How can I make a python numpy arange of datetime

Question

I have some input data, with timestamps in the input file in the form of hours from the date time specified in the filename.

This is a bit useless, so I need to convert it to python datetime.datetime objects, and then put it in a numpy array. I could write a for loop, but I'd like to do something like:

numpy.arange(datetime.datetime(2000, 1,1), datetime.datetime(2000, 1,2), datetime.timedelta(hours=1))

which throws a TypeError.

Can this be done? I'm stuck with python 2.6 and numpy 1.6.1.

See also http://stackoverflow.com/questions/993358/creating-a-range-of-dates-in-python — nneonneo, Aug 27 '12 at 06:42

score 53 · Answer 1 · edited Sep 24 '18 at 17:58

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from datetime import datetime, timedelta

t = np.arange(datetime(1985,7,1), datetime(2015,7,1), timedelta(days=1)).astype(datetime)

The key point here is to use astype(datetime), otherwise the result will be datetime64.

edited Sep 24 '18 at 17:58

Seanny123

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answered Dec 17 '15 at 01:58

Arc Shinus

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TypeError: 'module' object is not callable. Any idea? what to import first? – weefwefwqg3 May 12 '18 at 20:07
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@weefwefwqg3 Your import is probably `import datetime` instead of `from datetime import datetime`, so `datetime` is not the class but the module itself. – Right leg Apr 30 '19 at 09:12
It's probably worth pointing out that `object` arrays will be quite a bit slower than `datetime64` arrays for most operations; you should probably prefer `datetime64` arrays over `object` (`datetime.datetime`) arrays, at least for NumPy 1.7+. – nneonneo Sep 21 '22 at 23:23

nneonneo · Accepted Answer · 2022-09-21T23:21:51.583

See NumPy Datetimes and Timedeltas. Since NumPy 1.7, you can represent datetimes in NumPy using the numpy.datetime64 type, which permits you to do ranges of values:

>>> np.arange(np.datetime64("2000-01-01"), np.datetime64("2000-01-02"), np.timedelta64(1, "h"))
array(['2000-01-01T00', '2000-01-01T01', '2000-01-01T02', '2000-01-01T03',
       '2000-01-01T04', '2000-01-01T05', '2000-01-01T06', '2000-01-01T07',
       '2000-01-01T08', '2000-01-01T09', '2000-01-01T10', '2000-01-01T11',
       '2000-01-01T12', '2000-01-01T13', '2000-01-01T14', '2000-01-01T15',
       '2000-01-01T16', '2000-01-01T17', '2000-01-01T18', '2000-01-01T19',
       '2000-01-01T20', '2000-01-01T21', '2000-01-01T22', '2000-01-01T23'],
      dtype='datetime64[h]')

For NumPy 1.6, which has a much less useful datetime64 type, you can use a suitable list comprehension to build the datetimes (see also Creating a range of dates in Python):

base = datetime.datetime(2000, 1, 1)
arr = numpy.array([base + datetime.timedelta(hours=i) for i in xrange(24)])

This produces

array([2000-01-01 00:00:00, 2000-01-01 01:00:00, 2000-01-01 02:00:00,
   2000-01-01 03:00:00, 2000-01-01 04:00:00, 2000-01-01 05:00:00,
   2000-01-01 06:00:00, 2000-01-01 07:00:00, 2000-01-01 08:00:00,
   2000-01-01 09:00:00, 2000-01-01 10:00:00, 2000-01-01 11:00:00,
   2000-01-01 12:00:00, 2000-01-01 13:00:00, 2000-01-01 14:00:00,
   2000-01-01 15:00:00, 2000-01-01 16:00:00, 2000-01-01 17:00:00,
   2000-01-01 18:00:00, 2000-01-01 19:00:00, 2000-01-01 20:00:00,
   2000-01-01 21:00:00, 2000-01-01 22:00:00, 2000-01-01 23:00:00], dtype=object)

If only I had numpy 1.7, this would be the answer. But it seems I have 1.6.1, so the example doesn't work. — Melanie, Aug 27 '12 at 07:02
And is also compatible with the datetime I need to output. Thanks! — Melanie, Aug 27 '12 at 07:10

score 23 · Answer 3 · answered Mar 01 '17 at 07:41

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With modern NumPy you can do this:

np.arange(np.datetime64('2017-01-01'), np.datetime64('2017-01-08'))

And it gives you:

array(['2017-01-01', '2017-01-02', '2017-01-03', '2017-01-04',
       '2017-01-05', '2017-01-06', '2017-01-07'], dtype='datetime64[D]')

answered Mar 01 '17 at 07:41

John Zwinck

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Can I put step here? like: a week, 10 days, a month, or a year? – weefwefwqg3 May 12 '18 at 20:22
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@weefwefwqg3: Sure. Use `np.timedelta64(...)`. – John Zwinck May 14 '18 at 00:58

score 10 · Answer 4 · answered Jun 27 '16 at 16:05

As noted in another answer, for Numpy > 1.7, you can use Numpy's built-in datetime capability. The examples in the Numpy documentation don't include using np.arange with steps, so here's one:

timearray = np.arange('2000-01-01', '2000-01-02',np.timedelta64(1,'h'), dtype='datetime64')

Numpy sets the dtype of this result to datetime64[h]. You can set this explicitly to some smaller unit of time with dtype='datetime64[m]'.

In version 1.8.1 (and I expect earlier), trying to add an offset to that result array that is smaller than an hour will have no effect.

timearray += np.timedelta64(10,'s') does not change timearray
timearray2 = timearray + np.timedelta64(10,'s') will add 10 seconds to timearray and converts the dtype of timearray2 to datetime64[s]

score 9 · Answer 5 · answered Aug 27 '12 at 08:39

9

Note that @nneonneo solution can be simplified in

result = first_date + np.arange(24) * datetime.timedelta(hours=1)

thanks to NumPy array manipulations. The result array has then a dtype=object.

For more complex ranges, you might be interested in the scikits.timeseries package (no longer maintained) or better, the pandas package that reimplemented most of the ideas of scikits.timeseries. Both packages support older versions of NumPy (1.5, 1.6...)

answered Aug 27 '12 at 08:39

Pierre GM

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Thanks - it looks like I should have been using pandas for the entire task. Next time :-) – Melanie Aug 28 '12 at 05:46

How can I make a python numpy arange of datetime

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