add option to retrieve origin data from point cloud generation

[dmitrishastin]

Type

• Bug fix (non-breaking change which fixes an issue): Fixes #
• New feature (non-breaking change which adds functionality). Resolves #
• Breaking change (fix or feature that would cause existing functionality to not work as expected) Resolves #

Motivation and Context

Record the origin of points in the data used for generating those points. In my context, this is is a triangular mesh where I wish to record triangle ID and UV coordinates per point, to then relate a number functions on the same surface to the sampled point cloud.

Tried to code such that this pull request is: (1) non-breaking / does not slow existing functionality, (2) easily scalable so other origin data types (e.g., voxel ID / relative coordinates) can be added.

However, I am not fluent in C++ and not sure if the proposed implementation is reasonable. Further, I wasn't sure this suggested functionality will be considered useful / accepted. As such, I only added the option to record origin data data in

• PointCloud::operator+=
• TriangleMesh::SamplePointsUniformlyImpl
• TriangleMesh::SamplePointsUniformly
• TriangleMesh::SamplePointsPoissonDisk

Checklist:

• I have run python util/check_style.py --apply to apply Open3D code style
  to my code.
• This PR changes Open3D behavior or adds new functionality.
  • Both C++ (Doxygen) and Python (Sphinx / Google style) documentation is
    updated accordingly.
  • I have added or updated C++ and / or Python unit tests OR included test
    results (e.g. screenshots or numbers) here.
• I will follow up and update the code if CI fails.
• For fork PRs, I have selected Allow edits from maintainers.

Description

import open3d as o3d
import numpy as np
cube = o3d.geometry.TriangleMesh.create_box()                                                
pts = cube.sample_points_uniformly(4000)     
pts.origin_data()
{'tri_id': [], 'tri_uv': std::vector<Eigen::Vector2d> with 0 elements.
Use numpy.asarray() to access data.}

pts.has_origin_data()                                                                                            
False
                                                                                                                
pts = cube.sample_points_uniformly(4000, record_origin=True)
pts.has_origin_data()
True

pts.origin_data().keys()
dict_keys(['tri_id', 'tri_uv'])

np.asarray(pts.origin_data()['tri_uv'])                                                                          
array([[0.65370755, 0.25589466],                                                                                            
       [0.81755558, 0.14071095],                                                                                            
       [0.5513747 , 0.32234846],                                                                                            
       ...,                                                                                                                 
       [0.96641095, 0.02235785],                                                                                            
       [0.29790422, 0.39937283],                                                                                            
       [0.70699849, 0.17731053]])

v = np.asarray(cube.vertices) 
t = np.asarray(cube.triangles)
tidx = np.array(pts.origin_data()['tri_id']) # simple list
uv = np.asarray(pts.origin_data()['tri_uv'])
w = 1 - np.sum(uv, axis = 1)
c = v[t[tidx, 0], :] * uv[:, 0][:, None] + v[t[tidx, 1], :] * uv[:, 1][:, None] + v[t[tidx, 2], :] * w[:, None] # point coordinates from barycentric coordinates
np.allclose(c, np.asarray(pts.points))
True

[update-docs]

Thanks for submitting this pull request! The maintainers of this repository would appreciate if you could update the CHANGELOG.md based on your changes.

[dmitrishastin]

Afraid I am not certain regarding the Codacy findings above (under tests) - I thought I applied logic similar to other Boolean flags (e.g., use_triangle_normal), but it does not seem to like it. Will be grateful for advice.