similar to: Timestamps and manipulations

Hi All, I've been trying to compare if the previous value in a variable is equal to a binary value..(i.e i want to check if the last event was a yes or no) i've been trying to write some code for this, but it seems overly elaborate, can anyone suggest a better / shorter / neater way? The below doesn't quite work but shows my idea of splitting by the factor id, then creating a new

Hi All, I seem to be having a few troubles with aggregating data back onto the the dataframe, I want to take the max value of a user, and then apply this max value back against all id's that match (i.e a one to many matching) Can anyone offer any advice? is there a better way of doing this? Dummy data and code are listed below:- dataset is called Mcookie user_id c_we_conversion 1

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")

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:

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:

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

I have just become painfully aware that objects of class "difftime", generated by the difference of two POSIXct objects, carry a "units" attribute, which flashes by when the object is printed, for example. The pain was occasioned when I tried to turn these objects into numberic objects for use elsewhere as a covariate. as.numeric(difftime object) simply turns off the units

On the help page "DateTimeClasses {base}" it says: "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." However, > x<-Sys.time(); y<-Sys.time()+3600 > diff<-y-x > x; y; diff [1] "2005-11-23 19:58:20 GMT" [1] "2005-11-23 20:58:20 GMT" Time difference of 1 hours

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

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",

Hi I just started using RGui.exe under widnows. I have a text file containing date arranged in columns and rows, each column has the same format, each row with different formats. 3 of the columns are something like this 1/12/2006 3:59:45 PM I need to calculate the different in seconds between 2 selected periods using their row’s index My solution: Read the file in a data frame and

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

Colleagues, I submitted this several days ago and no one responded, so I am trying again, trying a different subject line: I just encountered some unexpected behavior of difftime in relationship to the change from daylight savings to standard time. My understanding is that DST and ST take effect at 2AM. However, the results below suggests that R (version 2.8.0 in OS X) implements the

Hi list, Could someone explain to me why the following result is not a integer? > difftime(strptime("24NOV2004", format="%d%b%Y"), strptime("13MAY2004", >format="%d%b%Y"), units="days") Time difference of 195.0417 days I'm using R2.12.0 on WinXP. Thanks! ...Tao

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

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

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

In some cases, incorrect results are produced by the code below intended to calculate the number of days between 2 dates. The year in question was a leap year. Note the results for 2004-04-04 and 2004-04-05 are the same! They should be 37 and 38 respectively. > as.integer(as.POSIXct("2004-04-02") - as.POSIXct("2004-02-27")) [1] 35 >

Dear All, what determines if a difference between POSIXct objects gets expressed in days or seconds? In the following example, it's sometimes seconds, sometimes days. as.POSIXct('2009-09-01') - as.POSIXct(NA) Time difference of NA secs c(as.POSIXct('2009-09-01'), as.POSIXct(NA)) - c(as.POSIXct('2009-09-01'), as.POSIXct('2009-08-31')) Time differences in

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"