NumPy array initialization (fill with identical values)

Question

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.

Asked By: max

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Source

Answer 1

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.

Answered By: Paul

Answer 2

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.

Answered By: Rolf Bartstra

Answer 3

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

Answered By: Yariv

Answer 4

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.

Answered By: Rolf Bartstra

Answer 5

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).

Answered By: Eric O Lebigot

Answer 6

I had np.array(n * [value]) in mind, but apparently that is slower than all other suggestions for large enough n. The best in terms of readability and speed is

np.full(n, 3.14)

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

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,
)

Answered By: Nico Schlömer

Answer 7

without numpy

>>>[2]*3
[2, 2, 2]

Answered By: tnusraddinov

Answer 8

You can also use np.broadcast_to.

To create an array of shape (dimensions) s and of value v, you can do (in your case, the array is 1-D, and s = (n,)):

a = np.broadcast_to(v, s).copy()

if a only needs to be read-only, you can use the following (which is way more efficient):

a = np.broadcast_to(v, s)

The advantage is that v can be given as a single number, but also as an array if different values are desired (as long as v.shape matches the tail of s).

Bonus: if you want to force the dtype of the created array:

a = np.broadcast_to(np.asarray(v, dtype), s).copy()

Answered By: User9123

Answer 9

We could also write

v=7
n=5
a=np.linspace(v,v,n)

Answered By: Gabriel Soranzo

NumPy array initialization (fill with identical values)

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