生成每月的最後一天的序列兩年

Question

我用lubridate和揣摩，這將是如此容易

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

但看看一個得到什麼。這一切都搞砸了！

[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" 

然後，我讀了如何lubridate迎合現象，如間隔，持續時間和週期。所以，我知道一個月實際上是由（365 * 4 + 1）/ 48 = 30.438天定義的天數。所以我試圖變得聰明，並重寫它爲

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

但是，只是給了一個錯誤。

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

Answer 1

是的，你找到了正確的把戲：從下個月第一天回來。

這裏是作爲一襯墊在基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（而被罰款封裝）不需要用於這樣的簡單的任務。另外，它現有基函數的超載仍然讓我覺得有些危險......

Answer 2

令人吃驚的是打字了一個問題的重點創作能量。我想我找出了答案。我不妨將它發佈在這裏，以便發現自己浪費時間的下一個可憐的靈魂。

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

只需指定下個月的第一天，從生成一個序列，但減去1天，從它獲得對上一個月的最後一天。

[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" 

順便說一句，我該如何擺脫煩人的「UTC」指定。時區在需要時是救生員。其餘時間他們是一個滋擾。