To expand slightly on what Milan said, while you can pass DateTime an array
of strings and a format code, in some cases the conversion function you
want to use may not accept arrays. In this case, you can do the same thing
with array comprehension:
df1 = DataFrame(V1 = DateTime[DateTime(d, "m/d/y H:M") for d in ["4/5/2002
04:20", "4/5/2002 04:25"]])
On Thursday, May 26, 2016 at 6:55:42 AM UTC-7, Milan Bouchet-Valat wrote:
>
> Le jeudi 26 mai 2016 à 06:15 -0700, akrun a écrit :
> > I am using the DataFrames package. I find it difficult to convert to
> > DateTime.
> >
> > println(DateTime("4/5/2002 04:20", "m/d/y H:M"))
> > gives output
> >
> > 2002-04-05T04:20:00
> >
> > However, if I try
> >
> > df1 = DataFrame(V1 = ["4/5/2002 04:20", "4/5/2002 04:25"])
> > println(DateTime(df1[:V1]))
> > gives
> >
> > ArgumentError: Delimiter mismatch. Couldn't find first
> > delimiter, "-", in date string
> > in parse at dates/io.jl:152
> >
> >
> > Is there any workaround?
> This isn't specific to data frames. You also get this with
> V1 = ["4/5/2002 04:20", "4/5/2002 04:25"]
> DateTime(V1)
>
> Anyway, you need to pass the format as in your first example:
> DateTime(V1, "m/d/y H:M")DateTime(df1[:V1], "m/d/y H:M")
>
>
> Regards
>
