Question

我使用了lubridate，并认为这很容易

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

但看看会得到什么。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"

然后我读到了lubridate如何迎合间隔，持续时间和周期等现象。那么，好吧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将其重写为

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

但这只是一个错误。

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

Answer 1

是的，你找到了正确的伎俩：从下个月的第一天回来一天。

以下是基础R中的单行：

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>

因此，对于像这样的简单任务，不需要lubridate（虽然是一个很好的包）。此外，它的现有基础功能的超载仍然让我觉得有些危险......

Answer 2

如何输入一个问题来集中创造力，这真是令人惊讶。我想我找到了答案。我也可以把它发布在这里，为下一个发现自己在浪费时间的可怜的灵魂。

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

只需指定下个月的第一天并生成一个序列，但从中减去1天即可获得前一个月的最后一天。

[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，我如何摆脱讨厌的“UTC”指定。在需要时，时区可以节省生命。其余的时间他们都很讨厌。

生成两个月中最后一天的序列

2 个答案: