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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Walter Roberson
Walter Roberson il 12 Mar 2021
S = 'CM.SJC..HHE.D.2015.213.172556.SAC';
parts = regexp(S,'\.','split');
year = str2double(parts{6});
doy = str2double(parts{7});
hour = str2double(parts{8}(1:2));
minute = str2double(parts{8}(3:4));
second = str2double(parts{8}(5:6));
output = datetime(year, 1, 1, hour, minute, second, 'Format', 'yyyy M d hh:mm:ss', 'timezone','utc') + days(doy);
output
output = datetime
2015 8 2 05:25:56
days(datetime(2015,7,26)-datetime(2015,1,1))
ans = 206
Could you confirm that you really do have Julian days? Julian Days are number of days since a particular date around 4700 BCE. As you can see, there is a difference of exactly one week between your intended output and a calculation based upon day of the year.
  1 Commento
Daphne Sagel Aguilar
Daphne Sagel Aguilar il 12 Mar 2021
Thank you very much, you have given us a plus in our work. We were working with the length of the SAC file, and with this code we can divide it into parts, excellent.

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Più risposte (1)

Peter Perkins
Peter Perkins il 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 Commento
Walter Roberson
Walter Roberson il 23 Mar 2021
The regexp version is not so bad
t1 = 'CM.SJC..HHE.D.2015.213.172556.SAC'
t1 = 'CM.SJC..HHE.D.2015.213.172556.SAC'
t3 = regexp(t1, '\d[\d.]+', 'match')
t3 = 1×1 cell array
{'2015.213.172556.'}
d3 = datetime(t3, 'InputFormat', 'uuuu.DDD.HHmmss.')
d3 = datetime
01-Aug-2015 17:25:56

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