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我正在编写一个代码，它使用两个时间戳向量来为我提供特定过程所需的时间（只是 difftime）。

但在周末，时间应该停止。因此，如果产品在星期五进货并在星期二离开，我必须减去周末时间。所以我需要一种可能性来确定产品处理时间何时经过周末，周末会自动从时差中减去。

这些产品的识别问题比减法更重要。

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2 回答 2

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is.weekend从chron包装中使用。

days<-seq.Date(as.Date("01-01-2017",format= "%d-%m-%Y"),as.Date("01-04-2017",format= "%d-%m-%Y"),by = "days")   
library(chron)  
  days[is.weekend(days)]
  [1] "2017-01-01" "2017-01-07" "2017-01-08" "2017-01-14" "2017-01-15" "2017-01-21" "2017-01-22"
  [8] "2017-01-28" "2017-01-29" "2017-02-04" "2017-02-05" "2017-02-11" "2017-02-12" "2017-02-18"
 [15] "2017-02-19" "2017-02-25" "2017-02-26" "2017-03-04" "2017-03-05" "2017-03-11" "2017-03-12"
 [22] "2017-03-18" "2017-03-19" "2017-03-25" "2017-03-26" "2017-04-01"

  weekdays.Date(days[is.weekend(days)])
  [1] "Sunday"   "Saturday" "Sunday"   "Saturday" "Sunday"   "Saturday" "Sunday"   "Saturday"
  [9] "Sunday"   "Saturday" "Sunday"   "Saturday" "Sunday"   "Saturday" "Sunday"   "Saturday"
 [17] "Sunday"   "Saturday" "Sunday"   "Saturday" "Sunday"   "Saturday" "Sunday"   "Saturday"
 [25] "Sunday"   "Saturday"
于 2018-07-30T14:07:03.787 回答
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该功能weekdays可能在这里有所帮助：

weekdays(as.Date("2018-07-30"))
# [1] "monday"

因此，如果您传递日期对象的向量，您将能够识别周六和周日。

例如：

which(weekdays(as.Date(c("2018-07-30", "2018-07-29", "2018-07-28", "2018-07-27"))) %in% c("saturday", "sunday"))
[1] 2 3
于 2018-07-30T13:59:53.963 回答