Calculating the difference between two two-digit years - r

Question

Is there any easy way in R to calculate the difference between two columns of two-digit years (just years, no months/days as it's unnecessary here) in order to produce a column of ages?
I've fairly new to this and have been playing with 'if' statements and algebra without success.
The data looks like this, but larger:
dat <- data.frame(year1=c("98","99","00","01","02"),
year2=c("03","04","05","06","07"))

Answer 1

You could use strptime() with the format %y:
dat <- data.frame(year1=c("98","99","00","01","02"),
year2=c("03","04","05","06","07"),
stringsAsFactors = F) # You might want to use this as a default!
dat$year1 <- strptime(dat$year1, format = "%y")
dat$year2 <- strptime(dat$year2, format = "%y")
as.vector(difftime(dat$year2,
dat$year1,
units = "days"))/365.242
4.999311 5.002163 4.999425 4.999425 4.999425

Answer 2

Format to a date, format back to a number, take the difference:
do.call(`-`, lapply(dat[1:2], function(x)
as.numeric(format(as.Date(x, format="%y"), "%Y"))))
#[1] -5 -5 -5 -5 -5
This may hit cases where it doesn't work if you have old dates in the early 1900's. As per ?strptime:
‘%y’ Year without century (00-99). On input, values 00 to 68 are
prefixed by 20 and 69 to 99 by 19 - that is the behaviour
specified by the 2004 and 2008 POSIX standards, but they do
also say ‘it is expected that in a future version the default
century inferred from a 2-digit year will change’.

Answer 3

df$age <- ifelse(df$year2 < df$year1, df$year2 - df$year1 + 100, df$year2 -df$year1)
should work under the assumption year2 is some kind of current year and year1 is the year of birth and there are no people born before 1918.
Example:
df <- data.frame(year1 = sample(18:99, 1000, replace = T),
year2 = sample(1:99, 1000, replace = T))
> head(df)
year1 year2
1 27 88
2 41 55
3 90 36
4 81 93
5 56 60
6 27 61
df$age <- ifelse(df$year2 < df$year1, df$year2 - df$year1 + 100, df$year2 -df$year1)
> head(df)
year1 year2 age
1 73 88 15
2 50 17 67
3 47 41 94
4 54 43 89
5 36 82 46
6 62 85 23
With your data example:
dat <- data.frame(year1=c("98","99","00","01","02"),
year2=c("03","04","05","06","07"))
dat$age <- ifelse(as.numeric(as.character(dat$year2)) < as.numeric(as.character(dat$year1)),
as.numeric(as.character(dat$year2)) - as.numeric(as.character(dat$year1)) + 100,
as.numeric(as.character(dat$year2)) - as.numeric(as.character(dat$year1)))
> dat
year1 year2 age
1 98 03 5
2 99 04 5
3 00 05 5
4 01 06 5
5 02 07 5

Answer 4

one method is to use as.Date with a dplyr chain:
dat %>%
mutate(year1 = as.Date(year1, format = "%y"),
year2 = as.Date(year2, format = "%y")) %>%
mutate(age = year2 - year1)
which returns:
year1 year2 age
1 1998-10-26 2003-10-26 1826 days
2 1999-10-26 2004-10-26 1827 days
3 2000-10-26 2005-10-26 1826 days
4 2001-10-26 2006-10-26 1826 days
5 2002-10-26 2007-10-26 1826 days
p.s. it assumes default day and month for both columns, but it assumes same value for both, so does not affect the difference calculation.