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

Question

I have two dates let´s say 14.01.2013 and 26.03.2014.

I would like to get the difference between those two dates in terms of weeks(?), months(in the example 14), quarters(4) and years(1).

Do you know the best way to get this?

Answer 1

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?

Answer 2

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.

Answer 3

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.

Answer 4

I just wrote this for another question, then stumbled here.

library(lubridate)

#' Calculate age
#' 
#' By default, calculates the typical "age in years", with a
#' \code{floor} applied so that you are, e.g., 5 years old from
#' 5th birthday through the day before your 6th birthday. Set
#' \code{floor = FALSE} to return decimal ages, and change \code{units}
#' for units other than years.
#' @param dob date-of-birth, the day to start calculating age.
#' @param age.day the date on which age is to be calculated.
#' @param units unit to measure age in. Defaults to \code{"years"}. Passed to \link{\code{duration}}.
#' @param floor boolean for whether or not to floor the result. Defaults to \code{TRUE}.
#' @return Age in \code{units}. Will be an integer if \code{floor = TRUE}.
#' @examples
#' my.dob <- as.Date('1983-10-20')
#' age(my.dob)
#' age(my.dob, units = "minutes")
#' age(my.dob, floor = FALSE)
age <- function(dob, age.day = today(), units = "years", floor = TRUE) {
    calc.age = interval(dob, age.day) / duration(num = 1, units = units)
    if (floor) return(as.integer(floor(calc.age)))
    return(calc.age)
}

Usage examples:

my.dob <- as.Date('1983-10-20')

age(my.dob)
# [1] 31

age(my.dob, floor = FALSE)
# [1] 31.15616

age(my.dob, units = "minutes")
# [1] 16375680

age(seq(my.dob, length.out = 6, by = "years"))
# [1] 31 30 29 28 27 26

Answer 5

Here the still lacking lubridate answer (although Gregor's function is built on this package)

The lubridate timespan documentation is very helpful for understanding the difference between periods and duration. I also like the lubridate cheatsheet and this very useful thread

library(lubridate)

dates <- c(dmy('14.01.2013'), dmy('26.03.2014'))

span <- dates[1] %--% dates[2] #creating an interval object

#creating period objects 
as.period(span, unit = 'year') 
#> [1] "1y 2m 12d 0H 0M 0S"
as.period(span, unit = 'month')
#> [1] "14m 12d 0H 0M 0S"
as.period(span, unit = 'day')
#> [1] "436d 0H 0M 0S"

Periods do not accept weeks as units. But you can convert durations to weeks:

as.duration(span)/ dweeks(1)
#makes duration object (in seconds) and divides by duration of a week (in seconds)
#> [1] 62.28571

^{Created on 2019-11-04 by the reprex package (v0.3.0)}

Answer 6

Here's a solution:

dates <- c("14.01.2013", "26.03.2014")

# Date format:
dates2 <- strptime(dates, format = "%d.%m.%Y")

dif <- diff(as.numeric(dates2)) # difference in seconds

dif/(60 * 60 * 24 * 7) # weeks
[1] 62.28571
dif/(60 * 60 * 24 * 30) # months
[1] 14.53333
dif/(60 * 60 * 24 * 30 * 3) # quartes
[1] 4.844444
dif/(60 * 60 * 24 * 365) # years
[1] 1.194521

Answer 7

This is a simple way to find out the difference in years with the lubridate package:

as.numeric(as.Date("14-03-2013", format = "%d-%m-%Y") %--% as.Date("23-03-2014", format = "%d-%m-%Y"), "years")

This returns 1.023956

You can use floor() if you don't want the decimals.

Answer 8

try this for a months solution

StartDate <- strptime("14 January 2013", "%d %B %Y") 
EventDates <- strptime(c("26 March 2014"), "%d %B %Y") 
difftime(EventDates, StartDate) 

Answer 9

A more "precise" calculation. That is, the number of week/month/quarter/year for a non-complete week/month/quarter/year is the fraction of calendar days in that week/month/quarter/year. For example, the number of months between 2016-02-22 and 2016-03-31 is 8/29 + 31/31 = 1.27586

explanation inline with code

#' Calculate precise number of periods between 2 dates
#' 
#' @details The number of week/month/quarter/year for a non-complete week/month/quarter/year 
#'     is the fraction of calendar days in that week/month/quarter/year. 
#'     For example, the number of months between 2016-02-22 and 2016-03-31 
#'     is 8/29 + 31/31 = 1.27586
#' 
#' @param startdate start Date of the interval
#' @param enddate end Date of the interval
#' @param period character. It must be one of 'day', 'week', 'month', 'quarter' and 'year'
#' 
#' @examples 
#' identical(numPeriods(as.Date("2016-02-15"), as.Date("2016-03-31"), "month"), 15/29 + 1)
#' identical(numPeriods(as.Date("2016-02-15"), as.Date("2016-03-31"), "quarter"), (15 + 31)/(31 + 29 + 31))
#' identical(numPeriods(as.Date("2016-02-15"), as.Date("2016-03-31"), "year"), (15 + 31)/366)
#' 
#' @return exact number of periods between
#' 
numPeriods <- function(startdate, enddate, period) {

    numdays <- as.numeric(enddate - startdate) + 1
    if (grepl("day", period, ignore.case=TRUE)) {
        return(numdays)

    } else if (grepl("week", period, ignore.case=TRUE)) {
        return(numdays / 7)
    }

    #create a sequence of dates between start and end dates
    effDaysinBins <- cut(seq(startdate, enddate, by="1 day"), period)

    #use the earliest start date of the previous bins and create a breaks of periodic dates with
    #user's period interval
    intervals <- seq(from=as.Date(min(levels(effDaysinBins)), "%Y-%m-%d"), 
        by=paste("1",period), 
        length.out=length(levels(effDaysinBins))+1)

    #create a sequence of dates between the earliest interval date and last date of the interval
    #that contains the enddate
    allDays <- seq(from=intervals[1],
        to=intervals[intervals > enddate][1] - 1,
        by="1 day")

    #bin all days in the whole period using previous breaks
    allDaysInBins <- cut(allDays, intervals)

    #calculate ratio of effective days to all days in whole period
    sum( tabulate(effDaysinBins) / tabulate(allDaysInBins) )
} #numPeriods

Please let me know if you find more boundary cases where the above solution does not work.

Answer 10

Hoping I can add something new to the table that's simple, intuitive and efficient.

Using time_diff() from my package timeplyr, one can easily calculate the time difference between any 2 date/datetime vectors.

# remotes::install_github("NicChr/timeplyr")
library(timeplyr)
library(lubridate)

x <- dmy("14-01-2013")
y <- dmy("26-03-2014")

time_diff(x, y, "days")
#> [1] 436
time_diff(x, y, "hours")
#> [1] 10464
time_diff(x, y, "weeks")
#> [1] 62.28571
time_diff(x, y, "months")
#> [1] 14.3871
time_diff(x, y, "years")
#> [1] 1.194521

^{Created on 2023-04-10 with reprex v2.0.2}

You can specify exotic units as well.

.extra_time_units
#>  [1] "fortnights" "quarters"   "semesters"  "olympiads"  "lustrums"  
#>  [6] "decades"    "indictions" "scores"     "centuries"  "milleniums"

time_diff(x, y, "fortnight")
#> [1] 31.14286
time_diff(x, y, "quarter")
#> [1] 4.788889
time_diff(x, y, "decade")
#> [1] 0.1193866

Choose between lubridate durations and periods.

time_diff(x, y, dmonths(1))
#> [1] 14.32444
time_diff(x, y, months(1))
#> [1] 14.3871

And even choose multi-units, as it's vectorised.

time_diff(x, y, time_by = list("days" = -10:10))
#>  [1]  -43.60000  -48.44444  -54.50000  -62.28571  -72.66667  -87.20000
#>  [7] -109.00000 -145.33333 -218.00000 -436.00000        Inf  436.00000
#> [13]  218.00000  145.33333  109.00000   87.20000   72.66667   62.28571
#> [19]   54.50000   48.44444   43.60000