Question

我使用以下代码：

dates<-seq(as.Date("1991/1/4"),as.Date("2010/3/1"),"days")

但是，我想只有工作日，怎么办呢？（假设1991/1/4是星期一，我想排除：1991/6/4和1991/7/4。每周一次。）

感谢您的帮助。

Answer 1

您输入的日期不正确。为了使用1991/1/4星期一隐含的YYYY / DD / MM输入模式，你需要在as.Date中有一个格式字符串。

因此，假设您要排除周末的完整解决方案是：

 X <- seq( as.Date("1991/1/4", format="%Y/%m/%d"), as.Date("2010/3/1", format="%Y/%m/%d"),"days")
weekdays.X <- X[ ! weekdays(X) %in% c("Saturday", "Sunday") ]  
        # negation easier since only two cases in exclusion
        # probably do not want to print that vector to screen.
str(weekdays.X)

关于你的评论，我无法重现。我明白了：

> table(weekdays(weekdays.X) )

   Friday    Monday  Thursday   Tuesday Wednesday 
     1000      1000       999       999       999

Answer 2

这对你有用吗？（注意，它需要安装timeDate包）

# install.packages('timeDate')
require(timeDate)

# A ’timeDate’ Sequence
tS <- timeSequence(as.Date("1991/1/4"), as.Date("2010/3/1"))
tS

# Subset weekdays
tW <- tS[isWeekday(tS)]; tW
dayOfWeek(tW)

Answer 3

我在查找工作日功能时遇到了这个问题，并且由于OP要求＆＃34;工作日＆＃34;而不是＆＃34;工作日＆＃34;和timeDate也有isBizday功能，这个答案使用了它。

# A timeDate Sequence
date.sequence <- timeSequence(as.Date("1991-12-15"), as.Date("1992-01-15"));  # a short example period with three London holidays
date.sequence;

# holidays in the period
years.included <- unique( as.integer( format( x=date.sequence, format="%Y" ) ) );
holidays <- holidayLONDON(years.included)  #  (locale was not specified by OP in question nor in profile, so this assumes for example: holidayLONDON; also supported by timeDate are: holidayNERC, holidayNYSE, holidayTSX & holidayZURICH)

# Subset business days
business.days <- date.sequence[isBizday(date.sequence, holidays)]; 
business.days

R时间序列数据，每日只有工作日

3 个答案: