我使用lubridate并認為這會很容易ymd("2010-01-31")+months(0:23)但是，看看有什么。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，我閱讀了潤滑如何滿足間隔，持續時間和周期等現象。所以，好的，我意識到一個月實際上是（365 * 4 + 1）/ 48 = 30.438天定義的天數。所以我試圖變得聰明，并將其重寫為ymd("2010-01-31")+ as.period(months(0:23))但這只是一個錯誤。Error in as.period.default(months(0:23)) : (list) object cannot be coerced to type 'double'
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明月笑刀無情
TA貢獻1828條經驗 獲得超4個贊
令人驚訝的是，輸入問題會集中精力發揮創造力。我想我已經解決了。我不妨在這里張貼給下一個發現自己在浪費時間的可憐的靈魂。
ymd("2010-02-01")+ months(0:23)-days(1)
只需指定下個月的第一天并從中生成一個序列，但從中減去1天即可得到上個月的最后一天。
[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"
順便說一句，我如何擺脫討厭的“ UTC”稱號。時區在需要時可以節省生命。其余的時間他們很討厭。
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