Having had a similar problem with data from 1800 to now, this worked for me:
data2$date=as.character(data2$date)
lct <- Sys.getlocale("LC_TIME");
Sys.setlocale("LC_TIME","C")
data2$date<- as.Date(data2$date, format = "%Y %m %d") # and it works
The question did not state precisely what output is expected but assuming that for month you want the month number (January = 1) and for the year you want the numeric 4 digit year then assuming that we have just run the code in the question:
cycle(date1)
## [1] 3
as.integer(date1)
## [1] 2012
The lubridate package is amazing for this kind of thing:
> require(lubridate)
> month(date1)
[1] 3
> year(date1)
[1] 2012
You can use format
:
library(zoo)
x <- as.yearmon(Sys.time())
format(x,"%b")
[1] "Mar"
format(x,"%Y")
[1] "2012"
For large vectors:
y = as.POSIXlt(date1)$year + 1900 # x$year : years since 1900
m = as.POSIXlt(date1)$mon + 1 # x$mon : 0–11
I know the OP is using zoo
here, but I found this thread googling for a standard ts
solution for the same problem. So I thought I'd add a zoo
-free answer for ts
as well.
# create an example Date
date_1 <- as.Date("1990-01-01")
# extract year
as.numeric(format(date_1, "%Y"))
# extract month
as.numeric(format(date_1, "%m"))
Source: Stackoverflow.com