' and 'glm'", so it's not suitable for the aovlist that aov() with Error()-term returns. How can I compute Type III SS for such objects? kind regards, Paul -- Paul Lemmens NICI, University of Nijmegen ASCII Ribbon Campaign /"\ Montessorilaan 3 (B.01.05) Against HTML Mail \ / NL-6525 HR Nijmegen X The Netherlands / \ Phonenumber +31-24-...

Hi Ruser I'm trying to replicate some SAS code. I have to add a spline to my longitudinal spaghetti plot. I have the plot, but I can't add the spline, a overall trend line. In the SAS code they use the command 'I=SM50S' and I would prefer something similar. I?m using R 2.10.1 on windows XP? I have made this working example. tolerance.pp <-

..., so this should range from 350 - 450. On the x-axis I want two ticks for each level (sc and oc) of the 'cat' factor. In the plot, three lines should be displayed, for each level of the factor 'loc', pref. including a legend. How do I do this??? Please send a CC to my address (p.lemmens at nici.kun.nl), as I'm not yet subscribed to the list. thank you for your help! Paul -- -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", o...

...even with options(digits=20), but *still* the test fails. IMHO it's not my code and neither the distribution I use to test, but still, can you point out an obvious failure of my programming or is it indeed something of R that I don't yet grasp? thank you for your help, Paul -- Paul Lemmens NICI, University of Nijmegen ASCII Ribbon Campaign /"\ Montessorilaan 3 (B.01.03) Against HTML Mail \ / NL-6525 HR Nijmegen X The Netherlands / \ Phonenumber +31-24-...

...an I have the same amount of protection as I previously had with dovecot 1.x over inetd ? BTW: I tried to post on the mailinglist but it seems I can't access them it the moment, altough I did subscribe and login. Is there anything wrong ? Thanks for your reply ! -- With best regards, Kees Lemmens. ----------------------------------------------------------------------- Delft Institute of Applied Mathematics (DIAM), Faculty Electrical Engineering, Mathematics and Computer Science, Delft University of Technology, Delft, The Netherlands,

Hello,   is there a simple way to give all combinations for a given vector:   v<-c("a","b","c")   combination(v,v) becomes "aa","ab","ac","bb","bc","cc'   combination(v,v,v) becomes "aaa","aab","aac","abb",......     [[alternative HTML version deleted]]

Is there a modulus operator in R? Anna

> -----Original Message----- > From: Richard A. O'Keefe [mailto:ok at cs.otago.ac.nz] <snip> > The very existence of an "is.na<-" which accepts a logical > vector containing FALSE as well as TRUE ... And don't forget this is not the only usage of is.na<-. In fact it is designed to take any valid indexing value. For example: > a<-1:10 >

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I'm puzzled by the following problem, which appears when attempting to run an analysis on part of a dataset: If I try: csubset <- dat$Diagnosis==0 cont <- lme(fixed=cform, random = ~1|StudyName, data=dat,subset=csubset,na.action=na.omit) Then I get: Error in eval(expr, envir, enclos) : Object "csubset" not found But if I do

...(a, a, b, b, c, c, a, b, c), or even c(3, 3, 1, 1, 2, 2, 3, 1, 2). So the fact that participant (index) 1, 2, and 8 belong to each other is key. I am most puzzled about how to do the matching / find the similarity between each column and gold standard. Thank you for your time! best regards, Paul Lemmens

Hallo! Does anyone know a possibility to perform a sample size estimation for an ANCOVA? Would be great! Thanks Karl

Hi! I have a 3d array: > dim(ib5km15.dbc) [1] 190 241 19 and a set of positions to extract: > ib5km.lincol.random[1:3,] [,1] [,2] [1,] 78 70 [2,] 29 213 [3,] 180 22 Geting the values of a 2D array for that set of positions would be: > ima <- ib5km15.dbc[,,1] > ima[ib5km.lincol.random[1:10,]] but don't find the way for the case of the 3D array: >