- When computing date vectors, MATLAB sets month values less than 1 to 1. Day values, D, less than 1 are set to the last day of the previous month minus D. However, if 0 ≤ DateNumber < 1, then datevec(DateNumber) returns a date vector of the form [0 0 0 H MN S], where H, MN, and S are hours, minutes, and seconds, respectively.
Calculate difference between two times?...something it's wrong here¡
7 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Rapala
am 17 Sep. 2014
Beantwortet: Star Strider
am 17 Sep. 2014
I would be very grateful if someone could help me with the following problem, I try to subtract dates, but clearly the result is not what I expect....look¡
>>date =
2012 9 25 0 0 0
2012 9 23 21 0 0
>>d=datenum(date)
d =
1.0e+05 *
7.3514
7.3514
>>difference=d(1)-d(2);
ans =
1.1250
>>datestr(difference,'yyyy mm dd HH MM SS')
ans =
0000 01 01 03 00 00 % should be 0000 00 01 03 00 00
Thanks
0 Kommentare
Akzeptierte Antwort
Star Strider
am 17 Sep. 2014
Limitations
This does not appear in the datestr documentation, but I would be surprised if it does not apply there as well, since both datestr and datevec give the same results to your calculations.
0 Kommentare
Weitere Antworten (1)
José-Luis
am 17 Sep. 2014
Bearbeitet: José-Luis
am 17 Sep. 2014
What makes you think it should be 0000 00 01 03 00 00?
From the documentation:
A serial date number represents a calendar date as the number of days that has passed since a fixed base date. In MATLAB, serial date number 1 is January 1, 0000.
If you subtract two datenumbers, the result will not be the time elapsed between the two dates.
datevec(723300 - 723297)
If you want the elapsed time:
etime(datevec(723300),datevec(723297)) %in seconds
3 Kommentare
Siehe auch
Kategorien
Mehr zu Calendar finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!