similar to: Obtaining the day of any given date

I know that this is correct: library(chron) x = dates("01-03-04", format="d-m-y", out.format="day mon year") print(x) It gives me the string "01 Mar 2004" which is correct. I also know that I can say: print(day.of.week(3,1,2004)) in which case he says 1, for today is monday. My question is: How do I combine these two!? :-) I have a

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

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]]

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,

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) +

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 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 <-

Dear all, I have a question related to the POSIXlt function in R. I have a set of dates and times, for exmaple: startx <- as.POSIXct("2011-01-01 00:00:00") finx <- as.POSIXct("2011-12-31 00:00:00") daysx<- seq(startx, finx, by="24 hours") I want to change the dates of all the days falling on a Saturday to the next working day (i.e. Monday). So I

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

Hi all, Is there a simple function already implemented for getting the ISO weeks of a Date object? I couldn't find one, and so wrote my own function to do it, but would appreciate a pointer to the "default" way. If a function is not yet implemented, could the code below be of interest to submit to CRAN? Best Regards, Gustaf --------------------

Thanks Mike. My data has longitude and latitude coords and I used distAB {clim.pact} then mst {ape} to calculate my minimum spanning tree. The nodes are telecoms sites from all over Australia. My goal is to determine the minimum cost of linking them via cabling, and I'm starting by calculating the distance "as the crow flies", but will probably eventually need to calculate the

Dear All At institutes where there are PC labs used by Masters and undergraduate students and software is maintained /managed by a central IT services, do you always update R every time a new version is released - around beginning of April and October? Do you also ask them to install all patch releases as they turn up? Currently our IT department update once a year before the start of the Autumn

Hi I have just upgraded from R2.6.0 to R2.7.1 (running on Windows) and a part of my code that previously ran ok now gives an error. The following is a simple example to demonstrate my problem. > a <- array(c(1,2,3,4,5,6,rep(NA,6)),dim=c(6,2)) > apply(a,2,sd,na.rm=T) In R2.6.0 this gives (which is what I would like) [1] 1.870829 NA In R2.7.1 it gives the following error

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 I was wondering if someone could help me with an rpart problem. I can see that cross-validation is the default for tree selection in rpart -- has a bootstrap method been implemented anywhere? I think this is a different thing to 'bagging' or 'boosting' -- I still want 'one' tree at the end, I just would like it chosen using a bootstrap method. Any ideas??? Thanks

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" # =