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

问题

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 straightaway?

I can use wday() in the lubridate package and then determine if returned value is 01 or 07, but anything else more straightforward?

x <- seq(Sys.Date()-10, Sys.Date(), by = 1)
x[lubridate::wday(x) %in% c(1, 7)]

回答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"

回答2:

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

回答3:

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"

回答4:

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.

回答5:

You could use isWeekend from package timeDate. Hard to do more straightforward :). wday specify which days should be considered as weekdays. By default from Mondays to Fridays.

> today <- isWeekend(Sys.Date(), wday = 1:5)
if (as.logical(today)){
  print("YES")
} else print("NO")

From the documentation :

## Dates in April, currentYear:
   currentYear = getRmetricsOptions("currentYear")
   tS = timeSequence(
      from = paste(currentYear, "-03-01", sep = ""),
      to = paste(currentYear, "-04-30", sep = ""))
   tS

## Subset of Weekends:
   isWeekend(tS)
   tS[isWeekend(tS)]

It works also with isWeekday.

回答6:

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

来源：https://stackoverflow.com/questions/26441700/how-to-determine-if-date-is-a-weekend-or-not-not-using-lubridate

标签

datetime

lubridate