43

我使用 lubridate 并认为这很容易

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

但看看一个人得到了什么。一切都乱了套！

 [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))

但这只是给出了一个错误。

Error in as.period.default(months(0:23)) : 
  (list) object cannot be coerced to type 'double'
4

2 回答 2

97

是的，您找到了正确的窍门：从下个月的第一天开始。

这是作为基础 R 中的单行代码：

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>

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

于 2011-11-30T23:29:45.643 回答
20

输入问题如何集中创造力是令人惊讶的。我想我想出了答案。我不妨把它贴在这里，给下一个发现自己在浪费时间的可怜人。

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

只需指定下个月的第一天并从中生成一个序列，但从中减去 1 天即可得到上个月的最后一天。

[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，我如何摆脱讨厌的“UTC”名称。时区在需要时可以挽救生命。其余的时间他们都是讨厌的。

于 2011-11-30T22:29:24.527 回答