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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T09:48:27+00:00 2026-06-14T09:48:27+00:00

I’m working with monthly data and have a character vector of dates, formatted: Sep/2012

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I’m working with monthly data and have a character vector of dates, formatted:

Sep/2012
Aug/2012
Jul/2012

and so on, back to 1981. I’ve tried using

as.Date(dates, "%b/%Y")

where %b represents month abbreviations, but this only returns NAs. What am I doing wrong?

Note: I already found a workaround using gsub() to add “01/” in front of each entry, like so:

01/Sep/2012
01/Aug/2012
01/Jul/2012

Then as.Dates() works, but this seems a little inelegant, and isn’t strictly accurate anyway.

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  1. Editorial Team
    Editorial Team
    2026-06-14T09:48:28+00:00Added an answer on June 14, 2026 at 9:48 am

    You are looking for as.yearmon() in the zoo package. Given your dates

    dates <- c("Sep/2012","Aug/2012","Jul/2012")
    

    we load the package and convert to the "yearmon" class

    require(zoo)
    dates1 <- as.yearmon(dates, format = "%b/%Y")
    dates1
    

    Which gives

    R> dates1
    [1] "Sep 2012" "Aug 2012" "Jul 2012"
    

    You can coerce to an object of class "Date" using the as.Date() method

    R> as.Date(dates1)
    [1] "2012-09-01" "2012-08-01" "2012-07-01"
    

    Which would be a simpler way of getting the thing you did via gsub(). There is a frac argument which controls how far through the month the day component should be:

    R> as.Date(dates1, frac = 0.5)
    [1] "2012-09-15" "2012-08-16" "2012-07-16"
    

    But that may ont be sufficient for you.

    If you really only want the dates stored as you have them, then they aren’t really dates but if you are happy to work within the zoo package then the "yearmon" class can be used as an index for a zoo object which is a time series.

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