我已经通过Gmsh在FiPy中定义了两个网格,并希望找到这两个网格之间的交界处的节点。在FiPy中有没有办法做到这一点?
siliconGeometry = '''
SetFactory("OpenCASCADE");
//set node spacing
ns = 1e-1;
ns2 = 1e-2;
x1 = 0;
y1 = 0;
x2 = 1;
y2 = 0.5;
Point(1) = {x1, y1, 0, ns};
Point(2) = {x2, y1, 0, ns};
Point(3) = {x2, y2, 0, ns2};
Point(4) = {x1, y2, 0, ns2};
Line(1) = {1, 2};
Line(2) = {2, 3};
Line(3) = {3, 4};
Line(4) = {4, 1};
Curve Loop(1) = {1, 2, 3, 4};
Plane Surface(1) = {1};
Physical Surface("Silicon") = {1};
'''
oxideGeometry = '''
SetFactory("OpenCASCADE");
//set node spacing
ns = 1e-1;
ns2 = 1e-2;
x1 = 0;
y1 = 0.5;
x2 = 1;
y2 = 1;
Point(5) = {x1, y1, 0, ns2};
Point(6) = {x2, y1, 0, ns2};
Point(7) = {x2, y2, 0, ns};
Point(8) = {x1, y2, 0, ns};
Line(5) = {5, 6};
Line(6) = {6, 7};
Line(7) = {7, 8};
Line(8) = {8, 5};
Curve Loop(2) = {5, 6, 7, 8};
Plane Surface(2) = {2};
Physical Surface("Oxide") = {2};
m0 = Gmsh2D(siliconGeometry)
m1 = Gmsh2D(oxideGeometry)我想得到mesh m0和m1交界处的所有节点(或线)。
发布于 2021-08-05 22:11:28
FiPy有一个名为nearest的函数,它可以获取两个向量的最接近的值。因此,要同时从m0和m1获取重叠的ID,请使用
from fipy import Gmsh2D
from fipy.tools.numerix import nearest
import numpy as np
siliconGeometry = '''
...
'''
oxideGeometry = '''
...
'''
m0 = Gmsh2D(siliconGeometry)
m1 = Gmsh2D(oxideGeometry)
near_ids = nearest(m0.vertexCoords, m1.vertexCoords)
mask = np.all(np.isclose(m0.vertexCoords[:, near_ids], m1.vertexCoords), axis=0)
m1_close_ids = np.arange(len(m1.vertexCoords[0]))[mask]
m0_close_ids = near_ids[mask]
for i in range(len(m0_close_ids)):
m0_id = m0_close_ids[i]
m1_id = m1_close_ids[i]
print()
print(f'm0: {m0_id}, {m0.vertexCoords[:, m0_id]}')
print(f'm1: {m1_id}, {m1.vertexCoords[:, m1_id]}')我们使用nearest获取m0的near_ids,然后使用np.isclose检查节点是否接近。使用mask,我们可以在两个网格之间构建相应的关闭ID。
请注意,使用Scipy的KDTree查找最接近的值可能要快得多。然而,对于这个小问题,这并不重要,但对于一个非常大的网格,我认为这将是重要的。事实上,FiPy真的应该在内部使用它(目前还没有)。
此外,此处的另一个调整是,您知道上面和下面重叠,因此可以简单地请求m0.facesTop和m1.facesBottom遮罩,以将问题减少到仅重叠的面,但这依赖于网格的先验知识。
发布于 2021-08-10 23:07:47
谢谢@wd15。下面是我所做的:
m0 = Gmsh2D(siliconGeometry)
m1 = Gmsh2D(oxideGeometry)
m0_faces = m0.exteriorFaces
m0_vertices = numerix.unique(m0.faceVertexIDs[..., m0_faces].flatten()) m0_vertexCoords = m0.vertexCoords[..., m0_vertices ]
m1_faces = m1.exteriorFaces
m1_vertices = numerix.unique(m1.faceVertexIDs[..., m1_faces].flatten()) m1_vertexCoords = m1.vertexCoords[..., m0_vertices ]
surfVertexID = numerix.nearest(m0_vertexCoords , m1_vertexCoords)https://stackoverflow.com/questions/68596013
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