我有一个三维网格坐标

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

和我需要个指数点（即X [I]，Y [J]，Z [K]）内的位置（X0，Y0，Z0）的某些半径R。 N_i可能相当大。我可以做一个简单的循环来找到我所需要的。

and I need to index points (i.e. x[i],y[j],z[k]) within some radius R of a position (x0,y0,z0). N_i can be quite large. I can do a simple loop to find what I need

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))

但相当缓慢。是否有一个更自然的/更快的方法做这种类型的操作与numpy的？

but this quite slow. Is there a more natural/ faster way to do this type of operation with numpy?

推荐答案

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]

相反，最后两行，你应该能够做到：

Instead of the last two lines you should be able to do:

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

如果你不想叫空间：

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