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 would be much slower.v is None, and also
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).
help(numpy.full) in a Python shell. I am also surprised that it is not in the web documentation.
np.fill() does not exist and should be arr.fill()), with a difference of about 10 %. If the difference was bigger, I would raise an issue in the NumPy bug tracker. :) I prefer more explicit and clearer code, for such a small difference in executing time, so I go with np.full() all the time.
np.full() is the way to go, I observed under some circumstances that np.zeros() + val can be faster (e.g. for very large arrays in a HPC).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
np.full() would be useful. On my machine, with NumPy 1.8.1, it is about 15% slower than the less direct fill() version (which is unexpected, as full() has the potential of going slightly faster).fill() is the fastest solution. The multiplication solution is much slower.10000 instead of 1e4 makes a noticeable difference, for some reason (full() is almost 50% slower, with 1e4).
full(), it runs considerably slower when the datatype isn't explicitly a float. Otherwise, it's comparable (but slightly slower) with the best methods here.full(100000, 5), full(100000, 5, dtype=float), full(100000, 5, dtype=int) and a =np.empty(100000); a.fill(5) all take about the same time on my machine (with no caching: %timeit -r1 -n1 …) (NumPy 1.11.2).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.
fill, I saw that repeat suits my needs even better.a[:]=v is actually faster overall than the fill?
fill.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,
)
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.
without numpy
>>>[2]*3
[2, 2, 2]
[v] * n would be more directly relevant to the OP question.
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.
We could also write
v=7
n=5
a=np.linspace(v,v,n)
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()
a = np.zeros(n)in the loop is faster thana.fill(0). This is counter to what I expected since I thoughta=np.zeros(n)would need to allocate and initialize new memory. If anyone can explain this, I would appreciate it.v * ones(n)is still horrible, as it uses the expensive multiplication. Replace*with+though, andv + zeros(n)turns out to be surprisingly good in some cases (stackoverflow.com/questions/5891410/…).var = np.empty(n)and then to fill it with 'var[:] = v'. (btw,np.full()is as fast as this)