similar to: time zones and the chron to POSIXct conversion

Dear R list, I noticed the following 'problem' when changing the format of dates created with seq.dates() (from the Chron library) using as.POSIXct() (R 1.8.0 on OSX 10.2.8): > datesTest<-seq.dates(from="10/01/1952", length=3, by="days"); > datesTest [1] 10/01/52 10/02/52 10/03/52 # Now changing the format to show year as 1952. >

Dear R People: Here is an interesting question: >library(chron) >xt <- seq.dates(from="01/01/2004",by="days",length=5) >xt [1] 01/01/04 01/02/04 01/03/04 01/04/04 01/05/04 > #Fine so far >as.POSIXct(xt) [1] "2003-12-31 18:00:00 Central Standard Time" [2] "2004-01-01 18:00:00 Central Standard Time" [3] "2004-01-02 18:00:00 Central

Dear R-helpers, I have a problem converting an object made with the 'chron' function to a POSIXct object: # Make date based on DOY dat <- chron(dates=232, origin.=c(month=1, day=1, year=2008)) dat #[1] 08/20/08 # Converting to POSIXct uses current timezone (Sydney): as.POSIXct(dat) #[1] "2008-08-20 10:00:00 EST" # Setting GMT timezone has no effect? as.POSIXct(dat,

Hi list, I have some questions regarding 1) conversion of date + time characters to chron 2) formatting chron object printing Regarding (1), Gabor's Rnews 2004 4/1 article has been indispensible, but I often work with files where dates and times are contained in a single field. In this case, I would like to control input/output of chron objects when each observation of date and time is

Dear All- I am trying to merge two data files - they have different date formats and different times zones. I need to match up the date/time of the datasets and then invoke a conditional statement, such as: if dataC$mph is >= 12 then keep dataM$co23 for the corresponding time/date stamp. snippets of data files: *dataC.txt* LST in mph Deg DegF DegF2 % volts Deg

Running R 2.13.1 on Windows XP. I would like to get week of the year (1-52) for each date. library(chron) dts <- dates(c("02/27/92", "02/27/92", "01/14/92","02/28/92", "02/01/92")) dts dts.chron <- as.chron(dts) dts.chron class(dts.chron) # all of these component extractions work: months(dts.chron) weekdays(dts.chron) years(dts.chron)

the date were created with chron with this argument format=c(dates="Y/m/d", times="H:M:S")) so I have the dates being displayed as (10/06/22 12:00:00) I would like to have them displayed as "2010-06-22 12:00:00" or "%Y-%m-%d %H:%M:%S" and then I can convert these for mergeing with another data frame x <- (structure(c(14464, 14464.0104166667,

Hi everybody How do I do to calculate the median and average of a colum of time data like this: "8:50:10". I also need to plot the time difference between two colums Thanks a lot -- View this message in context: http://n4.nabble.com/median-for-time-data-tp932287p932287.html Sent from the R help mailing list archive at Nabble.com.

Hi R users, I have a maybe small problem which I cannot solve by myself. I want to convert "chron" "dates" "times" (04/30/06 11:35:00) to a number with as.POSIXct. The Problem is that I can't choose different timezones. I always get "CEST" and not "UTC" what I need. date = as.POSIXct(y,tz="UTC") "2006-04-30 11:35:00

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 Gabor and everybody; Thanks Gabor, with the alternative step you've told me the problem is resolved. Comparing the two procedures: Extract from the source 'character' data: > rain$ts[2039:2046] [1] "25/03/2000 22:00:00 UTC" "25/03/2000 23:00:00 UTC" [3] "26/03/2000 00:00:00 UTC" "26/03/2000 01:00:00 UTC" [5] "26/03/2000 02:00:00

Dear R users, I have the following script to create bins of specified time intervals bin_end=60/bin_size bin_size=bin_size*100 h=seq(070000,180000,by=10000) breaks=c() for (i in h) { for (j in 0:(bin_end-1)) { value=i+(bin_size)*j breaks=append(breaks,value) } } I would like to plot then using the time as x-axis. I tried the following prova=zoo(myseries,times(breaks)) but of

Hi. This feels like a bug to me, or at least an undocumented feature, but I thought I'd see what people here thought of it. Consider a POSIXlt object like this one: > a <- as.POSIXlt ("2011-01-23 12:45:45") > class (a) [1] "POSIXlt" "POSIXt" Fine. Now, if I do some arithmetic on that object, the result is converted to POSIXct. > class (a

Full_Name: Xiao Gang FAN Version: 2.4.0 OS: Windows Submission from: (NULL) (159.50.101.9) > library(chron) > as.Date(as.POSIXct(strptime("1994-01-24","%Y-%m-%d"))) [1] "1994-01-23"

Hi! Is not there an standard R function to retrieve the day of the year (since 1st Jan of the same year)? I know I can make my own using julian, but find it weird that having days(), months() etc doy() does not exist as an standard function. Also, is the following not a bit inconsistent? > a <- chron("20100506",format="ymd") > a [1] 100506 > years(a) [1] 2010

Hi all; I've already send a similar e-mail to the list and Prof. Brian Ripley answered me but my doubts remain unresolved. Thanks for the clarification, but perhaps I wasn't clear enough in posting my questions. I've got a postgres database which I read into R. The first column is Timestamp with timezone, and my data are already in UTC format. An 'printed' extract of R

Hello r-help, I'm trying to split a character vector of dates of the form "dd-mmm-yy" into component days, months and years. But > library(chron) > testdate<-as.POSIXlt(strptime("17-Sep-98","%d-%b-%y")) > testdate [1] "1998-09-17" > months(testdate) [1] "September" # which is fine, but for days and years I get >

The problem with POSIXt is that you must consider timezones and daylight vs. standard time issues even if you don't want to. This violates modularity (viz. your routines becomes coupled to unrelated information) and leads to subtle errors where different routines are assuming different time zones. The problem is that the time, date, day of the week, month, etc. of a date depend on its

Hi, I'm a bit confused how julian() works. If I understand right, it returns the number of days since the origin. I have a vector: > SLDATX[1:10] [1] "1986-01-06" "1986-01-17" "1986-02-02" "1986-02-04" [5] "1986-02-04" "1986-02-21" "1986-03-06" "1986-03-25" [9] "1986-04-06"

Dear list, this might be an easy one, but I could figure out a solution (or how to google the right term). Is there any way to test whether all elements of a vector are identical? For numeric vectors I would use sum(diff(vector)==0)==0 but I have character vectors. Any Ideas? Cheers Jannis