Calculating Ages in R

Question

Hello everyone. I have two data frames in R. One frame has a persons year of birth:

YEAR
/1931
/1924

and then another column shows a more recent time.

RECENT
09/08/2005
11/08/2005

What I want to do is subtract the years so that I can calculate their age in number of years, however I am not sure how to approach this. Any help please?

Answer 1

You can do some formating:

as.numeric(format(as.Date("01/01/2010", format="%m/%d/%Y"), format="%Y")) - 1930

With your data:

> yr <- c(1931, 1924)
> recent <- c("09/08/2005", "11/08/2005")
> as.numeric(format(as.Date(recent, format="%m/%d/%Y"), format="%Y")) - yr
[1] 74 81

Since you have your data in a data.frame (I'll assume that it's called df), it will be more like this:

as.numeric(format(as.Date(df$recent, format="%m/%d/%Y"), format="%Y")) - df$year

Answer 2

Based on the previous answer, convert your columns to date objects and subtract. Some conversion of types between character and numeric is necessary:

> foo=data.frame(RECENT=c("09/08/2005","11/08/2005"),YEAR=c("/1931","/1924"))
> foo
      RECENT  YEAR
1 09/08/2005 /1931
2 11/08/2005 /1924
> foo$RECENTd = as.Date(foo$RECENT, format="%m/%d/%Y")
> foo$YEARn = as.numeric(substr(foo$YEAR,2,999))
> foo$AGE = as.numeric(format(foo$RECENTd,"%Y")) - foo$YEARn
> foo
      RECENT  YEAR    RECENTd YEARn AGE
1 09/08/2005 /1931 2005-09-08  1931  74
2 11/08/2005 /1924 2005-11-08  1924  81

Note I've assumed you have that slash in your year column.

Also, tip for when asking questions about dates is to include a day that is past the twelfth so we know if you are a month/day/year person or a day/month/year person.

Answer 3

Given the data in your example:

> m <- data.frame(YEAR=c("/1931", "/1924"),RECENT=c("09/08/2005","11/08/2005"))
> m
   YEAR     RECENT
1 /1931 09/08/2005
2 /1924 11/08/2005

Extract year with the strptime function:

> strptime(m[,2], format = "%m/%d/%Y")$year - strptime(m[,1], format = "/%Y")$year
[1] 74 81

Answer 4

You can solve this with the lubridate package.

> library(lubridate)

I don't think /1931 is a common date class. So I'll assume all the entries are character strings.

> RECENT <- data.frame(recent = c("09/08/2005", "11/08/2005"))
> YEAR <- data.frame(year = c("/1931", "/1924"))

First, let's notify R that the recent dates are dates. I'll assume the dates are in month/day/year order, so I use mdy(). If they're in day/month/year order just use dmy().

> RECENT$recent <- mdy(RECENT$recent)
      recent
1 2005-09-08
2 2005-11-08

Now, lets turn the years into numbers so we can do some math with them.

> YEAR$year <- as.numeric(substr(YEAR$year, 2, 5))

Now just do the math. year() extracts the year value of the RECENT dates.

> year(RECENT$recent) - YEAR
  year
1   74
2   81

p.s. if your year entries are actually full dates, you can get the difference in years with

> YEAR1 <- data.frame(year = mdy("01/08/1931","01/08/1924"))
> as.period(RECENT$recent - YEAR1$year, units = "year")
[1] 74 years and 8 months   81 years and 10 months

Answer 5

You can even use simple string manipulation :

m <- data.frame(YEAR=c("/1931", "/1924"),
  RECENT=c("09/08/2005","11/08/2005"),
  stringsAsFactors=F)

m$YEAR <- as.numeric(gsub("/","",m$YEAR))
m$recentyear <- as.numeric(substr(m$RECENT,7,10))

m$age <-  m$recentyear - m$YEAR

If you have birth dates, you can take that into account. This code gives the correct age, give or take a day or two due to the hack for calculating the age.

require(chron)
m <- data.frame(YEAR=c("11/2/1931", "17/11/1924"),
  RECENT=c("09/08/2005","11/08/2005"),
  stringsAsFactors=F)

m$YEAR <- chron(m$YEAR,format="d/m/y")
m$RECENT <- chron(m$RECENT,format="d/m/y")

m$age <-  floor((m$RECENT - m$YEAR)/365.25)

There are multiple ways of dealing with dates. I personally like the chron() library the best, but that's a matter of taste. All other options mentioned here are equally valid.