我想从面部数组中删除重复的面部 - 我尝试了下面的一些代码,但我不确定如何完成它。
首先我惊讶地发现:
new THREE.Vector3(0,0,0) == new THREE.Vector3(0,0,0)
产生 false(我希望它产生 True),而且下面的代码也产生 false(我再次希望它产生 True)。
var triangleGeometry = new THREE.Geometry();
triangleGeometry.vertices.push(new THREE.Vector3( 0.0, 1.0, 0.0));
triangleGeometry.vertices.push(new THREE.Vector3(-1.0, -1.0, 0.0));
triangleGeometry.vertices.push(new THREE.Vector3( 1.0, -1.0, 0.0));
triangleGeometry.faces.push(new THREE.Face3(0, 1, 2));
var triangleGeometry2 = new THREE.Geometry();
triangleGeometry2.vertices.push(new THREE.Vector3( 0.0, 1.0, 0.0));
triangleGeometry2.vertices.push(new THREE.Vector3(-1.0, -1.0, 0.0));
triangleGeometry2.vertices.push(new THREE.Vector3( 1.0, -1.0, 0.0));
triangleGeometry2.faces.push(new THREE.Face3(0, 1, 2));
triangleGeometry2.faces[0] === triangleGeometry.faces[0] - yields false
至于我的代码来确定面孔是否已经在面孔数组中,我编写了以下内容:
function faceInArray(arrayOfFaces,face)
{ // https://stackoverflow.com/questions/29759480/how-to-customize-object-equality-for-javascript-set
// Determine whether a face is in an array of faces
// The ES6 Set object does not have any compare methods or custom compare extensibility.
// For this reason this function will be called before adding an face to an array of faces
// to ensure that duplicate faces are not placed in an array
for(let i = 0; i < arrayOfFaces.length; i++)
{
vertexaFaceFromArray = buildingGeometry.vertices[arrayOfFaces[i].a]
vertexbFaceFromArray = buildingGeometry.vertices[arrayOfFaces[i].b]
vertexcFaceFromArray = buildingGeometry.vertices[arrayOfFaces[i].c]
vertexaFace = buildingGeometry.vertices[face.a]
vertexbFace = buildingGeometry.vertices[face.b]
vertexcFace = buildingGeometry.vertices[face.c]
// Compare the vertices in each face I'm not sure how to do this?
}
}
现在我不确定如何从这里继续,因为简单地检查 vertex1 == vertex2 不起作用,正如我在第一个代码块中演示的那样。比较时我真的需要提取每张脸的 x、y 和 z 坐标吗?此外,顶点的顺序重要吗?
最佳答案
这不起作用的原因:new THREE.Vector3(0,0,0) == new THREE.Vector3(0,0,0)
在这种情况下, ==
检查两个值是否是对同一对象的引用。但是你的向量是不同的对象,它们恰好具有相同的 x、y 和 z 值。您应该在 Vector3 上使用 Three.js equals
函数:
new THREE.Vector3(0,0,0).equals(new THREE.Vector3(0,0,0))
所以你的函数可以像这样工作:
function faceInArray(arrayOfFaces, face) {
for(let i = 0; i < arrayOfFaces.length; i++) {
vertexaFaceFromArray = buildingGeometry.vertices[arrayOfFaces[i].a]
vertexbFaceFromArray = buildingGeometry.vertices[arrayOfFaces[i].b]
vertexcFaceFromArray = buildingGeometry.vertices[arrayOfFaces[i].c]
vertexaFace = buildingGeometry.vertices[face.a]
vertexbFace = buildingGeometry.vertices[face.b]
vertexcFace = buildingGeometry.vertices[face.c]
if (vertexaFaceFromArray.equals(vertexaFace) &&
vertexbFaceFromArray.equals(vertexbFace) &&
vertexcFaceFromArray.equals(vertexcFace)) {
return true;
}
}
return false;
}
但是,当然,这仅检查每个面的顶点顺序是否与输入面完全相同。这取决于您要使用它的用途,但原则上面 (1, 2, 3) 与面 (2, 3, 1) 和 (3, 1, 2) 相同。
此外,如果你的面是两侧的,那么它也与顶点的任何顺序相同。 IE。 (3,2,1),(2,1,3)和(1,3,2)。因此,您可能需要扩展代码以另外检查这些情况。
关于javascript - 从面部数组中删除重复的面部 - Three.js,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46259217/