similar to: Most efficient way to check the length of a variable mentioned in a formula.

On 01/04/2015 1:35 PM, Gabriel Becker wrote: > Joris, > > > The second argument to evalq is envir, so that line says, roughly, "call > environment() to generate me a new environment within the environment > defined by data". I think that's not quite right. environment() returns the current environment, it doesn't create a new one. It is evalq() that created

Dear list members, I'm a bit confused about the evaluation of expressions using with() or within() versus subset() and transform(). I always teach my students to use with() and within() because of the warning mentioned in the helppages of subset() and transform(). Both functions use nonstandard evaluation and are to be used only interactively. I've never seen that warning on the help

Dear all, I'm updating a package regarding a new type of models, and I'm looking to extend the formula interface with two functions (L() and R() ) for construction of these models. I want to use as much of the formula interface as possible, and hoped to do something similarly to I(). I know the I() function does nothing more than add the class "AsIs". I've been browsing the

On 01/04/2015 2:33 PM, Joris Meys wrote: > Thank you for the insights. I understood as much from the code, but I > can't really see how this can cause a problem when using with() or > within() within a package or a function. The environments behave like > I would expect, as does the evaluation of the arguments. The second > argument is supposed to be an expression, so I

Dear, I'm following the r tag on stackoverflow.com, and couldn't but notice there are quite some questions popping up that deal with scoping in relation to custom functions. I grinded my teeth on it already, and I have absolutely no clue what goes wrong. The general pattern is as follows : ff <- function(x){ y <- some_value some_function(y) } > ff(x) Error in eval(expr,

Hi all, I noticed some very odd behaviour in the termplot function of the stats package due to the following lines : 18. if (is.null(data)) 19. data <- eval(model$call$data, envir) This one will look in the global environment, and renders the two lines after this 20. if (is.null(data)) 21. data <- mf completely obsolete. If nothing is found, an error is returned. If

I've noticed that I get a warning message every time a package masks some functions from Hmisc. The warning message says : Warning message: In identical(get(., i), get(., lib.pos)) : ignoring non-pairlist attributes This happens with eg: library(plyr) library(xtable) I think I've seen this passing by before, but I'm not sure any more. Just thought I'd mention it. Cheers Joris

Joris, the point is that 'z' is NOT used as a predictor in the model. Therefore it should not affect predictions. Also, I find it suspicious that the error only occurs when the response variable conitains missings and 'z' is unique (I have tested several other cases to confirm this). -Mark Op vr 16 mrt. 2018 om 13:03 schreef Joris Meys <jorismeys at gmail.com>: >

?typeof? is your friend here: > typeof(`[`) [1] "special" > typeof(mc[[1]]) [1] "symbol" > typeof(mc2[[1]]) [1] "special" so mc[[1]] is a symbol, and thus not a primitive. - Lukas > On 28 Mar 2017, at 14:46, Michael Lawrence <lawrence.michael at gene.com> wrote: > > There is a difference between the symbol and the function (primitive >

Hi, I would like to have an index for a column in a matrix encoded in a cell of the same matrix. For example: x = matrix(c(11,12,13,1, 21,22,23,3, 31,32,33,2),byrow=T,ncol=4) In this case, column 4 is the index. I then access the column specified in the index by: > for (i in 1:3) print(x[i,x[i,4]]) [1] 11 [1] 23 [1] 32 > > for (i in 1:3) {x[i,x[i,4]] <- x[i,x[i,4]] + 5} > x

On 31/01/2018 6:33 AM, Joris Meys wrote: > 3. given your criticism, I'd like your opinion on where I can improve > the documentation of https://github.com/CenterForStatistics-UGent/pim. > I'm currently busy updating the help files for a next release on CRAN, > so your input is more than welcome. After this invitation I sent some private comments to Joris. I would say his

Dear gurus, I was utterly surprised to learn that one of my examples illustrating the need of match.fun() doesn't give me the expected result. center <- function(x,FUN) FUN(x) center(1:10, mean) mean <- 4 center(1:10, mean) Used to give me the error message "could not find function FUN". Now it just works, even though I didn't expect it to. I believe this is at least

I would like to prepare the data for barplot. But I only have the data frame now. x1=rnorm(10,mean=2) x2=rnorm(20,mean=-1) x3=rnorm(15,mean=3) data=data.frame(x1,x2,x3) If there a way to put data within a specific range? The expected result is as follows: range x1 x2 x3 -10-0 2 5 1 (# points in this

Apparently, as.POSIXlt takes one o'clock as the start of the day : > as.POSIXlt(0,origin="1970-01-01") [1] "1970-01-01 01:00:00 CET" > as.POSIXlt(0,origin="1970-01-01 00:00:00") [1] "1970-01-01 01:00:00 CET" > as.POSIXlt(0,origin="1970-01-01 23:59:59") [1] "1970-01-02 00:59:59 CET" Cheers -- Joris Meys Statistical

Dear Mr. or Ms.,   I used the R-software to run the zero-inflatoin negative binomial model (zeroinfl()) .   Firstly, I introduced one dummy variable to the model as an independent variable, and I got the estimators of parameters. But the results are not satisfied to me. So I introduced three dummy variables to the model. but I could not get the results. And the error message is

Dear all, I want to apply a weighted median on a huge dataset, and I remember a function from the package R.basic that could do this using an internal sorting algorithm qsort. This speeded things up quite a bit. Alas, I can't find that package anywhere anymore. There is a weighted.median function in the package limma too, but I didn't use that before. Anybody who knows what happened to

Dear all, Before I file this as a bug, I wanted to check if I didn't miss something. The help page of lag() says that the function returns a time series object. It actually does return something that looks like a ts object (the attribute tsp is set). But when using a vector, the class "ts" is not added to the result: > avec <- 1:10 > lag(avec) [1] 1 2 3 4 5 6 7 8

Seriously, if a method gives a wrong result, it's wrong. line() does NOT implement the algorithm of Tukey, even not after the patch. We're not discussing Excel here, are we? The method of Tukey is rather clear, and it is NOT using the default quantile definition from the quantile function. Actually, it doesn't even use quantiles to define the groups. It just says that the groups

Dear dev team, I was sorry to see the announcement of Duncan about his retirement from maintaining the R Windows build and Rtools. Duncan, thank you incredibly much for your 15 years of devotion and your impressive contribution to the R community as a whole. Thinking about the future, I wondered whether there were plans for the succession of Duncan. Is it the intention to continue providing

Hi, I want to give a summary or anova for "arima" model in R, as "summary", and "anova" for "lm". As including various intervention factors in arima(xreg = ) part, I want to assess the significancy of thse factors. I can do it using interrupted analysis of time series by linear regression, but want to see whether arima model works for the data first.