Question

在计算工作日的日期差异时遇到问题，即在Excel中排除网络日期功能等周末。

这是我的数据。

e <- structure(list(date.pr = structure(c(15909, 15933, 16517, 15961, 15974, 15978), class = "Date"), date.po = structure(c(15909, 15933, 15954, 15961, 15974, 15978), class = "Date")), .Names = c("date.1", "date.2"), class = c("tbl_df", "data.frame"), row.names = c(NA, -6L))

找到此任务的“bizdays”包。哪个适合这个。

> bizdays(e$date.2,e$date.1)
[1]   0   0 563   0   0   0

但我的数据包含date.2在date.1之前的情况。

e2 <- structure(list(date.pr = structure(c(15909, 15933, 16517, 15961, 5974, 15978, 15978), class = "Date"), date.po = structure(c(15909, 15933, 15954, 15961, 15974, 15978, 15979), class = "Date")), .Names = c("date.1", "date.2"), class = c("tbl_df", "data.frame"), row.names = c(NA, -7L))

现在它出现以下错误：

> cal <- Calendar(holidaysANBIMA, weekdays=c("saturday","sunday"))
> bizdays(e2$date.2,e2$date.1,cal)
Error in bizdays.Date(e2$date.2, e2$date.1, cal) : 
  All from dates must be greater than all to dates.

我在考虑使用ifelse（）逻辑，但它给了我同样的错误。

> ifelse(e2$date.2 < e2$date.1, NA, bizdays(e2$date.2,e2$date.1,cal))
Error in bizdays.Date(e2$date.2, e2$date.1, cal) : 
  All from dates must be greater than all to dates.

帮助表示赞赏。

Answer 1

Nweekdays()功能改编自@J。韩元。解决方案Calculate the number of weekdays between 2 dates in R

此修改后的功能考虑了正面或负面的日期差异，而上述链接已接受正日期差异的解决方案。

library("dplyr")

e2 <- structure(list(date.pr = structure(c(16524, 16524, 16507, 16510, 16510, 16524, 16510, 5974), class = "Date"), 
                     date.po = structure(c(16524, 16525, 16510, 16517, 16524, 16510, 16531, 15974), class = "Date")), 
                .Names = c("date.1", "date.2"), class = c("tbl_df", "data.frame"), row.names = c(NA, -8L))

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))
  })


> 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 [8 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 2015-03-13 2015-03-16 Friday   Monday                 3               1
4 2015-03-16 2015-03-23 Monday   Monday                 7               5
5 2015-03-16 2015-03-30 Monday   Monday                14              10
6 2015-03-30 2015-03-16 Monday   Monday                14              10
7 2015-03-16 2015-04-06 Monday   Monday                21              15
8 1986-05-11 2013-09-26 Sunday Thursday             10000            7143

> e2 %>% mutate(ndays_no_wkends = Nweekdays(date.1, date.2))

Source: local data frame [8 x 3]

      date.1     date.2 ndays_no_wkends
      (date)     (date)           (dbl)
1 2015-03-30 2015-03-30               0
2 2015-03-30 2015-03-31               1
3 2015-03-13 2015-03-16               1
4 2015-03-16 2015-03-23               5
5 2015-03-16 2015-03-30              10
6 2015-03-30 2015-03-16              10
7 2015-03-16 2015-04-06              15
8 1986-05-11 2013-09-26            7143

R：在工作日找到差异

1 个答案: