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Dear all, I have found the following (for me) incomprehensible behaviour of ISOdate (POSIXct): > ISOdate(1900,6,16) [1] "1900-06-15 14:00:00 Westeurop?ische Sommerzeit" > ISOdate(1950,6,16) [1] "1950-06-16 14:00:00 Westeurop?ische Sommerzeit" Note that in the first case I get the 15th of June back, not the 16th as I would have expected! This happened...

Dear R-people! I am using R 1.8.0, under Windows XP. While using ISOdate() and strptime(), I noticed the following behaviour when "wrong" arguments (e.g., months>12) are given to these functions: > ISOdate(year=2003,month=2,day=20) #ok [1] "2003-02-20 13:00:00 Westeurop?ische Normalzeit" > ISOdate(year=2003,month=2,day=30) #wrong day, bu...

hi, I've troubles with some difftime objects. e.g. ISOdate(2001, 4, 26) - ISOdate(2001, 2, 26) - 2 works, telling me "Time difference of 57 days". But when I'd like to add days, such as ISOdate(2001, 4, 26) - ISOdate(2001, 2, 26) + 2 the function gives me an error. Function "as.COMDate.chron" of the Rbloomberg package doesn...

...(for me at least ;-)). -Heinrich. > -----Urspr?ngliche Nachricht----- > Von: Prof Brian Ripley [mailto:ripley at stats.ox.ac.uk] > Gesendet: Samstag, 22. November 2003 20:56 > An: RINNER Heinrich > Cc: r-help at stat.math.ethz.ch; Simon.Fear at synequanon.com > Betreff: Re: [R] ISOdate() and strptime() > > > Confirmation that this *is* an OS-specific problem: A professional > implementation of the POSIX standard (Solaris) gets all of > these correct. > > Your so-called OS lacks any implementation of strptime, so we > borrowed one > from glibc....

Hi, Given a date, how do I get the last date of that month? I have data in the form YYYYMM, that I've read as a date using > x$Date <- as.Date(ISOdate(substr(x$YearEnd,1,4),substr(x$YearEnd,5,6),1)) But this gives the first day of the month. To get the last day of the month, I tried > as.Date(as.yearmon(x$Date,frac=0)) But I don't get the last day of the month here. (Tried frac=1 too.) I then add a month to the date, substract one...

...;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 fails for objects of class POSIXct. > class(ISOdate(2002, 5, 21)) [1] "POSIXt" "POSIXct" > I(ISOdate(2002, 5, 21)) Error in names(x) : evaluation is nested too deeply: infinite recursion? > Interestingly enough, it produces an (incorrect?) answer for objects of class POSIXlt > class(as.POSIXlt(ISOdate(2002, 5, 21))...

...ur kind inputs. Ogbos data <- read.table("2005daily.txt", col.names = c("year", "month", "day", "counts")) new.century <- data$year < 50 data$year <- ifelse(new.century, data$year + 2000, data$year + 1900) data$date <- as.Date(ISOdate(data$year, data$month, data$day)) x1 = data$date y = data$counts

I use POSIXct for datetimes. Is thee a timeAlign function that I can use where : align by year direction -1 ==> start of this year direction 1 ==> start of next year align by week direction -1 ==> date on last sunday direction 1 ==> date on next sunday align by day direction -1 ==> time at past midnight direction 1 ==> time at this comming

Hi Ogbos, You can just use ISOdate. If you pass more values, it will process them: ISOdate(2018,01,22) [1] "2018-01-22 12:00:00 GMT" > ISOdate(2018,01,22,18,17) [1] "2018-01-22 18:17:00 GMT" Add something like: if(is.null(data$hour),data$hour<-12 then pass data$hour as it will default to the same val...

I'd like to make a plot showing frequency of an event. The data is in a data from that includes Year, Month and Day (of month) fields, so I created a Date with ISOdate(Year, Month, Day, tz=''). I can plot frequencies for the year 2002 with > thisyear <- Date[Year==2002] > hist( thisyear, xaxt='n' ) > axis.POSIXct( 1, at=seq(min(thisyear), max(thisyear), by="month"), format='%b %Y' ) Now I...

...lmost works to just call `plot.' However if I do this while using the `ylab' parameter I get a warning message: parameter "ylab" couldn't be set in high-level plot() function Here is a function that demonstrates the behavior. ylabProblem <- function() { x <- ISOdate(2003, 4, 1:10) # POSIXct vector y <- rnorm(10) plot(x, y, ylab = 'I am y') } It works to invoke the low-level plotting routines by hand as follows: ylabNoProblem <- function() { x <- ISOdate(2003, 4, 1:10) # POSIXct vector y <- rnorm(...

Dear all I try to make hourly average by cut() function, which almost works as *I* expected. What puzled me is that if there is only one item at the end of your data it results in NA. Example will explain what I mean datum<-seq(ISOdate(2004,8,31), ISOdate(2004,9,1), "min") cut(datum[1370:1381],"hour", labels=F) [1] 1 1 1 1 1 1 1 1 1 1 1 NA cut(datum[1370:1382],"hour", labels=F) [1] 1 1 1 1 1 1 1 1 1 1 1 2 2 I do not understand why the last item in first call is NA. I found it only...

Hello out there.... Again I have a problem and I stuck... How can I sort a vector of dates? For example I have the vector a<-ISOdate(2001, 1, 1) + 70*86400*runif(10) How can this vector be sorted chronological? And what's the function I should work with to handle these entries? (in sense of: which(a>2001-01-04) or somehting like that) Thank you for helping M.Kirschbaum [[alternative HTML version deleted]]

In today's pre1.5.0 > ISOdate(2002,1,1) [1] "2002-03-01 04:00:00 PST" > ISOdate(2002,1,1)==ISOdate(2002,3,1) [1] TRUE It doesn't seem to happen for days other than 1/1 -thomas > version _ platform i686-pc-linux-gnu arch i686 os linux-gnu system i686, linux-gnu status Under deve...

Dear list: working with date/times I have come across a problem that ISOdate and ISOdatetime are too slow on large vectors of data. I was surprised just until I looked at the implementation and the man page: "ISOdatetime and ISOdate are convenience wrappers for strptime". In other terms, they convert data to character representation first in order to create a P...

...uot;, col.names = c("year", "month", "day", "counts")) Using the formula below, I converted the data to as date and plotted. new.century <- data$year < 70 data$year <- ifelse(new.century, data$year + 2000, data$year + 1900) data$date <- as.Date(ISOdate(data$year, data$month, data$day)) The form of the data is: 16 1 19 9078 16 1 20 9060 16 1 21 9090 16 1 22 9080 16 1 23 9121 16 1 24 9199 16 1 25 9289 16 1 26 9285 16 1 27 9245 16 1 28 9223 16 1 29 9298 16 1 30 9327 16 1 31 9365 Now, I wish to include time (hour) in my data. The new data is of...

Comparison of Dovecot, Uwash, Courier, Cyrus and M-Box: http://www.isode.com/whitepapers/mbox-benchmark.html

Dear Brian and other R-developers, I have to say that I don't understand why what I wrote should have caused any offence. A smile was what I was hoping for. You know I devote more time than I am supposed to, to support R and its users, in partial repayment of my immeasurable debt to all the Developers. It's not much, it's sometimes misguided (I later discover), and my resources

Hi, I have a data.frame like this: y <- rnorm(60) lev <- gl(3,20, labels=paste("lev", 1:3, sep="")) date1 <- as.Date(seq(ISOdate(2007,9,1), ISOdate(2007,11,5), by=60*60*24)) date1 <- date1[-c(3,4,15,34,38,40)] df <- data.frame(lev=lev, date1=date1, y=y) I would like to produce a new data.frame with missing days in df$date1 in each df$lev, like this: lev date1 1 lev1 2007-09-03 2 lev1 2007-09...

Hi, I am new to R, coming from a few years using Stata. I've been twisting my brain and checking several R and S references over the last few days to try to solve this data management problem: I have a data set with a unique patient identifier that is repeated along multiple rows, a variable with month of patient encounter, and a continous variable for cost of individual encounters. The data