python - NumPy array initialization (fill with identical values)

285
votes

I need to create a NumPy array of length n, each element of which is v.

Is there anything better than:

a = empty(n)
for i in range(n):
    a[i] = v

I know zeros and ones would work for v = 0, 1. I could use v * ones(n), but it won't work when v is None, and also would be much slower.

7
pythonarraysnumpy
On my computer, for the 0 case, using a = np.zeros(n) in the loop is faster than a.fill(0). This is counter to what I expected since I thought a=np.zeros(n) would need to allocate and initialize new memory. If anyone can explain this, I would appreciate it. – user3731622
You cannot put Nonein a numpy array, since the cells are created with a specific data type while None has it's own type and is in fact a pointer. – Camion
@Camion Yeah I know now :) Of course v * ones(n) is still horrible, as it uses the expensive multiplication. Replace * with + though, and v + zeros(n) turns out to be surprisingly good in some cases (stackoverflow.com/questions/5891410/…). – max
max, instead of creating an array with zeros before adding v, it is even faster to create it empty with var = np.empty(n) and then to fill it with 'var[:] = v'. (btw, np.full() is as fast as this) – Camion

7 Answers

398
votes

NumPy 1.8 introduced np.full(), which is a more direct method than empty() followed by fill() for creating an array filled with a certain value:

>>> np.full((3, 5), 7)
array([[ 7.,  7.,  7.,  7.,  7.],
       [ 7.,  7.,  7.,  7.,  7.],
       [ 7.,  7.,  7.,  7.,  7.]])

>>> np.full((3, 5), 7, dtype=int)
array([[7, 7, 7, 7, 7],
       [7, 7, 7, 7, 7],
       [7, 7, 7, 7, 7]])

This is arguably the way of creating an array filled with certain values, because it explicitly describes what is being achieved (and it can in principle be very efficient since it performs a very specific task).

102
votes

Updated for Numpy 1.7.0:(Hat-tip to @Rolf Bartstra.)

a=np.empty(n); a.fill(5) is fastest.

In descending speed order:

%timeit a=np.empty(10000); a.fill(5)
100000 loops, best of 3: 5.85 us per loop

%timeit a=np.empty(10000); a[:]=5 
100000 loops, best of 3: 7.15 us per loop

%timeit a=np.ones(10000)*5
10000 loops, best of 3: 22.9 us per loop

%timeit a=np.repeat(5,(10000))
10000 loops, best of 3: 81.7 us per loop

%timeit a=np.tile(5,[10000])
10000 loops, best of 3: 82.9 us per loop
69
votes

I believe fill is the fastest way to do this.

a = np.empty(10)
a.fill(7)

You should also always avoid iterating like you are doing in your example. A simple a[:] = v will accomplish what your iteration does using numpy broadcasting.

27
votes

I had 

numpy.array(n * [value])

in mind, but apparently that is slower than all other suggestions for large enough n.

Here is full comparison with perfplot (a pet project of mine).

enter image description here

The two empty alternatives are still the fastest (with NumPy 1.12.1). full catches up for large arrays.

Code to generate the plot:

import numpy as np
import perfplot


def empty_fill(n):
    a = np.empty(n)
    a.fill(3.14)
    return a


def empty_colon(n):
    a = np.empty(n)
    a[:] = 3.14
    return a


def ones_times(n):
    return 3.14 * np.ones(n)


def repeat(n):
    return np.repeat(3.14, (n))


def tile(n):
    return np.repeat(3.14, [n])


def full(n):
    return np.full((n), 3.14)


def list_to_array(n):
    return np.array(n * [3.14])


perfplot.show(
    setup=lambda n: n,
    kernels=[empty_fill, empty_colon, ones_times, repeat, tile, full, list_to_array],
    n_range=[2 ** k for k in range(27)],
    xlabel="len(a)",
    logx=True,
    logy=True,
)
17
votes

Apparently, not only the absolute speeds but also the speed order (as reported by user1579844) are machine dependent; here's what I found:

a=np.empty(1e4); a.fill(5) is fastest;

In descending speed order:

timeit a=np.empty(1e4); a.fill(5) 
# 100000 loops, best of 3: 10.2 us per loop
timeit a=np.empty(1e4); a[:]=5
# 100000 loops, best of 3: 16.9 us per loop
timeit a=np.ones(1e4)*5
# 100000 loops, best of 3: 32.2 us per loop
timeit a=np.tile(5,[1e4])
# 10000 loops, best of 3: 90.9 us per loop
timeit a=np.repeat(5,(1e4))
# 10000 loops, best of 3: 98.3 us per loop
timeit a=np.array([5]*int(1e4))
# 1000 loops, best of 3: 1.69 ms per loop (slowest BY FAR!)

So, try and find out, and use what's fastest on your platform.

9
votes

without numpy

>>>[2]*3
[2, 2, 2]
8
votes

You can use numpy.tile, e.g. :

v = 7
rows = 3
cols = 5
a = numpy.tile(v, (rows,cols))
a
Out[1]: 
array([[7, 7, 7, 7, 7],
       [7, 7, 7, 7, 7],
       [7, 7, 7, 7, 7]])

Although tile is meant to 'tile' an array (instead of a scalar, as in this case), it will do the job, creating pre-filled arrays of any size and dimension.