從r出生日期到年齡段更改一列

Question

我首次使用data.table。

我在我的表中有一個約40萬年齡的專欄。我需要將它們從出生日期轉換爲年齡。

這樣做的最好方法是什麼？

Answer 1

從this blog entry的評論，我發現eeptools包中的age_calc函數。它處理邊緣情況（閏年等），檢查輸入並看起來相當健壯。

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

爲您的數據

yourdata$age <- floor(age_calc(yourdata$birthdate, units = "years")) 

假設你想在整數歲。

Answer 2

假設你有一個data.table，你可以做如下：

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() 

方法1：使用 「as.period」 從lubriate包

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 

選項2：如果你不知道就像選項1的格式一樣，你可以這樣做：

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 

相信選項2應該是更可取的。

Answer 3

我一直在想這件事，並且對迄今爲止的兩個答案感到不滿。我喜歡使用lubridate，正如@KFB所做的那樣，但我也希望事情能夠很好地包裝在一個函數中，就像我在使用eeptools包的答案中一樣。因此，這裏的使用lubridate區間方法與一些不錯的選擇包裝函數：

#' 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 = interval(dob, age.day)/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 

Answer 4

我並不高興與任何反應的，當涉及到計算時代幾個月或幾年，當處理閏年時，所以這是我使用lubridate軟件包的功能。

基本上，它會將from和to之間的間隔切分爲（最多）年度塊，然後調整該塊是否爲閏年的時間間隔。總間隔是每個塊的年齡的總和。

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 
} 

Answer 5

我喜歡做這個使用lubridate包，借用我原來在另一post遇到的語法。

有必要根據R日期對象標準化輸入日期，最好使用lubridate::mdy()或lubridate::ymd()或相似的函數（如適用）。您可以使用​​函數生成描述兩個日期之間所用時間的間隔，然後使用duration()函數來定義如何將該間隔「切塊」。

我總結了簡單的情況下，計算從下面兩個日期的時代，使用最新的語法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") 

年齡可能向下調整至最接近的整數完全使用基本R'地板（）`函數，如下所示：

df$Calc_AgeF <- floor(df$Calc_Age) 

可替換地，在基R中的digits=參數round()函數可用於舍向上或向下，並指定小數的確切數目在返回值中，像這樣：

df$Calc_Age2 <- round(df$Calc_Age, digits = 2) ## 2 decimals 
df$Calc_Age0 <- round(df$Calc_Age, digits = 0) ## nearest integer 

值得注意的是，一旦輸入日期通過以上（即，​​和duration()功能）中所述的計算步驟中通過，返回值將是數字，並不再在R.日期對象這是顯著而lubridate::floor_date()嚴格限於日期時間對象。

無論輸入日期是否出現在data.table或data.frame對象中，上述語法都適用。