How to Get the Same Weekday Last Year Given any Given Year? - r

Question

I would like to get the same day last year given any year. How can I best do this in R. For example, given Sunday 2010/01/03, I would like to obtain the Sunday of the same week the year before.
# "Sunday"
weekdays(as.Date("2010/01/03", format="%Y/%m/%d"))
# "Saturday"
weekdays(as.Date("2009/01/03", format="%Y/%m/%d"))

Answer 1

To find the same weekday one year ago, simply subtract 52 weeks or 364 days from the given date:
d <- as.Date("2010-01-03")
weekdays(d)
#[1] "Sunday"
d - 52L * 7L
#[1] "2009-01-04"
weekdays(d - 52L * 7L)
#[1] "Sunday"
Please note that the calendar year has 365 days (or 366 days in a leap year) which is one or two days more than 52 weeks. So, the calendar date of the same weekday one year ago moves on by one or two days. (Or, it explains why New Year's Eve is always on a different weekday.)

Answer 2

Using lubridate the following formula will give you the corresponding weekday in the same week in the previous year:
as.Date(dDate - 364 - ifelse(weekdays( dDate - 363) == weekdays( dDate ), 1, 0))
Where dDate is some date, i.e. dDate <- as.Date("2016-02-29"). The ifelse accounts for leap years.

Answer 3

Here's a simple algorithm. subtract 365 days from the day of interest. Adjust that day to the closest matching day of the week using the Tableau code below (easily translatable into other languages). This is equivalent to the rule in the table below (with 1 = Monday and 7 = Sunday). Basically you adjust day - 365 to be on the correct day of the week either in the same week if that moves <= 3 days otherwise you use the matching weekday from the previous/next week. It'll choose whichever leads to the least difference in terms of # of days.
[day prior year raw] = [day] - 365
[matching day prior year] =
if abs(datepart('weekday',[day]) - datepart('weekday',[day prior year raw]))<= 3
then [day prior year raw]+datepart('weekday',[day]) - datepart('weekday',[day prior year raw])
else [day prior year raw]+(if datepart('weekday',[day]) > datepart('weekday',[day prior year raw])
then -7+(datepart('weekday',[day]) - datepart('weekday',[day prior year raw]))
else 7+(datepart('weekday',[day]) - datepart('weekday',[day prior year raw])) end
)
end)

Answer 4

Look at ?years in package lubridate. This creates a period object which correctly spans a period, across leap years.
> library(lubridate)
> # set the reference date
> d1 = as.Date("2017/01/03", format="%Y/%m/%d")
>
> # verify across years and leap years
> d1 - years(1)
[1] "2016-01-03"
> d1 - years(2)
[1] "2015-01-03"
> d1 - years(3)
[1] "2014-01-03"
> d1 - years(4)
[1] "2013-01-03"
> d1 - years(5)
[1] "2012-01-03"
>
> weekdays(d1 - years(1))
[1] "Sunday"
> weekdays(d1 - years(2))
[1] "Saturday"
>
> # feb 29 on year period in yields NA
> ymd("2016/02/29") - years(1)
[1] NA
>
> # feb 29 in a non-leap year fails to convert
> ymd("2015/02/29") - years(1)
[1] NA
Warning message:
All formats failed to parse. No formats found.
>
> # feb 29, leap year with 4 year period works.
> ymd("2016/02/29") - years(4)
[1] "2012-02-29"
>