similar to: New package on CRAN: remindR

It is in the doBy package. Thanks From: Bert Gunter [mailto:bgunter.4567 at gmail.com] Sent: Tuesday, February 13, 2018 4:32 PM To: Pius Mwansa <pmwansa at shaw.ca> Cc: R-help <r-help at r-project.org> Subject: Re: [R] LSmeans and lsmeans Always cc the list unless there is good reason to keep your reply private. There is no LSmeans() function in the lsmeans package.

I suspect akshay is (or was? Not sure) unclear about what braces do. They are not closures... they create an expression that wraps multiple expressions into one expression... they are a little more like parentheses than closures. They are not intrinsically associated with creation of environments for holding variables. Functions are closures... they run in a call-specific environment that

A cursory reading indicates that they are identical; but others more knowledgeable than I need to confirm or deny this. -- Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Tue, Feb 13, 2018 at 3:38 PM, Pius Mwansa <pmwansa at

It should be written more like this: ```R z <- data.frame(a = 1:3, b = letters[1:3]) z |> names() |> _[2] <- "foo" z ``` Regards, Iris On Sat, Jul 20, 2024 at 4:47?PM Bert Gunter <bgunter.4567 at gmail.com> wrote: > > With further fooling around, I realized that explicitly assigning my > last "solution" 'works'; i.e. > >

Since these are color strings, you can use functions in the grDevices package (other others) to manipulate them. E.g., you can convert them to various color spaces and perhaps use the mean in one of those spaces as your 'average color'. > myColors <- c(One="#FF7C00",Two="#00BF40",Three="#FFFF00") > col2rgb(myColors) One Two Three red

I still need the output to match my requiremnt in my original post. With decision rules "clusters" and probability attached to them. The examples are sort of similar. You just provided links to general info about trees. Sent from my Verizon, Samsung Galaxy smartphone<div> </div><div> </div><!-- originalMessage --><div>-------- Original message

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

Oh thanks for that clarification Bert! Hope you enjoyed your coffee! I ended up just using the transform argument in the ddply function. It worked and it repeated, then I called a mode function in another call to ddply that summarised. Kinda hacky but oh well! On Tue, Apr 19, 2016 at 12:31 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > ... and I'm getting another cup of

Hi, Jumping into this thread mainly on the point of the mode of the distribution, while also supporting Bert's comments below on theory. If the vector 'x' that is being passed to this function is an integer vector, then a tabulation of the integers can yield a 'mode', presuming of course that there is only one unique mode. You may have to decide how you want to handle a

I think Bert Gunter is right, but do you want partial matches (not found by match), and how robust do you want the code to be? f <- function(?) { pos <- match('a', ...names()) if (is.na(pos)) stop("a is required.") ?elt(pos) } Incidentally, what is the best way to extract the expression without evaluating it? g <- function(...) { pos <-

Tjats great that you are familiar and thanks for responding. Have you ever done what I am referring to? I have alteady spent time going through links and tutorials about decision trees and random forrests and have even used them both before. Mike On Apr 13, 2016 5:32 PM, "Sarah Goslee" <sarah.goslee at gmail.com> wrote: It sounds like you want classification or regression trees.

??? IQR returns a single number. > IQR(rnorm(10)) [1] 1.090168 To your 2nd response: "I could have used average, min, max, they all would have returned the same thing., " I can only respond: huh?? Are all your values identical? You really need to provide a small reproducible example as requested by the posting guide -- I certainly don't get it, and I'm done guessing.

Hi, Here is what I am doing notGroupedAll <- ddply(data ,~groupColumn ,summarise ,col1_mean=mean(col1) ,col2_mode=Mode(col2) #Function I wrote for getting the mode shown below ,col3_Range=myIqr(col3) ) groupedAll <- ddply(data ,~groupColumn ,summarise

Hi, To clarify the default behavior that Boris is referencing below, note the definition of the 'bin' argument to the tabulate() function: bin: a numeric vector ***(of positive integers)***, or a factor. Long vectors are supported. I added the asterisks for emphasis. This is also noted in the examples used for the function in ?tabulate at the bottom of the help page. The second

Thank you for your effort Bert.., I knew what is the problem now, the values (1,2,3) were only an example. The values I have are 0 , 1, 2 . Tabulate () function seem to ignore calculating the frequency of 0 values and this is my exact problem as the frequency of 0 values should also be calculated for the maf to be calculated correctly. ________________________________ From: Bert Gunter

Dear @William<mailto:wdunlap at tibco.com>, thanks for the feedback. I have tested it on the larger dataset and noticed that it created two variables, pf_raw and pf_curated. The output we were looking for, was one that takes the variable pf_mcl in curated dataset and replaces pf_mcl in matching rows within the raw dataset. @Eric<mailto:ericjberger at gmail.com>?s solution was able to

hmmm... But note that you still used the nested assignment, names()[2] <- "foo", to circumvent R's pipe limitations, which is exactly what Iris's solution avoids. So I think I was overawed by your cleverness ;-) Best, Bert On Sun, Jul 21, 2024 at 8:01?AM Bert Gunter <bgunter.4567 at gmail.com> wrote: > > Wow! > Yes, this is very clever -- way too clever for

For the record please re-read my original message. It is clear, concise, polite and thankful for future help. I received a reply "Google it!". Thank you! Thank you Jeff for your links. I am aware of them. However, they do not point to an R package for GP for binary classification which produces prediction intervals. It seems that r-help is not as it was before. Wish you all the

Thanks, Calum. That was exactly what Duncan Murdoch proposed earlier in this thread, except, of course, he had to explicitly write the function first. -- Bert On Sun, Jul 21, 2024 at 8:12?AM CALUM POLWART <polc1410 at gmail.com> wrote: > > The tidy solution is rename > > literally: > > z |> rename(foo = 2) > > Or you could do it with other functions > > z

> On Mar 5, 2018, at 3:04 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > But of course the whole point of additivity is to decompose the combined effect as the sum of individual effects. Agreed. Furthermore your encoding of the treatment assignments has the advantage that the default treatment contrast for A+B will have a statistical estimate associated with it. That was a