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I'm trying to write an R function to calculate the number of weekdays between two dates. For example, something like Nweekdays('01/30/2011','02/04/2011') that would return 5.

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Update: I was wondering if anyone could think of a way to vectorize this, so that it'll work on 2 columns of dates.

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8 Answers 8

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Date1 <- as.Date("2011-01-30") Date2 <- as.Date("2011-02-04") sum(!weekdays(seq(Date1, Date2, "days")) %in% c("Saturday", "Sunday"))

EDIT: This can be vectorized:

Dates1 <- as.Date("2011-01-30") + rep(0, 10) Dates2 <- as.Date("2011-02-04") + seq(0, 9) Nweekdays <- Vectorize(function(a, b) sum(!weekdays(seq(a, b, "days")) %in% c("Saturday", "Sunday"))) Nweekdays(Dates1, Dates2)
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10 Comments

Thank you, that's very elegant
@J. Winchester Can you think of an easy way to vectorize this function so it works on more than one date at a time?
@Zach. Since there is no loop or apply, isn't it already vectorized? You could write the day check into a separate function, then apply that function to all your dates, but that seems like a needless complication.
@Zach: Never mind, it took me some mental digesting to remember what vectorize means. Edited the answer.
Will that work in a non-English locale? You need to get the words for Saturday and Sunday in the current locale and check for those.
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These modified functions takes into account of date differences of either positive or negative, whereas the accepted solution accounts for positive date difference.

library("dplyr") e2 <- structure(list(date.pr = structure(c(16524, 16524, 16524, 16524, 16524, 16524, 16524, 16524, 16524, 16524, 16545, 5974), class = "Date"), date.po = structure(c(16524, 16525, 16526, 16527, 16528, 16529, 16530, 16531, 16538, 16545, 16524, 15974), class = "Date")), .Names = c("date.1", "date.2"), class = c("tbl_df", "data.frame"), row.names = c(NA, -12L))

1. Locale Dependent Solution: Nweekdays() function is adapted from @J. Won.'s solution. It works for locale = "English_United States.1252"

Nweekdays <- Vectorize( function(a, b) { ifelse(a < b, return(sum(!weekdays(seq(a, b, "days")) %in% c("Saturday", "Sunday")) - 1), return(sum(!weekdays(seq(b, a, "days")) %in% c("Saturday", "Sunday")) - 1)) })

a. English Locale

> Sys.setlocale(category="LC_ALL", locale = "English_United States.1252") [1] "LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252" > Sys.getlocale() [1] "LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252" > e2 %>% mutate(wkd1 = format(date.1, "%A"), wkd2 = format(date.2, "%A"), ndays_with_wkends = ifelse((date.2 > date.1), (date.2 - date.1), (date.1 - date.2)), ndays_no_wkends = Nweekdays(date.1, date.2)) Source: local data frame [12 x 6] date.1 date.2 wkd1 wkd2 ndays_with_wkends ndays_no_wkends (date) (date) (chr) (chr) (dbl) (dbl) 1 2015-03-30 2015-03-30 Monday Monday 0 0 2 2015-03-30 2015-03-31 Monday Tuesday 1 1 3 2015-03-30 2015-04-01 Monday Wednesday 2 2 4 2015-03-30 2015-04-02 Monday Thursday 3 3 5 2015-03-30 2015-04-03 Monday Friday 4 4 6 2015-03-30 2015-04-04 Monday Saturday 5 4 7 2015-03-30 2015-04-05 Monday Sunday 6 4 8 2015-03-30 2015-04-06 Monday Monday 7 5 9 2015-03-30 2015-04-13 Monday Monday 14 10 10 2015-03-30 2015-04-20 Monday Monday 21 15 11 2015-04-20 2015-03-30 Monday Monday 21 15 12 1986-05-11 2013-09-26 Sunday Thursday 10000 7143

b. Chinese Locale

> Sys.setlocale(category="LC_ALL", locale = "chinese") [1] "LC_COLLATE=Chinese (Simplified)_People's Republic of China.936;LC_CTYPE=Chinese (Simplified)_People's Republic of China.936;LC_MONETARY=Chinese (Simplified)_People's Republic of China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People's Republic of China.936" > Sys.getlocale() [1] "LC_COLLATE=Chinese (Simplified)_People's Republic of China.936;LC_CTYPE=Chinese (Simplified)_People's Republic of China.936;LC_MONETARY=Chinese (Simplified)_People's Republic of China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People's Republic of China.936" > e2 %>% mutate(wkd1 = format(date.1, "%A"), wkd2 = format(date.2, "%A"), ndays_with_wkends = ifelse((date.2 > date.1), (date.2 - date.1), (date.1 - date.2)), ndays_no_wkends = Nweekdays(date.1, date.2)) Source: local data frame [12 x 6] date.1 date.2 wkd1 wkd2 ndays_with_wkends ndays_no_wkends (date) (date) (chr) (chr) (dbl) (dbl) 1 2015-03-30 2015-03-30 ÐÇÆÚÒ» ÐÇÆÚÒ» 0 0 2 2015-03-30 2015-03-31 ÐÇÆÚÒ» ÐÇÆÚ¶þ 1 1 3 2015-03-30 2015-04-01 ÐÇÆÚÒ» ÐÇÆÚÈý 2 2 4 2015-03-30 2015-04-02 ÐÇÆÚÒ» ÐÇÆÚËÄ 3 3 5 2015-03-30 2015-04-03 ÐÇÆÚÒ» ÐÇÆÚÎå 4 4 6 2015-03-30 2015-04-04 ÐÇÆÚÒ» ÐÇÆÚÁù 5 5 7 2015-03-30 2015-04-05 ÐÇÆÚÒ» ÐÇÆÚÈÕ 6 6 8 2015-03-30 2015-04-06 ÐÇÆÚÒ» ÐÇÆÚÒ» 7 7 9 2015-03-30 2015-04-13 ÐÇÆÚÒ» ÐÇÆÚÒ» 14 14 10 2015-03-30 2015-04-20 ÐÇÆÚÒ» ÐÇÆÚÒ» 21 21 11 2015-04-20 2015-03-30 ÐÇÆÚÒ» ÐÇÆÚÒ» 21 21 12 1986-05-11 2013-09-26 ÐÇÆÚÈÕ ÐÇÆÚËÄ 10000 10000

2. Locale Independent Solution: Nweekdays() function is adapted from @Sacha Epskamp's solution. It works for all locales, however @Sacha Epskamp used c(0,6) to weed out the weekends, which is different from this solution which uses c(2,3) to extract out weekends.

Nweekdays <- Vectorize( function(a, b) { return(sum(!(((as.numeric(b:a)) %% 7) %in% c(2,3))) - 1) # 2: Saturday and 3: Sunday })

a. English Locale 

> Sys.setlocale(category="LC_ALL", locale = "English_United States.1252") [1] "LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252" > Sys.getlocale() [1] "LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252" > e2 %>% mutate(wkd1 = format(date.1, "%A"), wkd2 = format(date.2, "%A"), ndays_with_wkends = ifelse((date.2 > date.1), (date.2 - date.1), (date.1 - date.2)), ndays_no_wkends = Nweekdays(date.1, date.2)) Source: local data frame [12 x 6] date.1 date.2 wkd1 wkd2 ndays_with_wkends ndays_no_wkends (date) (date) (chr) (chr) (dbl) (dbl) 1 2015-03-30 2015-03-30 Monday Monday 0 0 2 2015-03-30 2015-03-31 Monday Tuesday 1 1 3 2015-03-30 2015-04-01 Monday Wednesday 2 2 4 2015-03-30 2015-04-02 Monday Thursday 3 3 5 2015-03-30 2015-04-03 Monday Friday 4 4 6 2015-03-30 2015-04-04 Monday Saturday 5 4 7 2015-03-30 2015-04-05 Monday Sunday 6 4 8 2015-03-30 2015-04-06 Monday Monday 7 5 9 2015-03-30 2015-04-13 Monday Monday 14 10 10 2015-03-30 2015-04-20 Monday Monday 21 15 11 2015-04-20 2015-03-30 Monday Monday 21 15 12 1986-05-11 2013-09-26 Sunday Thursday 10000 7143

b. Chinese Locale

> Sys.setlocale(category="LC_ALL", locale = "chinese") [1] "LC_COLLATE=Chinese (Simplified)_People's Republic of China.936;LC_CTYPE=Chinese (Simplified)_People's Republic of China.936;LC_MONETARY=Chinese (Simplified)_People's Republic of China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People's Republic of China.936" > Sys.getlocale() [1] "LC_COLLATE=Chinese (Simplified)_People's Republic of China.936;LC_CTYPE=Chinese (Simplified)_People's Republic of China.936;LC_MONETARY=Chinese (Simplified)_People's Republic of China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People's Republic of China.936" > e2 %>% mutate(wkd1 = format(date.1, "%A"), wkd2 = format(date.2, "%A"), ndays_with_wkends = ifelse((date.2 > date.1), (date.2 - date.1), (date.1 - date.2)), ndays_no_wkends = Nweekdays(date.1, date.2)) Source: local data frame [12 x 6] date.1 date.2 wkd1 wkd2 ndays_with_wkends ndays_no_wkends (date) (date) (chr) (chr) (dbl) (dbl) 1 2015-03-30 2015-03-30 ÐÇÆÚÒ» ÐÇÆÚÒ» 0 0 2 2015-03-30 2015-03-31 ÐÇÆÚÒ» ÐÇÆÚ¶þ 1 1 3 2015-03-30 2015-04-01 ÐÇÆÚÒ» ÐÇÆÚÈý 2 2 4 2015-03-30 2015-04-02 ÐÇÆÚÒ» ÐÇÆÚËÄ 3 3 5 2015-03-30 2015-04-03 ÐÇÆÚÒ» ÐÇÆÚÎå 4 4 6 2015-03-30 2015-04-04 ÐÇÆÚÒ» ÐÇÆÚÁù 5 4 7 2015-03-30 2015-04-05 ÐÇÆÚÒ» ÐÇÆÚÈÕ 6 4 8 2015-03-30 2015-04-06 ÐÇÆÚÒ» ÐÇÆÚÒ» 7 5 9 2015-03-30 2015-04-13 ÐÇÆÚÒ» ÐÇÆÚÒ» 14 10 10 2015-03-30 2015-04-20 ÐÇÆÚÒ» ÐÇÆÚÒ» 21 15 11 2015-04-20 2015-03-30 ÐÇÆÚÒ» ÐÇÆÚÒ» 21 15 12 1986-05-11 2013-09-26 ÐÇÆÚÈÕ ÐÇÆÚËÄ 10000 7143
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5

I wrote this one, but the other answer is better :)

Nweekdays <- function(a,b) { dates <- as.Date(as.Date(a,"%m/%d/%y",origin="1900-01-01"):as.Date(b,"%m/%d/%y",origin="1900-01-01"),origin="1900-01-01") days <- format(dates,"%w")[c(-1,-length(dates))] return(sum(!days%in%c(0,6))) } Nweekdays('01/30/2011','02/04/2011') [1] 3

EDIT: Calculates how many weekdays are in between of the two specified days.

Edit:

Taking J. Winchesters advice, the function could be streamlined as:

 Nweekdays <- function(a,b) { dates <- as.numeric((as.Date(a,"%m/%d/%y")):(as.Date(b,"%m/%d/%y"))) dates <- dates[- c(1,length(dates))] return(sum(!dates%%7%in%c(0,6))) }

Some results:

> Nweekdays('01/30/2011','02/04/2011') [1] 4 > > Nweekdays('01/30/2011','01/30/2011') [1] 0 > > Nweekdays('01/30/2011','01/25/2011') [1] 3

Note that this is locale independent. (On that topic, how do I change locale anyway?)

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I started out along those lines, this should be faster than mine if you streamlined it. Converting your date sequence to a numeric vector lets you go modulo %% 7 and drop all the weekend days (eg, 2:3 if your origin is the default).
4

Working this out using lubridate you can create a function like:

library(lubridate) WorkingDays_function <- function(StartDate,EndDate){ startDate <- dmy(StartDate) endDate <- dmy(EndDate) #Now build a sequence between the dates: myDates <-seq(from = startDate, to = endDate, by = "days") #Week starts on Sunday (1) so to exclude Sun (1) and Sat (7) #use > 1 & < 7 working_days <- sum(wday(myDates)>1 & wday(myDates)<7) print(working_days) } WorkingDays_function("11/07/2019","20/07/2019")
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2

J. Win.'s answer is good, but can be quite a bit faster with lubridate. 

require(lubridate) count_weekdays<- Vectorize(function(from,to) sum(!wday(seq(from, to, "days")) %in% c(1,7)))

Here are time results from my machine:

> v1<- seq(from = ymd(19000101), to = ymd(20000101), by='month') > v2<- seq(from = ymd(20000101), to = ymd(21000101), by='month') > require(tictoc) > tic(); out<- Nweekdays(v1,v2); toc(); 293.06 sec elapsed > tic(); out<- count_weekdays(v1,v2); toc(); 9.95 sec elapsed

About 30x faster. Meaningful if your doing a lot of periods.

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1

I use the following approach - first a helper: 

weekDays <- function(UPPER = TRUE) { days <- c('MONDAY', 'TUESDAY', 'WEDNESDAY', 'THURSDAY', 'FRIDAY', 'SATURDAY', 'SUNDAY') if(!UPPER) return(.Internal(tolower(days))) days }

... and now the main function:

NumWeekDays <- function(dd, Xdays = c('saturday', 'sunday')) { # a function to count the number of non-Xdays in a month # > # first check if Xdays is of correct format stopifnot( all(.Internal(tolower(Xdays)) %in% weekDays(UP = FALSE))) # > # a helper function to find the number of non-X days between two dates NonXDays <- function(startDate, endDate, Xdays) { sum(!(.Internal(tolower(weekdays(seq(startDate, endDate, 'day')))) %in% .Internal(tolower(Xdays)))) } startDate <- as.Date(as.yearmon(index(dd)), frac = 0) endDate <- as.Date(as.yearmon(index(dd)), frac = 1) vapply(1:nrow(dd), FUN = function(i) NonXDays(startDate[i], endDate[i], Xdays = c('saturday', 'sunday')), FUN.VALUE = numeric(1)) }

Example: 

set.seed(1) dx <- apply.monthly(xts(rnorm(600), order.by = Sys.Date() + 1:600), mean) R> NumWeekDays(dx) [1] 23 21 22 23 20 23 22 20 22 22 21 22 23 21 22 22 21 23 21 21
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0

This should be faster

wkDaysCnt = function(endDate, startDate) { startDays = c("Monday"=5, "Tuesday"=4, "Wednesday"=3, "Thursday"=2, "Friday"=1, "Saturday"=0, "Sunday"=0) endDays = c("Monday"=1, "Tuesday"=2, "Wednesday"=3, "Thursday"=4, "Friday"=5, "Saturday"=5, "Sunday"=5) tDays = as.numeric(difftime(endDate, startDate, units="days"))+1 sdc1 = startDays[weekdays(startDate)] sdc2 = ifelse((tDays >= sdc1), sdc1, tDays) edc = endDays[weekdays(endDate)] * (tDays > sdc1+2) ret = sdc2 + edc + (5 * round((tDays - sdc2 - edc - 1) / 7) ) - 1 ret[ret < 0] = 0 names(ret) = NULL return(ret) }
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Surprised that bizdays has not been added here (note this returns the number of business days between two dates):

library(bizdays) # create a calendar cal <- create.calendar( name = "MyCalendar", weekdays = c("sunday", "saturday"), # non-work days holidays = "2023-02-20", # add holidays if you need them start.date = "2023-01-01", end.date = "2023-12-31" ) bizdays(from = "2023-02-01", to = "2023-02-28", "MyCalendar") [1] 18

There are additional options in ?create.calendar if your calendar start and end dates do not fall on business days.

This function is already vectorized:

data.frame(start_date = c("2023-02-01", "2023-05-01"), end_date = c("2023-02-28", "2023-05-31")) |> dplyr::mutate(biz = bizdays(from = start_date, to = end_date, "MyCalendar")) # start_date end_date biz #1 2023-02-01 2023-02-28 18 #2 2023-05-01 2023-05-31 22

If you are using the same calendar object repeatedly, you can set it as the default so you do not need to specify it each time in bizdays():

bizdays.options$set(default.calendar = "MyCalendar") bizdays(from = "2023-02-01", to = "2023-02-28") [1] 18 bizdays(from = "2023-05-01", to = "2023-05-31") [1] 22
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