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
>>>          


        

Answer 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

Answer 2

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)

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

Answer 4

If you don't mind None, you can use:

a = np.empty(3, dtype=object)

Answer 5

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+