Sklearn kNN usage with a user defined metric

Question

Currently I\'m doing a project which may require using a kNN algorithm to find the top k nearest neighbors for a given point, say P. im using python, sklearn package to do

Answer 1

A small addition to the previous answer. How to use a user defined metric that takes additional arguments.

>>> def mydist(x, y, **kwargs):
...     return np.sum((x-y)**kwargs["metric_params"]["power"])
...
>>> X = np.array([[-1, -1], [-2, -1], [-3, -2], [1, 1], [2, 1], [3, 2]])
>>> Y = np.array([-1, -1, -2, 1, 1, 2])
>>> nbrs = KNeighborsClassifier(n_neighbors=4, algorithm='ball_tree',
...            metric=mydist, metric_params={"power": 2})
>>> nbrs.fit(X, Y)
KNeighborsClassifier(algorithm='ball_tree', leaf_size=30,                                                                                                                                                          
       metric=<function mydist at 0x7fd259c9cf50>, n_neighbors=4, p=2,
       weights='uniform')
>>> nbrs.kneighbors(X)
(array([[  0.,   1.,   5.,   8.],
       [  0.,   1.,   2.,  13.],
       [  0.,   2.,   5.,  25.],
       [  0.,   1.,   5.,   8.],
       [  0.,   1.,   2.,  13.],
       [  0.,   2.,   5.,  25.]]),
 array([[0, 1, 2, 3],
       [1, 0, 2, 3],
       [2, 1, 0, 3],
       [3, 4, 5, 0],
       [4, 3, 5, 0],
       [5, 4, 3, 0]]))

Answer 2

You pass a metric as metric param, and additional metric arguments as keyword paramethers to NN constructor:

>>> def mydist(x, y):
...     return np.sum((x-y)**2)
...
>>> X = np.array([[-1, -1], [-2, -1], [-3, -2], [1, 1], [2, 1], [3, 2]])

>>> nbrs = NearestNeighbors(n_neighbors=4, algorithm='ball_tree',
...            metric='pyfunc', func=mydist)
>>> nbrs.fit(X)
NearestNeighbors(algorithm='ball_tree', leaf_size=30, metric='pyfunc',
         n_neighbors=4, radius=1.0)
>>> nbrs.kneighbors(X)
(array([[  0.,   1.,   5.,   8.],
       [  0.,   1.,   2.,  13.],
       [  0.,   2.,   5.,  25.],
       [  0.,   1.,   5.,   8.],
       [  0.,   1.,   2.,  13.],
       [  0.,   2.,   5.,  25.]]), array([[0, 1, 2, 3],
       [1, 0, 2, 3],
       [2, 1, 0, 3],
       [3, 4, 5, 0],
       [4, 3, 5, 0],
       [5, 4, 3, 0]]))