CODEWITHSUNDEEP
julia
codedancer
codedancer

Reputation: 1634

Convert date time format "1/22/20" into "2020-01-22" in Julia

I am wondering what the best way to convert date time format "1/22/20" into "2020-01-22". I tried DateTime.(d1.date, "m/d/y") and I feel I was missing something in the output. Any suggestion?

julia> first(d1, 10)
10×2 DataFrame
 Row │ date     value
     │ String   Int64
─────┼────────────────
   1 │ 1/22/20      1
   2 │ 1/23/20      1
   3 │ 1/24/20      2
   4 │ 1/25/20      2
   5 │ 1/26/20      5
   6 │ 1/27/20      5
   7 │ 1/28/20      5
   8 │ 1/29/20      6
   9 │ 1/30/20      6
  10 │ 1/31/20      8

julia> DateTime.(d1.date, "m/d/y")
519-element Vector{DateTime}:
 0020-01-22T00:00:00
 0020-01-23T00:00:00
 0020-01-24T00:00:00
 0020-01-25T00:00:00
 0020-01-26T00:00:00
 0020-01-27T00:00:00
 0020-01-28T00:00:00
 0020-01-29T00:00:00
 0020-01-30T00:00:00
 0020-01-31T00:00:00
 0020-02-01T00:00:00
 0020-02-02T00:00:00
 0020-02-03T00:00:00

Upvotes: 3

Views: 473

Answers (1)

BallpointBen
BallpointBen

Reputation: 13820

The DateTime constructor converts a string to a datetime. To convert back to a string (of a different format), use Dates.format:

Dates.format.(DateTime.(d1.date, dateformat"m/d/y"), dateformat"yyyy-mm-dd")

Since your years only have two digits, they're interpreted as year 20 instead of 2020. To fix this, you can add Dates.Year(2000) to the dates.

Dates.format.(DateTime.(d1.date, dateformat"m/d/y") .+ Dates.Year(2000), dateformat"yyyy-mm-dd")

You should make sure to use the @dateformat_str macro to construct DateFormats whenever possible because the conversion of String to DateFormat is quite expensive, and using a macro to do it means it happens a single time, at compilation time (when the macro is expanded) instead of once per DataFrame row.

Upvotes: 3

Related Questions