- 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¡
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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
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Star Strider
il 17 Set 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.
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José-Luis
il 17 Set 2014
Modificato: José-Luis
il 17 Set 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
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