r - Split date into different columns for year, month and day -

i have zoo objects like:

head(obs)

      index pp 1932-01-01  0 1932-01-02  0.2 1932-01-03  0 

and want split index 3 columns (years, months , days in separate columns) can analyses per day of month using ddply.

i don't know if makes difference dates created using:

dates <- as.date(cet[,1], "%d-%m-%y") obs <- xts(cet[,2], dates) 

where cet original file dates in column 1 , pp in column 2.

thanks helping!

1) columns. can use lubridate's year/month/day or chron's month.day.year:

1a) columns via lubridate

library(zoo) z <- zoo(1:1000, as.date("1932-01-01") + 0:999)  library(lubridate) tt <- time(z) zz <- cbind(z, year = year(tt), month = month(tt), day = day(tt)) 

1b) columns via chron

library(zoo) z <- zoo(1:1000, as.date("1932-01-01") + 0:999)  library(chron) zz <- with(month.day.year(time(z)), zoo(cbind(z, day, month, year))) 

2) aggregate. however, not need create columns in first place. can use aggregate.zoo directly original zoo object, z, using lubridate or chron or using yearmon zoo depending on want do:

2a) aggregate using lubridate

library(zoo) z <- zoo(1:1000, as.date("1932-01-01") + 0:999)  library(lubridate) aggregate(z, day, mean) aggregate(z, month, mean) aggregate(z, year, mean) 

2b) aggregate using chron

library(zoo) z <- zoo(1:1000, as.date("1932-01-01") + 0:999)  library(chron) mdy <- month.day.year(time(z))  aggregate(z, mdy$day, mean) aggregate(z, mdy$month, mean) aggregate(z, mdy$year, mean)  # or ct <- as.chron(time(z))  aggregate(z, days(ct), mean) aggregate(z, months(ct), mean) aggregate(z, years(ct), mean)  # days(ct) , years(ct) can # shortened days , years within above context # (and work months except out of order) aggregate(z, days, mean) aggregate(z, years, mean) 

2c) aggregate using yearmon

if wish summarize each year/month rather lumping january months together, february months together, etc. need neither chron nor lubridate rather can use zoo's yearmon:

library(zoo) z <- zoo(1:1000, as.date("1932-01-01") + 0:999)  aggregate(z, yearmon, mean)