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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,

Regarding 1 and 2, please read the Posting Guide mentioned at the bottom of every R-help post. R does not equal statistics... and education about statistics is way too ambitious to include in this mailing list that is about a tool that happens to be useful for statisticians. There are forums online that do cater to statistical methods (e.g. Cross Validated or many results from a search engine)...

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

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

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

Lisa, this seems to be fairly straight forward to do in R and I'm happy to help you get started. However, please be aware that you do have to have knowledge of statistics to do the analysis/modeling. Rolf, Jeff, I do appreciate your view that this is not a R probelm. It's more a 'how to use R' / 'help me get started' problem. The posting guidelines point to "Usenet

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

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

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

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

I have done a scatter plot in R. I want to insert the coefficient of determination R^2 = 0.62 as a text in the plot. I have tried to write R^2 but could not produce R2. I would appreciate it if someone could help me with the syntax. I have tried: expression(paste("", R^2,"=", 0.62)), but it did not produce R squared, rather it gave me error messages. Thanks. Jibrin Alhassan

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"]

Hans, It is a good question albeit R made a conscious decision to have indices that correspond to things like row numbers and thus start with 1. Some others have used a start of zero but often for reasons more related to making use of all combinations of the implementation of integers on many machines where starting with 1 would only allow use of the 255 of the 256 combinations available in 8

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])} >

Suppose I have the following list: test <- list(a=3,b=5,c=11) I'm trying to figure out how to extract the characters to the left of the equal sign (i.e., I want to extract a list of the variable names, a, b and c. I've tried the permutations I know of involving sub - things like sub("\\=.*", "", test), but no matter what I try, sub keeps returning (3, 5, 11).

Hello Peter, Unless I too misunderstand your point, negative indices for removal do work with the Oarray package (though -0 doesn't work to remove the 0th element, since -0 == 0 -- perhaps what you meant): > library(Oarray) > v <- Oarray(1:10, offset=0) > v [0,] [1,] [2,] [3,] [4,] [5,] [6,] [7,] [8,] [9,] 1 2 3 4 5 6 7 8 9 10 > dim(v)

Dear Brian, As Duncan mentioned, the terms type-I, II, and III sums of squares originated in SAS. The type-II and III SSs computed by the Anova() function in the car package take a different computational approach than in SAS, but in almost all cases produce the same results. (I slightly regret using the "type-*" terminology for car::Anova() because of the lack of exact