Pairwise difference of datetime vectors (i.e. pdist for time data)

I have two datetime vectors of differing lengths (500x1 and 15000x1) and was wondering how to produce a 500x15000 matrix of the differences between these times. I've previously used pdist2() to answer this question for double vectors, but can't find the function to do it for datetimes.

Answer 1

[Peter Perkins]

pdist is probably much more than you actually need, assuming that by "distance" you mean a simple subtraction. pdist is designed for pairwise diatances between vectors, using one of several distance measures. It will do what you want, but is kind of overkill. I think what you are looking for is what's referred to as "implicit expansion", a more elegant behavior that addresses the same cases as bsxfun used to be used for:

>> (1:3) - (1:4)'
ans =
     0     1     2
    -1     0     1
    -2    -1     0
    -3    -2    -1

I forget when this was added; a few years ago IIRC. In any case, you are right that datetime does not yet support it. But it's easy to do by hand, using ndgrid:

>> d1 = datetime(2019,7,1,0:4,0,0);
>> d2 = datetime(2019,7,1,0:3,0,30);
>> [D1,D2] = ndgrid(d1,d2)
D1 = 
  5×4 datetime array
   01-Jul-2019 00:00:00   01-Jul-2019 00:00:00   01-Jul-2019 00:00:00   01-Jul-2019 00:00:00
   01-Jul-2019 01:00:00   01-Jul-2019 01:00:00   01-Jul-2019 01:00:00   01-Jul-2019 01:00:00
   01-Jul-2019 02:00:00   01-Jul-2019 02:00:00   01-Jul-2019 02:00:00   01-Jul-2019 02:00:00
   01-Jul-2019 03:00:00   01-Jul-2019 03:00:00   01-Jul-2019 03:00:00   01-Jul-2019 03:00:00
   01-Jul-2019 04:00:00   01-Jul-2019 04:00:00   01-Jul-2019 04:00:00   01-Jul-2019 04:00:00
D2 = 
  5×4 datetime array
   01-Jul-2019 00:00:30   01-Jul-2019 01:00:30   01-Jul-2019 02:00:30   01-Jul-2019 03:00:30
   01-Jul-2019 00:00:30   01-Jul-2019 01:00:30   01-Jul-2019 02:00:30   01-Jul-2019 03:00:30
   01-Jul-2019 00:00:30   01-Jul-2019 01:00:30   01-Jul-2019 02:00:30   01-Jul-2019 03:00:30
   01-Jul-2019 00:00:30   01-Jul-2019 01:00:30   01-Jul-2019 02:00:30   01-Jul-2019 03:00:30
   01-Jul-2019 00:00:30   01-Jul-2019 01:00:30   01-Jul-2019 02:00:30   01-Jul-2019 03:00:30
>> delta = D1 - D2
delta = 
  5×4 duration array
   -00:00:30   -01:00:30   -02:00:30   -03:00:30
    00:59:30   -00:00:30   -01:00:30   -02:00:30
    01:59:30    00:59:30   -00:00:30   -01:00:30
    02:59:30    01:59:30    00:59:30   -00:00:30
    03:59:30    02:59:30    01:59:30    00:59:30

That makes two temporary arrays as big as the result, but however you do this, you will end up with a 15000x500 array as your result (unless I've misunderstood).

[dpb]

Good point, Peter! I didn't really think about the calculation itself as just thinking of how to get around the input data type mismatch for OP...

Answer 2

[dpb]

Presuming not excessive precision, one could convert to datenum, compute distances and then return to datetime

Seems like a feature enhancement.