python - 计算点间距不均匀的 3D 梯度

Question

我目前有一个由几百万个不均匀分布的粒子组成的体积，每个粒子都有一个属性(对于那些好奇的人来说是潜在的)，我想为其计算局部力(加速度)。

np.gradient 仅适用于均匀间隔的数据，我在这里查看:Second order gradient in numpy需要插值但我无法在 Numpy 中找到 3D 样条实现。

一些将产生代表性数据的代码:

import numpy as np    
from scipy.spatial import cKDTree

x = np.random.uniform(-10, 10, 10000)
y = np.random.uniform(-10, 10, 10000)
z = np.random.uniform(-10, 10, 10000)
phi = np.random.uniform(-10**9, 0, 10000)

kdtree = cKDTree(np.c_[x,y,z])
_, index = kdtree.query([0,0,0], 32) #find 32 nearest particles to the origin
#find the gradient at (0,0,0) by considering the 32 nearest particles?  

(我的问题和Function to compute 3D gradient with unevenly spaced sample locations很相似，但是好像没有解决办法，所以我想我再问一次。)

如有任何帮助，我们将不胜感激。

Answer 1

这是一个 Julia 实现，可以完成您的要求

using NearestNeighbors

n = 3;
k = 32; # for stability use  k > n*(n+3)/2

# Take a point near the center of cube
point = 0.5 + rand(n)*1e-3;
data = rand(n, 10^4);
kdtree = KDTree(data);
idxs, dists = knn(kdtree, point, k, true);

# Coords of the k-Nearest Neighbors
X = data[:,idxs];

# Least-squares recipe for coefficients
 C = point * ones(1,k); # central node
dX = X - C;  # diffs from central node
 G = dX' * dX;
 F =  G .* G;
 v = diag(G);
 N = pinv(G) * G;
 N = eye(N) - N;
 a =  N * pinv(F*N) * v;  # ...these are the coeffs

# Use a temperature distribution of  T = 25.4 * r^2
# whose analytical gradient is   gradT = 25.4 * 2*x
X2 = X .* X;
C2 = C .* C;
T  = 25.4 * n * mean(X2, 1)';
Tc = 25.4 * n * mean(C2, 1)'; # central node
dT = T - Tc;       # diffs from central node

y = dX * (a .* dT);   # Estimated gradient
g = 2 * 25.4 * point; # Analytical

# print results
@printf "Estimated  Grad  = %s\n" string(y')
@printf "Analytical Grad  = %s\n" string(g')
@printf "Relative Error   = %.8f\n" vecnorm(g-y)/vecnorm(g)

这种方法有大约 1% 的相对误差。这是几次运行的结果...

Estimated  Grad  = [25.51670916224472 25.421038632006926 25.6711949674633]
Analytical Grad  = [25.41499027802736 25.44913042322385  25.448202594123806]
Relative Error   = 0.00559934

Estimated  Grad  = [25.310574056859014 25.549736360607493 25.368056350800604]
Analytical Grad  = [25.43200914200516  25.43243178887198  25.45061497749628]
Relative Error   = 0.00426558

更新
我不太了解 Python，但这里有一个似乎有效的翻译

import numpy as np
from scipy.spatial import KDTree

n = 3;
k = 32;

# fill the cube with random points
data = np.random.rand(10000,n)
kdtree = KDTree(data)

# pick a point (at the center of the cube)
point = 0.5 * np.ones((1,n))

# Coords of k-Nearest Neighbors
dists, idxs = kdtree.query(point, k)
idxs = idxs[0]
X = data[idxs,:]

# Calculate coefficients
C = (np.dot(point.T, np.ones((1,k)))).T # central node
dX= X - C                    # diffs from central node
G = np.dot(dX, dX.T)
F = np.multiply(G, G)
v = np.diag(G);
N = np.dot(np.linalg.pinv(G), G)
N = np.eye(k) - N;
a = np.dot(np.dot(N, np.linalg.pinv(np.dot(F,N))), v)  # these are the coeffs

#  Temperature distribution is  T = 25.4 * r^2
X2 = np.multiply(X, X)
C2 = np.multiply(C, C)
T  = 25.4 * n * np.mean(X2, 1).T
Tc = 25.4 * n * np.mean(C2, 1).T # central node
dT = T - Tc;       # diffs from central node

# Analytical gradient ==>  gradT = 2*25.4* x
g = 2 * 25.4 * point;
print( "g[]: %s" % (g) )

# Estimated gradient
y = np.dot(dX.T, np.multiply(a, dT))
print( "y[]: %s,   Relative Error = %.8f" % (y, np.linalg.norm(g-y)/np.linalg.norm(g)) )

更新#2
我想我可以使用格式化的 ASCII 而不是 LaTeX 来写一些易于理解的东西......

`Given a set of M vectors in n-dimensions (call them b_k), find a set of`coeffs (call them a_k) which yields the best estimate of the identity`matrix and the zero vector``                                 M` (1) min ||E - I||,  where  E = sum  a_k b_k b_k`     a_k                        k=1``                                 M` (2) min ||z - 0||,  where  z = sum  a_k b_k`     a_k                        k=1```Note that the basis vectors {b_k} are not required`to be normalized, orthogonal, or even linearly independent.``First, define the following quantities:``  B             ==> matrix whose columns are the b_k`  G = B'.B      ==> transpose of B times B`  F = G @ G     ==> @ represents the hadamard product`  v = diag(G)   ==> vector composed of diag elements of G``The above minimizations are equivalent to this linearly constrained problem``  Solve  F.a = v`  s.t.   G.a = 0``Let {X} denote the Moore-Penrose inverse of X.`Then the solution of the linear problem can be written:``  N = I - {G}.G       ==> projector into nullspace of G`  a = N . {F.N} . v``The utility of these coeffs is that they allow you to write`very simple expressions for the derivatives of a tensor field.```Let D be the del (or nabla) operator`and d be the difference operator wrt the central (aka 0th) node,`so that, for any scalar/vector/tensor quantity Y, we have:`  dY = Y - Y_0``Let x_k be the position vector of the kth node.`And for our basis vectors, take`  b_k = dx_k  =  x_k - x_0.``Assume that each node has a field value associated with it` (e.g. temperature), and assume a quadratic model [about x = x_0]` for the field [g=gradient, H=hessian, ":" is the double-dot product]``     Y = Y_0 + (x-x_0).g + (x-x_0)(x-x_0):H/2`    dY = dx.g + dxdx:H/2`   D2Y = I:H            ==> Laplacian of Y```Evaluate the model at the kth node ``    dY_k = dx_k.g  +  dx_k dx_k:H/2``Multiply by a_k and sum``     M               M                  M`    sum a_k dY_k =  sum a_k dx_k.g  +  sum a_k dx_k dx_k:H/2`    k=1             k=1                k=1``                 =  0.g   +  I:H/2`                 =  D2Y / 2``Thus, we have a second order estimate of the Laplacian``                M`   Lap(Y_0) =  sum  2 a_k dY_k`               k=1```Now play the same game with a linear model`    dY_k = dx_k.g``But this time multiply by (a_k dx_k) and sum``     M                    M`    sum a_k dx_k dY_k =  sum a_k dx_k dx_k.g`    k=1                  k=1``                      =  I.g`                      =  g```In general, the derivatives at the central node can be estimated as``           M`    D#Y = sum  a_k dx_k#dY_k`          k=1``           M`    D2Y = sum  2 a_k dY_k`          k=1`` where`   # stands for the {dot, cross, or tensor} product`       yielding the {div, curl,  or grad} of Y` and`   D2Y stands for the Laplacian of Y`   D2Y = D.DY = Lap(Y)