Convert julian date SAC to specific format date
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I need convert: CM.SJC..HHE.D.2015.213.172556.SAC to specific format date, examples: 2015 7 26
Please
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More Answers (1)
Peter Perkins
on 23 Mar 2021
I'm gonna suggest another solution that might be simpler to follow.
First thing is that this isn't a Julian Date in the strictly correct sense of the word. In many fields though, "Julian day/date" means "day of year". That's what you have.
datetime text conversion will handle literals in a timestamp (the double-quoted format contains single-quote-escaped literals), and will handle day of year ("D", not "d"). So this
>> t1 = 'CM.SJC..HHE.D.2015.213.172556.SAC'
t1 =
'CM.SJC..HHE.D.2015.213.172556.SAC'
>> d1 = datetime(t1,'InputFormat',"'CM.SJC..HHE.D.'uuuu.DDD.HHmmss'.SAC'")
would be the way to go. Unfortunately, there's a fairly recent (and VERY narrowly-scoped) bug in parsing timestamps that causes that to error (thanks, Walter, for reporting it). That will be fixed, but for now, there are two work-arounds. One is to lower-case everything:
>> t2 = lower(t1)
t2 =
'cm.sjc..hhe.d.2015.213.172556.sac'
d2 = datetime(t2,'InputFormat',"'cm.sjc..hhe.d.'uuuu.DDD.HHmmss'.sac'")
d2 =
datetime
01-Aug-2015 17:25:56
The other is to peel off the literals. In versions since R2016b it's easiest to use some string functions rather than regexp:
>> t3 = extractAfter(extractBefore(t1,".SAC"),"CM.SJC..HHE.D.")
t3 =
'2015.213.172556'
d3 = datetime(t3,'InputFormat',"uuuu.DDD.HHmmss")
d3 =
datetime
01-Aug-2015 17:25:56
The above shows scalars, but of course it's completely vectorized.
1 Comment
The regexp version is not so bad
t1 = 'CM.SJC..HHE.D.2015.213.172556.SAC'
t3 = regexp(t1, '\d[\d.]+', 'match')
d3 = datetime(t3, 'InputFormat', 'uuuu.DDD.HHmmss.')
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