similar to: Generic Function read?

?.S3methods f <- function()(2) > length(.S3methods(f)) [1] 0 > length(.S3methods(print)) [1] 206 There may be better ways, but this is what came to my mind. -- Bert On Wed, Mar 8, 2023 at 11:09?AM Leonard Mada via R-help < r-help at r-project.org> wrote: > Dear R-Users, > > I want to change the args() function to return by default the arguments > of the default

If one makes the reasonable assumption that Pct is much larger than Cutoff, sorting Cutoff is the expensive part e.g O(nlog2(n) for Quicksort (n = length Cutoff). I believe looping is O(n^2). Jeff's approach using findInterval may be faster. Of course implementation details matter. -- Bert On Mon, Oct 16, 2023 at 4:41?AM Leonard Mada <leo.mada at syonic.eu> wrote: > > Dear

Dear Jason, The code could look something like: dummyData = data.frame(Tract=seq(1, 10, by=1), ?? ?Pct = c(0.05,0.03,0.01,0.12,0.21,0.04,0.07,0.09,0.06,0.03), ?? ?Totpop = c(4000,3500,4500,4100,3900,4250,5100,4700,4950,4800)) # Define the cutoffs # - allow for duplicate entries; by = 0.03; # by = 0.01; cutoffs <- seq(0, 0.20, by = by) # Create a new column with cutoffs dummyData$Cutoff

Dear R-Users, I want to change the args() function to return by default the arguments of the default generic function: args = function(name, default = TRUE) { ?? ?# TODO: && is.function.generic(); ?? ?if(default) { ?? ???? fn = match.call()[[2]]; ?? ???? fn = paste0(as.character(fn), ".default"); ?? ???? name = fn; ?? ?} ?? ?.Internal(args(name)); } Is there a nice way

My mistake! It does actually something else, which is incorrect. One could still use (although the code is more difficult to read): subset(tmp <- table(sample(1:10, 100, T)), tmp > 10) Sincerely, Leonard On 10/21/2023 10:26 PM, Leonard Mada wrote: > Dear List Members, > > There was recently an issue on R-devel (which I noticed only very late): >

On 2024-09-05 4:23 p.m., Leo Mada via R-help wrote: > Dear R Users, > > Is this desired behaviour? > I presume it's a bug. > > atan(1i) > # 0+Infi > > tan(atan(1i)) > # 0+1i > > atan(1i) / 5 > # NaN+Infi There's no need to involve atan() and tan() in this: > (0+Inf*1i)/5 [1] NaN+Infi Why do you think this is a bug? Duncan Murdoch

Dear all, I am new R user and I am sure that this question has been asked quite often and I have also googled it and read about it! I understood that in order to read excel sheet into R you need to open it and saved it as csv or text, is this true? or you can use read.delim2 and read.csv2 to do this without the following error > dat <- read.csv2(file="C:\\Dokumente und

Dear Akshay, The best response was given by Andrew. "{...}" is not a closure. This is unusual for someone used to C-type languages. But I will try to explain some of the rationale. In the case that "{...}" was a closure, then external variables would need to be explicitly declared before the closure (in order to reuse those values): intermediate = c() { ??? intermediate

atan(1i) -> 0 + Inf i complex(1/5) -> 0.2 + 0i atan(1i) -> (0 + Inf i) * (0.2 + 0i) -> 0*0.2 + 0*0i + Inf i * 0.2 + Inf i * 0i infinity times zero is undefined -> 0 + 0i + Inf i + NaN * i^2 -> 0 + 0i + Inf i - NaN -> NaN + Inf i I am not sure how complex arithmetic could arrive at another answer. I advise against messing with infinities... use atan2() if you don't

Dear Shadee, If you have a data.frame with the following columns: n = 100; # population size x = data.frame( ??????Sex = sample(c("M","F"), n, T), ??????Country = sample(c("AA", "BB", "US"), n, T), ??????Income = as.factor(sample(1:3, n, T)) ) # Dummy variable ONE = rep(1, nrow(x)) r = aggregate(ONE ~ Sex + Income + Country, length, data = x)

Dear Bert, These behave like real divisions/multiplications: complex(re=Inf, im = Inf) * 5 # Inf+Infi complex(re=-Inf, im = Inf) * 5 # -Inf+Infi The real division / multiplication should be faster and also is well behaved. I was expecting R to do the real division/multiplication on a complex number. Which R actually does for these very particular cases; but not when only Im(x) is Inf.

> complex(real = 0, imaginary = Inf) [1] 0+Infi > Inf*1i [1] NaN+Infi >> complex(real = 0, imaginary = Inf)/5 [1] NaN+Infi See the Note in ?complex for the explanation, I think. Duncan can correct if I'm wrong. -- Bert On Thu, Sep 5, 2024 at 3:20?PM Leo Mada <leo.mada at syonic.eu> wrote: > Dear Bert, > > These behave like real divisions/multiplications: >

Dear Rich, It depends how the data is generated. Although I am not an expert in ecology, I can explain it based on a biomedical example. Certain variables are generated geometrically (exponentially), e.g. MIC or Titer. MIC = Minimum Inhibitory Concentration for bacterial resistance Titer = dilution which still has an effect, e.g. serially diluting blood samples; Obviously, diluting the

Dear members of the R Development Team, I am looking for people with a deep understanding of R internals to assist in bridging R to OpenOffice. While R is a state of the art statistical environment, less experienced users often find it difficult to work with R. Therefore, I believe that a bridge between R and a spreadsheet program will make this transition less painful. I sincerely believe

? Sat, 1 Jun 2024 16:16:23 +0000 Leo Mada via R-help <r-help at r-project.org> ?????: > When highlighting pdf-documents with Microsoft Edge, the bounding box > is sometimes misplaced, and quite ugly so. It also lacks the ability > to draw lines or arrows. > > On the other hand, I did not get used to Acrobat Reader: it usually > involves much more effort to add specific

Dear list members, Every "modern" OS comes with dozens of useless fonts, so that the current font drop-down list in most programs is overcrowded with fonts one never will use. Selecting a useful font becomes a nightmare. In an attempt to ease the selection of useful fonts, I began looking into sorting fonts using some statistical techniques. I summed my ideas on the OpenOffice.org

Dear R Users, Is this desired behaviour? I presume it's a bug. atan(1i) # 0+Infi tan(atan(1i)) # 0+1i atan(1i) / 5 # NaN+Infi There were some changes in handling of complex numbers. But it looks like a bug. Sincerely, Leonard [[alternative HTML version deleted]]

Perhaps > Inf*1i [1] NaN+Infi clarifies why it is *not* a bug. (Boy, did that jog some long dusty math memories :-) ) -- Bert On Thu, Sep 5, 2024 at 2:48?PM Duncan Murdoch <murdoch.duncan at gmail.com> wrote: > On 2024-09-05 4:23 p.m., Leo Mada via R-help wrote: > > Dear R Users, > > > > Is this desired behaviour? > > I presume it's a bug. > >

Dear Jason, I do not think that the solution based on aggregate offered by GPT was correct. That quasi-solution only aggregates for every individual level. As I understand, you want the cumulative sum. The idea was proposed by Bert; you need only to sort first based on the cutoff (e.g. using an ordered factor). And then only extract the last value for each level. If Pct is unique, than you

Dear List Members, There was recently an issue on R-devel (which I noticed only very late): https://stat.ethz.ch/pipermail/r-devel/2023-October/082943.html It is possible to use subset as well, almost as initially stated: subset(table(sample(1:5, 100, T)), table > 10) # Error in table > 10 : #? comparison (>) is possible only for atomic and list types subset(table(sample(1:5, 100,