Reputation: 61
From date 10/31/2018, I want to obtain 09/30/2018.
I tried:
PREV.PROD.DATE<-seq.Date(from=as.Date(chron("10/31/2018")),length=2,by="-1 months")
but it returns:
"2018-10-31" "2018-10-01"
How can I obtain 09/30 instead of 10/01?
Notes: I would like to avoid to use an external package, and I would like the solution to work for any end of month date.
Upvotes: 0
Views: 1808
Reputation: 145775
I like to floor
the date - making it the first day of its month, and then subtract 1 to make it the last day of the previous month:
x = as.Date("2018-10-31")
library(lubridate)
floor_date(x, unit = "months") - 1
# [1] "2018-09-30"
Here's a version without using other packages:
as.Date(format(x, "%Y-%m-01")) - 1
# [1] "2018-09-30"
Upvotes: 1
Reputation: 160447
The integer units of Date
are days, so you can do:
seq.Date(from=as.Date("2018-10-31"), length=2, by="-1 months") - c(0,1)
# [1] "2018-10-31" "2018-09-30"
If you want arbitrary prior-last-date:
(d <- as.POSIXlt(Sys.Date()))
# [1] "2018-11-08 UTC"
d$mday <- 1L
as.Date(d) - 1
# [1] "2018-10-31"
Replace Sys.Date()
with whatever single date you have. If you want to vectorize this:
(ds <- Sys.Date() + c(5, 20, 50))
# [1] "2018-11-13" "2018-11-28" "2018-12-28"
lapply(as.POSIXlt(ds), function(a) as.Date(`$<-`(a, "mday", 1L)) - 1L)
# [[1]]
# [1] "2018-10-31"
# [[2]]
# [1] "2018-10-31"
# [[3]]
# [1] "2018-11-30"
Upvotes: 3
Reputation: 883
I don't use function chron
. But I think the function duration
from lubridate
helps you.
You don't need use floor_date
to confuse you.
library(lubridate)
#>
#> 载入程辑包:'lubridate'
#> The following object is masked from 'package:base':
#>
#> date
library(tidyverse)
better_date <-
str_split("10/31/2018",'/') %>%
.[[1]] %>%
.[c(3,1,2)] %>%
str_flatten(collapse = '-') %>%
as.Date()
(better_date - duration('1 months')) %>%
as.Date()
#> [1] "2018-09-30"
Created on 2018-11-09 by the reprex package (v0.2.1)
Upvotes: 0