performing outer addition with numpy

Question

Sorry if this is a silly question but I am just getting started with python/numpy and I’m really not sure of the most efficient ways to go about things. I’m putting together a demo N-body simulator for some students, but for now, I am computing the force between particles by looping over the positions of those particles which is predictably as slow as molasses. Basically, given a vector x[i], I would like to compute:

n[i] = sum from j = 0 to n-1, j != i of (x[i]-x[j])^-2,

using numpy functions rather than looping. If there is a way to perform outer addition/multiplication:

m[i,j] = x[i]-x[j],

m[i,j] = x[i]*x[j],

I could use that to do the computation.

Asked By: Base

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Answer 1

Many of numpy’s basic operators such as np.add, np.subtract, np.multiply etc. are known as universal functions (ufuncs), and have (amongst other things) an .outer method:

import numpy as np
a = np.arange(3)
b = np.arange(5)
c = np.add.outer(a, b)
print(repr(c))
# array([[0, 1, 2, 3, 4],
#        [1, 2, 3, 4, 5],
#        [2, 3, 4, 5, 6]])

Another very powerful technique for doing this sort of thing is broadcasting:

print(repr(a[:, None] + b[None, :]))
# array([[0, 1, 2, 3, 4],
#        [1, 2, 3, 4, 5],
#        [2, 3, 4, 5, 6]])

Indexing with None (or alternatively, with np.newaxis) inserts a new dimension, so a[:, None] has shape (3, 1) and b[None, :] has shape (1, 5). Broadcasting “expands” the result along the singleton dimensions, so that it has shape (3, 5).

Answered By: ali_m

Answer 2

All universal functions that take two input arguments have an attribute outer:

x = np.array([1, 2, 3])
np.subtract.outer(x, x)

gives:

array([[ 0, -1, -2],
       [ 1,  0, -1],
       [ 2,  1,  0]])

and

np.multiply.outer(x, x)

results in:

array([[1, 2, 3],
       [2, 4, 6],
       [3, 6, 9]])

Answered By: Mike Müller

performing outer addition with numpy

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