similar to: Variable names with spaces in lm

Hi all, I just cannot think of how to do it: I want to take the first variable (column) of a data frame and regress it against all other variables. bla <- function (dat) { reg <- lm(whateverthefirstofthevariablenamesis ~., data=dat) return(reg) } What kind of function do I have to take instead of the whateverthefirstofthevariablenamesis, eval(), substitute(), get(), ... to

Full_Name: G. Grothendieck Version: R version 2.1.0, 2005-05-14 OS: Windows XP Submission from: (NULL) (216.59.254.207) This code: > plot(1:10) > text(5,5,lab=expression(italic(22*"33"))) has the effect of italicizing 33 (which is a character string) but not 22 (which is not). I would have thought that both, not just 33, would be italicized. I had previously posted about this

Hi, if we have a file called Rscript.R that contains the following, for example: x <- 1:100 outfile = "Rscript.Rout" sink(outfile) print(x) and then we run >> source("Rscript.R") we get an output file called Rscript.Rout - great! Is there an internal variable, something like .Platform, that holds the script name when it is being executed? I would like to use

many languages have shorthands for that operation like: variable += 1 or ++variable is there something like that in R ? -- View this message in context: http://r.789695.n4.nabble.com/increase-or-decrease-variable-by-1-tp3076390p3076390.html Sent from the R help mailing list archive at Nabble.com.

One can perform a for loop without indices over the columns of a dataframe like this: for( v in df ) ... some statements involving v ... Is there some way to do this for rows other than using indices: for( i in 1:nrow(df) ) ... some statements involving df[i,] ... If the dataframe had only numeric entries I could transpose it and then do it over columns but what about the general case?

The first component name has backticks around it and the second does not. Though not wrong, it seems inconsistent. list(a = 1, b = 2) ## $`a` ## [1] 1 ## ## $b ## [1] 2 R.version.string ## [1] "R version 3.5.1 Patched (2018-07-02 r74950)" -- Statistics & Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com

Hello, I have 2 variables in my sample Data: Data$A and Data$B Variable Data$A can assume values: 1, 2, 3, and 4. Variable Data$B identifies my cases and can assume values: 1 and 2. I need to recode my variable Data$A into a new variable Data$new such that: People who are Data[Data$B %in% 1, ] are recoded like this: Value on Data$A Value on Data$new 1 +1 2

I did not write that. On Jan 29, 2008 9:05 AM, S Ellison <S.Ellison at lgc.co.uk> wrote: > >>> "Gabor Grothendieck" <ggrothendieck at gmail.com> 29/01/2008 12:35:27 > >>> > > As is common in human affairs, even > > the illusion of understanding is preferred to a lofty digression > upon > > why the audience does not understand.

If you object to names(x)[2]<- ... then use replace: z |> list(x = _) |> within(replace(names(x), 2, "foo")) |> _$x On Sun, Jul 21, 2024 at 11:10?AM Bert Gunter <bgunter.4567 at gmail.com> wrote: > > hmmm... > But note that you still used the nested assignment, names()[2] <- > "foo", to circumvent R's pipe limitations, which is exactly

Until recently R 2.1.0 was called rw2010dev. I just visited http://cran.r-project.org/bin/windows/base/rdevel.html and noticed its now called rw2010alpha, not rw2010dev. I would like to use it without downloading all the libraries over again. Can I - just rename my rw2010dev folder (Windows XP) to rw2010alpha and then install rw2010alpha to that? or - install it in a separate

What a delicious example! (I'm taking the liberty of sharing it with r-devel, since it raises some good issues.) You have two questions, presumably: 1 - how could the order of the setMethod calls make a difference in the results? 2 - what's causing the infinite loop & how could it be avoided, reliably? Second question first. The danger sign is the "vector" method:

When I alter the levels of a factor, why does it alter the names too? f <- factor(c(A="one",B="two",C="one",D="one",E="three"), levels=c("one","two","three")) names(f) -- gives [1] "A" "B" "C" "D" "E" levels(f) <-

If I issue the command > update.packages() it wants to update 'gregmisc' but then if I do it again right afterwards it still wants to update 'gregmisc'. It should have updated everything the first time and should not be trying to update anything the second time. Any comments? Here is the transcript of a session (R version at end): > update.packages() gregmisc :

Wow! Yes, this is very clever -- way too clever for me -- and meets my criteria for a solution. I think it's also another piece of evidence of why piping in base R is not suited for complex/nested assignments, as discussed in Deepayan's response. Maybe someone could offer a better Tidydata piping solution just for completeness? Best, Bert On Sun, Jul 21, 2024 at 7:48?AM Gabor

Hi I'm using R 2.0 in SuSE 9.2. When I plot data as a boxplot, the boxes appear on the plot in alphabetical order (of group) rather than the order in which they appear in the data. So far, the only thing I can do to fix this is to prefix the group labels with a,b,c...etc to trick R into plotting them in the right order. Can sorting be turned off? How should I address this sensibly? Thanks

If a path name ends in slash then file.exists says it does not exist. I would have expected these to all return TRUE. > file.exists("/Program Files") [1] TRUE > file.exists("/Program Files/") [1] FALSE > file.exists(normalizePath("/Program Files/")) [1] FALSE > R.version.string [1] "R version 3.0.2 Patched (2013-11-25 r64299)" I am using

Dear All, The following code should return 1, but it returns 0: source("http://r-bc.googlecode.com/svn/trunk/R/bc.R") bc("9 % 2") Do you confirm this bug? Paul

Also why should that difference result in different behavior? On Tue, Nov 14, 2023 at 9:38?AM Gabor Grothendieck <ggrothendieck at gmail.com> wrote: > > In that case identical should be FALSE but it is TRUE > > identical(a1, a2) > ## [1] TRUE > > > On Tue, Nov 14, 2023 at 8:58?AM Deepayan Sarkar > <deepayan.sarkar at gmail.com> wrote: > > > >

Hi, Before I reinvent the wheel I wanted to kindly ask you for your opinion if there is a simple way to do it. I want to merge a larger number of tables from different data sources in R and the matching criterium are country names. The tables are of different size and sometimes the country names do differ slightly. Has anyone done this or any recommendation on what commands I

On Tue, 14 Nov 2023 at 09:41, Gabor Grothendieck <ggrothendieck at gmail.com> wrote: > > Also why should that difference result in different behavior? That's justifiable, I think; consider: > d1 = data.frame(a = 1:4) > d2 = d3 = data.frame(b = 1:2) > row.names(d3) = c("a", "b") > data.frame(d1, d2) a b 1 1 1 2 2 2 3 3 1 4 4 2 > data.frame(d1,