similar to: cut POSIX results in NA - bug?

Background: OS: Linux Mandrake 10.1 release: R 2.0.0 editor: GNU Emacs 21.3.2 front-end: ESS 5.2.3 --------------------------------- Colleagues I have a wind speed time series with a normal frequency distribution and a spike in the 5 metres/second bin. The most likely explanation is that the instrument was returning duplicate values at this speed. To check this, I want to extract all the unique

Hi R, I have an excel file in which the third column is "date" and others are "character" and "numeric". Number of columns are 12 If I use this to read the file in R: x = read.xls("D:\\file.xls") The problem is that my date column is read in julian dates. So I am using: x = read.xls("D:\\file.xls",

Hi there =20 I use Gnu R sometimes at work.=20 =20 Unfortunately we use Windows and R has problems with getting the calendar week number with the first week as the one which has at least 4 days. A colleague told me that he has the same problem.=20 =20 Below you can see what I use now, for finding the right week.=20 =20 =20 Kind regards,=20 Samuel Andreas Meichtry =20 =20

I don't use SPSS but following through on your detective work can provide the likely answer. First note that both date numbers are evenly divisible by the number of seconds in a day, i.e. 24*60*60. This suggests that these numbers are seconds since some origin. Since we know "2003/02/11" corresponds to 13264300800 we deduce that the origin must be spss.orig <-

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

I have a monthly price index series x, the related return series y = diff(log(x)) and a POSIXlt date-time variable dp. I would like to apply annual blocks to compute for example annual block maxima and mean of y. When studying the POSIX classes, in the first stage of the learning curve, I computed the maximum drawdown of x: > mdd <- maxdrawdown(x) > max.dd <- mdd$maxdrawdown > from

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

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

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

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'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 have a number of plots that I'm trying to do using the lattice package, unfortunately, they involve Posix values. A small sample of the data is as follows: Time TC.1 <Snipped> 1 2003-04-24 13:29:15 26.61 2 2003-04-24 13:29:30 26.48 3 2003-04-24 13:30:00 26.53 4 2003-04-24 13:30:30 27.85 <Snipped> where str(data) `data.frame': 415 obs. of

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 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 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)

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

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