How to determine if date is a weekend or not (not using lubridate)

Question

I have a vector of date objects (yyyy-mm-dd) and I want to determine if any of them are on weekend or not. Is there a function that can determine this straighta

Answer 1

I put @AnandaMahto's suggestion here rather than a comment:

library(chron)
x <- seq(Sys.Date()-10, Sys.Date(), by = 1)
x[is.weekend(x)]

## [1] "2014-10-11" "2014-10-12" "2014-10-18"

Answer 2

The wday in both lubridate and data.table (yes, data.table has pretty much everything but the kitchen sink :-) both do a variation on:

as.POSIXlt(x, tz = tz(x))$wday + 1 # lubridate
as.POSIXlt(x)$wday + 1L            # data.table

So you could, in theory, just do:

as.POSIXlt("2014-10-18")$wday + 1
## [1] 7

and then test for the weekend days as other answer(s) do.

Answer 3

After looking a lot about this topic, I found this solution particularly simple using the package RQuantLib

install.packages("RQuantLib")
library(RQuantLib)
isBusinessDay(calendar="WeekendsOnly", dates=yourdatesofinterest)

You can modify this code with different calendars to add to the weekends different sets of holidays in different countries (below just an example, but they have many more).

isBusinessDay(calendar="UnitedStates", dates=yourdatesofinterest)
isBusinessDay(calendar="UnitedStates/Settlement", dates=yourdatesofinterest)
isBusinessDay(calendar="UnitedStates/NYSE", dates=yourdatesofinterest)
isBusinessDay(calendar="Sweden", dates=yourdatesofinterest)
isBusinessDay(calendar="Mexico", dates=yourdatesofinterest)

I hope it helps somebody

Answer 4

Using lubridate to avoid other packages, you can use:

is_weekday = function(timestamp){
  lubridate::wday(timestamp, week_start = 1) < 6
}

Answer 5

You can use the base R function weekdays().

x <- seq(Sys.Date() - 10, Sys.Date(), by = 1)
weekdays(x, abbr = TRUE)
# [1] "Wed" "Thu" "Fri" "Sat" "Sun" "Mon" "Tue" "Wed" "Thu" "Fri" "Sat"
x[grepl("S(at|un)", weekdays(x))]
# [1] "2014-10-11" "2014-10-12" "2014-10-18"

As far as lubridate goes, wday() has a label argument. When set to TRUE, the (abbreviated) day names are returned instead of numbers. Use the abbr argument to change to full names.

library(lubridate)
wday(x, label = TRUE)
# [1] Wed   Thurs Fri   Sat   Sun   Mon   Tues  Wed   Thurs Fri   Sat  
# Levels: Sun < Mon < Tues < Wed < Thurs < Fri < Sat

Answer 6

Another approach could be to use format and %u, which gives a number for the day of the week, starting with "1" representing "Monday".

With that, you can do:

x <- seq(as.Date("2014-10-18")-10, Sys.Date(), by = 1)
format(x, "%u") %in% c(6, 7)
#  [1] FALSE FALSE FALSE  TRUE  TRUE FALSE FALSE FALSE FALSE FALSE  TRUE
x[format(x, "%u") %in% c(6, 7)]
# [1] "2014-10-11" "2014-10-12" "2014-10-18"