我使用lubridate并认为这会很容易
ymd("2010-01-31")+months(0:23)
但是，看看有什么。都搞砸了!
[1] "2010-01-31 UTC" "2010-03-03 UTC" "2010-03-31 UTC" "2010-05-01 UTC" "2010-05-31 UTC" "2010-07-01 UTC" "2010-07-31 UTC" "2010-08-31 UTC" "2010-10-01 UTC"
[10] "2010-10-31 UTC" "2010-12-01 UTC" "2010-12-31 UTC" "2011-01-31 UTC" "2011-03-03 UTC" "2011-03-31 UTC" "2011-05-01 UTC" "2011-05-31 UTC" "2011-07-01 UTC"
[19] "2011-07-31 UTC" "2011-08-31 UTC" "2011-10-01 UTC" "2011-10-31 UTC" "2011-12-01 UTC" "2011-12-31 UTC"
然后，我阅读了润滑如何满足间隔，持续时间和周期等现象。因此，好的，我意识到一个月实际上是(365 * 4 + 1)/ 48 = 30.438天定义的天数。所以我试图变得聪明，并将其重写为
ymd("2010-01-31")+ as.period(months(0:23))
但这只是一个错误。
最佳答案
是的，您找到了正确的把戏:从下个月的第一天开始。
这是基数R中的单线:
R> seq(as.Date("2010-02-01"), length=24, by="1 month") - 1
[1] "2010-01-31" "2010-02-28" "2010-03-31" "2010-04-30" "2010-05-31"
[6] "2010-06-30" "2010-07-31" "2010-08-31" "2010-09-30" "2010-10-31"
[11] "2010-11-30" "2010-12-31" "2011-01-31" "2011-02-28" "2011-03-31"
[16] "2011-04-30" "2011-05-31" "2011-06-30" "2011-07-31" "2011-08-31"
[21] "2011-09-30" "2011-10-31" "2011-11-30" "2011-12-31"
R>
因此，不需要lubridate，它对于这样的简单任务是不需要的(虽然是一个很好的软件包)。另外，它对现有基本功能的重载仍然令我感到有些危险...
关于r - 生成两年中每月最后一天的序列，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/8333838/