计算R中2个日期之间的工作日数

Question

计算R中2个日期之间的工作日数

我正在尝试编写一个R函数来计算两个日期之间的工作日数.例如,Nweekdays('01/30/2011','02/04/2011')等于5.

与此问题类似.谢谢!

/编辑:@J.温彻斯特的答案很棒,但我想知道是否有人能想出一种方法来对此进行矢量化,以便它可以在两列日期上工作.谢谢!/编辑2:再次感谢!

Answer 1

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

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编辑:和扎克说,让那里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)

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Answer 2

Sat*_*ish 6

这些修改后的函数考虑了正数或负数的日期差异,而接受的解决方案考虑了正数日期差异.

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))

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1. Locale Dependent Solution: Nweekdays()功能改编自@J.赢了.解决方案.它适用于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))
  })

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一个.英语区域

> 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

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湾中国区域

> 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

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2. Locale Independent Solution: Nweekdays()功能改编自@Sacha Epskamp的解决方案.它适用于所有语言环境,但@Sacha Epskamp过去常常c(0,6)在周末淘汰,这与c(2,3)用于提取周末的解决方案不同 .

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

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一个.英语区域

> 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

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湾中国区域

> 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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Answer 3

Sac*_*amp 5

我写了这个,但另一个答案更好:)

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

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编辑:计算两个指定日期之间的工作日数.

编辑:

根据J. Winchesters的建议,该功能可以简化为:

    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)))
}

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一些结果:

> 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

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请注意,这与区域设置无关.(关于该主题,我如何更改语言环境？)

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