Is there a better way to broadcast arrays?

Question

I want to broadcast an array b to the shape it would take if it were in an arithmetic operation with another array a.

For example, if a.shape = (3,3) and b was a scalar, I want to get an array whose shape is (3,3) and is filled with the scalar.

One way to do this is like this:

>>> import numpy as np
>>> a = np.arange(9).reshape((3,3))
>>> b = 1 + a*0
>>> b
array([[1, 1, 1],
       [1, 1, 1],
       [1, 1, 1]])

Although this works practically, I can't help but feel it looks a bit weird, and wouldn't be obvious to someone else looking at the code what I was trying to do.

Is there any more elegant way to do this? I've looked at the documentation for np.broadcast, but it's orders of magnitude slower.

In [1]: a = np.arange(10000).reshape((100,100))

In [2]: %timeit 1 + a*0
10000 loops, best of 3: 31.9 us per loop

In [3]: %timeit bc = np.broadcast(a,1);np.fromiter((v for u, v in bc),float).reshape(bc.shape)
100 loops, best of 3: 5.2 ms per loop

In [4]: 5.2e-3/32e-6
Out[4]: 162.5

Answer 1

If you just want to fill an array with a scalar, fill is probably the best choice. But it sounds like you want something more generalized. Rather than using broadcast you can use broadcast_arrays to get the result that (I think) you want.

>>> a = numpy.arange(9).reshape(3, 3)
>>> numpy.broadcast_arrays(a, 1)[1]
array([[1, 1, 1],
       [1, 1, 1],
       [1, 1, 1]])

This generalizes to any two broadcastable shapes:

>>> numpy.broadcast_arrays(a, [1, 2, 3])[1]
array([[1, 2, 3],
       [1, 2, 3],
       [1, 2, 3]])

It's not quite as fast as your ufunc-based method, but it's still on the same order of magnitude:

>>> %timeit 1 + a * 0
10000 loops, best of 3: 23.2 us per loop
>>> %timeit numpy.broadcast_arrays(a, 1)[1]
10000 loops, best of 3: 52.3 us per loop

But scalars, fill is still the clear front-runner:

>>> %timeit b = numpy.empty_like(a, dtype='i8'); b.fill(1)
100000 loops, best of 3: 6.59 us per loop

Finally, further testing shows that the fastest approach -- in at least some cases -- is to multiply by ones:

>>> %timeit numpy.broadcast_arrays(a, numpy.arange(100))[1]
10000 loops, best of 3: 53.4 us per loop
>>> %timeit (1 + a * 0) * numpy.arange(100)
10000 loops, best of 3: 45.9 us per loop
>>> %timeit b = numpy.ones_like(a, dtype='i8'); b * numpy.arange(100)
10000 loops, best of 3: 28.9 us per loop

Answer 2

The fastest and cleanest solution I know is:

b_arr = numpy.empty(a.shape)  # Empty array
b_arr.fill(b)  # Filling with one value