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
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)
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.
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.
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.
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
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)
Source: Stackoverflow.com