search for: wdays

Hi all, Is there a simple function already implemented for getting the ISO weeks of a Date object? I couldn't find one, and so wrote my own function to do it, but would appreciate a pointer to the "default" way. If a function is not yet implemented, could the code below be of interest to submit to CRAN? Best Regards, Gustaf --------------------

Hi All, Two years back, I posted a small function for getting the ISO 8601 defined week number of a date (such as the week number used in all Swedish calendars), in a os-independent manner. I've since discovered an inaccuracy in that code, and so I thought I'd repost the corrected version. Hopefully this will come in handy for someone searching the mailing list archives in the future.

Wanted to raise two questions: 1. Is bugs.r-project.org down? I haven't been able to reach it for two or three days: ``` ping bugs.r-project.org PING rbugs.research.att.com (207.140.168.137): 56 data bytes Request timeout for icmp_seq 0 Request timeout for icmp_seq 1 Request timeout for icmp_seq 2 Request timeout for icmp_seq 3 Request timeout for icmp_seq 4 Request timeout for icmp_seq 5

Hey guys, I was playing with the cron trigger, and I noticed that wday is not honored. Looking at the code confirms this. Has anyone else seen this? Thanks Adam -------------- next part -------------- An HTML attachment was scrubbed... URL: http://rubyforge.org/pipermail/backgroundrb-devel/attachments/20071217/44d3a8a6/attachment.html

Hello, I need to create a dataframe containing all possible combinations of three variables: SITE (101,102,103,104), WDAY (MON,TUE,WED,THR,FRI), and TOD (MORN, AFTN). There should be a total of 40 unique combinations in my dataframe. I used expand.grid() successfully(?) to create my dataframe, but then when I went to order it by SITE, the resultant dataframe only contained four rows, one for each

date <- as.POSIXlt(Sys.time()) #present date for (i in 1:difftime(as.POSIXlt(Sys.Date()),"2007-10-01")) if (date$wday != 0 & date$wday != 6) {print(date);assign("date", (date-86400))} else (assign("date", (date-86400))) I am trying to print dates from present day to a day in the past, but omitting weekends. I am not doing something right, but can't

Hi, I have a dataframe column of the form v<-c("Fri Feb 05 20:00:01.43000 2010","Fri Feb 05 20:00:02.274000 2010","Fri Feb 05 20:00:02.274000 2010","Fri Feb 05 20:00:06.34000 2010") I need to convert this to datetime form. I did the following.. lapply(v,function(x){strptime(x, "%a %b %d %H:%M:%OS %Y")}) This gives me a list that looks like

Dear all, I have a question related to the POSIXlt function in R. I have a set of dates and times, for exmaple: startx <- as.POSIXct("2011-01-01 00:00:00") finx <- as.POSIXct("2011-12-31 00:00:00") daysx<- seq(startx, finx, by="24 hours") I want to change the dates of all the days falling on a Saturday to the next working day (i.e. Monday). So I

Have you noticed any problems with big dates (>=1/1/2040) in R? Here is the bit of code that I'm having trouble with: > test.date <- strptime("1/1/2040",format="%m/%d/%Y") > > unlist(test.date) sec min hour mday mon year wday yday isdst 0 0 0 1 0 140 0 0 0 > > date.plus.one <- as.POSIXct(test.date) +

Dear R-users, I use unlist of POSIX dates to extract the year, hour etc. With that I can search for files in my database which are in the form 'yyyymmddhh_synops.txt' However, I get stucked during midnight where unlist just gives NA's. The script is given below, the problem accurs at acc.period[16] (midnight). However when I write out the character, unlist works well. But

I have the code for computing a weekly event: (start_date..end_date).select{|d| d.wday == wday} but I am wondering if anyone has any ideas for using this to figure out a recurring event for every other week. Is it possible to somehow specify every other within the .select method? Any other implementation ideas? --~--~---------~--~----~------------~-------~--~----~ You received this message

Gabor, Thanks much. Your solution is elegant. My overall scheme is to take present date, and check whether it is a weekend, if not, then create a string based on the date, to concatenate into a url link for download.file( ). The files I need to download have a part which is in the format: mmddyy. I am working to make myself a system to connect to exchanges, and download end of day files from

I am trying to download a bunch of files from a server, for which I am using download.file( ) within a for loop. The script is working fine except until download.file hits a URL which has no file, at which point it exits. I want to change this behavior to simple log the failure and maintain state within the for loop and iterate to next. I read about try / tryCatch but am having trouble

John Putz wrote: > Thanks. > > */ > email: /**/johnputz3655@yahoo.com/* <mailto:johnputz3655@yahoo.com> > */home: 206-632-6522 > cell: 206-910-5229/* > > > ----- Original Message ---- > From: Joe W. Byers <ecjbosu@aol.com> > To: John Putz <johnputz3655@yahoo.com> > Cc: r-sig-finance@stat.math.ethz.ch > Sent: Friday, December 7, 2007

Can someone, please, explain the difference is results below (notice the isdst value) > unlist(as.POSIXlt('2005-7-1')) sec min hour mday mon year wday yday isdst 0 0 0 1 6 105 5 181 1 > unlist(as.POSIXlt(as.Date('2005-7-1'))) sec min hour mday mon year wday yday isdst 0 0 0 1 6 105 5 181 0

Hi, Is it possible to create a ts() object, whose data is daily based BUT measured only on working days? In other words, suppose I have a data set with 255 observations, measured from 29 June 1959 to 30 June 1960. How would I create such a data? I tried something like: ts(c(...), start(1959, 180)) but I'm not sure what to use for frequency. In other words I don't know how to

Hey, I need to get the number of years (as a number) between two dates. Here is what I have as a helper. # Returns the number of years between now and the specified date. def years_ago(date) dateDifference = DateTime.now - date results =Date.day_fraction_to_time(dateDifference) return results[0] / 24 / 365; end I''m sure there is a better way to do this. What is it?

Full_Name: Petr Simecek Version: 2.5.1, 2.6.1 OS: Windows XP Submission from: (NULL) (195.113.231.2) Several times I have experienced that a length of a POSIXt vector has not been computed right. Example: tv<-structure(list(sec = c(50, 0, 55, 12, 2, 0, 37, NA, 17, 3, 31 ), min = c(1L, 10L, 11L, 15L, 16L, 18L, 18L, NA, 20L, 22L, 22L ), hour = c(12L, 12L, 12L, 12L, 12L, 12L, 12L, NA, 12L,

x = structure(list(date = structure(list(sec = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), min = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), hour = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), mday = c(1L, 2L, 3L, 4L, 5L, 8L, 9L, 10L, 11L, 12L, 15L, 16L, 17L, 18L, 19L, 22L, 23L, 24L,

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