similar to: vector of factors to POSIXlt

Hello, I have a data.frame > names(popA) [1] "Year" "Series" "Age" "WM" "WF" "HM" "HF" "BM" [9] "BF" "IM" "IF" "AM" "AF" "Yr" how do i loop over a subset of variables in this frame to create a vector of

looking for some thoughts on incorporating R functionality to create histograms of data stored in an informix db. im gonna write the app in .Net and will use a managed provider to access the data. what R libs might I have to package in the assemblies? (sorry my Q is general as Ive only just looked at wanting this yet) Thanks. Amer.

Hi, how it's possible to extract the year and the number of days from Julian date. i'm little confused about the last two functions and ?years . EDATE comes from sqlQuery with as.is=T EDATE <- as.POSIXlt(datvears$ENROLLDAY) Many thanks, Christian > EDATE[1:5] [1] "2000-06-30 11:25:01" "2000-06-30 11:39:55" "2000-06-30 12:11:11" [4]

How can I extract only the time component from an POSIXlt object? For example if I try the following it still returns both the date and time... >as.POSIXlt(tr.date[1]) [1] "2010-10-18 21:46:53" >as.POSIXlt(tr.date[1],"%H:%M:%S") [1] "2010-10-18 21:46:53" round and trunc don't help... is there an "as.Time" equivalent to as.Date ? Thanks,

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

I tried the code that Richard O'Keefe posted last week, to wit: library(chron) ymd.to.POSIXlt <- function (y, m, d) as.POSIXlt(chron(julian(y=y, x=m, d=d))) n <- 100000 y <- sample(1970:2004, n, replace=TRUE) m <- sample(1:12, n, replace=TRUE) d <- sample(1:28, n, replace=TRUE) system.time(ymd.to.POSIXlt(y, m, d)) [1] 8.78 0.10

Dear R users, Given this data: x <- seq(1,100,1) dx <- as.POSIXct(x*900, origin="2007-06-01 00:00:00") dfx <- data.frame(dx) Now to play around for example: subset(dfx, dx > as.POSIXct("2007-06-01 16:00:00")) Ok. Now for some reason I want to extract the datapoints between hours 10:00:00 and 14:00:00, so I thought well: subset(dfx, dx >

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

Should I be able to use axis() on a barplot? i have a data.frame, the first 3 values of which are: > c[1:3,] median mean A1 56.5 58.50000 A61 73.0 73.00000 A62 63.0 63.00000 > str(c) `data.frame': 19 obs. of 2 variables: $ median: num 56.5 73 63 161 51 55 44.5 22 54 49 ... $ mean : num 58.5 73.0 63.0 161.0 47.5 ... if I do barplot(median) and then try

Hey R folks, i found some strange (to me) behaviour with chron to POSIXct conversion. The two lines of code result in two different results, on ewith the correct time zone, one without: library(chron) as.POSIXct(chron('12/12/2000'), tz = 'UTC') as.POSIXlt(chron('12/12/2000'), tz = 'UTC') Only the code below would give me a POSIXct object with the correct time

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. >

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

Hi list, I'm trying to turn a date into something productive. (Not what you may be thinking....) I want three functions so I could take a "date" object and get the day of week, month, and year from it. xx <- as.Date("2006-01-05") month(xx) equal 1 day(xx) equal 5 year(xx) equal 2006 I'm aware of the weekdays() and months() functions in the base package. But

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"

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

Full_Name: Alec Stephenson Version: 1.6.0 OS: linux Submission from: (NULL) (148.88.138.5) Appears to be a sign error in as.POSIX(lt/ct) > library("chron") The following is fine, with default origin. > tmp <- chron(1:2, origin = c(1,1,1970)) > as.POSIXlt(tmp) [1] "1970-01-02 01:00:00 GMT" "1970-01-03 01:00:00 GMT" These are not. > tmp <-

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

I've always found the chron library to be useful for tasks like this: > x <- round(runif(10)*100000, digits=0) > y <- as.Date(x, origin="1970-01-01") > library(chron) > days(y) [1] 7 25 26 25 10 24 1 31 12 8 31 Levels: 1 < 2 < 3 < 4 < 5 < 6 < 7 < 8 < 9 < 10 < 11 < 12 < 13 < ... < 31 Notice that it returns the days as

Hello!! I need some help! I tried everything, but nothing worked! I have a vector c with dates in it, in the format "2004-04-01" and i need to convert it in the form "04/01/2004" or "01/04/2004" ! How can i do that?? Thanks Bye Martina --

Hola a todos: Debe de ser una tontería, pero no consigo saber porque la siguiente linea no devuelve la fecha actual: as.Date(as.numeric(Sys.time())) He hecho esa prueba porque no consigo pasar un numero convertido a partir de una fecha y modificado a fecha de nuevo. Gracias por adelantado. Un saludo, Alberto. [[alternative HTML version deleted]]