How can I avoid that this statement gets NA for september when run on the 31.08.2022?
format(lubridate::as_date(as.character(lubridate::now() + months(-1:2))), "%m.%Y")
[1] "07.2022" "08.2022" NA "10.2022"
lubridate month vector generation with NA on 31.08.2022
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2 Answers
I found Add and subtract months to a date without exceeding the last day of the new month
format(lubridate::as_date(as.character(lubridate::now() %m+% months(-1:2))), "%m.%Y")
[1] "07.2022" "08.2022" "09.2022" "10.2022"
Use dmonths instead of months, I also removed some redundant code.
This will do:
format(now() + dmonths(-1:2), "%m.%Y")
# [1] "07.2022" "08.2022" "09.2022" "10.2022"
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