NumPy array initialization (fill with identical values)
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Chapters
00:00 Numpy Array Initialization (Fill With Identical Values)
00:26 Answer 1 Score 77
00:45 Answer 2 Score 119
01:12 Answer 3 Score 20
01:45 Accepted Answer Score 561
02:12 Thank you
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Full question
https://stackoverflow.com/questions/5891...
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Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...
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Tags
#python #arrays #numpy
#avk47
ACCEPTED ANSWER
Score 561
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).
ANSWER 2
Score 119
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
ANSWER 3
Score 77
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.
ANSWER 4
Score 20
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.