Question

我有一个坐标为

的三维网格

x = linspace(0, Lx, Nx)
y = linspace(0, Ly, Ny)
z = linspace(0, Lz, Nz)

我需要在位置（x0，y0，z0）的某个半径R内索引点（即x [i]，y [j]，z [k]）。 N_i可能非常大。我可以做一个简单的循环来找到我需要的东西

points=[]
i0,j0,k0 = floor( (x0,y0,z0)/grid_spacing )
Nr = (i0,j0,k0)/grid_spacing + 2
for i in range(i0-Nr, i0+Nr):
    for j in range(j0-Nr, j0+Nr):
        for k in range(k0-Nr, k0+Nr):
            if norm(array([i,j,k])*grid_spacing - (x0,y0,k0)) < cutoff:
                points.append((i,j,k))

但这很慢。是否有一种更自然/更快捷的方式来进行numpy这种类型的操作？

Answer 1

这个怎么样：

import scipy.spatial as sp
x = np.linspace(0, Lx, Nx)
y = np.linspace(0, Ly, Ny)
z = np.linspace(0, Lz, Nz)

#Manipulate x,y,z here to obtain the dimensions you are looking for

center=np.array([x0,y0,z0])

#First mask the obvious points- may actually slow down your calculation depending.
x=x[abs(x-x0)<cutoff]
y=y[abs(y-y0)<cutoff]
z=z[abs(z-z0)<cutoff]


#Generate grid of points
X,Y,Z=np.meshgrid(x,y,z)
data=np.vstack((X.ravel(),Y.ravel(),Z.ravel())).T

distance=sp.distance.cdist(data,center.reshape(1,-1)).ravel()
points_in_sphere=data[distance<cutoff]

而不是最后两行你应该能够做到：

tree=sp.cKDTree(data)
mask=tree.query_ball_point(center,cutoff)
points_in_sphere=data[mask]

如果您不想呼叫空间：

distance=np.power(np.sum(np.power(data-center,2),axis=1),.5)
points_in_sphere=data[distance<cutoff]

在numpy中索引3d网格数据的球形子集

1 个答案: