r - 生成两年内每月最后一天的序列

Question

我使用 lubridate 并认为这会很容易

ymd("2010-01-31")+months(0:23)

但是看看会得到什么。一切都乱了!

 [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 lubridate 如何迎合间隔、持续时间和周期等现象。所以，好吧，我意识到一个月实际上是由 (365*4+1)/48 = 30.438 天定义的天数。所以我尝试变得聪明并将其重写为

ymd("2010-01-31")+ as.period(months(0:23))

但这只是给出了一个错误。

Error in as.period.default(months(0:23)) : 
  (list) object cannot be coerced to type 'double'

Answer 1

是的，您找到了正确的技巧:从下个月的第一天回溯一天。

这是基本 R 中的一行:

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> 

所以不需要lubridate像这样的简单任务不需要它(虽然是一个很好的包)。另外，它对现有基本函数的重载仍然让我觉得有些危险......