Is there a way to convert time difference in terms of years and months

Question

Is there a way to convert below time difference in terms of years and months?

as.Date(Sys.Date()) - as.Date("2015-08-20")
Time difference of 2753 days

Expected output (Since it app.. around 7 years and 5 months?

7.5

Does this answer your question? [Get date difference in years (floating point)](https://stackoverflow.com/questions/15569333/get-date-difference-in-years-floating-point) — GuedesBF, Mar 04 '23 at 18:36
Did you mean years and fraction of a year as there is about 7.5 years between those dates but there is not 7 years and 5 months? — G. Grothendieck, Mar 05 '23 at 14:53

score 0 · Answer 1 · answered Mar 04 '23 at 17:43

You may try the below code

library(lubridate)

months <- time_length(as.numeric(difftime(as.Date(Sys.Date()) , as.Date("2015-08-20"), units = 'days')), 'month')

years <- time_length(as.numeric(difftime(as.Date(Sys.Date()) , as.Date("2015-08-20"), units = 'days')), 'year')

Allan Cameron · Answer 2 · 2023-03-04T18:15:43.140

A trick using only base R is to add the difftime on to the first of January in the year 0 AD (this exists in R even if it didn't technically exist in reality).

The resulting date is the number of years, months and days of the difftime (with an extra day, which is easy to subtract)

(as.POSIXct('0000-01-01') + (as.Date(Sys.Date()) - as.Date("2015-08-20"))) |>
  as.character() |>
  strsplit('-') |>
  getElement(1) |>
  as.numeric() |>
  setNames(c('years', 'months', 'days')) - c(0, 0, 1)
#>  years months   days 
#>      7      7     15

^{Created on 2023-03-04 with reprex v2.0.2}

Is there a way to convert time difference in terms of years and months

2 Answers2