使用python中的内置函数查找3d距离

Question

I have 6 lists storing x,y,z coordinates of two sets of positions (3 lists each). I want to calculate the distance between each point in both sets. I have written my own distance function but it is slow. One of my lists has about 1 million entries. I have tried cdist, but it produces a distance matrix and I do not understand what it means. Is there another inbuilt function that can do this?

Answer 1

If possible, use the numpy module to handle this kind of things. It is a lot more efficient than using regular python lists.

I am interpreting your problem like this

1. You have two sets of points
2. Both sets have the same number of points ( N )
3. Point k in set 1 is related to point k in set 2. If each point is the coordinate of some object, I am interpreting it as set 1 containing the initial point and set 2 the point at some other time t.
4. You want to find the distance d(k) = dist(p1(k), p2(k)) where p1(k) is point number k in set 1 and p2(k) is point number k in set 2.

Assuming that your 6 lists are x1_coords , y1_coords , z1_coords and x2_coords , y2_coords , z2_coords respectively, then you can calculate the distances like this

import numpy as np
p1 = np.array([x1_coords, y1_coords, z1_coords])
p2 = np.array([x2_coords, y2_coords, z2_coords])

squared_dist = np.sum((p1-p2)**2, axis=0)
dist = np.sqrt(squared_dist)

The distance between p1(k) and p2(k) is now stored in the numpy array as dist[k] .

As for speed: On my laptop with a "Intel(R) Core(TM) i7-3517U CPU @ 1.90GHz" the time to calculate the distance between two sets of points with N=1E6 is 45 ms.

Answer 2

Although this solution uses numpy , np.linalg.norm could be another solution.

Say you have one point p0 = np.array([1,2,3]) and a second point p1 = np.array([4,5,6]) . Then the quickest way to find the distance between the two would be:

dist = np.linalg.norm(p0 - p1)

Answer 3

# Use the distance function in Cartesian 3D space:
# Example
import math     
def distance(x1, y1, z1, x2, y2, z2):
d = 0.0
d = math.sqrt((x2 - x1)**2 + (y2 - y1)**2 + (z2 - z1)**2)
return d

Answer 4

您可以使用math.dist(A, B) ，其中 A 和 B 是坐标数组