CODEWITHSUNDEEP
math3dasciimeshmetrics
Mec-Eng
Mec-Eng

Reputation: 199

Alternative metric for Hausdorff distance

For my project, I need to measure the distance between two 3D meshes based on OBJ-Files. I have to implement two different metrics and compare them. In the course of my literature research, I have so far found only the Hausdorff distance as a metric. Apparently, the Hausdorff distance can be used to calculate the distance of 3D meshes.

Is there an adequate alternative for the Hausdorff distance?

This topic is similiar to mine, but i want to implement two different metrics. Measure distance between meshes

Upvotes: 0

Views: 1151

Answers (1)

Rafael Valero
Rafael Valero

Reputation: 2816

Many. Depends on your case.

Hausdorff distance is "it is the greatest of all the distances from a point in one set to the closest point in the other set." from wikipedia

Consider the below example of two sets (u and v) in 2 dimensions:

from scipy.spatial.distance import directed_hausdorff
import numpy as np


u = np.array([(1.0, 0.0),
              (0.0, 1.0),
              (-1.0, 0.0),
              (0.0, -1.0)])
v = np.array([(2.0, 0.0),
              (0.0, 2.0),
              (-2.0, 0.0),
              (0.0, -4.0)])

print(directed_hausdorff(u, v))
(2.23606797749979, 3, 0)

Depending on the group you have: 2.23606797749979 or 3.

Going back to the definition I can easily reproduce that results using euclidian distance.

print(euclidean_distances(u, v).min(axis = 0).max(axis = 0))
print(euclidean_distances(u, v).min(axis = 1).max(axis = 0))
3.0
2.23606797749979

Let have a look to all the distances between all the points of the two sets:

print(euclidean_distances(u, v))
[[1.         2.23606798 3.         4.12310563]
 [2.23606798 1.         2.23606798 5.        ]
 [3.         2.23606798 1.         4.12310563]
 [2.23606798 3.         2.23606798 3.        ]]

As you can see the sortest distance is 1 and the longest 5 for instance. I could formalize that as follow:

print(np.max(euclidean_distances(u, v)))
print(np.min(euclidean_distances(u, v)))
5
1

I could take the average, too:

print(np.mean(euclidean_distances(u, v)))
2.603913694764629

As you see, you have different alternatives there.

Upvotes: 1

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