Convert string "date code" for month and year to proper date type

New to Julia
1

I have a data frame with where one of the columns contains a string “date code”. An example of this code is as follows:

“04_22”

which indicates the 4th month of the year 2022.

“11_23”

would indicate the 11th month of 2023.

Rather than working with this code in string, I would like to convert the column of string code to a column of proper date types. How can that be done?

2 
julia> dStr = "04_22"
"04_22"

julia> m, y = split(dStr, '_')
2-element Vector{SubString{String}}:
 "04"
 "22"

julia> DateTime(2000 + parse(Int, y), parse(Int, m), 1)
2022-04-01T00:00:00
3

Also:

vd = ["04_22","11_23"]
vd .= replace.(vd, "_" => "-20")
Date.(vd, "mm-yyyy")
4

There is no need to use split/parse/replace here, just parse directly with a DateFormat (and then add 2000 years):

julia> Date("11_23", dateformat"mm_yy") + Year(2000)
2023-11-01
5 Likes
5

That is nice. Would you know why the dot is not needed to broadcast the sum of the 2000 years with the vector?

vd = ["04_22","11_23"]
Date.(vd, "mm_yy") .+ Year(2000)
6

Yes, that is not intended, but something that was forgotten to deprecate for Julia 1.0, see Dates/src/deprecated.jl.

2 Likes
7

That is very nice and intuitive. Suppose I knew that it was not the first of the month that the code indicated, but always the last day of the month, what would your solution look like then?

8

Maybe you could do:

Date.(vd, "mm_yy") .+ Year(2000) .+ Month(1) .- Day(1)
1 Like
9

An alternative for the last day of the month is

@. Dates.lastdayofmonth(Date(vd, "mm_yy") + Year(2000))
1 Like