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?
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)
np.dot: delta = a-b; dist = np.dot(delta, delta); dist = np.sqrt(dist)
dot vectorizes like that; it computes matrix products for 2-d inputs. You could probably do something with einsum, but I don't know the Einstein summation convention, so it's hard for me to give answers using it.inner1d that does it: import numpy.core.umath_tests as ut; delta = a-b; dist = np.sqrt(dnp.inner1d(delta, delta)). Alternatively dist = np.sqrt(np.einsum('ij, ij->i', delta, delta)).hypot is another valid alternative
a, b = randn(10, 2), randn(10, 2)
ahat, bhat = (a - b).T
r = hypot(ahat, bhat)
Result of timeits 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
try adding [:, np.newaxis, :] to the first parameter
np.linalg.norm(grid[:, np.newaxis, :] - scenario.target, axis=-1)
ref Numpy Broadcast to perform euclidean distance vectorized
