如何在 Python 中为 3D 顶点列表创建索引列表？

Question

我有这组 3D 顶点。每条线保存 2 个不同点的坐标，因此每条线就像 x1,y1,z1,x2,y2,z2。

0.142000E+03 -.600000E+03 0.262790E+04 -.142000E+03 -.600000E+03 0.262790E+04
-.984000E+02 -.599999E+03 0.258430E+04 0.984000E+02 -.599999E+03 0.258430E+04
0.984000E+02 -.599999E+03 0.258430E+04 0.984000E+02 -.599999E+03 0.238750E+04
0.984000E+02 -.599999E+03 0.258430E+04 -.984000E+02 -.599999E+03 0.258430E+04
-.984000E+02 -.599999E+03 0.238750E+04 0.984000E+02 -.599999E+03 0.238750E+04
-.142000E+03 0.600000E+03 0.262800E+04 0.142000E+03 0.600000E+03 0.262800E+04
0.984000E+02 0.599999E+03 0.258430E+04 -.984000E+02 0.599999E+03 0.258430E+04
0.142000E+03 0.600000E+03 0.262800E+04 0.142000E+03 0.600000E+03 0.234400E+04

我想首先分离每个三元组的顶点，然后将所有三元组合并到一个列表 A 中。然后，我想从这个列表中创建一个由唯一三元组组成的列表 B，并为每个三元组提供一个索引。从那里，我想根据唯一的三元组列表 B 创建 A 中每个元素的索引列表 C。我本质上是想制作一个 .off 文件，其中索引组构成一张脸。

我在Python中尝试了这段代码：

import re

def takeNumFaces(filepath):
    with open(filepath, 'r') as file:
        lines = file.readlines()
        second_line = lines[1].strip()
        numbers = second_line.split()
        faces = int(numbers[1])
    return(faces)

def takeNumPoints(filepath):
    with open(filepath, 'r') as file:
        lines = file.readlines()
        second_line = lines[1].strip()
        numbers = second_line.split()
        points = int(numbers[2])
    return(points)

geofile = "C:/mygeofile.geo"

geofaces = takeNumFaces(geofile)
facepoints = takeNumPoints(geofile)
#print(geofaces, facepoints)

def gatherVertices(filepath):
    vertices = []
    unique_vert = []
    index = 0
    with open(filepath, 'r') as file:
        lines = file.readlines()
        for line in lines[2 : 2*geofaces + 2]:
            parts = re.findall(r'[-0]\.\d+E[+-]\d{2}', line)
            # print(parts)
            for i in range(0, len(parts), 3):
                vertex = [parts[i], parts[i+1], parts[i+2]]
                #print(vertex)
                vertices.append(vertex)
        
    return vertices
                
vertices = gatherVertices(geofile)
#print(vertices)
unique_vertices = set(tuple(vertex) for vertex in vertices)
unique_vertices_list = [list(vertex) for vertex in unique_vertices]
# print(len(unique_vertices), len(vertices))
# for vertex in unique_vertices_list:
#     print(vertex)

def getVertexIndices(vertices1, vertices2):
    # Convert vertices2 to tuples (since unique_vertices_list is already list of lists)
    unique_vertices_tuples = [tuple(vertex) for vertex in vertices2]
    
    # Create a mapping of vertices to indices
    vertex_to_index = {tuple(vertex): index for index, vertex in enumerate(unique_vertices_tuples)}
    # print(vertex_to_index)
    # Get the index of each vertex in vertices1 in vertices2
    index_list = [vertex_to_index[tuple(vertex)] for vertex in vertices1]

    return index_list

index_list = getVertexIndices(vertices, unique_vertices_list)
# for index in index_list:
#     print(index)
# print(len(index_list))

# num = 2000
# if num in index_list:
#     print(f"{num} is in the list")
# else:
#     print(f"{num} is not in the list")

def groupFaces(indices):
    faces = [indices[i:i+4] for i in range(0, len(indices), 4)]
    return faces

face_list = groupFaces(index_list)
# for face in face_list:
#     print(face)

off_file = "C:/myPrivateFilePath"

object_dimension = 3

groupsize = facepoints // object_dimension 

def write_off(offpath):
    with open(offpath, 'w') as file:
        file.write("OFF\n")
        file.write(f"{len(vertices)} {geofaces} 0\n")

        for vertex in vertices:
            file.write(" ".join(vertex) + "\n")

        for face in face_list:
            line = f"{groupsize} " + " ".join(map(str, face)) + "\n"
            file.write(line)


write_off(off_file)

我设法制作了一个三元组列表，例如 [0.142000E+03, -.600000E+03, 0.262790E+04], [-.142000E+03, -.600000E+03, 0.262790E+04] 等。出于某种原因，我的索引被打乱了，我得到了一个以 :

开头的 index_list

4 547 1181 1335 867
4 1181 2243 182 1335
4 2243 787 2275 182
4 787 547 867 2275

所以第一组甚至不是由前 4 个顶点三元组组成的。我应该获取顶点中每个元素（顶点三元组）的索引，然后将其索引（根据 unique_vertices_list）提供给index_list。

如何确保顶点三元组的索引按顺序出现（即顶点的元素 0 是唯一列表的元素 0，因此 index_list 的元素 0 是 0）？

Answer 1

看起来使用

set

来获取唯一的顶点是问题的核心。

set

未订购。这是另一个讨论这个问题的问题。答案中有很多您可能不需要的细节，但您可能可以从最上面的答案中获取代码来开始。

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