生成每月的最后一天的序列两年

Question

我用lubridate和揣摩，这将是如此容易

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

但看看一个得到什么。这一切都搞砸了！

[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，我读了如何lubridate迎合现象，如间隔，持续时间和周期。所以，我知道一个月实际上是由（365 * 4 + 1）/ 48 = 30.438天定义的天数。所以我试图变得聪明，并重写它为

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

但是，只是给了一个错误。

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

Answer 1

是的，你找到了正确的把戏：从下个月第一天回来。

这里是作为一衬垫在基R：

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其中（而被罚款封装）不需要用于这样的简单的任务。另外，它现有基函数的超载仍然让我觉得有些危险......

Answer 2

令人吃惊的是打字了一个问题的重点创作能量。我想我找出了答案。我不妨将它发布在这里，以便发现自己浪费时间的下一个可怜的灵魂。

ymd("2010-02-01")+ months(0:23)-days(1) 

只需指定下个月的第一天，从生成一个序列，但减去1天，从它获得对上一个月的最后一天。

[1] "2010-01-31 UTC" "2010-02-28 UTC" "2010-03-31 UTC" "2010-04-30 UTC" "2010-05-31 UTC" "2010-06-30 UTC" "2010-07-31 UTC" "2010-08-31 UTC" "2010-09-30 UTC" 
[10] "2010-10-31 UTC" "2010-11-30 UTC" "2010-12-31 UTC" "2011-01-31 UTC" "2011-02-28 UTC" "2011-03-31 UTC" "2011-04-30 UTC" "2011-05-31 UTC" "2011-06-30 UTC" 
[19] "2011-07-31 UTC" "2011-08-31 UTC" "2011-09-30 UTC" "2011-10-31 UTC" "2011-11-30 UTC" "2011-12-31 UTC" 

顺便说一句，我该如何摆脱烦人的“UTC”指定。时区在需要时是救生员。其余时间他们是一个滋扰。