Claudiu - 2 months ago 10
Python Question

# How to efficiently apply an operator to the cartesian product of two arrays?

I have

`a = array([1, 2, 3, 4, 5])`
and
`b = array([10, 20, 30, 40, 50])`
. I want:

``````array([[ -9, -19, -29, -39, -49],
[ -8, -18, -28, -38, -48],
[ -7, -17, -27, -37, -47],
[ -6, -16, -26, -36, -46],
[ -5, -15, -25, -35, -45]])
``````

What's the most efficient way to do this? I have

``````np.transpose([a]) - np.tile(b, (len(a), 1))
``````

However I wonder if there's a more efficient way than this if
`a`
is very large, which wouldn't require copying
`b`
so much (or vice versa).

Some NumPy functions, like `np.subtract`, `np.add`, `np.multiply`, `np.divide`, `np.logical_and`, `np.bitwise_and`, etc. have an "outer" method which can be used to create the equivalent of "multiplication tables":

``````In [76]: np.subtract.outer(a, b)
Out[76]:
array([[ -9, -19, -29, -39, -49],
[ -8, -18, -28, -38, -48],
[ -7, -17, -27, -37, -47],
[ -6, -16, -26, -36, -46],
[ -5, -15, -25, -35, -45]])
``````

``````In [96]: a[:, None] - b
Out[96]:
array([[ -9, -19, -29, -39, -49],
[ -8, -18, -28, -38, -48],
[ -7, -17, -27, -37, -47],
[ -6, -16, -26, -36, -46],
[ -5, -15, -25, -35, -45]])
``````

The performance of the two is about the same:

``````In [98]: a = np.tile(a, 1000)

In [99]: b = np.tile(b, 1000)

In [100]: %timeit a[:, None] - b
10 loops, best of 3: 88.3 ms per loop

In [101]: %timeit np.subtract.outer(a, b)
10 loops, best of 3: 87.8 ms per loop

In [102]: %timeit np.transpose([a]) - np.tile(b, (len(a), 1))
10 loops, best of 3: 120 ms per loop
``````