R:取TS对象中月份的名称
假设您有以下时间序列。月刊R:取TS对象中月份的名称,r,date,R,Date,假设您有以下时间序列。月刊 require(forecast) a<-ts(seq(1:50),frequency = 12) Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 1 1 2 3 4 5 6 7 8 9 10 11 12 2 13 14 15 16 17 18 19 20 21 22 23 24 3 25 26 27 28 29 30 31 32
require(forecast)
a<-ts(seq(1:50),frequency = 12)
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
1 1 2 3 4 5 6 7 8 9 10 11 12
2 13 14 15 16 17 18 19 20 21 22 23 24
3 25 26 27 28 29 30 31 32 33 34 35 36
4 37 38 39 40 41 42 43 44 45 46 47 48
5 49 50
我也试过了。但是没有运气。有什么建议吗
最好的 1)月份名称试试以下方法:
> month.abb[cycle(a)]
[1] "Jan" "Feb" "Mar" "Apr" "May" "Jun" "Jul" "Aug" "Sep" "Oct" "Nov" "Dec"
[13] "Jan" "Feb" "Mar" "Apr" "May" "Jun" "Jul" "Aug" "Sep" "Oct" "Nov" "Dec"
[25] "Jan" "Feb" "Mar" "Apr" "May" "Jun" "Jul" "Aug" "Sep" "Oct" "Nov" "Dec"
[37] "Jan" "Feb" "Mar" "Apr" "May" "Jun" "Jul" "Aug" "Sep" "Oct" "Nov" "Dec"
[49] "Jan" "Feb"
2)yearmon如果真正需要的是年和月,那么这将给出一个动物园“yearmon”
类结果:
library(zoo)
a <- ts(1:50, start = 2000, freq = 12)
as.yearmon(time(a))
或本月最后一天:
as.Date(as.yearmon(time(a)), frac = 1)
更新添加了(2)和(3)。多大的漏洞!:)。然后我使用as.Date()来更改格式。谢谢如果这是你想要的,而不仅仅是这个月,那么我添加了更多的选择。
as.Date(as.yearmon(time(a)))
as.Date(as.yearmon(time(a)), frac = 1)