Finding 3d distances using an inbuilt function in python
Solution 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
- You have two sets of points
- Both sets have the same number of points (
N
) - Point
k
in set 1 is related to pointk
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. - You want to find the distance
d(k) = dist(p1(k), p2(k))
wherep1(k)
is point numberk
in set 1 andp2(k)
is point numberk
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.
Solution 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)
Solution 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
Abhinav Kumar
I am a graduate student (Physics) at University of California, Merced.
Updated on October 31, 2021Comments
-
Abhinav Kumar over 2 years
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?
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RattleyCooper about 8 yearsThis doesn't answer the original question. What if you do not have access to numpy?
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Steven C. Howell about 5 yearsIf you do not have NumPy, you would do something like this: stackoverflow.com/a/51665185/3585557. You certainly would want to use NumPy array if you could because lists are not a great data structure for this type of calculation.
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karel over 2 yearsYou certainly would want to use a NumPy array if you could because lists are not a great data structure for this type of calculation.