Triple for-loop in a vector

Question

I have a simple numpy array (3xN) like:

v = np.array([[-3.33829, -3.42467, -3.53332],
   [-2.67681, -2.6082 , -3.49502],
   [-3.49497, -2.73177, -2.61499],
   [-2.76056, -3.57753, -2.67334],
   [-1.96801, -3.47521, -3.51974],
   [-1.25571, -2.69451, -3.45554],
   [-1.94568, -2.59504, -2.72568],
   [-1.28991, -3.47927, -2.73176],
   [-0.51201, -3.50684, -3.40448],
   [ 0.22398, -2.70244, -3.43421]])

Here, N = 10, but it is much larger than here (+500) in my real case. Each row is a point – Euclidean coordinate.

I would like to carry out:

where i, j and k indicate different rows from v.

How can I implement it on Python in a fast way?

Asked By: pmoreira

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Source

Answer 1

You can do this using numpy broadcasting operations:

diffs = ((v[:, None] - v) ** 2).sum(-1)
d = np.exp(diffs + diffs[:, None]).sum((0, 1))
print(d)

# [3.08316899e+11 2.37020625e+07 4.05357364e+12 8.22697743e+08
#  8.85209202e+04 2.55340202e+05 7.33879459e+04 1.88175133e+05
#  8.10134295e+08 6.62122925e+12]

Even for an array of size 500, the result is computed in just a few seconds:

%%time
v = np.random.rand(500, 3)
diffs = np.sum((v[:, None] - v) ** 2, -1)
d = np.exp(diffs + diffs[:, None]).sum((0, 1))

# CPU times: user 2.74 s, sys: 5.5 ms, total: 2.75 s
# Wall time: 2.75 s

Answered By: jakevdp

Answer 2

IIUC, the equation suggests pairwise vector differences, and not squared distance between vectors.

The pairwise difference between N vectors will be N*N vectors.

Finally, I would assume since you are only reducing over j and k axes, the output vector is (10,3) and not (10,). Do correct me if I am wrong.

import numpy as np

d = np.exp(((v[:,None]-v)**2)[:,None] + ((v[:,None]-v)**2)).sum((0,1))
print(d)

#### Stepwise breakdown
#v                           #i,3 -> 10,3
#diff = (v[:,None]-v)**2     #j,i,3 -> 10,10,3
#power = diff[:,None]+diff   #k,j,i,3 -> 10,10,10,3
#exp = np.exp(power)         #k,j,i,3 -> 10,10,10,3
#d = np.sum(exp,(1,2))       #i,3 -> 10,3

array([[4.38558108e+11, 2.11224470e+02, 2.08153285e+02],
       [6.10332697e+09, 2.42309774e+02, 2.00079357e+02],
       [1.37237360e+12, 2.11552094e+02, 2.32739462e+02],
       [9.98934092e+09, 2.51158071e+02, 2.16562340e+02],
       [1.77827910e+08, 2.22151678e+02, 2.05163797e+02],
       [1.91234145e+08, 2.19457894e+02, 1.92858561e+02],
       [1.63391357e+08, 2.46419838e+02, 2.04498335e+02],
       [1.67512751e+08, 2.23119070e+02, 2.03232700e+02],
       [8.45322705e+09, 2.30065042e+02, 1.85024981e+02],
       [1.14468558e+12, 2.17683864e+02, 1.89388595e+02]])

Benchmark –

%%timeit
np.exp(((v[:,None]-v)**2)[:,None] + ((v[:,None]-v)**2)).sum((0,1))
# 21.2 s ± 3.27 s per loop (mean ± std. dev. of 7 runs, 1 loop each)

Answered By: Akshay Sehgal

Triple for-loop in a vector

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