CODEWITHSUNDEEP
rdate
Sanoj
Sanoj

Reputation: 1467

Extract year from date string in this format Dec-06, Jan-90, Feb-76 etc in R

I have a data frame in R. It has a column containing dates in this format Dec-06, Jan-90, Feb-76 etc. They are strings. How can I extract year section of it in this format: 2006, 1990, 1976 etc? I want to discard month segment and treat it as distance on year portion of it and treat this column as continuous variable for my logistic regression.

I tried several Date format package provided in R like POSIX, lubridate etc but was not able to extract.

Any idea?

Upvotes: 1

Views: 366

Answers (4)

akrun
akrun

Reputation: 886948

Here is another option with zoo

library(zoo)
data.table::year(as.yearmon("Dec-06", "%b-%y"))
#[1] 2006

Or as @G.Grothendieck mentioned, as.integer returns the year

as.integer(as.yearmon("Dec-06", "%b-%y"))
#[1] 2006

Upvotes: 0

PKumar
PKumar

Reputation: 11128

Using lubridate , it is easy, year function is a part of lubridate:

library(lubridate)
dat <- data.frame(x=c("Mar-06","Jan-90","May-76"))
dat$date <- as.POSIXlt(paste0("01-",tolower(dat$x)),format="%d-%b-%y",origin="1970-01-01")
dat$year <- year(dat$date)

Answer:

> dat
       x       date year
1 Mar-06 2006-03-01 2006
2 Jan-90 1990-01-01 1990
3 May-76 1976-05-01 1976

Upvotes: 1

d.b
d.b

Reputation: 32548

format(as.Date(gsub(".*-","","Dec-06"), format = "%y"), "%Y")
#[1] "2006"

OR

library(lubridate)
format(myd(paste("Dec-06","-01",sep="")), "%Y")
#[1] "2006"

Upvotes: 4

Ronak Shah
Ronak Shah

Reputation: 388817

We convert the string into a Date class and then extract only the year from it. 

format(as.Date(paste0("01-", x), "%d-%b-%y"), "%Y")

#[1] "2006" "1990" "1976"

data

x <- c("Dec-06", "Jan-90", "Feb-76 ")

Upvotes: 3

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