Home/ Questions/Ways to resolve the error - error in chartodate(x) : character string is not in a standard unambiguous format
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Damien Wilson
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Asked: 2022-05-12T08:03:36+00:00 In:

Ways to resolve the error – error in chartodate(x) : character string is not in a standard unambiguous format

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I am tired of fixing the problem: error in chartodate(x) : character string is not in a standard unambiguous format in the r; even if I get the reference from another forum, it still returns an error:

$ R --vanilla
 
 > as.Date("01 Jan 2000")
 Error in charToDate(x) :
  character string is not in a standard unambiguous format

To identity the problem, I will show you the detail here:

> as.Date("01/01/2000")
 [1] "0001-01-20"
> sessionInfo()
 R version 2.15.2 (2012-10-26)
 Platform: x86_64-w64-mingw32/x64 (64-bit)
 
 locale:
 [1] LC_COLLATE=English_United Kingdom.1252
 [2] LC_CTYPE=English_United Kingdom.1252
 [3] LC_MONETARY=English_United Kingdom.1252
 [4] LC_NUMERIC=C
 [5] LC_TIME=English_United Kingdom.1252
 
 attached base packages:
 [1] stats graphics grDevices utils datasets methods base

How do I do that? Could you support me in improving this problem?

error in chartodate(x)
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2 Answers

  1. Best Answer
    Expert

    The cause: This behavior is documented. ?as.Date:

    format: A character string. If not specified, it will try '"%Y-%m-%d"' then '"%Y/%m/%d"' on the first non-'NA' element, and give an error if neither works.

    as.Date("01 Jan 2000") returns an error as the format of as.Date("01 Jan 2000") is not one of the ones listed above. as.Date("01/01/2000") returns an error because the date is not in one of these formats.

    Solution: I interpret “standard unambiguous” to refer to “ISO-8601” (even if as.Date isn’t as strict as ISO-8601, since “%m/%d/%Y” isn’t ISO-8601).

    This error can be fixed by specifying the date format you received. Use the Details section of ?strptime to do this. You must ensure that both the order and any separators match the format of your in string. You should also be careful if your data contains day/month names or abbreviations. The conversion will depend upon your location

    • 12

  2. In other words, is there a better solution than needing to specify the format?

    Yes, it is (ie in late 2016), thanks anytime::anydate from anytime.

    For some examples, see the following: 

    R> anydate(c("01 Jan 2000", "01/01/2000", "2015/10/10"))
[1] "2000-01-01" "2000-01-01" "2015-10-10"
R>

    These are unambiguous, and should work as you have already stated. They do, thanks to anydate(). Without a format.

    • 22

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