Calculate the number of weekdays between 2 dates in R
I'm trying to write an R function to calculate the number of weekdays between two dates. For example, Nweekdays('01/30/2011','02/04/2011') would equal 5.
Similar to this question. Thanks!
/edit: @J. Winchester's answer is great, but I was wondering if anyone could think of a way to vectorize this, so that it'll work on 2 columns of dates. Thanks! /edit 2: Thanks again!
Date1 <- as.Date("2011-01-30")
Date2 <- as.Date("2011-02-04")
sum(!weekdays(seq(Date1, Date2, "days")) %in% c("Saturday", "Sunday"))
EDIT: And Zach said, let there be Vectorize
:)
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.
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?)