Create numpy matrix filled with NaNs

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Chapters
00:00 Question
00:31 Accepted answer (Score 381)
01:38 Answer 2 (Score 258)
02:02 Answer 3 (Score 77)
02:56 Answer 4 (Score 27)
03:33 Thank you

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Full question
https://stackoverflow.com/questions/1704...

Answer 1 links:
[numpy.full]: https://docs.scipy.org/doc/numpy/referen...

Answer 2 links:
[image]: https://i.stack.imgur.com/oGR81.png

Answer 3 links:
[mailing list thread]: http://www.mail-archive.com/numpy-discus...

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Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...

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Tags
#python #numpy

#avk47

ACCEPTED ANSWER

Score 408

You rarely need loops for vector operations in numpy. You can create an uninitialized array and assign to all entries at once:

>>> a = numpy.empty((3,3,))
>>> a[:] = numpy.nan
>>> a
array([[ NaN,  NaN,  NaN],
       [ NaN,  NaN,  NaN],
       [ NaN,  NaN,  NaN]])

I have timed the alternatives a[:] = numpy.nan here and a.fill(numpy.nan) as posted by Blaenk:

$ python -mtimeit "import numpy as np; a = np.empty((100,100));" "a.fill(np.nan)"
10000 loops, best of 3: 54.3 usec per loop
$ python -mtimeit "import numpy as np; a = np.empty((100,100));" "a[:] = np.nan" 
10000 loops, best of 3: 88.8 usec per loop

The timings show a preference for ndarray.fill(..) as the faster alternative. OTOH, I like numpy's convenience implementation where you can assign values to whole slices at the time, the code's intention is very clear.

Note that ndarray.fill performs its operation in-place, so numpy.empty((3,3,)).fill(numpy.nan) will instead return None.

ANSWER 2

Score 299

Another option is to use numpy.full, an option available in NumPy 1.8+

a = np.full([height, width, 9], np.nan)

This is pretty flexible and you can fill it with any other number that you want.

ANSWER 3

Score 28

Are you familiar with numpy.nan?

You can create your own method such as:

def nans(shape, dtype=float):
    a = numpy.empty(shape, dtype)
    a.fill(numpy.nan)
    return a

Then

nans([3,4])

would output

array([[ NaN,  NaN,  NaN,  NaN],
       [ NaN,  NaN,  NaN,  NaN],
       [ NaN,  NaN,  NaN,  NaN]])

I found this code in a mailing list thread.

ANSWER 4

Score 14

You can always use multiplication if you don't immediately recall the .empty or .full methods:

>>> np.nan * np.ones(shape=(3,2))
array([[ nan,  nan],
       [ nan,  nan],
       [ nan,  nan]])

Of course it works with any other numerical value as well:

>>> 42 * np.ones(shape=(3,2))
array([[ 42,  42],
       [ 42,  42],
       [ 42, 42]])

But the @u0b34a0f6ae's accepted answer is 3x faster (CPU cycles, not brain cycles to remember numpy syntax ;):

$ python -mtimeit "import numpy as np; X = np.empty((100,100));" "X[:] = np.nan;"
100000 loops, best of 3: 8.9 usec per loop
(predict)laneh@predict:~/src/predict/predict/webapp$ master
$ python -mtimeit "import numpy as np; X = np.ones((100,100));" "X *= np.nan;"
10000 loops, best of 3: 24.9 usec per loop