mirror of
https://github.com/wassname/simpeg.git
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Rowan Cockett
289 lines
8.4 KiB
Python
289 lines
8.4 KiB
Python
import numpy as np
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from scipy import sparse as sp
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from matutils import mkvc, ndgrid, sub2ind
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from sputils import sdiag
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def exampleLomGird(nC, exType):
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assert type(nC) == list, "nC must be a list containing the number of nodes"
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assert len(nC) == 2 or len(nC) == 3, "nC must either two or three dimensions"
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exType = exType.lower()
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possibleTypes = ['rect', 'rotate']
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assert exType in possibleTypes, "Not a possible example type."
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if exType == 'rect':
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return ndgrid([np.cumsum(np.r_[0, np.ones(nx)/nx]) for nx in nC], vector=False)
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elif exType == 'rotate':
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if len(nC) == 2:
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X, Y = ndgrid([np.cumsum(np.r_[0, np.ones(nx)/nx]) for nx in nC], vector=False)
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amt = 0.5-np.sqrt((X - 0.5)**2 + (Y - 0.5)**2)
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amt[amt < 0] = 0
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return X + (-(Y - 0.5))*amt, Y + (+(X - 0.5))*amt
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elif len(nC) == 3:
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X, Y, Z = ndgrid([np.cumsum(np.r_[0, np.ones(nx)/nx]) for nx in nC], vector=False)
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amt = 0.5-np.sqrt((X - 0.5)**2 + (Y - 0.5)**2 + (Z - 0.5)**2)
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amt[amt < 0] = 0
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return X + (-(Y - 0.5))*amt, Y + (-(Z - 0.5))*amt, Z + (-(X - 0.5))*amt
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def volTetra(xyz, A, B, C, D):
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"""
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Returns the volume for tetrahedras volume specified by the indexes A to D.
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:param numpy.array xyz: X,Y,Z vertex vector
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:param numpy.array A,B,C,D: vert index of the tetrahedra
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:rtype: numpy.array
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:return: V, volume of the tetrahedra
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Algorithm http://en.wikipedia.org/wiki/Tetrahedron#Volume
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.. math::
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V = {1 \over 3} A h
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V = {1 \over 6} | ( a - d ) \cdot ( ( b - d ) ( c - d ) ) |
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"""
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AD = xyz[A, :] - xyz[D, :]
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BD = xyz[B, :] - xyz[D, :]
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CD = xyz[C, :] - xyz[D, :]
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V = (BD[:, 0]*CD[:, 1] - BD[:, 1]*CD[:, 0])*AD[:, 2] - (BD[:, 0]*CD[:, 2] - BD[:, 2]*CD[:, 0])*AD[:, 1] + (BD[:, 1]*CD[:, 2] - BD[:, 2]*CD[:, 1])*AD[:, 0]
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return V/6
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def indexCube(nodes, gridSize, n=None):
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"""
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Returns the index of nodes on the mesh.
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Input:
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nodes - string of which nodes to return. e.g. 'ABCD'
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gridSize - size of the nodal grid
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n - number of nodes each i,j,k direction: [ni,nj,nk]
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Output:
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index - index in the order asked e.g. 'ABCD' --> (A,B,C,D)
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TWO DIMENSIONS::
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node(i,j) node(i,j+1)
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A -------------- B
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| cell(i,j) |
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| I |
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D -------------- C
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node(i+1,j) node(i+1,j+1)
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THREE DIMENSIONS::
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node(i,j,k+1) node(i,j+1,k+1)
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E --------------- F
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/| / |
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/ | / |
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/ | / |
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node(i,j,k) node(i,j+1,k)
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A -------------- B |
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| H ----------|---- G
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| /cell(i,j) | /
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| / I | /
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| / | /
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D -------------- C
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node(i+1,j,k) node(i+1,j+1,k)
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"""
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assert type(nodes) == str, "Nodes must be a str variable: e.g. 'ABCD'"
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assert type(gridSize) == np.ndarray, "Number of nodes must be an ndarray"
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nodes = nodes.upper()
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# Make sure that we choose from the possible nodes.
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possibleNodes = 'ABCD' if gridSize.size == 2 else 'ABCDEFGH'
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for node in nodes:
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assert node in possibleNodes, "Nodes must be chosen from: '%s'" % possibleNodes
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dim = gridSize.size
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if n is None:
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n = gridSize - 1
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if dim == 2:
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ij = ndgrid(np.arange(n[0]), np.arange(n[1]))
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i, j = ij[:, 0], ij[:, 1]
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elif dim == 3:
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ijk = ndgrid(np.arange(n[0]), np.arange(n[1]), np.arange(n[2]))
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i, j, k = ijk[:, 0], ijk[:, 1], ijk[:, 2]
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else:
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raise Exception('Only 2 and 3 dimensions supported.')
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nodeMap = {'A': [0, 0, 0], 'B': [0, 1, 0], 'C': [1, 1, 0], 'D': [1, 0, 0],
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'E': [0, 0, 1], 'F': [0, 1, 1], 'G': [1, 1, 1], 'H': [1, 0, 1]}
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out = ()
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for node in nodes:
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shift = nodeMap[node]
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if dim == 2:
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out += (sub2ind(gridSize, np.c_[i+shift[0], j+shift[1]]).flatten(), )
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elif dim == 3:
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out += (sub2ind(gridSize, np.c_[i+shift[0], j+shift[1], k+shift[2]]).flatten(), )
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return out
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def faceInfo(xyz, A, B, C, D, average=True, normalizeNormals=True):
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"""
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function [N] = faceInfo(y,A,B,C,D)
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Returns the averaged normal, area, and edge lengths for a given set of faces.
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If average option is FALSE then N is a cell array {nA,nB,nC,nD}
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Input:
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xyz - X,Y,Z vertex vector
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A,B,C,D - vert index of the face (counter clockwize)
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Options:
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average - [true]/false, toggles returning all normals or the average
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Output:
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N - average face normal or {nA,nB,nC,nD} if average = false
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area - average face area
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edgeLengths - exact edge Lengths, 4 column vector [AB, BC, CD, DA]
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see also testFaceNormal testFaceArea
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@author Rowan Cockett
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Last modified on: 2013/07/26
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"""
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assert type(average) is bool, 'average must be a boolean'
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assert type(normalizeNormals) is bool, 'normalizeNormals must be a boolean'
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# compute normal that is pointing away from you.
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#
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# A -------A-B------- B
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# | |
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# | |
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# D-A (X) B-C
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# | |
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# | |
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# D -------C-D------- C
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AB = xyz[B, :] - xyz[A, :]
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BC = xyz[C, :] - xyz[B, :]
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CD = xyz[D, :] - xyz[C, :]
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DA = xyz[A, :] - xyz[D, :]
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def cross(X, Y):
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return np.c_[X[:, 1]*Y[:, 2] - X[:, 2]*Y[:, 1],
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X[:, 2]*Y[:, 0] - X[:, 0]*Y[:, 2],
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X[:, 0]*Y[:, 1] - X[:, 1]*Y[:, 0]]
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nA = cross(AB, DA)
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nB = cross(BC, AB)
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nC = cross(CD, BC)
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nD = cross(DA, CD)
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length = lambda x: np.sqrt(x[:, 0]**2 + x[:, 1]**2 + x[:, 2]**2)
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normalize = lambda x: x/np.kron(np.ones((1, x.shape[1])), mkvc(length(x), 2))
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if average:
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# average the normals at each vertex.
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N = (nA + nB + nC + nD)/4 # this is intrinsically weighted by area
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# normalize
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N = normalize(N)
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else:
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if normalizeNormals:
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N = [normalize(nA), normalize(nB), normalize(nC), normalize(nD)]
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else:
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N = [nA, nB, nC, nD]
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# Area calculation
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#
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# Approximate by 4 different triangles, and divide by 2.
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# Each triangle is one half of the length of the cross product
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#
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# So also could be viewed as the average parallelogram.
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#
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# TODO: This does not compute correctly for concave quadrilaterals
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area = (length(nA)+length(nB)+length(nC)+length(nD))/4
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return N, area
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def inv3X3BlockDiagonal(a11, a12, a13, a21, a22, a23, a31, a32, a33):
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""" B = inv3X3BlockDiagonal(a11, a12, a13, a21, a22, a23, a31, a32, a33)
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inverts a stack of 3x3 matrices
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Input:
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A - a11, a12, a13, a21, a22, a23, a31, a32, a33
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Output:
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B - inverse
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"""
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a11 = mkvc(a11)
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a12 = mkvc(a12)
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a13 = mkvc(a13)
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a21 = mkvc(a21)
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a22 = mkvc(a22)
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a23 = mkvc(a23)
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a31 = mkvc(a31)
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a32 = mkvc(a32)
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a33 = mkvc(a33)
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detA = a31*a12*a23 - a31*a13*a22 - a21*a12*a33 + a21*a13*a32 + a11*a22*a33 - a11*a23*a32
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b11 = +(a22*a33 - a23*a32)/detA
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b12 = -(a12*a33 - a13*a32)/detA
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b13 = +(a12*a23 - a13*a22)/detA
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b21 = +(a31*a23 - a21*a33)/detA
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b22 = -(a31*a13 - a11*a33)/detA
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b23 = +(a21*a13 - a11*a23)/detA
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b31 = -(a31*a22 - a21*a32)/detA
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b32 = +(a31*a12 - a11*a32)/detA
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b33 = -(a21*a12 - a11*a22)/detA
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B = sp.vstack((sp.hstack((sdiag(b11), sdiag(b12), sdiag(b13))),
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sp.hstack((sdiag(b21), sdiag(b22), sdiag(b23))),
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sp.hstack((sdiag(b31), sdiag(b32), sdiag(b33)))))
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return B
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def inv2X2BlockDiagonal(a11, a12, a21, a22):
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""" B = inv2X2BlockDiagonal(a11, a12, a21, a22)
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Inverts a stack of 2x2 matrices by using the inversion formula
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inv(A) = (1/det(A)) * cof(A)^T
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Input:
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A - a11, a12, a13, a21, a22, a23, a31, a32, a33
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Output:
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B - inverse
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"""
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a11 = mkvc(a11)
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a12 = mkvc(a12)
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a21 = mkvc(a21)
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a22 = mkvc(a22)
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# compute inverse of the determinant.
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detAinv = 1./(a11*a22 - a21*a12)
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b11 = +detAinv*a22
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b12 = -detAinv*a12
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b21 = -detAinv*a21
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b22 = +detAinv*a11
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B = sp.vstack((sp.hstack((sdiag(b11), sdiag(b12))),
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sp.hstack((sdiag(b21), sdiag(b22)))))
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return B
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