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?

Solution 1

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

Solution 2

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+

Solution 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

Solution 4

Just for future reference, the multiplication by np.nan only works because of the mathematical properties of np.nan. For a generic value N, one would need to use np.ones() * N mimicking the accepted answer, however, speed-wise, this is not a terribly good choice.

Best choice would be np.full() as already pointed out, and, if that is not available for you, np.zeros() + N seems to be a better choice than np.ones() * N, while np.empty() + N or np.empty() * N are simply not suitable. Note that np.zeros() + N will also work when N is np.nan.

%timeit x = np.full((1000, 1000, 10), 432.4)
8.19 ms ± 97.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit x = np.zeros((1000, 1000, 10)) + 432.4
9.86 ms ± 55.1 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit x = np.ones((1000, 1000, 10)) * 432.4
17.3 ms ± 104 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit x = np.array([432.4] * (1000 * 1000 * 10)).reshape((1000, 1000, 10))
316 ms ± 37.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

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Comments

[Davidmh]

This is the correct way to do it. If you are in older versions, you would need to do np.zeros(3) + value

[norok2]

Note however that this will not work for generic values, for which @JoshAdel answer is better for NumPy 1.8+, while for earlier version np.zeros(shape) + value should be used. See my answer for timings.