概述
x = linspace(0,Lx,Nx) y = linspace(0,Ly,Ny) z = linspace(0,Lz,Nz)
我需要在一个位置(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,z0)/grid_spacing )
Nr = (i0,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,k0)) < cutoff:
points.append((i,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？
import scipy.spatial as sp x = np.linspace(0,Nx) y = np.linspace(0,Ny) z = np.linspace(0,Nz) #Manipulate x,y,z here to obtain the dimensions you are looking for center=np.array([x0,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,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]
如果您不想调用空间：
distance=np.power(np.sum(np.power(data-center,2),axis=1),.5) points_in_sphere=data[distance<cutoff]
总结
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