Parse ISO 8601 date-time in format YYYY-MM-DDTHH-MM-SSZ

Question

I have a large dataframe with time stamps that look like this:

"2019-05-15T01:42:15.072Z"

It resembles a ISO 8601 combined date and tim

Answer 1

You can simply parse the timestamp by specifying the format in as.POSIXct (or strptime)

as.POSIXct("2019-05-15T01:42:15.072Z", format = "%Y-%m-%dT%H:%M:%OSZ")
#[1] "2019-05-15 01:42:15 AEST"

Explanation:

%Y, %m and %d denote the year (with century), month and day; %H, %M and %OS denote the hours, minutes and seconds (including milliseconds). The T and Z are simply added to the format string, because

Any character in the format string not part of a conversion specification is interpreted literally

See ?strptime for the different conversion specifications.