Best way to initialize and fill an numpy array?

Question

I want to initialize and fill a numpy array. What is the best way?

This works as I expect:

>>> import numpy as np
>>> np.empty(3)
array([ -1.28822975e-231,  -1.73060252e-077,   2.23946712e-314])

But this doesn't:

>>> np.empty(3).fill(np.nan)
>>> 

Nothing?

>>> type(np.empty(3))
<type 'numpy.ndarray'>

It seems to me that the np.empty() call is returning the correct type of object, so I don't understand why .fill() is not working?

Assigning the result of np.empty() first works fine:

>>> a = np.empty(3)
>>> a.fill(np.nan)
>>> a
array([ nan,  nan,  nan])

Why do I need to assign to a variable in order to use np.fill()? Am I missing a better alternative?

Answer 1

You could also try:

In [79]: np.full(3, np.nan) Out[79]: array([ nan,  nan,  nan]) 

The pertinent doc:

Definition: np.full(shape, fill_value, dtype=None, order='C') Docstring: Return a new array of given shape and type, filled with `fill_value`. 

Although I think this might be only available in numpy 1.8+

Answer 2

np.fill modifies the array in-place, and returns None. Therefor, if you're assigning the result to a name, it gets a value of None.

An alternative is to use an expression which returns nan, e.g.:

a = np.empty(3) * np.nan

Answer 3

I find this easy to remember:

numpy.array([numpy.nan]*3)

Out of curiosity, I timed it, and both @JoshAdel's answer and @shx2's answer are far faster than mine with large arrays.

In [34]: %timeit -n10000 numpy.array([numpy.nan]*10000)
10000 loops, best of 3: 273 µs per loop

In [35]: %timeit -n10000 numpy.empty(10000)* numpy.nan
10000 loops, best of 3: 6.5 µs per loop

In [36]: %timeit -n10000 numpy.full(10000, numpy.nan)
10000 loops, best of 3: 5.42 µs per loop