factor to date returns NA

Question

I am sorry but I struggle with this:

mydate <- factor("2016-10-25")
as.Date(mydate, format = "%Y-%M-%D")

it returns NA. Any ideas? Thanks!

Try `as.Date(mydate, format = "%Y-%m-%d")`. Some more info on the formats here: https://www.statmethods.net/input/dates.html — AntoniosK, Nov 09 '18 at 11:53
Similar answer here: [link]https://stackoverflow.com/questions/17496358/r-help-converting-factor-to-date — Jamie_D, Nov 09 '18 at 11:53
um `as.Date(mydate)` works _just fine_ since that's the default ISO format it expects — hrbrmstr, Nov 09 '18 at 12:23
Suggesting the use of `lubridate` for this example is crazy talk. It's the default ISO format and I don't think anyone rly wants to take **1250** microseconds to do the conversion (for _one_ element) when the built-in `as.Date()` does it in **30** microseconds. — hrbrmstr, Nov 09 '18 at 12:25

score 3 · Accepted Answer · answered Nov 09 '18 at 12:00

3

You need to use small letters for month and day ("%Y-%m-%d") instead of capital letters ("%Y-%M-%D").

mydate <- factor("2016-10-25")
as.Date(mydate, format = "%Y-%m-%d")

"2016-10-25"

answered Nov 09 '18 at 12:00

Can H.

well I tried this - I think but yes this works now. Thanks! – cs0815 Nov 09 '18 at 12:01
You might want to also add the output of `microbenchmark::microbenchmark(base = as.Date(mydate), lubr = lubridate::ymd(mydate))` to this answer to show the stark contrast to `lubridate` (also the `format` string is not necessary since it'll try the ISO format one by default) – hrbrmstr Nov 09 '18 at 12:28

score 2 · Answer 2 · answered Nov 09 '18 at 12:14

2

You can use lubridate as follows:

    mydate <- factor("2016-10-25")
require(lubridate)
ymd(mydate)

answered Nov 09 '18 at 12:14

NelsonGon

2 Answers2