On Apr 21, 2010, at 9:40 AM, Michael Hosack wrote:
Thank you David, this approach is a start in the right direction but
it does
not yield the needed results. I need new week numbers to only begin
on Saturdays. The only exception will be for the first date (May 01)
which will start week 1 on a different day of the week depending on
the year. The proceeding Saturday will begin Week #2. The approach you
provided does not end Week 1 on a Friday and does not end subsequent
week #'s on Fridays.
My solution should be adjustable for any day-of-week ending number.
Why don't you provide a minimal example for testing and show how apply
my solution fails? Your original version was definitely not "minimal."
--
David.
Thanks again,
Mike
----------------------------------------
CC: mhosa...@hotmail.com; r-help@r-project.org
From: dwinsem...@comcast.net
To: dwinsem...@comcast.net
Subject: Re: [R] Assigning Week Numbers
Date: Tue, 20 Apr 2010 16:03:09 -0400
On Apr 20, 2010, at 2:55 PM, David Winsemius wrote:
On Apr 20, 2010, at 1:59 PM, Michael Hosack wrote:
R experts,
How could I extract the week number from a date vector (in Date
class)
such that week numbering (week 1...2...) begins (May 01) and ends
(October 31) on the same specific dates each year? Week numbering
must conform to the following day numbering format
(Sat=1,Sun=2,Mon=3.....Fri=7).
This means that new weeks must begin on Saturdays, and end on
Fridays
(except for the first date of May 01, which always begins week 1;
week 2
begins on the proceeding Saturday). This needs to be applicable
across years
to work effectively. I have tried using both vectorized and loop
approaches with
no success.
Modulo arithmetic will work if you first convert the difftime object
to integer:
And you can use teh chron function day.of.week to return an offset
for
the first week
weekdays(as.Date("2013-05-01") )
[1] "Wednesday"
day.of.week(2013, 5, 1)
[1] 5
SCHEDULE3$wkoffset <- as.integer(SCHEDULE3$DATE. -
as.Date("2013-05-01")- day.of.week(2013, 5, 1) ) %/% 7
range(SCHEDULE3$wkoffset)
[1] -1 25
# So you need a further adjustment of 2 to set the starting week
number to "1":
SCHEDULE3$wkoffset <- 2+as.integer(SCHEDULE3$DATE. -
as.Date("2013-05-01")- day.of.week(2013, 5, 1) ) %/% 7
range(SCHEDULE3$wkoffset)
[1] 1 27
SCHEDULE3$wkoffset <- as.integer(SCHEDULE3$DATE. -
as.Date("2010-05-01") ) %/% 7
str(head(SCHEDULE3))
'data.frame': 6 obs. of 9 variables:
$ DATE. :Class 'Date' int [1:6] 15826 15826 15826 15826 15826 15826
$ YEAR : num 2013 2013 2013 2013 2013 ...
$ MONTH : num 5 5 5 5 5 5
$ DAY : num 1 1 1 1 1 1
$ DOW : chr "Wed" "Wed" "Wed" "Wed" ...
$ SITE : num 101 101 102 102 103 103
$ TOD : Factor w/ 2 levels "MORN","AFTN": 1 2 1 2 1 2
$ DOW. : num 5 5 5 5 5 5
$ wkoffset: num 156 156 156 156 156 156
range(SCHEDULE3$DATE.)
[1] "2013-05-01" "2013-10-31"
range(SCHEDULE3$wkoffset)
[1] 156 182
I am including a bit of old Systat code that does the trick simply
and concisely.
If anyone knows an analogous method in R please let me know. My R
dataframe contains
all the variables and data in the Systat temp file.
Use sched3.t
Save sched4.t
Hold
By mm dd
If bof then let week=1
Else if bog and DOW$="SAT" then let week = week + 1
Run
Thank you,
Mike
SCHEDULE3 <-
structure(list(DATE. = structure(c(15826L, 15826L, 15826L, 15826L,
snipped the rather larger than minimal example.
David Winsemius, MD
West Hartford, CT
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