NumPy array initialization (fill with identical values)
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
Solution 1:
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).
Solution 2:
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