convert difftime time to years, months and days - r

Question

How can I accurately convert the products (units is in days) of the difftime below to years, months and days?
difftime(Sys.time(),"1931-04-10")
difftime(Sys.time(),"2012-04-10")
This does years and days but how could I include months?
yd.conv<-function(days, print=TRUE){
x<-days*0.00273790700698851
x2<-floor(x)
x3<-x-x2
x4<-floor(x3*365.25)
if (print) cat(x2,"years &",x4,"days\n")
invisible(c(x2, x4))
}
yd.conv(difftime(Sys.time(),"1931-04-10"))
yd.conv(difftime(Sys.time(),"2012-04-10"))
I'm not sure how to even define months either. Would 4 weeks be considered a month or the passing of the same month day. So for the later definition of a month if the initial date was 2012-01-10 and the current 2012-05-31 then we'd have 0 years, 5 months and 21 days. This works well but what if the original date was on the 31st of the month and the end date was on feb 28 would this be considered a month?
As I wrote this question the question itself evolved so I'd better clarify:
What would be the best (most logical approach) to defining months and then how to find diff time in years, months and days?

Answer 1

If you're doing something like
difftime(Sys.time(), someDate)
It comes as implied that you must know what someDate is. In that case, you can convert this to a POSIXct class object that gives you the ability to extract temporal information directly (package chron offers more methods, too). For instance
as.POSIXct(c(difftime(Sys.time(), someDate, units = "sec")), origin = someDate)
This will return your desired date object. If you have a timezone tz to feed into difftime, you can also pass that directly to the tz parameter in as.POSIXct.
Now that you have your date object, you can run things like months(.) and if you have chron you can do years(.) and days(.) (returns ordered factor).
From here, you could do more simple math on the difference of years, months, and days separately (converting to appropriate numeric representations). Of course, convert someDate to POSIXct will be required.
EDIT: On second thought, months(.) returns a character representation of the month, so that may not be efficient. At least, it'll require a little processing (not too difficult) to give a numeric representation.

Answer 2

R has not implemented these features out of ignorance. difftime objects are transitive. A 700 day difference on any arbitrary start-date can yield a differing number of years depending on whether there was a leap year or not. Similarly for months, they take between 28-31 days.
For research purposes, we use these units a lot (months and years) and pragmatically, we define a year as 365.25 days and a month as 365.25/12 = 30.4375 days.
To do arithmetic on a given difftime, you must convert this value to numeric using as.numeric(difftime.obj) which is, in default, days so R stops spouting off the units.

Answer 3

You can not simply convert a difftime to month, since the definition of months depends on the absolute time at which the difftime has started.
You'll need to know the start date or the end date to accurately tell the number of months.
You could then, e.g., calculate the number of months in the first year of your timespan, the number of month in the last your of the timespan, and add the number of years between times 12.

Answer 4

Hmm. I think the most sensible would be to look at the various units themselves. So compare the day of the month first, then compare the month of the year, then compare the year. At each point, you can introduce a carry to avoid negative values.
In other words, don't work with the product of difftime, but recode your own difftime.