Create numpy matrix filled with NaNs

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]])

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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.

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.

I compared the suggested alternatives for speed and found that, for large enough vectors/matrices to fill, all alternatives except val * ones and array(n * [val]) are equally fast.

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Code to reproduce the plot:

import numpy
import perfplot

val = 42.0


def fill(n):
    a = numpy.empty(n)
    a.fill(val)
    return a


def colon(n):
    a = numpy.empty(n)
    a[:] = val
    return a


def full(n):
    return numpy.full(n, val)


def ones_times(n):
    return val * numpy.ones(n)


def list(n):
    return numpy.array(n * [val])


perfplot.show(
    setup=lambda n: n,
    kernels=[fill, colon, full, ones_times, list],
    n_range=[2 ** k for k in range(20)],
    logx=True,
    logy=True,
    xlabel="len(a)",
)