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

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Alex G 2019년 7월 1일
댓글: dpb 2019년 7월 3일
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.

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### 채택된 답변

Peter Perkins 2019년 7월 2일
편집: Peter Perkins 2019년 7월 2일
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).
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dpb 2019년 7월 3일
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...

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### 추가 답변 (1개)

dpb 2019년 7월 2일
Presuming not excessive precision, one could convert to datenum, compute distances and then return to datetime
Seems like a feature enhancement.

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