This object is in character format and i would like to avoid extracting the 10 by string based functions since it is very cumbersome. If there is a way through as.Date() and some particular format, i would be happy to use that
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This may help: https://stackoverflow.com/questions/6242955/converting-year-and-month-yyyy-mm-format-to-a-date – jjl Dec 01 '17 at 16:34
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Because 6,7 is not generic. Month could be 5 as well as 12 – Abhay Saini Dec 01 '17 at 16:38
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The least cumbersome would probably just be a regex that captures all digits from "-" to the end. – joran Dec 01 '17 at 16:40
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1library(stringr); str_extract("2010-10", pattern = "[0-1]?[0-9]$") – jjl Dec 01 '17 at 16:42
3 Answers
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I've really been enjoying anydate from the anytime package lately.
Try this:
library(anytime)
format(anydate("2017-10"), "%m")
tomasu
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I believe this question has been asked before on SO.
As an alternative to the anytime package you may use lubridate
library(lubridate)
format(ymd("2017-10", truncated = 1L), "%m")
or with base R
format(as.Date(paste0("2017-10", "-01")), "%m")
Uwe
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This converts it to a yearmon object (which represents a year and month with no day) and then extracts the month. See ?yearmon for more information.
library(zoo)
cycle(as.yearmon("2017-10"))
## [1] 10
G. Grothendieck
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