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How can I parse this date format? Should I change this colon to dot or maybe someone know better solution?
> x <- "2012.01.15 09:00:02:002"
> strptime(x, "%Y.%m.%d %H:%M:%S:%OS")
[1] "2012-01-15 09:00:02"
> strptime(x, "%Y.%m.%d %H:%M:%OS")
[1] "2012-01-15 09:00:02"
> x <- "2012.01.15 09:00:02.002"
> strptime(x, "%Y.%m.%d %H:%M:%OS")
[1] "2012-01-15 09:00:02.001"
200
To display the millisecond component of a DateTime valueParse(String) or DateTimeOffset. Parse(String) method. To extract the string representation of a time's millisecond component, call the date and time value's DateTime.
Usually we display time in in 12 hour format hh:mm:aa format (e.g. 12:30 PM) or 24 hour format HH:mm (e.g. 13:30), however sometimes we also want to show the milliseconds in the time. To show the milliseconds in the time we include “SSS” in the pattern which displays the Milliseconds.
There's a subtle distinction here that may be throwing you off.
As ?strptime notes:
for 'strptime' '%OS' will input seconds including fractional seconds.
To emphasize that a bit, %OS represents the seconds including fractional seconds --- not just the fractional part of the seconds: if the seconds value is 44.234, %OS or %OS3 represents 44.234, not .234
So the solution is indeed to substitute a . for that final :.
Here's one way you might do that:
x <- "2012.01.15 09:00:02:002"
strptime(gsub(":", ".", x), "%Y.%m.%d %H.%M.%OS")
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