similar to: Posix Problem, difftime

Displaying 20 results from an estimated 2000 matches similar to: "Posix Problem, difftime"

2007 Mar 22
2
R difftime function: How can we fix the difftime unit?
Hi, I am trying to take difference of two time objects. I want to fix the result's unit to minutes. How can I do that? Here is an example: > difftime(x, y) Time difference of 2.030720 hours > difftime(x, z) Time difference of 30.34672 mins where x = '2007-03-05 08:32:58' y = '2007-03-05 06:31:07' and z = '2007-03-05 08:02:37' How can I get answer
2010 Jun 22
1
New errors with difftime()-objects in 2.11.1 (was Re: Request: difftime method for cut())
On Thu, Jun 10, 2010 at 3:39 PM, Gustaf Rydevik <gustaf.rydevik at gmail.com> wrote: > Hi all, > > The recent change in 2.11 that made as.numeric() return false on > difftime-objects broke some of my code that calculated age classes of > individuals using cut(). While this was no big thing to fix for me, it > might be wise > to provide a cut.difftime method to ?stop
2010 Jun 22
1
New errors with difftime()-objects in 2.11.1 (was Re: Request: difftime method for cut())
On Thu, Jun 10, 2010 at 3:39 PM, Gustaf Rydevik <gustaf.rydevik at gmail.com> wrote: > Hi all, > > The recent change in 2.11 that made as.numeric() return false on > difftime-objects broke some of my code that calculated age classes of > individuals using cut(). While this was no big thing to fix for me, it > might be wise > to provide a cut.difftime method to ?stop
2006 Apr 03
1
weird "max" behavior for difftime class
If you apply the "max" function to a vector of class "difftime" with units="days", the returned value is in units of "seconds". Is this not a bug? At any rate it can lead to confusing results if one buries a call to "max" deep in some data analysis code. Details: > y<-structure(1, class = "difftime", units = "days")
2009 Feb 06
1
Operations on difftime (abs, /, c)
Since both comparison and negation are well-defined for time differences, I wonder why abs and division are not defined for class difftime. This behavior is clearly documented on the man page: "limited arithmetic is available on 'difftime' objects"; but why? Both are natural, semantically sound, and useful operations and I see no obvious reason that they should give an error:
2009 Feb 06
1
Operations on difftime (abs, /, c)
Since both comparison and negation are well-defined for time differences, I wonder why abs and division are not defined for class difftime. This behavior is clearly documented on the man page: "limited arithmetic is available on 'difftime' objects"; but why? Both are natural, semantically sound, and useful operations and I see no obvious reason that they should give an error:
2013 Jul 09
1
Is difftime a "class"
I am trying to write S4 methods with "difftime" in the signature but am being "informed" (? not a warning or error) that "difftime" is not a class. Nevertheless, dispatch takes place. Should I simply ignore that "information"? Here is a toy example: > setClass("foo", contains = "Date") > setMethod("+", c("foo",
2011 Oct 25
2
difftime producing NA values in R 2.12.2
R-listers, I have noticed several posts on issues with difftime producing NA's but they have been for older versions of R. Here's the issue associated with difftime that I am dealing with in R 2.12.2. > preciptime = strptime("01/10/2007 14:00",format="%m/%d/%Y %H:%M") > class(preciptime) [1] "POSIXlt" "POSIXt" > # Now using difftime, this
2018 Aug 01
1
RFC: make as.difftime more consistent or convenient
Hello! you, Emil Bode <emil.bode at dans.knaw.nl>, wrote on Tuesday, July 31, 2018 1:55 PM: > Some of the changes you're proposing could be made (with effort), but note that you're not > restricted to providing strings with a format. > What you're trying to do can be accomplished with as.difftime(12, units='weeks'), see also > ?as.difftime > > Or if
2010 Apr 05
1
using difftime()
I'm new to R and have the following problem with difftime: if I directly assign date/time strings in difftime I get the expected result: > a<-"2010-03-23 10:52:00" > a [1] "2010-03-23 10:52:00" > b<-"2010-03-23 11:53:00" > u2<-as.difftime(c(a,b), format ="%Y-%m-%d %H:%M:%S", units="mins") > u2 Time differences in mins
2007 Feb 21
1
Adding difftime objects to POSIXt objects
Hello, ?DateTimeClasses states that "one can add or subtract a number of seconds or a 'difftime' object from a date-time object, but not add two date-time objects." So, is the below expected behavior? > x <- Sys.time() > x [1] "2007-02-21 16:19:56 CST" > x + as.difftime("1","%H") [1] "2007-02-21 16:19:57 CST" Warning
2009 Sep 25
1
Collision between difftime and ggplot2.
It seems that there are several folks Out There with an itch to scratch with respect to difftimes. http://article.gmane.org/gmane.comp.lang.r.devel/19223/match=difftime http://article.gmane.org/gmane.comp.lang.r.devel/18441/match=difftime http://article.gmane.org/gmane.comp.lang.r.devel/10882/match=difftime http://article.gmane.org/gmane.comp.lang.r.devel/11675/match=difftime and I might be
2007 Oct 08
2
Incompatible methods ("-.POSIXt", "Ops.difftime") for "-"
Dear all, according to the Help-page of DateTimeClasses {base} I should be able to do time - z with time date-time objects z a numeric vector (in seconds) or an object of class "difftime". However, on R version 2.6.0 (Windows XP) I get > Sys.time() - as.difftime(c("0:3:20", "11:23:15")) Time differences in mins [1] 1191837998 1191837318
2002 Dec 04
2
difftime arithmetic (PR#2345)
Full_Name: Barry Rowlingson Version: 1.6.0 OS: RH8 i386 Submission from: (NULL) (148.88.136.205) Strange things happen if I premultiply a difftime() object with a number. Example: > d1 <- difftime(Sys.time(),Sys.time()) > d2 <- 1 * difftime(Sys.time(),Sys.time()) > d3 <- difftime(Sys.time(),Sys.time()) * 1 > d1 Time difference of 0 secs - thats fine > d2 [1] 0
2012 Mar 19
1
diff(time) vs. difftime?
I just encountered another RTFM problem: With diff(as.POSIXct(...), ...) I was unable to control the units of the results. Examples: > (d.d <- diff(as.POSIXct(c('2012-12-12', '2012-12-13')))) Time difference of 1 days > (d.h <- diff(as.POSIXct(c('2012-12-12 08:00', '2012-12-12 09:00')))) Time difference of 1 hours > (d.m <-
2011 Aug 14
2
Trouble: Time Difference with difftime
Hello all!!! I want to measure the duration of events (given a start and an end time). The catch is that I require the output in calender days. This means: 02-Jan-2011 00:01:00 minus 01-Jan-2011 23:59:00 should be 1 day (although the real time difference is only 2 minutes) My data is the following head(episode.ct) [1] "2009-07-13 13:37:20 CEST" "2009-07-14 07:29:20 CEST"
2008 Dec 09
3
difftime
Hi. I'm trying to take the difference in days between two times. Can you point out what's wrong, or suggest a different function? When I try the following code, The following code works fine: a <- strptime(1911100807,format="%Y%m%d%H",tz="GMT") b <- strptime(1911102718,format="%Y%m%d%H",tz="GMT") x <- difftime(b, a,
2008 May 21
3
Converting a 'difftime' to integer - How to???
I want to find the DOY (Day of Year) of some dates. I think to substract the date 1. January from the data to achive this. Something like: > d <- as.Date("2006-03-13") - as.Date("2006-01-01") +1 > d Time difference of 72 days So far so good. But d is a 'difftime' object. How do I get an Integer value from that? I tried severel things, incuding the
2008 May 14
1
Time differences (as.difftime?) issue
Dear all, I have a vector generated using the function strptime: > my.dt [1] "2004-04-19 08:35:00 W. Europe Daylight Time" "2004-04-19 09:35:00 W. Europe Daylight Time" "2004-04-19 11:35:00 W. Europe Daylight Time" [4] "2004-04-19 13:35:00 W. Europe Daylight Time" "2004-04-20 07:50:00 W. Europe Daylight Time" > class(my.dt) [1]
2009 Mar 17
1
Mean of difftime vectors : "code infelicity" or intended behaviour ?
Dear list, "+" (and "-") being defined for difftime class, I expected mean() to return something sensible. This is only half-true : > mean(c(1:5, 5:1),na.rm=TRUE) [1] 3 > mean(as.difftime(c(1:5, 5:1),unit="mins"),na.rm=TRUE) Time difference of 3 mins Fine so far. However : > mean(c(1:5, NA,5:1),na.rm=TRUE) [1] 3 > mean(as.difftime(c(1:5,