change a column from birth date to age in r - r

Question

I am using data.table for the first time.
I have a column of about 400,000 ages in my table. I need to convert them from birth dates to ages.
What is the best way to do this?

Answer 1

I've been thinking about this and have been dissatisfied with the two answers so far. I like using lubridate, as #KFB did, but I also want things wrapped up nicely in a function, as in my answer using the eeptools package. So here's a wrapper function using the lubridate interval method with some nice options:
#' Calculate age
#'
#' By default, calculates the typical "age in years", with a
#' \code{floor} applied so that you are, e.g., 5 years old from
#' 5th birthday through the day before your 6th birthday. Set
#' \code{floor = FALSE} to return decimal ages, and change \code{units}
#' for units other than years.
#' #param dob date-of-birth, the day to start calculating age.
#' #param age.day the date on which age is to be calculated.
#' #param units unit to measure age in. Defaults to \code{"years"}. Passed to \link{\code{duration}}.
#' #param floor boolean for whether or not to floor the result. Defaults to \code{TRUE}.
#' #return Age in \code{units}. Will be an integer if \code{floor = TRUE}.
#' #examples
#' my.dob <- as.Date('1983-10-20')
#' age(my.dob)
#' age(my.dob, units = "minutes")
#' age(my.dob, floor = FALSE)
age <- function(dob, age.day = today(), units = "years", floor = TRUE) {
calc.age = lubridate::interval(dob, age.day) / lubridate::duration(num = 1, units = units)
if (floor) return(as.integer(floor(calc.age)))
return(calc.age)
}
Usage examples:
> my.dob <- as.Date('1983-10-20')
> age(my.dob)
[1] 31
> age(my.dob, floor = FALSE)
[1] 31.15616
> age(my.dob, units = "minutes")
[1] 16375680
> age(seq(my.dob, length.out = 6, by = "years"))
[1] 31 30 29 28 27 26

Answer 2

From the comments of this blog entry, I found the age_calc function in the eeptools package. It takes care of edge cases (leap years, etc.), checks inputs and looks quite robust.
library(eeptools)
x <- as.Date(c("2011-01-01", "1996-02-29"))
age_calc(x[1],x[2]) # default is age in months
[1] 46.73333 224.83118
age_calc(x[1],x[2], units = "years") # but you can set it to years
[1] 3.893151 18.731507
floor(age_calc(x[1],x[2], units = "years"))
[1] 3 18
For your data
yourdata$age <- floor(age_calc(yourdata$birthdate, units = "years"))
assuming you want age in integer years.

Answer 3

Assume you have a data.table, you could do below:
library(data.table)
library(lubridate)
# toy data
X = data.table(birth=seq(from=as.Date("1970-01-01"), to=as.Date("1980-12-31"), by="year"))
Sys.Date()
Option 1 : use "as.period" from lubriate package
X[, age := as.period(Sys.Date() - birth)][]
birth age
1: 1970-01-01 44y 0m 327d 0H 0M 0S
2: 1971-01-01 43y 0m 327d 6H 0M 0S
3: 1972-01-01 42y 0m 327d 12H 0M 0S
4: 1973-01-01 41y 0m 326d 18H 0M 0S
5: 1974-01-01 40y 0m 327d 0H 0M 0S
6: 1975-01-01 39y 0m 327d 6H 0M 0S
7: 1976-01-01 38y 0m 327d 12H 0M 0S
8: 1977-01-01 37y 0m 326d 18H 0M 0S
9: 1978-01-01 36y 0m 327d 0H 0M 0S
10: 1979-01-01 35y 0m 327d 6H 0M 0S
11: 1980-01-01 34y 0m 327d 12H 0M 0S
Option 2 : if you do not like the format of Option 1, you could do below:
yr = duration(num = 1, units = "years")
X[, age := new_interval(birth, Sys.Date())/yr][]
# you get
birth age
1: 1970-01-01 44.92603
2: 1971-01-01 43.92603
3: 1972-01-01 42.92603
4: 1973-01-01 41.92329
5: 1974-01-01 40.92329
6: 1975-01-01 39.92329
7: 1976-01-01 38.92329
8: 1977-01-01 37.92055
9: 1978-01-01 36.92055
10: 1979-01-01 35.92055
11: 1980-01-01 34.92055
Believe Option 2 should be the more desirable.

Answer 4

I prefer to do this using the lubridate package, borrowing syntax I originally encountered in another post.
It's necessary to standardize your input dates in terms of R date objects, preferably with the lubridate::mdy() or lubridate::ymd() or similar functions, as applicable. You can use the interval() function to generate an interval describing the time elapsed between the two dates, and then use the duration() function to define how this interval should be "diced".
I've summarized the simplest case for calculating an age from two dates below, using the most current syntax in R.
df$DOB <- mdy(df$DOB)
df$EndDate <- mdy(df$EndDate)
df$Calc_Age <- interval(start= df$DOB, end=df$EndDate)/
duration(n=1, unit="years")
Age may be rounded down to the nearest complete integer using the base R 'floor()` function, like so:
df$Calc_AgeF <- floor(df$Calc_Age)
Alternately, the digits= argument in the base R round() function can be used to round up or down, and specify the exact number of decimals in the returned value, like so:
df$Calc_Age2 <- round(df$Calc_Age, digits = 2) ## 2 decimals
df$Calc_Age0 <- round(df$Calc_Age, digits = 0) ## nearest integer
It's worth noting that once the input dates are passed through the calculation step described above (i.e., interval() and duration() functions) , the returned value will be numeric and no longer a date object in R. This is significant whereas the lubridate::floor_date() is limited strictly to date-time objects.
The above syntax works regardless whether the input dates occur in a data.table or data.frame object.

Answer 5

I wanted an implementation that didn't increase my dependencies beyond data.table, which is usually my only dependency. The data.table is only needed for mday, which means day of the month.
Development function
This function is logically how I would think about someone's age. I start with [current year] - [brith year] - 1, then add 1 if they've already had their birthday in the current year. To check for that offset I start by considering month, then (if necessary) day of month.
Here is that step by step implementation:
agecalc <- function(origin, current){
require(data.table)
y <- year(current) - year(origin) - 1
offset <- 0
if(month(current) > month(origin)) offset <- 1
if(month(current) == month(origin) &
mday(current) >= mday(origin)) offset <- 1
age <- y + offset
return(age)
}
Production function
This is the same logic refactored and vectorized:
agecalc <- function(origin, current){
require(data.table)
age <- year(current) - year(origin) - 1
ii <- (month(current) > month(origin)) | (month(current) == month(origin) &
mday(current) >= mday(origin))
age[ii] <- age[ii] + 1
return(age)
}
Experimental function that uses strings
You could also do a string comparison on the month / day part. Perhaps there are times when this is more efficient, for example if you had the year as a number and the birth date as a string.
agecalc_strings <- function(origin, current){
origin <- as.character(origin)
current <- as.character(current)
age <- as.numeric(substr(current, 1, 4)) - as.numeric(substr(origin, 1, 4)) - 1
if(substr(current, 6, 10) >= substr(origin, 6, 10)){
age <- age + 1
}
return(age)
}
Some tests on the vectorized "production" version:
## Examples for specific dates to test the calculation with things like
## beginning and end of months, and leap years:
agecalc(as.IDate("1985-08-13"), as.IDate("1985-08-12"))
agecalc(as.IDate("1985-08-13"), as.IDate("1985-08-13"))
agecalc(as.IDate("1985-08-13"), as.IDate("1986-08-12"))
agecalc(as.IDate("1985-08-13"), as.IDate("1986-08-13"))
agecalc(as.IDate("1985-08-13"), as.IDate("1986-09-12"))
agecalc(as.IDate("2000-02-29"), as.IDate("2000-02-28"))
agecalc(as.IDate("2000-02-29"), as.IDate("2000-02-29"))
agecalc(as.IDate("2000-02-29"), as.IDate("2001-02-28"))
agecalc(as.IDate("2000-02-29"), as.IDate("2001-02-29"))
agecalc(as.IDate("2000-02-29"), as.IDate("2001-03-01"))
agecalc(as.IDate("2000-02-29"), as.IDate("2004-02-28"))
agecalc(as.IDate("2000-02-29"), as.IDate("2004-02-29"))
agecalc(as.IDate("2000-02-29"), as.IDate("2011-03-01"))
## Testing every age for every day over several years
## This test requires vectorized version:
d <- data.table(d=as.IDate("2000-01-01") + 0:10000)
d[ , b1 := as.IDate("2000-08-15")]
d[ , b2 := as.IDate("2000-02-29")]
d[ , age1_num := (d - b1) / 365]
d[ , age2_num := (d - b2) / 365]
d[ , age1 := agecalc(b1, d)]
d[ , age2 := agecalc(b2, d)]
d
Below is a trivial plot of ages as numeric and integer. As you can see the
integer ages are a sort of stair step pattern that is tangent to (but below) the
straight line of numeric ages.
plot(numeric_age1 ~ today, dt, type = "l",
ylab = "ages", main = "ages plotted")
lines(integer_age1 ~ today, dt, col = "blue")

Answer 6

I wasn't happy with any of the responses when it comes to calculating the age in months or years, when dealing with leap years, so this is my function using the lubridate package.
Basically, it slices the interval between from and to into (up to) yearly chunks, and then adjusts the interval for whether that chunk is leap year or not. The total interval is the sum of the age of each chunk.
library(lubridate)
#' Get Age of Date relative to Another Date
#'
#' #param from,to the date or dates to consider
#' #param units the units to consider
#' #param floor logical as to whether to floor the result
#' #param simple logical as to whether to do a simple calculation, a simple calculation doesn't account for leap year.
#' #author Nicholas Hamilton
#' #export
age <- function(from, to = today(), units = "years", floor = FALSE, simple = FALSE) {
#Account for Leap Year if Working in Months and Years
if(!simple && length(grep("^(month|year)",units)) > 0){
df = data.frame(from,to)
calc = sapply(1:nrow(df),function(r){
#Start and Finish Points
st = df[r,1]; fn = df[r,2]
#If there is no difference, age is zero
if(st == fn){ return(0) }
#If there is a difference, age is not zero and needs to be calculated
sign = +1 #Age Direction
if(st > fn){ tmp = st; st = fn; fn = tmp; sign = -1 } #Swap and Change sign
#Determine the slice-points
mid = ceiling_date(seq(st,fn,by='year'),'year')
#Build the sequence
dates = unique( c(st,mid,fn) )
dates = dates[which(dates >= st & dates <= fn)]
#Determine the age of the chunks
chunks = sapply(head(seq_along(dates),-1),function(ix){
k = 365/( 365 + leap_year(dates[ix]) )
k*interval( dates[ix], dates[ix+1] ) / duration(num = 1, units = units)
})
#Sum the Chunks, and account for direction
sign*sum(chunks)
})
#If Simple Calculation or Not Months or Not years
}else{
calc = interval(from,to) / duration(num = 1, units = units)
}
if (floor) calc = as.integer(floor(calc))
calc
}

Answer 7

(Sys.Date() - yourDate) / 365.25

Answer 8

A very simple way of calculating the age from two dates without using any additional packages probably is:
df$age = with(df, as.Date(date_2, "%Y-%m-%d") - as.Date(date_1, "%Y-%m-%d"))

Answer 9

Here is a (I think simpler) solution using lubridate:
library(lubridate)
age <- function(dob, on.day=today()) {
intvl <- interval(dob, on.day)
prd <- as.period(intvl)
return(prd#year)
}

Answer 10

Note that age_calc from the eeptools package in particular fails on cases with the year 2000 around birthdays.
Some examples that don't work in age_calc:
library(lubridate)
library(eeptools)
age_calc(ymd("1997-04-21"), ymd("2000-04-21"), units = "years")
age_calc(ymd("2000-04-21"), ymd("2019-04-21"), units = "years")
age_calc(ymd("2000-04-21"), ymd("2016-04-21"), units = "years")
Some of the other solutions also have some output that is not intuitive to what I would want for decimal ages when leap years are involved. I like #James_D 's solution and it is precise and concise, but I wanted something where the decimal age is calculated as complete years plus the fraction of the year completed from their last birthday to their next birthday (which would be out of 365 or 366 days depending on year). In the case of leap years I use lubridate's rollback function to use March 1st for non-leap years following February 29th. I used some test cases from #geneorama and added some of my own, and the output aligns with what I would expect.
library(lubridate)
# Calculate precise age from birthdate in ymd format
age_calculation <- function(birth_date, later_year) {
if (birth_date > later_year)
{
stop("Birth date is after the desired date!")
}
# Calculate the most recent birthday of the person based on the desired year
latest_bday <- ymd(add_with_rollback(birth_date, years((year(later_year) - year(birth_date))), roll_to_first = TRUE))
# Get amount of days between the desired date and the latest birthday
days_between <- as.numeric(days(later_year - latest_bday), units = "days")
# Get how many days are in the year between their most recent and next bdays
year_length <- as.numeric(days((add_with_rollback(latest_bday, years(1), roll_to_first = TRUE)) - latest_bday), units = "days")
# Get the year fraction (amount of year completed before next birthday)
fraction_year <- days_between/year_length
# Sum the difference of years with the year fraction
age_sum <- (year(later_year) - year(birth_date)) + fraction_year
return(age_sum)
}
test_list <- list(c("1985-08-13", "1986-08-12"),
c("1985-08-13", "1985-08-13"),
c("1985-08-13", "1986-08-13"),
c("1985-08-13", "1986-09-12"),
c("2000-02-29", "2000-02-29"),
c("2000-02-29", "2000-03-01"),
c("2000-02-29", "2001-02-28"),
c("2000-02-29", "2004-02-29"),
c("2000-02-29", "2011-03-01"),
c("1997-04-21", "2000-04-21"),
c("2000-04-21", "2016-04-21"),
c("2000-04-21", "2019-04-21"),
c("2017-06-15", "2018-04-30"),
c("2019-04-20", "2019-08-24"),
c("2020-05-25", "2021-11-25"),
c("2020-11-25", "2021-11-24"),
c("2020-11-24", "2020-11-25"),
c("2020-02-28", "2020-02-29"),
c("2020-02-29", "2020-02-28"))
for (i in 1:length(test_list))
{
print(paste0("Dates from ", test_list[[i]][1], " to ", test_list[[i]][2]))
result <- age_calculation(ymd(test_list[[i]][1]), ymd(test_list[[i]][2]))
print(result)
}
Output:
[1] "Dates from 1985-08-13 to 1986-08-12"
[1] 0.9972603
[1] "Dates from 1985-08-13 to 1985-08-13"
[1] 0
[1] "Dates from 1985-08-13 to 1986-08-13"
[1] 1
[1] "Dates from 1985-08-13 to 1986-09-12"
[1] 1.082192
[1] "Dates from 2000-02-29 to 2000-02-29"
[1] 0
[1] "Dates from 2000-02-29 to 2000-03-01"
[1] 0.00273224
[1] "Dates from 2000-02-29 to 2001-02-28"
[1] 0.9972603
[1] "Dates from 2000-02-29 to 2004-02-29"
[1] 4
[1] "Dates from 2000-02-29 to 2011-03-01"
[1] 11
[1] "Dates from 1997-04-21 to 2000-04-21"
[1] 3
[1] "Dates from 2000-04-21 to 2016-04-21"
[1] 16
[1] "Dates from 2000-04-21 to 2019-04-21"
[1] 19
[1] "Dates from 2017-06-15 to 2018-04-30"
[1] 0.8739726
[1] "Dates from 2019-04-20 to 2019-08-24"
[1] 0.3442623
[1] "Dates from 2020-05-25 to 2021-11-25"
[1] 1.50411
[1] "Dates from 2020-11-25 to 2021-11-24"
[1] 0.9972603
[1] "Dates from 2020-11-24 to 2020-11-25"
[1] 0.002739726
[1] "Dates from 2020-02-28 to 2020-02-29"
[1] 0.00273224
[1] "Dates from 2020-02-29 to 2020-02-28"
Error in age_calculation(ymd(test_list[[i]][1]), ymd(test_list[[i]][2])) :
Birth date is after the desired date!
As others have been saying, the trunc function is excellent to get integer age.

Answer 11

I realise there are a lot of answers but since I can't help myself, I might as well add to the discussion.
I'm building a package that's focused on dates and datetimes and in it I use a function called time_diff(). Here is a simplified version.
time_diff <- function(x, y, units, num = 1,
type = c("duration", "period"),
as_period = FALSE){
type <- match.arg(type)
units <- match.arg(units, c("picoseconds", "nanoseconds", "microseconds",
"milliseconds", "seconds", "minutes", "hours", "days",
"weeks", "months", "years"))
int <- lubridate::interval(x, y)
if (as_period || type == "period"){
if (as_period) int <- lubridate::as.period(int, unit = units)
unit <- lubridate::period(num = num, units = units)
} else {
unit <- do.call(get(paste0("d", units),
asNamespace("lubridate")),
list(x = num))
}
out <- int / unit
out
}
# Wrapper around the more general time_diff
age_years <- function(x, y){
trunc(time_diff(x, y, units = "years", num = 1,
type = "period", as_period = TRUE))
}
library(lubridate)
#>
#> Attaching package: 'lubridate'
#> The following objects are masked from 'package:base':
#>
#> date, intersect, setdiff, union
bday <- dmy("01-01-2000")
time_diff(bday, today(), "years", type = "period")
#> [1] 23.11233
leap1 <- dmy("29-02-2020")
leap2 <- dmy("28-02-2021")
leap3 <- dmy("01-03-2021")
# Many people might say this is wrong so use the more exact age_years
time_diff(leap1, leap2, "years", type = "period")
#> [1] 1
# age in years, accounting for leap years properly
age_years(leap1, leap2)
#> [1] 0
age_years(leap1, leap3)
#> [1] 1
# So to add a column of ages in years, one can do this..
library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#> filter, lag
#> The following objects are masked from 'package:base':
#>
#> intersect, setdiff, setequal, union
my_data <- tibble(dob = seq(bday, today(), by = "day"))
my_data <- my_data %>%
mutate(age_years = age_years(dob, today()))
slice_head(my_data, n = 10)
#> # A tibble: 10 x 2
#> dob age_years
#> <date> <dbl>
#> 1 2000-01-01 23
#> 2 2000-01-02 23
#> 3 2000-01-03 23
#> 4 2000-01-04 23
#> 5 2000-01-05 23
#> 6 2000-01-06 23
#> 7 2000-01-07 23
#> 8 2000-01-08 23
#> 9 2000-01-09 23
#> 10 2000-01-10 23
Created on 2023-02-11 with reprex v2.0.2