生成两个月中最后一天的序列

Question

我使用lubridate并认为这将是如此简单

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))

但那只是一个错误.

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.

这是作为基地R的单线程:

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,不需要使用lubridate(虽然是一个精美的包).此外,它现有基础功能的超载仍然让我感到有点危险......

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从中生成一个序列,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.其余时间他们是一个讨厌的人.