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