使用具有疏水功能的突变体计算年龄 [英] Calculating age using mutate with lubridate functions

问题描述

我想根据出生日期计算年龄。

I would like to calculate age based on birth date.

如果我使用lubridate，我只需运行以下内容，如高效准确的年龄计算（以年，月，或周）在R给出生日期和任意日期

If I use lubridate, I would just run the following as in Efficient and accurate age calculation (in years, months, or weeks) in R given birth date and an arbitrary date

as.period（new_interval（start = birthdate，end = givendate） ）$ year

但是，当我尝试在<$ c中使用 mutate $ c> dplyr 创建新变量，我遇到了一个错误。

However, when I tried to use mutate in dplyr to create the new variable, I ran into an error.

library(dplyr); library(lubridate)

birthdate <- ymd(c(NA, "1978-12-31", "1979-01-01", "1962-12-30"))
givendate <- ymd(c(NA, "2015-12-31", "2015-12-31", NA))

df <- data.frame(
    birthdate = birthdate,
    givendate = givendate)

以下工作虽然它提供了所有的日期和时间值。即年，月，日，小时，分和秒。

The following works though it gives all the date and time values. i.e. year, month, day, hour, minute and second.

df<-df %>% mutate(age=as.period(interval(start = birthdate, end = givendate)))

# df
#    birthdate  givendate                  age
# 1       <NA>       <NA>                 <NA>
# 2 1978-12-31 2015-12-31   37y 0m 0d 0H 0M 0S
# 3 1979-01-01 2015-12-31 36y 11m 30d 0H 0M 0S
# 4 1962-12-30       <NA>                 <NA>

以下内容不起作用：

df<-df %>% 
       mutate(age=as.period(interval(start = birthdate, end = givendate))$year)

它给出错误：

错误在mutate_impl（.data，dots）中：无效的下标类型'closure'

Error in mutate_impl(.data, dots) : invalid subscript type 'closure'

我以为可能是因为缺少的值。所以，我试过：

I thought it might be because of the missing values. So, I tried:

df<-df %>% 
   mutate(age=as.period(interval(start = birthdate, end = givendate))) %>% 
   mutate(age=if_else(!is.na(age),age$year,age))

它也会出现错误：

mutate_impl中的错误（.data，dots）：没有找到对象'age'

Error in mutate_impl(.data, dots) : object 'age' not found

推荐答案

code> do

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

---

由于 as .period 带有期间类，我们可能需要S4方法来提取它

---

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

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