Get the difference between dates in terms of weeks, months, quarters, and years

what about this:

# get difference between dates `"01.12.2013"` and `"31.12.2013"`

# weeks
difftime(strptime("26.03.2014", format = "%d.%m.%Y"),
strptime("14.01.2013", format = "%d.%m.%Y"),units="weeks")
Time difference of 62.28571 weeks

# months
(as.yearmon(strptime("26.03.2014", format = "%d.%m.%Y"))-
as.yearmon(strptime("14.01.2013", format = "%d.%m.%Y")))*12
[1] 14

# quarters
(as.yearqtr(strptime("26.03.2014", format = "%d.%m.%Y"))-
as.yearqtr(strptime("14.01.2013", format = "%d.%m.%Y")))*4
[1] 4

# years
year(strptime("26.03.2014", format = "%d.%m.%Y"))-
year(strptime("14.01.2013", format = "%d.%m.%Y"))
[1] 1

as.yearmon() and as.yearqtr() are in package zoo. year() is in package lubridate. What do you think?

For weeks, you can use function difftime:

date1 <- strptime("14.01.2013", format="%d.%m.%Y")
date2 <- strptime("26.03.2014", format="%d.%m.%Y")
difftime(date2,date1,units="weeks")
Time difference of 62.28571 weeks

But difftime doesn't work with duration over weeks.
The following is a very suboptimal solution using cut.POSIXt for those durations but you can work around it:

seq1 <- seq(date1,date2, by="days")
nlevels(cut(seq1,"months"))
15
nlevels(cut(seq1,"quarters"))
5
nlevels(cut(seq1,"years"))
2

This is however the number of months, quarters or years spanned by your time interval and not the duration of your time interval expressed in months, quarters, years (since those do not have a constant duration). Considering the comment you made on @SvenHohenstein answer I would think you can use nlevels(cut(seq1,"months")) - 1 for what you're trying to achieve.

All the existing answers are imperfect (IMO) and either make assumptions about the desired output or don't provide flexibility for the desired output.

Based on the examples from the OP, and the OP's stated expected answers, I think these are the answers you are looking for (plus some additional examples that make it easy to extrapolate).

(This only requires base R and doesn't require zoo or lubridate)

Convert to Datetime Objects

date_strings = c("14.01.2013", "26.03.2014")
datetimes = strptime(date_strings, format = "%d.%m.%Y") # convert to datetime objects

Difference in Days

You can use the diff in days to get some of our later answers

diff_in_days = difftime(datetimes[2], datetimes[1], units = "days") # days
diff_in_days
#Time difference of 435.9583 days

Difference in Weeks

Difference in weeks is a special case of units = "weeks" in difftime()

diff_in_weeks = difftime(datetimes[2], datetimes[1], units = "weeks") # weeks
diff_in_weeks
#Time difference of 62.27976 weeks

Note that this is the same as dividing our diff_in_days by 7 (7 days in a week)

as.double(diff_in_days)/7
#[1] 62.27976

Difference in Years

With similar logic, we can derive years from diff_in_days

diff_in_years = as.double(diff_in_days)/365 # absolute years
diff_in_years
#[1] 1.194406

You seem to be expecting the diff in years to be "1", so I assume you just want to count absolute calendar years or something, which you can easily do by using floor()

# get desired output, given your definition of 'years'
floor(diff_in_years)
#[1] 1

Difference in Quarters

# get desired output for quarters, given your definition of 'quarters'
floor(diff_in_years * 4)
#[1] 4

Difference in Months

Can calculate this as a conversion from diff_years

# months, defined as absolute calendar months (this might be what you want, given your question details)
months_diff = diff_in_years*12
floor(month_diff)
#[1] 14

I know this question is old, but given that I still had to solve this problem just now, I thought I would add my answers. Hope it helps.