similar to: calculating the number of days from dates

The interface for dates in R is a little confusing to me. I want to create a vector of Date objects from vectors of years, months, and days. One solution I found is: years <- c(1991, 1992) months <- c(1, 10) days <- c(1, 2) dates <- as.Date(ISOdate(years, months, days)) But, in this solution the ISOdate function converts the vectors into characters, which can cause serious

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

Hello, today I've updated on the newest R-Version. But sadly a function I needed didnt want to work: The input is e.g. days(as.Date("21-07-2005","%d-%m-%y")) the error is: Fehler in Math.Date(dts): floor nicht definiert f??r Date Objekte (Error in Math.Date(dts): floor not defined for date objects) Same for year. Only months gives me the correct output. In Version

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, 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

note: whole function is below- I am sure I am doing something silly. when I use it like USGS(input="precipitation") it is choking on the precip.1 <- subset(DF, precipitation!="NA") b <- ddply(precip.1$precipitation, .(precip.1$gauge_name), cumsum) DF.precip <- precip.1 DF.precip$precipitation <- b$.data part, but runs fine outside of the function: days=7

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

Hello R-experts! I am using R Version 1.7.1 (2003-06-16) on a Debian Linux box and I have discovered an odd result when plotting data involving dates. Please try this minimal example: a = seq(ISOdate(2000,1,1), ISOdate(2001,1,1), "months") b = 1:13 plot(a,b, col="red") What I get is a plot that looks as expected except the x-axis is mostly red. Can anyone reproduce this

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),

Look at the bottom of the message for my question #here is a little function that I wrote USGS <- function(input="discharge", days=7){ library(chron) library(gsubfn) #021973269 is the Waynesboro Gauge on the Savannah River Proper (SRS) #02102908 is the Flat Creek Gauge (ftbrfcms) #02133500 is the Drowning Creek (ftbrbmcm) #02341800 is the Upatoi Creek Near Columbus (ftbn) #02342500 is

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

Not sure if there is an R way to do this or a regular express way, but here is what I am trying to do. I've got lots of data where the format is HH:MM:SS, but I need to format it like HH:MM:00, i.e. round the second down to zero. What is the best way to do this? Thanks again. Jason

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"

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

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

I decided my earlier email on this topic was rather long and wordy; here's a condensed version. I am sitting at a Solaris computer in the US/Pacific timezone. I have a file of data having times that includes the following three values 2002-4-7 1:30:00 GMT 2002-4-7 2:30:00 GMT 2002-4-7 3:30:00 GMT I have not been able to find a way to correctly convert these to either of the POSIX

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

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

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

The most important thing is that Date objects by definition do not include time of day. You want to look at ISOdatetime() and as.POSIXct() instead. And beware daylight savings time issues. -pd On 18 Apr 2016, at 15:09 , Ogbos Okike <giftedlife2014 at gmail.com> wrote: > Dear All, > > I have a data set containing year, month, day and counts as shown below: > data <-