similar to: problem with unlist POSIX date at midnight

Hi, I have a dataframe column of the form v<-c("Fri Feb 05 20:00:01.43000 2010","Fri Feb 05 20:00:02.274000 2010","Fri Feb 05 20:00:02.274000 2010","Fri Feb 05 20:00:06.34000 2010") I need to convert this to datetime form. I did the following.. lapply(v,function(x){strptime(x, "%a %b %d %H:%M:%OS %Y")}) This gives me a list that looks like

Can someone, please, explain the difference is results below (notice the isdst value) > unlist(as.POSIXlt('2005-7-1')) sec min hour mday mon year wday yday isdst 0 0 0 1 6 105 5 181 1 > unlist(as.POSIXlt(as.Date('2005-7-1'))) sec min hour mday mon year wday yday isdst 0 0 0 1 6 105 5 181 0

Full_Name: Petr Simecek Version: 2.5.1, 2.6.1 OS: Windows XP Submission from: (NULL) (195.113.231.2) Several times I have experienced that a length of a POSIXt vector has not been computed right. Example: tv<-structure(list(sec = c(50, 0, 55, 12, 2, 0, 37, NA, 17, 3, 31 ), min = c(1L, 10L, 11L, 15L, 16L, 18L, 18L, NA, 20L, 22L, 22L ), hour = c(12L, 12L, 12L, 12L, 12L, 12L, 12L, NA, 12L,

Dear R users, Does anyone know if there's a median filter available in R? I have a considerable amount of images as 256 by 256 matrices and want to smooth them with a 5x5 median filter. Until now I'm not lucky in searching the R-help list/files. Thanks, Hanneke ir. J.M. (Hanneke) Schuurmans (PhD student) Department of Physical Geography Faculty of Geosciences, Utrecht University, The

Have you noticed any problems with big dates (>=1/1/2040) in R? Here is the bit of code that I'm having trouble with: > test.date <- strptime("1/1/2040",format="%m/%d/%Y") > > unlist(test.date) sec min hour mday mon year wday yday isdst 0 0 0 1 0 140 0 0 0 > > date.plus.one <- as.POSIXct(test.date) +

Dear all, I have just detected what seems a minor inconsistence with data types. If one unlists a POSIXlt time with GMT zone gets a numeric vector, since the POSIXlt list has no `zone` element, while if one unlists a POSIXlt time with a non GMT zone (also non specifying tz if the Sys.timezone is not GMT) gets a character vector due to including the `zone` element. > x <-

x = structure(list(date = structure(list(sec = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), min = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), hour = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), mday = c(1L, 2L, 3L, 4L, 5L, 8L, 9L, 10L, 11L, 12L, 15L, 16L, 17L, 18L, 19L, 22L, 23L, 24L,

The return value of Sys.time() today with a timezone of US/Eastern is unchanged between 4.0.3-patched and devel, but on devel the following test fails all.equal(x, as.POSIXlt(x)) with x = Sys.time() This means that devel does not complete make tests (failure on tests/reg-tests-2.R) It is entirely possible that it is an error on my end, I use export TZ="US/Eastern" but I have been

Here's a perl script that works when I run it manually. But when I run it via cron, it won't create the directory. But worse than that, an email isn't sent to the account running the job. So I'm not getting an error, but it does work when I run it manually. If I put an obvious error in the script, cron does generate an email giving me the STDERR. What could it be? === Al

I realize it's -12C and we're having the next best thing to a blizzard but why can I not melt this data frame. I am missing something terribly obvious but I just don't understand what the error message is saying. John Kane Kingston ON Canada Code and aata below #================================================# library(reshape2) melt(mydata, id.vars = c("date"))

We're almost at the stage of rolling out our new RH6.1/Samba server - one last item. We have some 60 users and 7 departments. Can I use/how do I use a logon script to log groups onto the server. I currently have 1 logon script per user, but a small change in 1 share means updating every script. Can I log them on by group ID but still map their personal shares with the %U flag so I then

After installing R-3.4.0 I ran 'make check' which halted here: $ > tail reg-tests-1d.Rout.fail -n 16 > ## format()ing invalid hand-constructed POSIXlt objects > d <- as.POSIXlt("2016-12-06"); d$zone <- 1 > tools::assertError(format(d)) > d$zone <- NULL > stopifnot(identical(format(d),"2016-12-06")) > d$zone <- "CET" # =

After installing R-3.4.0 I ran 'make check' which halted here: $ > tail reg-tests-1d.Rout.fail -n 16 > ## format()ing invalid hand-constructed POSIXlt objects > d <- as.POSIXlt("2016-12-06"); d$zone <- 1 > tools::assertError(format(d)) > d$zone <- NULL > stopifnot(identical(format(d),"2016-12-06")) > d$zone <- "CET" # =

# hi all, # I'm trying to plot temperatures by date in a trellis plot by their stations # I'm plotting the following data.frame library(lattice) h <- structure(list(station_name = structure(c(3L, 4L, 2L, 10L, 11L, 12L, 6L, 7L, 5L, 8L, 9L, 3L, 4L, 2L, 10L, 11L, 12L, 6L, 7L, 5L, 8L, 9L, 3L, 4L, 2L, 10L, 11L, 12L, 6L, 7L), .Label = c("Ashqelon", "Beer Sheva",

Yes, the potential issue I see is that make check fails when I explicitly set TZ. However, I set it to be the same as what the system reports when I login. Details: The system (RHEL) I am working on has $ strings /etc/localtime | tail -n 1 EST5EDT,M3.2.0,M11.1.0 $ date +%Z EDT $ echo $TZ US/Eastern On Fri, Oct 2, 2020 at 9:48 AM Sebastian Meyer <seb.meyer at fau.de> wrote: > Thank

I have a vector of dates. I wish to find the month end date for each. Any suggestions? e.g. For 12/15/05, I want 12/31/05, For 10/15/1995, I want 10/31/1995, etc __________________________________________________ [[alternative HTML version deleted]]

Hello! I've been trying to get this right for quite a while now and fear there is an easy solution I just don't see. I did not have this problem in Linux, and I searched r-help and Google but did not find a solution, but of course I am grateful for and resources I might not have found our not understood yet. I try to parse a time stamp with time zone. I essentially just want to parse the

Although the documentation is somewhat sketchy, I() can be used to create objects of class AsIs: > I("a") [1] "a" attr(,"class") [1] "AsIs" "character" > I(4) [1] 4 attr(,"class") [1] "AsIs" "numeric" > I(4 + 0i) [1] 4+0i attr(,"class") [1] "AsIs" "complex" > This

Can anyone point me to a doc or a script that shows how to generate netlogon script on the fly and supports users having multiple groups? I tried the Perl script referenced at <http://us2.samba.org/samba/docs/man/AdvancedNetworkManagement.html> but I believe it only supports the user being a member of one group. -- Andrew Gaffney

> x<-as.POSIXct("1970-1-1", tz="UTC")-.5 > y<-as.POSIXct("1970-1-1", tz="UTC")+.5 > x==y [1] FALSE # of course but x and y "appear" to be the same when formatted, even with extra precision: > format(x, format="%Y-%m-%d %H:%M:%OS2") [1] "1970-01-01 00:00:00.50" > format(y, format="%Y-%m-%d