Python中的平均最近坐标

Question

This is the continuation of my previous question. I have now a sorted list of coordinates in a Euclidean space. I want to average the closest coordinates in such a way that clustering works ie a whole cluster is averaged and returns one single point in Euclidean space. So, for example the list below

a = [[ 42, 206],[ 45,  40],[ 45, 205],[ 46,  41],[ 46, 205],[ 47,  40],[ 47, 202],[ 48,  40],[ 48, 202],[ 49,  38]]

will return avg_coordinates = [[47.0, 39.8], [45.6, 204.0]] . This is done by averaging first 5 closest points (or cluster) and then last 5 closest points. Right now I am using gradient approach that is I am looping through all coordinates and wherever the gradient is higher then some set threshold then I consider it another cluster of points (because list is already sorted). But problem arise when I have higher denominator then numerator in the gradient formula gradient = (y2-y1)/(x2-x1) which return a smaller value then threshold. So logically I am doing it wrong. Any good suggestions for doing this ? Please note I do not want to apply clustering.

Answer 1

Here's an approach -

thresh = 100 # Threshold for splitting, heuristically chosen for given sample

# Lex-sort of coordinates
b = a[np.lexsort(a.T)]

# Interval indices that partition the clusters
diff_idx = np.flatnonzero(np.linalg.norm(b[1:] - b[:-1],axis=1) > thresh)+1
idx = np.hstack((0, diff_idx, b.shape[0]))
sums = np.add.reduceat(b, idx[:-1])
counts = idx[1:] - idx[:-1]
out = sums/counts.astype(float)[:,None]

Sample input, output -

In [141]: a
Out[141]: 
array([[ 42, 206],
       [ 45,  40],
       [ 45, 205],
       [ 46,  41],
       [ 46, 205],
       [ 47,  40],
       [ 47, 202],
       [ 48,  40],
       [ 48, 202],
       [ 49,  38]])

In [142]: out
Out[142]: 
array([[  47. ,   39.8],
       [  45.6,  204. ]])

Answer 2

如果您乐意使用库而不是重新实现群集，则可以使用scikit-learn中的k-means ： http ：//scikit-learn.org/stable/modules/generated/sklearn.cluster.KMeans.html