pytorch3d/pytorch3d/loss/mesh_edge_loss.py
Patrick Labatut af93f34834 License lint codebase
Summary: License lint codebase

Reviewed By: theschnitz

Differential Revision: D29001799

fbshipit-source-id: 5c59869911785b0181b1663bbf430bc8b7fb2909
2021-06-22 03:45:27 -07:00

51 lines
1.9 KiB
Python

# Copyright (c) Facebook, Inc. and its affiliates.
# All rights reserved.
#
# This source code is licensed under the BSD-style license found in the
# LICENSE file in the root directory of this source tree.
import torch
def mesh_edge_loss(meshes, target_length: float = 0.0):
"""
Computes mesh edge length regularization loss averaged across all meshes
in a batch. Each mesh contributes equally to the final loss, regardless of
the number of edges per mesh in the batch by weighting each mesh with the
inverse number of edges. For example, if mesh 3 (out of N) has only E=4
edges, then the loss for each edge in mesh 3 should be multiplied by 1/E to
contribute to the final loss.
Args:
meshes: Meshes object with a batch of meshes.
target_length: Resting value for the edge length.
Returns:
loss: Average loss across the batch. Returns 0 if meshes contains
no meshes or all empty meshes.
"""
if meshes.isempty():
return torch.tensor(
[0.0], dtype=torch.float32, device=meshes.device, requires_grad=True
)
N = len(meshes)
edges_packed = meshes.edges_packed() # (sum(E_n), 3)
verts_packed = meshes.verts_packed() # (sum(V_n), 3)
edge_to_mesh_idx = meshes.edges_packed_to_mesh_idx() # (sum(E_n), )
num_edges_per_mesh = meshes.num_edges_per_mesh() # N
# Determine the weight for each edge based on the number of edges in the
# mesh it corresponds to.
# TODO (nikhilar) Find a faster way of computing the weights for each edge
# as this is currently a bottleneck for meshes with a large number of faces.
weights = num_edges_per_mesh.gather(0, edge_to_mesh_idx)
weights = 1.0 / weights.float()
verts_edges = verts_packed[edges_packed]
v0, v1 = verts_edges.unbind(1)
loss = ((v0 - v1).norm(dim=1, p=2) - target_length) ** 2.0
loss = loss * weights
return loss.sum() / N