Sklearn kNN用法与用户定义的指标 [英] Sklearn kNN usage with a user defined metric

问题描述

目前，我正在做一个项目，可能需要使用kNN算法来查找给定点的前k个最近的邻居，例如使用python和sklearn包的P. im来完成这项工作，但我们的预定义指标不是这些默认指标.因此我必须使用sklearn文档中的用户定义指标，可以在此处.

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 the job, but our predefined metric is not one of those default metrics. so I have to use the user defined metric, from the documents of sklearn, which can be find here and here.

似乎sklearn kNN的最新版本支持用户定义的指标，但是我找不到如何使用它:

It seems that the latest version of sklearn kNN support the user defined metric, but i cant find how to use it:

import sklearn
from sklearn.neighbors import NearestNeighbors
import numpy as np
from sklearn.neighbors import DistanceMetric
from sklearn.neighbors.ball_tree import BallTree
BallTree.valid_metrics

说我已经定义了一个名为mydist = max(x-y)的度量，然后使用DistanceMetric.get_metric使其成为DistanceMetric对象:

say i have defined a metric called mydist=max(x-y), then use DistanceMetric.get_metric to make it a DistanceMetric object:

dt=DistanceMetric.get_metric('pyfunc',func=mydist)

在文档中，该行应如下所示

from the document, the line should looks like this

nbrs = NearestNeighbors(n_neighbors=4, algorithm='auto',metric='pyfunc').fit(A)
distances, indices = nbrs.kneighbors(A)

但是我可以在哪里放置dt?谢谢

but where can i put the dt in? Thanks

推荐答案

您将指标作为metric参数传递，并将其他指标参数作为关键字参数传递给NN构造函数:

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]]))

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