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x[?V2?] would retain columns of x headed by V2. What I need is the opposite??I need a data grime with those columns excluded. Steven from iPhone > On Feb 13, 2023, at 9:33 AM, Rolf Turner <r.turner at auckland.ac.nz> wrote: > > ? >> On Sun, 12 Feb 2023 14:57:36 -0800 >> Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: >> >> x["V2"]

Please see fortunes::fortune(285). cheers, Rolf Turner -- Honorary Research Fellow Department of Statistics University of Auckland Stats. Dep't. (secretaries) phone: +64-9-373-7599 ext. 89622 Home phone: +64-9-480-4619

See fortunes::fortune(36). cheers, Rolf Turner -- Honorary Research Fellow Department of Statistics University of Auckland Stats. Dep't. (secretaries) phone: +64-9-373-7599 ext. 89622 Home phone: +64-9-480-4619

On Tue, 7 May 2024 06:34:50 -0400 Duncan Murdoch <murdoch.duncan at gmail.com> wrote: > On 07/05/2024 6:31 a.m., Iago Gin? V?zquez wrote: > > Thanks Duncan. > > > > I am currently on Windows. Is there any solution for it? > > Switch to Linux or MacOS? Fortune nomination! cheers, Rolf Turner -- Honorary Research Fellow Department of Statistics University of

On Sat, 28 Sep 2024 10:26:31 +0100 CALUM POLWART <polc1410 at gmail.com> wrote: > Avi > > I fear this was all a huge social experiment. > > Testing if a post titled "sexy way" would increase engagement... <SNIP> I conjecture that this conjecture was tongue-in-cheek. Be that as it were ??, let me assure everyone that such was not my intention. The usage

On Sun, 1 Dec 2024 21:43:45 -0800 Bert Gunter <bgunter.4567 at gmail.com> wrote: > Finally, my best advice would be to forget about SAS if you wish to > use R. Trying to translate SAS paradigms into R is the devil's work. cheers, Rolf Turner -- Honorary Research Fellow Department of Statistics University of Auckland Stats. Dep't. (secretaries) phone:

I have no real idea what you are trying to do, but if a table is what you want, you can probably get it using the table() function. Or, more likely, the xtabs() function. Using your example from an earlier post (adjusted to make it comprehensible to the human mind): set.seed(1000) time <- factor(rep(c("Pre","Post"),each=200)) treatment <-

Admit it, Rolf. Haven't you wondered if S, in a more private way, is sexier than R? OK, kidding aside, we have talked this to death. Just FYI, the conversation was stimulating for some of us and I have continued on my own and located functions I see as useful in the stringi and stringr packages to make my silly version ever less silly! LOL! -----Original Message----- From: Rolf Turner

Heh. Did anyone bring up negative indices yet? -pd > On 22 Apr 2024, at 10:46 , Rolf Turner <rolfturner at posteo.net> wrote: > > > See fortunes::fortune(36). > > cheers, > > Rolf Turner > > -- > Honorary Research Fellow > Department of Statistics > University of Auckland > Stats. Dep't. (secretaries) phone: > +64-9-373-7599

On Sun, 18 May 2025 13:40:31 -0400 paul zachos via R-help <r-help at r-project.org> wrote: > Dear R Community > > I am an R beginner > > I have a vector of ?1?s and ?0?s > > x > [1] 0 0 1 0 0 1 1 0 1 0 1 1 1 0 1 1 0 0 1 0 1 0 1 0 0 0 0 > [28] 0 1 1 0 1 0 1 1 0 0 0 0 1 0 1 1 0 0 1 0 0 1 0 0 0 1 1 > [55] 0 0 1 0 1 0 0 0 1 1 1 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0

Hi Rolf, No it is not. I don't know to which question did you want to respond ? I desribed everything in my first email and attached links from SO with pictures included, which are quite understandable. Cheers, Jacek wt., 27 lut 2024 o 02:29 Rolf Turner <rolfturner at posteo.net> napisa?(a): > > I have no real idea what you are trying to do, but if a table is > what you

Rolf, If it is homework, I have to wonder at what level the class is before any of the proposed answers would be considered acceptable by the instructor. For a true beginner who has not even learned that much of R is vectorized, we are back to perhaps using a manual loop with indices and ... I will say that these days, I see too many assignments handed in by novices, such as 12-year-old kids

You could have negative indices. There are two ways to do this. 1) provide a large offset. Offset <- 30 for (i in -29 to 120) { print(df[i+Offset])} 2) use absolute values if all indices are negative. for (i in -200 to -1) {print(df[abs(i)])} Tim -----Original Message----- From: R-help <r-help-bounces at r-project.org> On Behalf Of Peter Dalgaard via R-help Sent: Monday, April 22,

Hi all, I've just could test your suggestions on using dataedit... entries now, and indeed they work... partially. See, please, the next screenshot: https://ibb.co/Dkn2pVs dataedituser modifies the red borders dataedittext the yellow text dataeditfg... I do not know, specified to green the output of View(mtcars) does not change from that of the screenshot dataeditbg is the (almost) black

with the following simple data frame dd = structure(list(z = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L ), .Label = c("a", "b"), class = "factor"), x = c(0.3, 0.2, 0.1, 0, 0, 0, 0.2, 0.3)), .Names = c("z", "x"), row.names = c(NA, -8L), class = "data.frame") I would like know if it's possible to use model.matrix() to construct the

the version 2 parser thinks I have unnecessary braces, but I can't find any. False positive or am I missing something? If a false positive, is there any way to work around the warning? * checking Rd files against version 2 parser ... WARNING Warning: ./man/dbetabinom.Rd:32-34: Unnecessary braces at ?{p(x) = % (C(N,x)*Beta(N-x+theta*(1-p),x+theta*p))/% Beta(theta*(1-p),theta*p)}?

I have (toy example): x <- list(`1` = c(7, 13, 1, 4, 10), `2` = c(2, 5, 14, 8, 11), `3` = c(6, 9, 15, 12, 3)) and f <- factor(rep(1:3,5)) I want to create a vector v of length 15 such that the entries of v, corresponding to level l of f are the entries of x[[l]]. I.e. I want v to equal c(7, 2, 6, 13, 5, 9, 1, 14, 15, 4, 8, 12, 10, 11, 3) I can create v

I am having difficulty with evaluation/environment construction for a formula to be evaluated by model.matrix(). Basically, I want to construct a model matrix that first looks in "newdata" for the values of the model parameters, then in "object at data". Here's what I've tried: 1. model.matrix(~f,data=c(newdata,object at data)) -- fails because something (terms()?)

Is there a reason that Reduce() doesn't take a "..." argument that would allow arbitrary extra arguments to be passed through to the function? Here is a little example of how this would be convenient: z <- list( data.frame(state=c("California"), cases=0), data.frame(state=c("California","Massachusetts"),

Doesn't sound like you got the point. x[-1] normally removes the first element. With 0-based indices, this cannot work. - pd > On 22 Apr 2024, at 17:31 , Ebert,Timothy Aaron <tebert at ufl.edu> wrote: > > You could have negative indices. There are two ways to do this. > 1) provide a large offset. > Offset <- 30 > for (i in -29 to 120) { print(df[i+Offset])} >