similar to: Converting a dataframe column from string to datetime

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

Dear R-users, I use unlist of POSIX dates to extract the year, hour etc. With that I can search for files in my database which are in the form 'yyyymmddhh_synops.txt' However, I get stucked during midnight where unlist just gives NA's. The script is given below, the problem accurs at acc.period[16] (midnight). However when I write out the character, unlist works well. But

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,

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,

A quick fix for anyone running into this problem: load_schedule was crashing cron_trigger.rb#160, and I found that for triggers like "0 20 5 * * * *" it was putting in 32 for the date (even though its Mon Mar 31 04:16:21 BST 2008). Something to do with switching to day light savings I think, which concludes after 2am. I patched it by adding day = 31 if day == 32 on the line before.

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 <-

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

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"))

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" # =

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

# 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",

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]]

> is.na(strptime("5/2/1992", format="%m/%d/%Y")) [1] FALSE > is.na(strptime("5/3/1992", format="%m/%d/%Y")) [1] TRUE Any idea what's going on with this? Running strptime against all dates from around 1946, only 5/3/1992 was converted as "NA". Even stranger, it still seems to have a value associated with it (even though is.na thinks

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

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

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

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

Dear R-help subscribers Would you tell me how to apply SSfpl with binary data as below? Unfortunately, there is not the EXAMPLE in help(SSfpl) for binary data but for quantitative data(Chick). V1: dose V2: log-transformed dose V3: response (rate) V1 V2 V3 1 0.775 -0.2548922 0.1666667 2 5.000 1.6094379 0.8148148 3 10.000 2.3025851 0.5000000 4 20.000 2.9957323