Given a numpy array x of shape (m,) and a numpy array y of shape (m/n,), how do I multiply x by corresponding elements of y efficiently?
Here's my best attempt:
In [13]: x = np.array([1, 5, 3, 2, 9, 1])
In [14]: y = np.array([2, 4, 6])
In [15]: n = 2
In [16]: (y[:, np.newaxis] * x.reshape((-1, n))).flatten()
Out[16]: array([ 2, 10, 12, 8, 54, 6])
Your solution looks pretty good to me.
If you wanted to speed it up slightly, you could:
Use ravel() instead of flatten() (the former will return a view if possible, the latter always returns a copy).
Reshape x in Fortran order to avoid the overhead of another indexing operation on y (although subsequent timings suggest this speedup is negligible)
So rewritten the multiplication becomes:
>>> (x.reshape((2, -1), order='f') * y).ravel('f')
array([ 2, 10, 12, 8, 54, 6])
Timings:
>>> %timeit (y[:, np.newaxis] * x.reshape((-1, n))).flatten()
100000 loops, best of 3: 7.4 µs per loop
>>> %timeit (x.reshape((n, -1), order='f') * y).ravel('f')
100000 loops, best of 3: 4.98 µs per loop
x of length 1000000 and y of length 500000 suggests it's faster by a factor of 10 (much more that I'd imagined). Most of the speedup comes from using ravel().