Claudiu - 1 year ago 57

Python Question

I have

`a = array([1, 2, 3, 4, 5])`

`b = array([10, 20, 30, 40, 50])`

`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`

`b`

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Answer Source

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]])
```

or, using broadcasting:

```
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
```