用numpy计算欧氏距离

Question

我有一个点集,我将其坐标存储在三个不同的数组(xa,ya,za)中.现在,我想用另一个点集(xb,yb,zb)的所有点来计算此点集(xa [0],ya [0],za [0]等)的每个点之间的欧氏距离. )并且每次都在新阵列中存储最小距离.

假设xa.shape =(11,),ya.shape =(11,),za.shape =(11,).分别为xb.shape =(13,),yb.shape =(13,),zb.shape =(13,).我想要做的是每次使用一个xa [],ya [],za [],并计算它与xb,yb,zb的所有元素的距离,并在最后将最小值存储到xfinal中. shape =(11,)数组.

你认为这可能与numpy有关吗？

Answer 1

另一种解决方案是使用scipy的空间模块,特别是KDTree.

本课程从一组数据中学习,并可在给定新数据集的情况下进行查询:

from scipy.spatial import KDTree
# create some fake data
x = arange(20)
y = rand(20)
z = x**2
# put them togheter, should have a form [n_points, n_dimension]
data = np.vstack([x, y, z]).T
# create the KDTree
kd = KDTree(data)

现在,如果你有一个点,你可以通过这样做来询问壁橱点(或N个最近点)的距离和索引:

kd.query([1, 2, 3])
# (1.8650720813822905, 2)
# your may differs

或者,给定一系列职位:

#bogus position
x2 = rand(20)*20
y2 = rand(20)*20
z2 = rand(20)*20
# join them togheter as the input
data2 = np.vstack([x2, y2, z2]).T
#query them
kd.query(data2)

#(array([ 14.96118553,   9.15924813,  16.08269197,  21.50037074,
#    18.14665096,  13.81840533,  17.464429  ,  13.29368755,
#    20.22427196,   9.95286671,   5.326888  ,  17.00112683,
#     3.66931946,  20.370496  ,  13.4808055 ,  11.92078034,
#     5.58668204,  20.20004206,   5.41354322,   4.25145521]),
#array([4, 3, 2, 4, 2, 2, 4, 2, 3, 3, 2, 3, 4, 4, 3, 3, 3, 4, 4, 4]))