query(self, x, k=1, eps=0, p=2, distance_upper_bound=inf, workers=1)
An array of points to query.
Either the number of nearest neighbors to return, or a list of the k-th nearest neighbors to return, starting from 1.
Return approximate nearest neighbors; the kth returned value is guaranteed to be no further than (1+eps) times the distance to the real kth nearest neighbor.
Which Minkowski p-norm to use. 1 is the sum-of-absolute-values distance ("Manhattan" distance). 2 is the usual Euclidean distance. infinity is the maximum-coordinate-difference distance. A large, finite p may cause a ValueError if overflow can occur.
Return only neighbors within this distance. This is used to prune tree searches, so if you are doing a series of nearest-neighbor queries, it may help to supply the distance to the nearest neighbor of the most recent point.
Number of workers to use for parallel processing. If -1 is given all CPU threads are used. Default: 1.
The distances to the nearest neighbors. If x
has shape tuple+(self.m,)
, then d
has shape tuple+(k,)
. When k == 1, the last dimension of the output is squeezed. Missing neighbors are indicated with infinite distances. Hits are sorted by distance (nearest first).
If k=None
, then d
is an object array of shape tuple
, containing lists of distances. This behavior is deprecated and will be removed in SciPy 1.8.0, use query_ball_point
instead.
The index of each neighbor in self.data
. i
is the same shape as d. Missing neighbors are indicated with self.n
.
Query the kd-tree for nearest neighbors.
>>> import numpy as np
... from scipy.spatial import KDTree
... x, y = np.mgrid[0:5, 2:8]
... tree = KDTree(np.c_[x.ravel(), y.ravel()])
To query the nearest neighbours and return squeezed result, use
>>> dd, ii = tree.query([[0, 0], [2.2, 2.9]], k=1)
... print(dd, ii, sep='\n') [2. 0.2236068] [ 0 13]
To query the nearest neighbours and return unsqueezed result, use
>>> dd, ii = tree.query([[0, 0], [2.2, 2.9]], k=[1])
... print(dd, ii, sep='\n') [[2. ] [0.2236068]] [[ 0] [13]]
To query the second nearest neighbours and return unsqueezed result, use
>>> dd, ii = tree.query([[0, 0], [2.2, 2.9]], k=[2])
... print(dd, ii, sep='\n') [[2.23606798] [0.80622577]] [[ 6] [19]]
To query the first and second nearest neighbours, use
>>> dd, ii = tree.query([[0, 0], [2.2, 2.9]], k=2)
... print(dd, ii, sep='\n') [[2. 2.23606798] [0.2236068 0.80622577]] [[ 0 6] [13 19]]
or, be more specific
>>> dd, ii = tree.query([[0, 0], [2.2, 2.9]], k=[1, 2])See :
... print(dd, ii, sep='\n') [[2. 2.23606798] [0.2236068 0.80622577]] [[ 0 6] [13 19]]
The following pages refer to to this document either explicitly or contain code examples using this.
scipy.spatial._kdtree.KDTree.query
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