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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?
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
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.
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.
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.
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")
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
}
(Sys.Date() - yourDate) / 365.25
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"))
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)
}
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.
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
I'm working with a dataset that has a lot of timestamps. There are some invalid timestamps which I try to identify and set to NA. Because if_else() forces me to have the same data type in both arms, I'm using as.POSIXct(NA) to encode such missing values.
Interestingly, the results differ when I invert the test (and change the true and false argument) in if_else().
Here is some code to illustrate my problems:
x <- tibble(
A = parse_datetime("2020-08-18 19:00"),
B = if_else(TRUE, A, as.POSIXct(NA)),
C = if_else(FALSE, as.POSIXct(NA), A)
)
> x
# A tibble: 1 x 3
A B C
<dttm> <dttm> <dttm>
1 2020-08-18 19:00:00 2020-08-18 19:00:00 2020-08-18 21:00:00
Any idea, why C is two hours later?
Follow-up:
Based on the great answers below, I think a more readable solution should perhaps generate a missing datetime object with parse_datetime(NA_character_) and use this in the code instead of as.POSIXct().
R> NA_datetime_ <- parse_datetime(NA_character_)
R> x <- tibble(
A = parse_datetime("2020-08-18 19:00"),
B = if_else(TRUE, A, NA_datetime_),
C = if_else(FALSE, NA_datetime_, A)
)
R> map(x, lubridate::tz)
$A
[1] "UTC"
$B
[1] "UTC"
$C
[1] "UTC"
At First, you need to know that parse_datetime() returns a date-time object with an tzone attribute default to UTC. You can use lubridate::tz(x$A) and attributes(x$A) to check it.
From the document of if_else(), it said the true and false arguments must be the same type. All other attributes are taken from true. Hence, in part C of your tibble:
C = if_else(FALSE, as.POSIXct(NA), A)
as.POSIXct(NA) doesn't have a tzone attribute, so A's tzone is dropped and reset to the time zone of your region. Actually, C is not two hours later. The three columns have equal time but unequal time zones. To fix it, you can adjust as.POSIXct(NA) to own a tzone attribute, i.e. replace it with
as.POSIXct(NA_character_, tz = "UTC")
Note: You must use NA_character_ instead of NA because the tz argument in as.POSIXct() only works on character objects.
Finally, revise your code as
x <- tibble(
A = parse_datetime("2020-08-18 19:00"),
B = if_else(TRUE, A, as.POSIXct(NA_character_, tz = "UTC")),
C = if_else(FALSE, as.POSIXct(NA_character_, tz = "UTC"), A)
)
# # A tibble: 1 x 3
# A B C
# <dttm> <dttm> <dttm>
# 1 2020-08-18 19:00:00 2020-08-18 19:00:00 2020-08-18 19:00:00
Remember to check their time zones.
R > lubridate::tz(x$A)
[1] "UTC"
R > lubridate::tz(x$B)
[1] "UTC"
R > lubridate::tz(x$C)
[1] "UTC"
This is a timezone problem :
lubridate::tz(x$A)
[1] "UTC"
lubridate::tz(x$B)
[1] "UTC"
lubridate::tz(x$C)
[1] ""
This is due to the way if_else <- function (test, yes, no) works : it uses the attributes of the yes argument which for C is NA.
I need to sum the number of days across multiple intervals in the same observations. I have seen many different examples in StackOverflow about this task. Still, I cannot reproduce them using my data, because I have to do it in more than two overlapping times, and across multiple intervals.
library(lubridate)
library(dplyr)
a <- c(as_date(0), as_date(8), as_date(80),as_date(60))
b <-c(as_date(2), as_date(20), as_date(100),as_date(80))
c <-c(as_date(1), as_date(16), as_date(95),as_date(85))
d <- c(as_date(100), as_date(19), as_date(120),as_date(100))
e <-c(as_date(0), as_date(50), as_date(101),as_date(65))
f <- c(as_date(150), as_date(100), as_date(200),as_date(200))
df <- data.frame(int.1 = interval(a, b), int.2 = interval(c, d), int.3 = interval(e, f))
I can sum the total time between the intervals, but the time that overlaps is included:
df %>%
mutate(overlapping.time = int.1 %/% days(1) + int.2 %/% days(1) + int.3 %/% days(1))
int.1 int.2 int.3 overlapping.time
1 1970-01-01 UTC--1970-01-03 UTC 1970-01-02 UTC--1970-04-11 UTC 1970-01-01 UTC--1970-05-31 UTC 251
2 1970-01-09 UTC--1970-01-21 UTC 1970-01-17 UTC--1970-01-20 UTC 1970-02-20 UTC--1970-04-11 UTC 65
3 1970-03-22 UTC--1970-04-11 UTC 1970-04-06 UTC--1970-05-01 UTC 1970-04-12 UTC--1970-07-20 UTC 144
4 1970-03-02 UTC--1970-03-22 UTC 1970-03-27 UTC--1970-04-11 UTC 1970-03-07 UTC--1970-07-20 UTC 170
Below is a function overlapping_days(), which will take a set of interval columns and calculate the total amount of overlapping days. See inline comments for how it works. It covers intervals completely contained within another, partially overlapping, and makes no assumptions about the relationships between columns. Subtracting the result of the function from your previous calculation will get you what you want. Note that I modified the data I used a bit from what you originally posted.
library(lubridate)
#>
#> Attaching package: 'lubridate'
#> The following object is masked from 'package:base':
#>
#> date
library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:lubridate':
#>
#> intersect, setdiff, union
#> The following objects are masked from 'package:stats':
#>
#> filter, lag
#> The following objects are masked from 'package:base':
#>
#> intersect, setdiff, setequal, union
a <- c(as_date(0), as_date(1), as_date(80),as_date(60))
b <-c(as_date(20), as_date(22), as_date(100),as_date(80))
c <-c(as_date(1), as_date(16), as_date(95),as_date(85))
d <- c(as_date(3), as_date(19), as_date(120),as_date(100))
e <-c(as_date(0), as_date(50), as_date(101),as_date(65))
f <- c(as_date(150), as_date(100), as_date(200),as_date(200))
df <- data.frame(int.1 = interval(a, b), int.2 = interval(c, d), int.3 = interval(e, f))
overlapping_days <- function(...) {
# Collect the vectors passed into a list
ll <- list(...)
# Create all possible 2-combinations for the number of columns passed in.
combinations <- combn(length(ll), 2)
# Create a column for each combination, and a row for each element in the vectors.
overlaps <- matrix(data = 0, nrow = length(ll[[1]]), ncol = length(combinations))
# Loop through the combinations
iterations <- seq_len(ncol(combinations))
for (k in iterations) {
# I'll refer to each of these indices as intervals -- they each represent
# a vector passed in.
i <- combinations[1, k]
j <- combinations[2, k]
overlaps[,k] <- case_when(
# If the interval i is within interval j, add i to the overlap
ll[[i]] %within% ll[[j]] ~ ll[[i]] %/% days(1),
# If the interval j is within interval i, add j to the overlap
ll[[j]] %within% ll[[i]] ~ ll[[j]] %/% days(1),
# If they overlap, either int_start(i) < int_end(j), or int_start(j) < int_end(i)
# Calculate the appropriate difference -- these look backwards but
# are needed so a positive number is produced.
int_overlaps(ll[[i]], ll[[j]]) & int_start(ll[[j]]) < int_end(ll[[i]]) ~
int_start(ll[[j]]) %--% int_end(ll[[i]]) %/% days(1),
int_overlaps(ll[[j]], ll[[i]]) & int_start(ll[[i]]) < int_end(ll[[j]]) ~
int_start(ll[[i]]) %--% int_end(ll[[j]]) %/% days(1),
# If none of these are true, the intervals do not overlap and we add 0 to
# the overlap amount.
TRUE ~ 0
)
}
# Sum across rows to get the total number of overlapping days.
rowSums(overlaps)
}
df %>%
mutate(overlapping.time = int.1 %/% days(1) + int.2 %/% days(1) + int.3 %/% days(1), overlap = overlapping_days(int.1, int.2, int.3))
#> Note: method with signature 'Timespan#Timespan' chosen for function '%/%',
#> target signature 'Interval#Period'.
#> "Interval#ANY", "ANY#Period" would also be valid
#> int.1 int.2
#> 1 1970-01-01 UTC--1970-01-21 UTC 1970-01-02 UTC--1970-01-04 UTC
#> 2 1970-01-02 UTC--1970-01-23 UTC 1970-01-17 UTC--1970-01-20 UTC
#> 3 1970-03-22 UTC--1970-04-11 UTC 1970-04-06 UTC--1970-05-01 UTC
#> 4 1970-03-02 UTC--1970-03-22 UTC 1970-03-27 UTC--1970-04-11 UTC
#> int.3 overlapping.time overlap
#> 1 1970-01-01 UTC--1970-05-31 UTC 172 24
#> 2 1970-02-20 UTC--1970-04-11 UTC 74 3
#> 3 1970-04-12 UTC--1970-07-20 UTC 144 24
#> 4 1970-03-07 UTC--1970-07-20 UTC 170 30
From date 10/31/2018, I want to obtain 09/30/2018.
I tried:
PREV.PROD.DATE<-seq.Date(from=as.Date(chron("10/31/2018")),length=2,by="-1 months")
but it returns:
"2018-10-31" "2018-10-01"
How can I obtain 09/30 instead of 10/01?
Notes: I would like to avoid to use an external package, and I would like the solution to work for any end of month date.
The integer units of Date are days, so you can do:
seq.Date(from=as.Date("2018-10-31"), length=2, by="-1 months") - c(0,1)
# [1] "2018-10-31" "2018-09-30"
If you want arbitrary prior-last-date:
(d <- as.POSIXlt(Sys.Date()))
# [1] "2018-11-08 UTC"
d$mday <- 1L
as.Date(d) - 1
# [1] "2018-10-31"
Replace Sys.Date() with whatever single date you have. If you want to vectorize this:
(ds <- Sys.Date() + c(5, 20, 50))
# [1] "2018-11-13" "2018-11-28" "2018-12-28"
lapply(as.POSIXlt(ds), function(a) as.Date(`$<-`(a, "mday", 1L)) - 1L)
# [[1]]
# [1] "2018-10-31"
# [[2]]
# [1] "2018-10-31"
# [[3]]
# [1] "2018-11-30"
I like to floor the date - making it the first day of its month, and then subtract 1 to make it the last day of the previous month:
x = as.Date("2018-10-31")
library(lubridate)
floor_date(x, unit = "months") - 1
# [1] "2018-09-30"
Here's a version without using other packages:
as.Date(format(x, "%Y-%m-01")) - 1
# [1] "2018-09-30"
I don't use function chron. But I think the function duration from lubridate helps you.
You don't need use floor_date to confuse you.
library(lubridate)
#>
#> 载入程辑包：'lubridate'
#> The following object is masked from 'package:base':
#>
#> date
library(tidyverse)
better_date <-
str_split("10/31/2018",'/') %>%
.[[1]] %>%
.[c(3,1,2)] %>%
str_flatten(collapse = '-') %>%
as.Date()
(better_date - duration('1 months')) %>%
as.Date()
#> [1] "2018-09-30"
Created on 2018-11-09 by the reprex package (v0.2.1)
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?
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
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$birthdate, units = "years"))
assuming you want age in integer years.
Assume you have a data.table, you could do below:
library(data.table)
library(lubridate)
# toy data
X = data.table(birth=seq(from=as.Date("1970-01-01"), to=as.Date("1980-12-31"), by="year"))
Sys.Date()
Option 1 : use "as.period" from lubriate package
X[, age := as.period(Sys.Date() - birth)][]
birth age
1: 1970-01-01 44y 0m 327d 0H 0M 0S
2: 1971-01-01 43y 0m 327d 6H 0M 0S
3: 1972-01-01 42y 0m 327d 12H 0M 0S
4: 1973-01-01 41y 0m 326d 18H 0M 0S
5: 1974-01-01 40y 0m 327d 0H 0M 0S
6: 1975-01-01 39y 0m 327d 6H 0M 0S
7: 1976-01-01 38y 0m 327d 12H 0M 0S
8: 1977-01-01 37y 0m 326d 18H 0M 0S
9: 1978-01-01 36y 0m 327d 0H 0M 0S
10: 1979-01-01 35y 0m 327d 6H 0M 0S
11: 1980-01-01 34y 0m 327d 12H 0M 0S
Option 2 : if you do not like the format of Option 1, you could do below:
yr = duration(num = 1, units = "years")
X[, age := new_interval(birth, Sys.Date())/yr][]
# you get
birth age
1: 1970-01-01 44.92603
2: 1971-01-01 43.92603
3: 1972-01-01 42.92603
4: 1973-01-01 41.92329
5: 1974-01-01 40.92329
6: 1975-01-01 39.92329
7: 1976-01-01 38.92329
8: 1977-01-01 37.92055
9: 1978-01-01 36.92055
10: 1979-01-01 35.92055
11: 1980-01-01 34.92055
Believe Option 2 should be the more desirable.
I prefer to do this using the lubridate package, borrowing syntax I originally encountered in another post.
It's necessary to standardize your input dates in terms of R date objects, preferably with the lubridate::mdy() or lubridate::ymd() or similar functions, as applicable. You can use the interval() function to generate an interval describing the time elapsed between the two dates, and then use the duration() function to define how this interval should be "diced".
I've summarized the simplest case for calculating an age from two dates below, using the most current syntax in R.
df$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.
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")
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
}
(Sys.Date() - yourDate) / 365.25
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"))
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)
}
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.
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