Create end of the month date from a date variable - r

Question

I have a large data frame with date variables, which reflect first day of the month. Is there an easy way to create a new data frame date variable that represents the last day of the month?
Below is some sample data:
date.start.month=seq(as.Date("2012-01-01"),length=4,by="months")
df=data.frame(date.start.month)
df$date.start.month
"2012-01-01" "2012-02-01" "2012-03-01" "2012-04-01"
I would like to return a new variable with:
"2012-01-31" "2012-02-29" "2012-03-31" "2012-04-30"
I've tried the following but it was unsuccessful:
df$date.end.month=seq(df$date.start.month,length=1,by="+1 months")

Answer 1

To get the end of months you could just create a Date vector containing the 1st of all the subsequent months and subtract 1 day.
date.end.month <- seq(as.Date("2012-02-01"),length=4,by="months")-1
date.end.month
[1] "2012-01-31" "2012-02-29" "2012-03-31" "2012-04-30"

Answer 2

Here is another solution using the lubridate package:
date.start.month=seq(as.Date("2012-01-01"),length=4,by="months")
df=data.frame(date.start.month)
library(lubridate)
df$date.end.month <- ceiling_date(df$date.start.month, "month") - days(1)
df$date.end.month
[1] "2012-01-31" "2012-02-29" "2012-03-31" "2012-04-30"
This uses the same concept given by James above, in that it gets the first day of the next month and subtracts one day.
By the way, this will work even when the input date is not necessarily the first day of the month. So for example, today is the 27th of the month and it still returns the correct last day of the month:
ceiling_date(Sys.Date(), "month") - days(1)
[1] "2017-07-31"

Answer 3

Use timeLastDayInMonth from the timeDate package:
df$eom <- timeLastDayInMonth(df$somedate)

Answer 4

library(lubridate)
as.Date("2019-09-01") - days(1)
[1] "2019-08-31"
or
library(lubridate)
as.Date("2019-09-01") + months(1) - days(1)
[1] "2019-09-30"

Answer 5

A straightforward solution would be using the yearmonfunction with the argument frac=1 from the xts-package. frac is a number between 0 and 1 that indicates the fraction of the way through the period that the result represents.
as.Date(as.yearmon(seq.Date(as.Date('2017-02-01'),by='month',length.out = 6)),frac=1)
[1] "2017-02-28" "2017-03-31" "2017-04-30" "2017-05-31" "2017-06-30" "2017-07-31"
Or if you prefer “piping” using magrittr:
seq.Date(as.Date('2017-02-01'),by='month',length.out = 6) %>%
as.yearmon() %>% as.Date(,frac=1)
[1] "2017-02-28" "2017-03-31" "2017-04-30" "2017-05-31" "2017-06-30" "2017-07-31"

Answer 6

A function as below would do the work (assume dt is scalar) -
month_end <- function(dt) {
d <- seq(dt, dt+31, by="days")
max(d[format(d,"%m")==format(dt,"%m")])
}
If you have a vector of Dates, then do the following -
sapply(dates, month_end)

Answer 7

you can use timeperiodsR
date.start.month=seq(as.Date("2012-01-01"),length=4,by="months")
df=data.frame(date.start.month)
df$date.start.month
# install.packages("timeperiodsR")
pm <- previous_month(df$date.start.month[1]) # get previous month
start(pm) # first day of previous month
end(pm) # last day of previous month
seq(pm) # vector with all days of previous month

Answer 8

We can also use bsts::LastDayInMonth:
transform(df, date.end.month = bsts::LastDayInMonth(df$date.start.month))
# date.start.month date.end.month
# 1 2012-01-01 2012-01-31
# 2 2012-02-01 2012-02-29
# 3 2012-03-01 2012-03-31
# 4 2012-04-01 2012-04-30

Answer 9

tidyverse has added the clock package in addition to the lubridate package that has nice functionality for this:
library(clock)
date_build(2012, 1:12, 31, invalid = "previous")
# [1] "2012-01-31" "2012-02-29" "2012-03-31" "2012-04-30" "2012-05-31" "2012-06-30"
# [7] "2012-07-31" "2012-08-31" "2012-09-30" "2012-10-31" "2012-11-30" "2012-12-31"
The invalid argument specifies what to do with an invalid date (e.g. 2012-02-31). From the documentation:
"previous": The previous valid instant in time.
"previous-day": The previous valid day in time, keeping the time of
day.
"next": The next valid instant in time.
"next-day": The next valid day in time, keeping the time of day.
"overflow": Overflow by the number of days that the input is invalid
by. Time of day is dropped.
"overflow-day": Overflow by the number of days that the input is
invalid by. Time of day is kept.
"NA": Replace invalid dates with NA.
"error": Error on invalid dates.