在Numpy中，找到两个阵列中每对之间的欧几里德距离

Question

I have two arrays of 2D coordinate points (x,y)

a = [ (x1,y1), (x2,y2), ... (xN,yN) ]
b = [ (X1,Y1), (X2,Y2), ... (XN,YN) ]

How can I find the Euclidean distances between each aligned pairs (xi,yi) to (Xi,Yi) in an 1xN array?

The scipy.spatial.cdist function gives me distances between all pairs in an NxN array.

If I just use norm function to calculate the distance one by one it seems to be slow.

Is there a built in function to do this?

Answer 1

I'm not seeing a built-in, but you could do it yourself pretty easily.

distances = (a-b)**2
distances = distances.sum(axis=-1)
distances = np.sqrt(distances)

Answer 2

hypot is another valid alternative

a, b = randn(10, 2), randn(10, 2)
ahat, bhat = (a - b).T
r = hypot(ahat, bhat)

Result of timeit s between manual calculation and hypot :

Manual:

timeit sqrt(((a - b) ** 2).sum(-1))
100000 loops, best of 3: 10.3 µs per loop

Using hypot :

timeit hypot(ahat, bhat)
1000000 loops, best of 3: 1.3 µs per loop

Now how about some adult-sized arrays:

a, b = randn(1e7, 2), randn(1e7, 2)
ahat, bhat = (a - b).T

timeit -r10 -n3 hypot(ahat, bhat)
3 loops, best of 10: 208 ms per loop

timeit -r10 -n3 sqrt(((a - b) ** 2).sum(-1))
3 loops, best of 10: 224 ms per loop

Not much of a performance difference between the two methods. You can squeeze out a tiny bit more from the latter by avoiding pow :

d = a - b

timeit -r10 -n3 sqrt((d * d).sum(-1))
3 loops, best of 10: 184 ms per loop