Numpy - Finding matches across multiple co-ordinat

问题:

I'm using somoclu to produce an emergent Self-Organising Map of some data. Once I have the BMUs (Best Matching Units) I'm performing a Delaunay Triangulation on the co-ordinates of the BMUs in order to find each BMU's neighbours in the SOM.

In the following snippet of Python, is there a more Pythonic version of the a == c and b == d conditional? In other words, how can I compare bmu and point directly without splitting out the separate co-ordinates?

points = np.unique(np.array(som.bmus), axis = 0)
for idx, bmu in enumerate(som.bmus):
    a, b = bmu
    for point_idx, point in enumerate(points):
        c, d = point
        if a == c and b == d: # More Pythonic version of this line?
            print (idx, point_idx)
            break

回答1:

Approach #1

We are working with NumPy arrays, so we can leverage broadcasting for a vectorized solution -

ar = np.array(som.bmus)
points = np.unique(ar, axis = 0)

mask = (ar[:,0,None]==points[:,0]) & (ar[:,1,None]==points[:,1])
indices = np.argwhere(mask)

Approach #1-G

One more compact way to get mask, which covers for a generic no. of columns in ar, would be -

mask = (ar[:,None] == points).all(axis=2)

Approach #2

Another memory-efficient approach for generic no. of cols would be with views and np.searchsorted -

# https://stackoverflow.com/a/45313353/ @Divakar
def view1D(a, b): # a, b are arrays
    a = np.ascontiguousarray(a)
    b = np.ascontiguousarray(b)
    void_dt = np.dtype((np.void, a.dtype.itemsize * a.shape[1]))
    return a.view(void_dt).ravel(),  b.view(void_dt).ravel()

n = len(ar)
indices = np.empty((n,2),dtype=int)
indices[:,0] = np.arange(n)
a,b = view1D(ar, points) # ar, points from app#1
indices[:,1] = np.searchsorted(b, a)

回答2:

With numpy arrays, you can use np.array_equal. This tests for same shape and same elements.

But if your logic is as simple as the code you have, use @Divakar's vectorized solution.

points = np.unique(np.array(som.bmus), axis = 0)

for idx, bmu in enumerate(som.bmus):
    for point_idx, point in enumerate(points):
        if np.array_equal(bmu, point):
            print(idx, point_idx)
            break