similar to: Change in how R handles assignments?

Hi, I have a data.frame that I need to construct iteratively. At the moment, I'm doing: d<-data.frame(x=c(),y=c(),z=()); # {and, within some loop} d<-rbind(d,data.frame(x=newx,y=newy,z=newz); While this works, it is horribly verbose and probably not efficient, either. My real data.frame has, of course, many more columns, which can be of different modes. I vaguely recall that

Full_Name: D Kreil Version: 2.1.1 OS: HP-UX B.11.23 U ia64 0029870451 unlimited-user license Submission from: (NULL) (62.178.15.60) When the dependency readline is installed in a non-standard location the installation procedure fails. If using a non-standard installation location like --prefix=/bi/common --exec-prefix=/bi/arch the configure script default include and library paths should

Full_Name: D Kreil Version: 2.8.1 and 2.9.0 OS: Debian Linux Submission from: (NULL) (141.244.140.179) Group: Accuracy I understand that most floating point numbers are approximated due to their binary storage. On the other hand, I thought that modern math CPUs used guard digits to protect against trivial underflows. Not true? # integers, no problem > 1+1+1==3 [1] TRUE # binary floating

Full_Name: D Kreil Version: 2.8.1 and 2.9.0 OS: Debian Linux Submission from: (NULL) (141.244.140.179) Group: Accuracy I understand that most floating point numbers are approximated due to their binary storage. On the other hand, I thought that modern math CPUs used guard digits to protect against trivial underflows. Not true? # integers, no problem > 1+1+1==3 [1] TRUE # binary floating

Full_Name: D Kreil Version: 2.1.1 OS: HP-UX B.11.23 U ia64 0029870451 unlimited-user license Submission from: (NULL) (62.178.15.60) During compiliation, I get: ld: Unsatisfied symbol "finite" in file colors.o Fix: replace "finite" with R_FINITE in colors.c line 269. I have, this time, searched through the R-admin.html, but not found a mention. This is all a bit hard for me

Dear R experts, I have a problem that is a related to the question raised in this earlier post https://stat.ethz.ch/pipermail/r-help/2007-January/124064.html My situation is different in that I have only 2 predictors (coordinates x,y) for local regression but a number of global ("parametric") offsets that I need to consider. Essentially, I have a spatial distortion overlaid over a

Dear John, > I suggest that you look at the abilities of the mgcv package. > There are notes of mine at > > http://www.maths.anu.edu.au/%7Ejohnm/r-book/xtras/autosmooth.pdf > > that may help you get started. Thank?you very much for the suggestion and the link to your write-up, it was indeed very helpful! I have experimented with this library for a while now and am really happy

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

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" >

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

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()

I was surprised to find that calling str() on an environment can remove attributes from it, as in > a <- new.env() > class(a) <- "jens" > class(a) [1] "jens" > str(a) Class 'jens' length 0 <environment> > class(a) NULL then I found in the R-Language-Manual in chapter 2.1.10 Environments the sentence: "In particular, assigning

We have deprecated codes() in R-devel. One we started looking into this, we found that *all* the uses of codes() in the R sources and probably all the uses in CRAN packages were not what we think was intended. For an ordered factor, codes() is the same as unclass/as.vector/as.integer and so was unneeded. For an unordered factor, codes() does *not* retrieve the internal codes. It

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

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

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

Dear R developers! Considering I got no response or comments in the general r-help forum so far, perhaps my question is actually better suited for this list? I have added some more hopefully relevant technical details to my original post (edited below). Any comments gratefully received! Best regards, David Kreil. ---------- Dear R experts, I have a problem that is a related to the question

R 3.5.0 Is it intended that the Date method of as.POSIXct does not respect the tz parameter? I suggest changing as.POSIXct.Date to this: function (x, tz = "", ...) .POSIXct(unclass(x) * 86400, tz = tz) Currently, the best workaround seems to be using the character method if one doesn't want the default timezone (which is often an annoying DST timezone). This came up on