Efficient and Accurate Age Calculation (In Years, Months, or Weeks) in R Given Birth Date and an Arbitrary Date

Efficient and accurate age calculation (in years, months, or weeks) in R given birth date and an arbitrary date

Ok, so I found this function in another post:

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)
}

It was posted by @Jim saying "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".

It is indeed simpler and it does the trick I was looking for. On average, it is actually faster than the arithmetic method (about 75% faster).

mbm <- microbenchmark(
    arithmetic = (givendate - birthdate) / 365.25,
    lubridate = interval(start = birthdate, end = givendate) /
        duration(num = 1, units = "years"),
    eeptools = age_calc(dob = birthdate, enddate = givendate, 
                        units = "years"),
    age = age(from = birthdate, to = givendate),
    times = 1000
)
mbm
autoplot(mbm)

And at least in my examples it does not make any mistake (and it should not in any example; it's a pretty straightforward function using ifelses).

toy_df <- data.frame(
    birthdate = birthdate,
    givendate = givendate,
    arithmetic = as.numeric((givendate - birthdate) / 365.25),
    lubridate = interval(start = birthdate, end = givendate) /
        duration(num = 1, units = "years"),
    eeptools = age_calc(dob = birthdate, enddate = givendate,
                        units = "years"),
    age = age(from = birthdate, to = givendate)
)
toy_df[, 3:6] <- floor(toy_df[, 3:6])
toy_df

    birthdate  givendate arithmetic lubridate eeptools age
1  1978-12-30 2015-12-31         37        37       37  37
2  1978-12-31 2015-12-31         36        37       37  37
3  1979-01-01 2015-12-31         36        37       36  36
4  1962-12-30 2015-12-31         53        53       53  53
5  1962-12-31 2015-12-31         52        53       53  53
6  1963-01-01 2015-12-31         52        53       52  52
7  2000-06-16 2050-06-17         50        50       50  50
8  2000-06-17 2050-06-17         49        50       50  50
9  2000-06-18 2050-06-17         49        50       49  49
10 2007-03-18 2008-03-19          1         1        1   1
11 2007-03-19 2008-03-19          1         1        1   1
12 2007-03-20 2008-03-19          0         1        0   0
13 1968-02-29 2015-02-28         46        47       46  46
14 1968-02-29 2015-03-01         47        47       47  47
15 1968-02-29 2015-03-02         47        47       47  47

I do not consider it as a complete solution because I also wanted to have age in months and weeks, and this function is specific for years. I post it here anyway because it solves the problem for the age in years. I will not accept it because:

1. I would wait for @Jim to post it as an answer.
2. I will wait to see if someone else come up with a complete solution (efficient, accurate and producing age in years, months or weeks as desired).

How to calculate the mean age from the data stated as 2 digit years in R

We could use lubridate's make_date() to turn the individual columns into a date column and then calculate the age. I have shown here how you could take care of the missing 19/20 in birthyear, but you might need to tweak it for your data.

library(dplyr)
library(lubridate)

mydf |> 
    mutate(date = make_date(if_else(birthyear > 21, birthyear+1900, birthyear), birthmonth, birthday),
           age  = as.period(interval(date, today()))$year
    )

Output:

  ID birthday birthmonth birthyear       date age
1  A       12          8        79 1979-08-12  43
2  B       23         10        62 1962-10-23  59
3  C        2          3        66 1966-03-02  56
4  D       20          9        83 1983-09-20  38

And to get the mean age with summarise:

mydf |> 
    mutate(date = make_date(if_else(birthyear > 21, birthyear+1900, birthyear), birthmonth, birthday),
           age  = as.period(interval(date, today()))$year
    ) |>
    summarise(mean_age = mean(age))

Output:

  mean_age
1       49

Update: It can be non-trivial to get the right age calculation (fast), check e.g. Efficient and accurate age calculation (in years, months, or weeks) in R given birth date and an arbitrary date

Calculating age using mutate with lubridate functions

We can use do

df %>%
   mutate(age=as.period(interval(start = birthdate, end = givendate))) %>%
   do(data.frame(.[setdiff(names(.), "age")], 
       age = ifelse(!is.na(.$age), .$age$year, .$age)))
#    birthdate  givendate age
#1       <NA>       <NA>  NA
#2 1978-12-31 2015-12-31  37
#3 1979-01-01 2015-12-31  36
#4 1962-12-30       <NA>  NA

---

As the as.period comes with period class, we may need S4 methods to extract it

df %>% 
    mutate(age=as.period(interval(start = birthdate, end = givendate))) %>%
   .$age %>%
   .@year %>%
    mutate(df, age = .)
#  birthdate  givendate age
#1       <NA>       <NA>  NA
#2 1978-12-31 2015-12-31  37
#3 1979-01-01 2015-12-31  36
#4 1962-12-30       <NA>  NA

change a column from birth date to age in r

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.

Getting negative ages using lubridate to calculate age from birth date and current date

If you check the output of dmy function

head(df$DATE_OF_BIRTH)
#[1] "20/10/01" "15/04/88" "16/12/58" "15/10/91" "09/02/66" "02/07/03"

head(dmy(df$DATE_OF_BIRTH))
#[1] "2001-10-20" "1988-04-15" "2058-12-16" "1991-10-15" "2066-02-09" "2003-07-02"

R interprets years 00 - 68 as 2000 - 2068 and 69 - 99 as 1969 - 1999. Hence, 58 is considered as 2058, 66 is considered to 2066 but 88 is 1988.

From ?strptime

%y
Year without century (00–99). On input, values 00 to 68 are prefixed by 20 and 69 to 99 by 19 – that is the behaviour specified by the 2004 and 2008 POSIX standards, but they do also say ‘it is expected that in a future version the default century inferred from a 2-digit year will change

---

For negative values you can add 100 to them to get equivalent positive values

library(dplyr)
library(lubridate)

df %>%
  mutate(age = interval(start = dmy(DATE_OF_BIRTH), end = dmy('01/07/17')) / 
          duration(num = 1, units = "years"), 
          age = if_else(age < 0, age + 100, age))


#   DATE_OF_BIRTH       age
#1       20/10/01 15.706849
#2       15/04/88 29.230137
#3       16/12/58 58.512329
#4       15/10/91 25.728767
#5       09/02/66 51.356164
#6       02/07/03 14.008219
#7       20/08/96 20.876712
#....

To get difference between dates in years, you could also use interval like this

df %>%
  mutate(age = interval(dmy(DATE_OF_BIRTH), dmy('01/07/17')) / years(1),
         age = if_else(age < 0, age + 100, age))

Calculate age (with decimal places) in R

Is this OK?

round(as.numeric((Testday - DOB) / 365.25), 2)
#[1]  3.28 11.18  0.92

In a single chain get a max date - n years

We could use a lambda function

aus_livestock$Month |>
    max() |> 
    as_date() |> 
    (\(x) x - years(6))()
[1] "2012-12-01"

---

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