我第一次使用data.table。
我的表中有大约400,000个年龄数据,我需要将它们从出生日期转换为年龄。
最佳方法是什么?
我一直在思考这个问题并对迄今为止给出的两个答案不满意。我喜欢使用lubridate
,就像@KFB所做的那样,但我也希望以我使用eeptools
包的答案中一样,将所有内容都封装到一个函数中。因此,这里有一个使用lubridate间隔方法和一些不错选项的封装函数:
#' 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)
}
使用示例:
> 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
age(dob = as.Date(“1970-06-01”),age.day = as.Date(“2018-05-31”))
(一个人48岁生日前一天)应该返回47岁,但实际上它返回了48(在floor=FALSE
时为48.03014)。可能有更简洁的方法,但as.numeric(as.period(interval(as.Date("1970-06-01"), as.Date("2018-05-31"))), "years")
似乎更好(它返回47.9988)。 - Hobo从这篇博客文章的评论中,我找到了在eeptools
包中的age_calc
函数。它处理了边缘情况(闰年等),检查了输入并且看起来非常健壮。
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 224.83118age_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
针对您的数据
yourdata$age <- floor(age_calc(yourdata$birthdate, units = "years"))
假设你想以整数年龄为单位。
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()
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
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
yr = duration(num = 1, units = "years"); birth <- as.Date("1970-06-01"); age_as_at <- as.Date("2018-05-31"); interval(birth, age_as_at)/yr
应小于48。 - Hobodata.table
以外的依赖项,通常data.table
是我的唯一依赖项。 data.table
仅用于mday,即月份中的日期。
这个函数逻辑上是我如何考虑某人的年龄。我从[当前年份] - [出生年份] - 1开始,然后如果他们已经过了今年的生日,则加1。为了检查这个偏移量,我首先考虑月份,然后(如果必要)考虑日期。
以下是逐步实现的步骤:
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)
}
这是相同逻辑的重构和向量化:
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)
}
您也可以对月份/日期部分进行字符串比较。也许有时这更有效,例如如果年份是数字而出生日期是字符串。
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)
}
对矢量化的“生产”版本进行了一些测试:
## 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
plot(numeric_age1 ~ today, dt, type = "l",
ylab = "ages", main = "ages plotted")
lines(integer_age1 ~ today, dt, col = "blue")
我更倾向于使用lubridate
包,借鉴我最初在另一个post中遇到的语法来完成这项任务。
为了以R日期对象的形式标准化输入日期,最好使用lubridate::mdy()
或lubridate::ymd()
或类似的函数(如适用)。您可以使用interval()
函数生成描述两个日期之间经过的时间间隔的区间,然后使用duration()
函数定义应如何"切块"该区间。
下面概述了使用R中最新语法计算两个日期之间年龄的最简单情况。
df$DOB <- mdy(df$DOB)
df$EndDate <- mdy(df$EndDate)
df$Calc_Age <- interval(start= df$DOB, end=df$EndDate)/
duration(n=1, unit="years")
使用R语言的'floor()`函数,可以将年龄四舍五入到最接近的整数。
df$Calc_AgeF <- floor(df$Calc_Age)
或者,在基本的R round()
函数中,digits=
参数可用于向上或向下舍入,并指定返回值中的精确小数位数,如下所示:
df$Calc_Age2 <- round(df$Calc_Age, digits = 2) ## 2 decimals
df$Calc_Age0 <- round(df$Calc_Age, digits = 0) ## nearest integer
interval()
和duration()
函数),返回的值将是数值,不再是R中的日期对象。这很重要,因为lubridate::floor_date()
严格限于日期时间对象。data.table
还是data.frame
对象中出现,上述语法都适用。(Sys.Date() - yourDate) / 365.25
在计算月龄或年龄时,处理闰年的情况,我对所有的回答都不满意,所以这是我使用lubridate包编写的函数。
基本上,它将from
和to
之间的时间间隔切成(最多)每年一块,并根据该块是否为闰年调整时间间隔。总时间间隔是每个块的年龄之和。
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
}
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")
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)
}
输出:
[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!
正如其他人所说,trunc
函数非常适合获取整数年龄。
library(lubridate)
age <- function(dob, on.day=today()) {
intvl <- interval(dob, on.day)
prd <- as.period(intvl)
return(prd@year)
}
一种非常简单的计算两个日期之间年龄的方法,而不需要使用任何额外的包可能是:
df$age = with(df, as.Date(date_2, "%Y-%m-%d") - as.Date(date_1, "%Y-%m-%d"))