Question

假设我有两套要点：

>>> points1.shape
(10000, 3)
>>> points2.shape
(1529, 3)

我想在points1中某点的欧氏距离cutoff内找到points2的索引列表。我可以使用scipy.spatial.distance.cdist轻松完成此操作：

from scipy.spatial.distance import cdist
import numpy

indices = numpy.argwhere(cdist(points1, points2).min(axis=0) < cutoff)

然而，这似乎效率低下，因为我不需要知道彼此之间有多远，只要它们是否在一个截止距离内。 KDTree可以帮助解决这个问题吗？

Answer 1

以下是3个替代方案，一个使用cdist，两个使用scipy.spatial.cKDTree：

import itertools as IT
import numpy as np
import scipy.spatial as spatial
import scipy.spatial.distance as dist
np.random.seed(2016)
points1 = np.random.randint(100, size=(10**5, 3))
points2 = np.random.randint(100, size=(1529, 3))
cutoff = 5

def using_cdist(points1, points2, cutoff):
    indices = np.where(dist.cdist(points1, points2) <= cutoff)[0]
    indices = np.unique(indices)
    return indices

def using_kdtree(points1, points2, cutoff):
    # build the KDTree using the *smaller* points array
    tree = spatial.cKDTree(points2)
    groups = tree.query_ball_point(points1, cutoff)
    indices = np.unique([i for i, grp in enumerate(groups) if len(grp)])
    return indices

def using_kdtree2(points1, points2, cutoff):
    # build the KDTree using the *larger* points array
    tree = spatial.cKDTree(points1)
    groups = tree.query_ball_point(points2, cutoff)
    indices = np.unique(IT.chain.from_iterable(groups))
    return indices

cdist_result = using_cdist(points1, points2, cutoff)
kdtree_result = using_kdtree(points1, points2, cutoff)
kdtree_result2 = using_kdtree2(points1, points2, cutoff)
assert np.allclose(cdist_result, kdtree_result)
assert np.allclose(cdist_result, kdtree_result2)

在这3个备选方案中，using_kdtree2是最快的：

In [80]: %timeit using_kdtree3(points1, points2, cutoff)
10 loops, best of 3: 92.4 ms per loop

In [103]: %timeit using_kdtree(points1, points2, cutoff)
1 loops, best of 3: 938 ms per loop

In [104]: %timeit using_cdist(points1, points2, cutoff)
1 loops, best of 3: 1.51 s per loop

我对最快速度的直觉证明是完全错误的。一世以为使用较小的点阵列构建KDTree会是最快的。即使使用更大的点阵列构建KDTree也是如此有点慢，在较小的点数组上调用tree.query_ball_point 更快：

In [68]: %timeit tree = spatial.cKDTree(points2)
1000 loops, best of 3: 312 µs per loop

In [69]: %timeit tree = spatial.cKDTree(points1)
10 loops, best of 3: 45.7 ms per loop

In [66]: %timeit tree = spatial.cKDTree(points2); groups = tree.query_ball_point(points1, cutoff)
1 loops, best of 3: 933 ms per loop

In [67]: %timeit tree = spatial.cKDTree(points1); groups = tree.query_ball_point(points2, cutoff)
10 loops, best of 3: 89.3 ms per loop

请注意，使用

时存在一些问题

def orig(points1, points2, cutoff):
    return np.argwhere(dist.cdist(points1, points2).min(axis=0) < cutoff)

首先，通过调用min(axis=0)，如果points1中有两个点，则会丢失信息都在cutoff中的points2点内min。你只会获得索引最接近的一点。另一个问题是通过调用points2 0轴，剩下的全部是1轴，与orig相关联。所以 points2将索引返回points1，而不是library(tidyr) library(dplyr) # reshape data to long format td <- d %>% gather(key, value, value1:value4) # create a copy w/ different names for merging td2 <- td %>% select(id2 = id, key, value2 = value) # full outer join to produce one row per pair of IDs dd <- merge(td, td2, by = "key", all = TRUE) # the result dd %>% filter(id != id2) %>% group_by(id, id2) %>% summarise(all_less = !any(value >= value2)) %>% filter(all_less)。

Answer 2

一些想法（？）：

如果您不需要知道距离，可以通过比较距离的平方与阈值的平方来保存平方根的计算（mServiceRestartEmitter.take(1).subscribe(action);将计算平方根）。
第一次排除x坐标已经超过阈值的点，然后与y和z相同，将节省一些计算。特别是因为＆＃39;或＆＃39;在Python中是懒惰的（如果x已经足够远，它甚至不会检查y）。

用scipy在其他点的截止距离内找到点

2 个答案: