How to convert datetime to datestr including day of year
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I can convert a date string to a datetime object using datetime(), and convert that object back to a string using datestr(). However, my date includes a three-digit day of year field (day 179, say). I can use formatting to convert the day of year to the correct datetime object by overloading the day of month field, but I see no good way to reverse the process in datestr.
time_string = '22179114244.580';
date_time = datetime(time_string, 'InputFormat', 'uuDDDHHmmss.SSS')
date_time =
datetime 28-Jun-0022 11:42:44 % this is correct so far
formatOut = 'yy:ddd:HH:mm:SS.FFF';
datestr(date_time,formatOut)
ans =
'22:Tue:11:06:44.580' % WRONG
What format must I use to represent the day of year using datestr? Python uses %j to specifically represent the day of year, but I see no equivalent in Matlab.
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Accepted Answer
Steven Lord
on 25 Oct 2022
Looking at the table of format identifiers in the description of the Format property of datetime objects on the datetime documentation page, the identifiers D, DD, or DDD will do what you want if you convert the datetime into a string using string. You will also need to use s or ss for whole seconds and S, SS, etc. for fractional seconds.
time_string = '22179114244.580';
date_time = datetime(time_string, 'InputFormat', 'uuDDDHHmmss.SSS')
formatOut = 'yy:D:HH:mm:ss.SSS';
s = string(date_time, formatOut)
Let's check.
day(date_time, 'dayofyear')
Wikipedia confirms that June 28th is the 179th (or 180th in a leap year) day of the year.
3 Comments
Steven Lord
on 25 Oct 2022
That's pretty easy. I'm going to use 6 second increments to make it easier to see the effect, but you could use 0.6 second increments.
N = datetime('now')
sixSecondIncrements = N + seconds(0:6:30).'
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