datetime with nanosecond precision, but arithmetic with millisecond precision?

I need to process GPS-based time stamps that have 100ns precision. I have successfully read and stored them with this precision using the datetime data type. Now, I would like to do simple calculations, i.e. difference between time stamps. Any function for time arithmetic (diff, between) or direct subtraction of two datetime objects only provides me a result with millisecond precision, although my datetime data has 100ns precision.

How can I subtract two 100ns precision time stamps and still get a 100ns precision result in Matlab?

[dpb]

Help us out with attaching a dataset to play with but I believe calculations are in full precision but the default display for durations is msec. You can set .SSSS fractions but that's probably rather inconvenient.

I'm guessing you may need to write some specialty functions; the datetime stuff really isn't made for other than just human calendar scale stuff.

Answer 1

[Peter Perkins]

As Walter says, datetime and duration* should have no trouble with this. I suspect this is just a format issue:

>> d = datetime(2018,7,5,0,0,0,'Format','dd-MMM-yyyy HH:mm:ss.SSSSSSSSS') + milliseconds((0:5)'/1e4)
d = 
  6×1 datetime array
   05-Jul-2018 00:00:00.000000000
   05-Jul-2018 00:00:00.000000100
   05-Jul-2018 00:00:00.000000200
   05-Jul-2018 00:00:00.000000300
   05-Jul-2018 00:00:00.000000400
   05-Jul-2018 00:00:00.000000500
>> dt = diff(d);
>> dt.Format = 's'
dt = 
  5×1 duration array
   1e-07 sec
   1e-07 sec
   1e-07 sec
   1e-07 sec
   1e-07 sec

• duration will start having round-off issues at 100ns precision when representing elapsed times larger than about 14 years. Not sure if your data are like that.

Answer 2

[Walter Roberson]

The calculation does have nanosecond precision; however the default formats do not display it.

aaa = seconds(sort(rand(1,2)));
aaa.Format = 's'    %the default for duration objects created with seconds(), but used for emphasis
bbb = diff(aaa);
bbb.Format = 's'

aaa = 
  1×2 duration array
   0.0975404049994095 sec    0.278498218867048 sec
bbb = 
  duration
   0.180957813867639 sec
>> sym('0.278498218867048') - sym('0.0975404049994095')
ans =
0.1809578138676385

so the 15th digit is rounded in the display of bbb.