similar to: version comparison puzzle

? Sat, 27 Apr 2024 13:56:58 -0500 Jonathan Keane <jkeane at gmail.com> ?????: > In devel: > > max(numeric_version(c("1.0.1.100000000", "1.0.3.100000000", > "1.0.2.100000000"))) > [1] ?1.0.1.100000000? > > max(numeric_version(c("1.0.1.10000000", "1.0.3.10000000", > "1.0.2.10000000"))) > [1]

Hi I looked, but I didn't find it: I need an array [0 to 101] where each element is a list (the result of Kest in spatstat). I.e. I want to do: A[2] <- Kest(pp1) A[3] <- Kest(pp2) ... A[101] <- Kest(pp100) How can I create A ? Rainer -- Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation Biology (UCT) Department of Conservation Ecology University of Stellenbosch Matieland

I have a data frame (attached) that has interpolated EOT errors for each minute before flight landing. It also has the runway and an index for the flight: > > times[1:4,] time error runway flight 1 0 -0.02206235 04R 1 2 1 -0.07961631 04R 1 3 2 -0.13795380 04R 1 4 3 -0.20726073 04R 1 > > sapply(times, class) time error

Hi R, I have a list > class(pp2) [1] "list" > length(pp2) [1] 1244 It is in the below format RIC Trade.Date Close.Price Currency.Code Convertion.Rate New.Price ABCD.SZ 2008/02/29 15.30 CNY 0.1408 2.154240 ABCD.SZ 2008/01/31 15.27 CNY 0.1392 2.125584 ABCD.SZ 2007/12/31 14.88 CNY 0.1371 2.040048

I've noticed something in R devel which seems a little off and not the behavior I see in 4.4.0 or earlier versions. With numeric_versions that have long (>8 digit) final components max and min return the first element and not the max or min: In devel: > max(numeric_version(c("1.0.1.100000000", "1.0.3.100000000", "1.0.2.100000000"))) [1] ?1.0.1.100000000?

I am using ggplot2 to make a boxplot that overlays a scatterplot: pp = qplot(time, error, data=times, size=I(1), geom="jitter", main=title, ylab="Error (min)", xlab="Time before ON (min)", alpha=I(1/10), color=times$runway, ylim=c(-30,40)) pp2 = pp + with(times, facet_wrap(~ runway, ncol=2)) print(pp2 + geom_boxplot(alpha=.5,

Hello. I have n files in a directory: file1, ..., filen. I read them with the following commands: list=scan(file="list",what=list(nom="")) # in the file list, I have all the filenames. n=length(list[[1]]) for (i in 1:n){ aux <- paste("p",i,sep="") assign(aux, as.matrix(read.table(list[[1]][i]))) } R creates the matrices p1,p2,...,pn. I want

Hi everyone, ? I am trying to do a horizonplot using my own time series data. I know that there is a horizonplot function in latticeExtra, but on closer examination i think that the graph itself is slightly wrong (it displays some regions as triangles and i think they should be trapezoids, and the red regions (that are below the baseline) are displayed on top of the blue areas ? while i think

Hi I have two processes which take with a certain probability (p1 and p2) x number of years to complete (age1 and age2). As soon as thge first process is completed, the second one begins. I want to calculate the time it takes for the both processes to be completed. I have the following script which gives me the answer, butI think there must be a more elegant way of doing the calculations

Dear all, I'm looking for primary help at aggregating table() results and at writing a loop (if useful) My dataset ( http://goo.gl/gEPKW ) is composed of 23k rows, each one representing a point in the space of which we know the land cover over 10 years (column y01 to y10). I need to analyse it with a temporal sliding window of 5 years (y01 to y05, y02 to y06 and so forth) For each period

This patch set fixes a two year old bug in guestfish, namely that the 'glob' command does not expand /dev/* patterns. https://bugzilla.redhat.com/show_bug.cgi?id=635971 Rich.

Dear R users, I have a function (below) which encompasses several tests. However, when I run it, only the output of the last test is displayed. How can I ensure that the function root(var) will run and display the output from all tests, and not just the last one? Thank you, b. root <- function(var) { #---Phillips-Perron PP.test(var, lshort = TRUE) PP.test(var, lshort = FALSE)

Full_Name: Ed Borasky Version: R-beta 2.1.0 2005-04-08 OS: Linux 2.6.11 GCC 3.3.5 Submission from: (NULL) (24.21.57.139) I downloaded the latest R-beta tarball and did a build with the default options. OS is Linux 2.6.11 and compiler is GCC 3.3.5. "make check-all" failed with the following message: make[3]: Entering directory `/home/znmeb/R-beta/tests' running code in

Hi, I've implemented parallelization in one of my packages using the 'parallel' package -- many thanks for providing it! In my package I'm importing 'parallel' and so added it to the DESCRIPTION file's 'Import:' tag and also added a 'importFrom("parallel", ...)' statement in the NAMESPACE file. Parallelization works nicely, but my package

Good Morning We are planning a Windows 2003 migration project through out our company and we are currently using SAMBA on the following platforms. to communicate with our NT servers, MPE/ix 6.5 pp2 and 7.0 HP-UX 11i Linux RedHat 9 Can you please advise which SAMBA version should we install for each of the above O/S's and where we may download these from? Many Thanks David Jason Senior

Hello We are running samba on SGI/SUN/HP fileservers. We saw a problem on the SGI fileservers. After a while a lot of defunct smbd deamons were hanging and occupying the slots in the process table. We saw the problem with IRIX 5.2 and samba 1.9.16p11 and IRIX 6.2 and samba 1.9.17.p4 . Has anybody seen this problem before? This is a serious problem, because the smbd daemons are filling the

Muchas gracias, Fernando y Javier. > pami$PP <- as.numeric(as.character(pami$PP)) Warning message: NAs introduced by coercion Y entonces los datos que antes convertía en labels(pami$PP) (v.g. 753,2256 a 61343 o 62,7688 ? a 17390, que me pa eran enteros del 1 al n de clases como decía Fernando), los convertía en NA.- Luego de probar también con read.table, pasé a: > pami <-

I'm trying to obtain numerical derivative of a probability computed with mvtnorm with respect to its parameters using grad() and jacobian() from NumDeriv. To simplify the matter, here is an example: PP1 <- function(p){ thetac <- p thetae <- 0.323340333 thetab <- -0.280970036 thetao <- 0.770768082 ssigma <- diag(4) ssigma[1,2] <- 0.229502120

Dear R devs, I installed the recent devel R to test a package error when I intercept this warning when loading packages: ``` Warning in .make_numeric_version(x, strict, .standard_regexps()$valid_numeric_version) : invalid non-character version specification 'x' (type: double) ``` After a long debugging, I realize that `numeric_version` in base does not support numerical input x by

En general lo que yo uso en esos casos es as.numeric(as.character(X)) No se los términos correctos pero los factores aunque se muestren con los nombres de las diferentes clases, internamente son clases separadas que se nombran como enteros por ejemplo del 1 al n de clases. Cuando usas directamente as.nuemric sobre un factor, este toma los numeros de las clases y no el valor de clase. Fijate en