Calculate ages in R - r

Question

I have two data frames in R. One frame has a persons year of birth:
YEAR
/1931
/1924
and then another column shows a more recent time.
RECENT
09/08/2005
11/08/2005
What I want to do is subtract the years so that I can calculate their age in number of years, however I am not sure how to approach this. Any help please?

Answer 1

The following function takes a vectors of Date objects and calculates the ages, correctly accounting for leap years. Seems to be a simpler solution than any of the other answers.
age = function(from, to) {
from_lt = as.POSIXlt(from)
to_lt = as.POSIXlt(to)
age = to_lt$year - from_lt$year
ifelse(to_lt$mon < from_lt$mon |
(to_lt$mon == from_lt$mon & to_lt$mday < from_lt$mday),
age - 1, age)
}

Answer 2

You can solve this with the lubridate package.
> library(lubridate)
I don't think /1931 is a common date class. So I'll assume all the entries are character strings.
> RECENT <- data.frame(recent = c("09/08/2005", "11/08/2005"))
> YEAR <- data.frame(year = c("/1931", "/1924"))
First, let's notify R that the recent dates are dates. I'll assume the dates are in month/day/year order, so I use mdy(). If they're in day/month/year order just use dmy().
> RECENT$recent <- mdy(RECENT$recent)
recent
1 2005-09-08
2 2005-11-08
Now, lets turn the years into numbers so we can do some math with them.
> YEAR$year <- as.numeric(substr(YEAR$year, 2, 5))
Now just do the math. year() extracts the year value of the RECENT dates.
> year(RECENT$recent) - YEAR
year
1 74
2 81
p.s. if your year entries are actually full dates, you can get the difference in years with
> YEAR1 <- data.frame(year = mdy("01/08/1931","01/08/1924"))
> as.period(RECENT$recent - YEAR1$year, units = "year")
[1] 74 years and 8 months 81 years and 10 months

Answer 3

I use a custom function, see code below, convenient to use in mutate and quite flexible (you'll need the lubridate package).
Examples
get_age("2000-01-01")
# [1] 17
get_age(lubridate::as_date("2000-01-01"))
# [1] 17
get_age("2000-01-01","2015-06-15")
# [1] 15
get_age("2000-01-01",dec = TRUE)
# [1] 17.92175
get_age(c("2000-01-01","2003-04-12"))
# [1] 17 14
get_age(c("2000-01-01","2003-04-12"),dec = TRUE)
# [1] 17.92176 14.64231
Function
#' Get age
#'
#' Returns age, decimal or not, from single value or vector of strings
#' or dates, compared to a reference date defaulting to now. Note that
#' default is NOT the rounded value of decimal age.
#' #param from_date vector or single value of dates or characters
#' #param to_date date when age is to be computed
#' #param dec return decimal age or not
#' #examples
#' get_age("2000-01-01")
#' get_age(lubridate::as_date("2000-01-01"))
#' get_age("2000-01-01","2015-06-15")
#' get_age("2000-01-01",dec = TRUE)
#' get_age(c("2000-01-01","2003-04-12"))
#' get_age(c("2000-01-01","2003-04-12"),dec = TRUE)
get_age <- function(from_date,to_date = lubridate::now(),dec = FALSE){
if(is.character(from_date)) from_date <- lubridate::as_date(from_date)
if(is.character(to_date)) to_date <- lubridate::as_date(to_date)
if (dec) { age <- lubridate::interval(start = from_date, end = to_date)/(lubridate::days(365)+lubridate::hours(6))
} else { age <- lubridate::year(lubridate::as.period(lubridate::interval(start = from_date, end = to_date)))}
age
}

Answer 4

You can do some formating:
as.numeric(format(as.Date("01/01/2010", format="%m/%d/%Y"), format="%Y")) - 1930
With your data:
> yr <- c(1931, 1924)
> recent <- c("09/08/2005", "11/08/2005")
> as.numeric(format(as.Date(recent, format="%m/%d/%Y"), format="%Y")) - yr
[1] 74 81
Since you have your data in a data.frame (I'll assume that it's called df), it will be more like this:
as.numeric(format(as.Date(df$recent, format="%m/%d/%Y"), format="%Y")) - df$year

Answer 5

Given the data in your example:
> m <- data.frame(YEAR=c("/1931", "/1924"),RECENT=c("09/08/2005","11/08/2005"))
> m
YEAR RECENT
1 /1931 09/08/2005
2 /1924 11/08/2005
Extract year with the strptime function:
> strptime(m[,2], format = "%m/%d/%Y")$year - strptime(m[,1], format = "/%Y")$year
[1] 74 81

Answer 6

Based on the previous answer, convert your columns to date objects and subtract. Some conversion of types between character and numeric is necessary:
> foo=data.frame(RECENT=c("09/08/2005","11/08/2005"),YEAR=c("/1931","/1924"))
> foo
RECENT YEAR
1 09/08/2005 /1931
2 11/08/2005 /1924
> foo$RECENTd = as.Date(foo$RECENT, format="%m/%d/%Y")
> foo$YEARn = as.numeric(substr(foo$YEAR,2,999))
> foo$AGE = as.numeric(format(foo$RECENTd,"%Y")) - foo$YEARn
> foo
RECENT YEAR RECENTd YEARn AGE
1 09/08/2005 /1931 2005-09-08 1931 74
2 11/08/2005 /1924 2005-11-08 1924 81
Note I've assumed you have that slash in your year column.
Also, tip for when asking questions about dates is to include a day that is past the twelfth so we know if you are a month/day/year person or a day/month/year person.

Answer 7

I think this might be a bit more intuitive and requires no formatting or stripping:
as.numeric(as.Date("2002-02-02") - as.Date("1924-08-03")) / 365
gives output:
77.55342
Then you can use floor(), round(), or ceiling() to round to a whole number.

Answer 8

Really solid way that also supports vectors using the lubridate package:
age <- function(date.birth, date.ref = Sys.Date()) {
if (length(date.birth) > 1 & length(date.ref) == 1) {
date.ref <- rep(date.ref, length(date.birth))
}
date.birth.monthdays <- paste0(month(date.birth), day(date.birth)) %>% as.integer()
date.ref.monthdays <- paste0(month(date.ref), day(date.ref)) %>% as.integer()
age.calc <- 0
for (i in 1:length(date.birth)) {
if (date.birth.monthdays[i] <= date.ref.monthdays[i]) {
# didn't had birthday
age.calc[i] <- year(date.ref[i]) - year(date.birth[i])
} else {
age.calc[i] <- year(date.ref[i]) - year(date.birth[i]) - 1
}
}
age.calc
}
This also accounts for leap years. I just check if someone has had a birthday already.