search for: isodates

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 under R-1.7.1 on

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, but returns a value [1]

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't work for that reason. I'm

Thanks for this clarification. I have learned in the meantime that it is necessary to be very careful when using all these POSIX things. As another example, here is something that made me scratch my head just yesterday: When I create a sequence of days that happens to start before and ends in daylight savings time, I seem to lose a day: > seq(from = strptime("20030329",

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

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 Members, Compliments of the Season!! Below is a part of a code I use for Fourier analysis of signals. The code handles data with the format 05 01 01 8628 (year, month, day and count) 05 01 02 8589 (year, month, day and count) The sample data is attached as 2005daily.txt. I would like to adapt the code to handle data of the form: 05 01 01 00 4009

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 value as if you hadn't passed it. Jim On Mon, Jan 22, 2018 at 6:01

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),

I am using R-1.7.0 and have some data which consist of one vector of numbers and a second corresponding vector of dates belonging to the POSIXct class. I would like to plot the numbers against the dates. What is the best way to do this? It almost works to just call `plot.' However if I do this while using the `ylab' parameter I get a warning message: parameter "ylab"

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

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

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 development (unstable) major 1 minor 5.0 year 2002 month

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 POSIXlt object that is then

Dear All, I have a data set containing year, month, day and counts as shown below: data <- read.table("data.txt", 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)

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

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