faster way of getting difference between each element of 2 numpy arrays

Question

I have 2 numpy arrays from which I am trying to find the difference for each element pair and store the difference in a matrix.

Here is the code used by me:

for i in range(arr1):
    for j in range(arr2):
        data[i,j] = float(arr1[i])-float(arr2[j])

what can be done to optimize the speed of this loop?

you can use np.subtract.outer(arr1, arr2) or broadcasting or np.subtract(*np.ix_(arr1, arr2)); broadcasting: arr1[:, None] - arr2[None, :] — Paul Panzer
– Paul Panzer, Commented Apr 11, 2017 at 6:01
arr1_arr2_transformation = numpy.dstack(-arr1, arr2) then do numpy.sum(arr1_arr2_transformation) — FancyDolphin
– FancyDolphin, Commented Apr 11, 2017 at 6:09
@FancyDolphin are you sure? First of all numpy.dstack takes only one argument returns an array. Summin of it yields a scalar. I dont see how this is supposed to work. — greole
– greole, Commented Apr 11, 2017 at 6:58

greole · Accepted Answer · 2017-04-11 06:42:41Z

As pointed out in the comments there are several ways to reach your goal.

In [1]: import numpy as np
In [6]: a = np.random.rand(1000)
In [7]: b = np.random.rand(1000)

In [9]: %timeit a - b.reshape((-1,1))
100 loops, best of 3: 2.46 ms per loop

In [10]: %timeit np.subtract.outer(a, b)
100 loops, best of 3: 2.52 ms per loop

It seems that the reshape and the subtract.outer are comparable in speed. However it looks like you need to transpose the result in order to have identical results for both methods

In [18]: a - b.reshape((-1,1)) == np.subtract.outer(a, b).T

array([[ True,  True,  True, ...,  True,  True,  True],
       [ True,  True,  True, ...,  True,  True,  True],
       [ True,  True,  True, ...,  True,  True,  True],
       ..., 
       [ True,  True,  True, ...,  True,  True,  True],
       [ True,  True,  True, ...,  True,  True,  True],
       [ True,  True,  True, ...,  True,  True,  True]], dtype=bool)

Edit The second method proposed by @PaulPanzer seems to be the slowest.

In [27]: %timeit np.subtract(*np.ix_(a, b)); a[:, None] - b[None, :]
100 loops, best of 3: 4.99 ms per loop

deathracer · Accepted Answer · 2017-04-11 08:05:47Z

0

np.subtract.outer(arr1, arr2) helped me solved the problem.

Thanks everyone

answered Apr 11, 2017 at 8:05

deathracer

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