问题描述
给定出生日期和任意日期，我面临着计算年龄(以年，月或周为单位)的常见任务.问题是，我经常必须对许多条记录(> 3亿条记录)执行此操作，因此性能是此处的关键问题.
I am facing the common task of calculating the age (in years, months, or weeks) given the date of birth and an arbitrary date. The thing is that quite often I have to do this over many many records (>300 millions), so performance is a key issue here.
在SO和Google中快速搜索之后，我发现了3种替代方法:
After a quick search in SO and Google I found 3 alternatives:
- 常见算术过程(/365.25)(链接)
- 使用软件包
lubridate中的函数new_interval()和duration()(链接) 包 - 函数
age_calc()(链接，链接， 链接)
eeptools中的- A common arithmetic procedure (/365.25) (link)
- Using functions
new_interval()andduration()from packagelubridate(link) - Function
age_calc()from packageeeptools(link, link, link)
所以，这是我的玩具代码:
So, here's my toy code:
# Some toy birthdates
birthdate <- as.Date(c("1978-12-30", "1978-12-31", "1979-01-01",
"1962-12-30", "1962-12-31", "1963-01-01",
"2000-06-16", "2000-06-17", "2000-06-18",
"2007-03-18", "2007-03-19", "2007-03-20",
"1968-02-29", "1968-02-29", "1968-02-29"))
# Given dates to calculate the age
givendate <- as.Date(c("2015-12-31", "2015-12-31", "2015-12-31",
"2015-12-31", "2015-12-31", "2015-12-31",
"2050-06-17", "2050-06-17", "2050-06-17",
"2008-03-19", "2008-03-19", "2008-03-19",
"2015-02-28", "2015-03-01", "2015-03-02"))
# Using a common arithmetic procedure ("Time differences in days"/365.25)
(givendate-birthdate)/365.25
# Use the package lubridate
require(lubridate)
new_interval(start = birthdate, end = givendate) /
duration(num = 1, units = "years")
# Use the package eeptools
library(eeptools)
age_calc(dob = birthdate, enddate = givendate, units = "years")
让我们稍后再讨论准确性，并首先关注性能.这是代码:
Let's talk later about accuracy and focus first on performance. Here's the code:
# Now let's compare the performance of the alternatives using microbenchmark
library(microbenchmark)
mbm <- microbenchmark(
arithmetic = (givendate - birthdate) / 365.25,
lubridate = new_interval(start = birthdate, end = givendate) /
duration(num = 1, units = "years"),
eeptools = age_calc(dob = birthdate, enddate = givendate,
units = "years"),
times = 1000
)
# And examine the results
mbm
autoplot(mbm)
结果在这里:
底线:lubridate和eeptools函数的性能比算术方法差得多(/365.25至少快10倍).不幸的是，算术方法不够准确，我无法承受该方法会犯的一些错误.
Bottom line: performance of lubridate and eeptools functions is much worse than the arithmetic method (/365.25 is at least 10 times faster). Unfortunately, the arithmetic method is not accurate enough and I cannot afford the few mistakes that this method will make.
当我在一些文章中读到的时候，lubridate和eeptools不会犯这样的错误(尽管我没有看代码/了解更多有关那些函数的信息，以了解它们使用哪种方法)，这就是为什么我想要使用它们，但它们的性能对我的实际应用程序无效.
As I read on some posts, lubridate and eeptools make no such mistakes (though, I haven't looked at the code/read more about those functions to know which method they use) and that's why I wanted to use them, but their performance does not work for my real application.
对有效，准确地计算年龄的方法有何想法?
Any ideas on an efficient and accurate method to calculate the age?
糟糕，看来lubridate也会出错.显然，基于这个玩具示例，它比算术方法犯了更多的错误(请参见第3、6、9、12行). (我做错什么了吗?)
Ops, it seems lubridate also makes mistakes. And apparently based on this toy example, it makes more mistakes than the arithmetic method (see lines 3, 6, 9, 12). (am I doing something wrong?)
toy_df <- data.frame(
birthdate = birthdate,
givendate = givendate,
arithmetic = as.numeric((givendate - birthdate) / 365.25),
lubridate = new_interval(start = birthdate, end = givendate) /
duration(num = 1, units = "years"),
eeptools = age_calc(dob = birthdate, enddate = givendate,
units = "years")
)
toy_df[, 3:5] <- floor(toy_df[, 3:5])
toy_df
birthdate givendate arithmetic lubridate eeptools
1 1978-12-30 2015-12-31 37 37 37
2 1978-12-31 2015-12-31 36 37 37
3 1979-01-01 2015-12-31 36 37 36
4 1962-12-30 2015-12-31 53 53 53
5 1962-12-31 2015-12-31 52 53 53
6 1963-01-01 2015-12-31 52 53 52
7 2000-06-16 2050-06-17 50 50 50
8 2000-06-17 2050-06-17 49 50 50
9 2000-06-18 2050-06-17 49 50 49
10 2007-03-18 2008-03-19 1 1 1
11 2007-03-19 2008-03-19 1 1 1
12 2007-03-20 2008-03-19 0 1 0
13 1968-02-29 2015-02-28 46 47 46
14 1968-02-29 2015-03-01 47 47 47
15 1968-02-29 2015-03-02 47 47 47
推荐答案
好，所以我在另一个:
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)
}
@Jim发表的话说:以下函数采用Date对象的向量并计算年龄，正确地计算了leap年.似乎比其他任何一个答案都更简单."
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".
它确实更简单，并且可以实现我一直在寻找的窍门.平均而言，它实际上比算术方法要快(大约快75％).
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)
至少在我的示例中，它没有犯任何错误(并且在任何示例中都不应犯错；这是使用ifelse s的非常简单的函数).
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
我不认为这是一个完整的解决方案，因为我也想将年龄设在几个月和几周之内，并且此功能特定于几年.无论如何，我将其发布在这里，因为它解决了多年以来的问题.我不会接受，因为:
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:
- 我将等待@Jim将其发布为答案.
- 我将拭目以待，看看其他人是否提出了完整的解决方案(有效，准确并且可以根据需要以年，月或周为单位的年龄).
这篇关于在给定的出生日期和任意日期的情况下，有效且准确地计算出R的年龄(以年，月或周为单位)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！