search for: unclassing

Dear All, Assume I have a data.frame that contains also factors and I would like to get another data.frame containing the factors as numeric vectors, to apply functions like sapply(..., median) on them. I read the warning concerning as.numeric or unclass, but in my case this makes sense, because the factor levels are properly ordered. I can do it, if I write for each single column

> Henrik Bengtsson: > > I'm looking for a way to get the length of an object 'x' as given by > base data type without dispatching on class. The performance improvement you're looking for is implemented in the latest version of pqR (pqR-2016-10-24, see pqR-project.org), along with corresponding improvements in several other circumstances where unclass(x) does not

Have I been sleeping in class? rw1071 from CRAN, windows XP incidencia is made by a call to tapply > class(incidencia) [1] "array" > incidencia <- unclass(incidencia) > class(incidencia) [1] "array" Kjetil Halvorsen

In a real example I was trying to remove the class from the result of table, just because it was to be used as a building block for other things and a simple integer vector seemed likely to be most efficient. I'm puzzled as to why unclass doesn't work. > zed <- table(1:5) > class(zed) [1] "table" > class(unclass(zed)) [1] "array" >

In a real example I was trying to remove the class from the result of table, just because it was to be used as a building block for other things and a simple integer vector seemed likely to be most efficient. I'm puzzled as to why unclass doesn't work. > zed <- table(1:5) > class(zed) [1] "table" > class(unclass(zed)) [1] "array" >

Regarding the discussion of getting length(unclass(x)) without an unclassed version of x being created... There are already no copies done for length(unclass(x)) in pqR (current version of 2017-06-09 at pqR-project.org, as well as the soon-to-be-release new version). This is part of a more general facility for avoiding copies from unclass in other circumstances as well - eg,

I'm looking for a way to get the length of an object 'x' as given by base data type without dispatching on class. Something analogous to how .subset()/.subset2(), e.g. a .length() function. I know that I can do length(unclass(x)), but that will trigger the creation of a new object unclass(x) which I want to avoid because 'x' might be very large. Here's a dummy example

On 08/24/2018 07:55 PM, Henrik Bengtsson wrote: > Is there a low-level function that returns the length of an object 'x' > - the length that for instance .subset(x) and .subset2(x) see? An > obvious candidate would be to use: > > .length <- function(x) length(unclass(x)) > > However, I'm concerned that calling unclass(x) may trigger an > expensive copy

Is there a low-level function that returns the length of an object 'x' - the length that for instance .subset(x) and .subset2(x) see? An obvious candidate would be to use: .length <- function(x) length(unclass(x)) However, I'm concerned that calling unclass(x) may trigger an expensive copy internally in some cases. Is that concern unfounded? Thxs, Henrik

The solution below introduces a dependency on data.table, but otherwise it does what you need: --- # special method for Foo objects length.Foo <- function(x) { length(unlist(x, recursive = TRUE, use.names = FALSE)) } # an instance of a Foo object x <- structure(list(a = 1, b = list(b1 = 1, b2 = 2)), class = "Foo") # its length stopifnot(length(x) == 3L) # get its length as

On 09/05/2018 11:18 AM, I?aki Ucar wrote: > The bottomline here is that one can always call a base method, > inexpensively and without modifying the object, in, let's say, > *formal* OOP languages. In R, this is not possible in general. It > would be possible if there was always a foo.default, but primitives > use internal dispatch. > > I was wondering whether it would be

Full_Name: Bo Zhou Version: 2.6.1 (2007-11-26) OS: Windows XP Submission from: (NULL) (207.237.54.242) Hi, I found an arithmetic problem when I'm doing something with POSIXct The code to reproduce it is as follows (This is the recommended way of finding out time zone difference on R News 2004-1 Page 32 URL http://cran.r-project.org/doc/Rnews/Rnews_2004-1.pdf) a=Sys.time()

Hi Tomas, On 09/03/2018 11:49 AM, Tomas Kalibera wrote: > Please don't do this to get the underlying vector length (or to achieve > anything else). Setting/deleting attributes of an R object without > checking the reference count violates R semantics, which in turn can > have unpredictable results on R programs (essentially undebuggable > segfaults now or more likely later

The bottomline here is that one can always call a base method, inexpensively and without modifying the object, in, let's say, *formal* OOP languages. In R, this is not possible in general. It would be possible if there was always a foo.default, but primitives use internal dispatch. I was wondering whether it would be possible to provide a super(x, n) function which simply causes the

Please don't do this to get the underlying vector length (or to achieve anything else). Setting/deleting attributes of an R object without checking the reference count violates R semantics, which in turn can have unpredictable results on R programs (essentially undebuggable segfaults now or more likely later when new optimizations or features are added to the language). Setting attributes

The following command produces red axis line in a pairs plot: pairs(iris[1:4], main = "Anderson's Iris Data -- 3 species", pch = "+", col = c("red", "green3", "blue")[unclass(iris$Species)]) Trying to fool pairs in the following way produces the same plot as above: pairs(iris[1:4], main = "Anderson's Iris Data -- 3

More generally, I think one of the issues is that R is not yet able to decrement a reference count (or mark a 'shared' data object as 'unshared' after it knows only one binding to it exists). This means passing variables to R closures will mark that object as shared: x <- list() .Internal(inspect(x)) # NAM(1) identity(x) .Internal(inspect(x)) # NAM(3) I think

Hi, Encouraged by a tip from Simon Urbanek I tried to use the S3 machinery to write a faster version of the data.frame class. This quickly hits a snag: the "[.default"(x, i) for some reason cares about the dimensionality of x. In the end there is a full transcript of my R session. It includes the motivation for writing the class and the problems I have encountered. As a result I see

Hi All, I'm perplexed by the way the unclass function displays a factor whose labels have been swapped with the relevel function. I realize it won't affect any results and that the relevel did nothing useful in this particular case. I'm just doing it to learn ways to manipulate factors. The display of unclass leaves me feeling that the relevel had failed. I've checked three books

Buenas tardes a todos los participantes del foro. Me dirijo a vosotros porque estoy atascado con una duda de programación respecto al data frame: > dd # Data frame de 5 variables, leído de un archivo txt id sexo nacim origen final 1 1 0 02/09/1955 01/04/1985 01/02/2014 2 2 1 29/10/1951 15/08/1996 01/05/2009 3 3 0 30/10/1942 02/08/2000 01/02/2014 4 4 1