使用numpy/scipy从3D阵列计算等值面 [英] Using numpy/scipy to calculate iso-surface from 3D array

问题描述

我有一个3D numpy数组，其中包含给定函数的值.我想计算一个二维等值面，或一组代表该函数某些值的等值面.

I have a 3D numpy array that contains the values of a given function. I want to calculate a 2D iso-surface, or a set of iso-surfaces that represent certain values of this function.

在这种特定情况下，可以独立处理3D数组的每个1D列(column = myarray[i, j, :]).因此，我想知道的是函数等于某个值(例如myvalue)的最后一个索引位置(二维数组).

In this particular case, each 1D column (column = myarray[i, j, :]) of the 3D array can be treated independently. So what I would like to know are the last index positions (2D array) where the function is equal to a certain value, say myvalue.

一些(慢速)代码示例:

Some (slow) code to exemplify:

# myarray = 3D ndarray
import numpy as np
from scipy import interpolate

result = np.zeros(nx, ny)
z_values = np.arange(nz)

for i in range(nx):
    for j in range(ny):
        f = interpolate.interp1d(my_array[i, j], z_values)
        result[i, j] = f(myvalue)

我知道可以使用np.ndenumerate和其他技巧来加快速度，但是想知道是否已经有一种更简单的方法来进行这种等值面处理.我在ndimage或其他库中找不到任何内容.我知道mayavi2和vtk有很多用于处理等值面的工具，但我的目的不是可视化-我想对这些等值面值进行计算，而不是显示它们.另外，vtk的许多等值面方法似乎都涉及多边形等，而我需要的只是每个等值面值的二维数组.

I know this can be sped up a bit with np.ndenumerate and other tricks, but was wondering if there is already a simpler way of doing this kind of iso-surface. I couldn't find anything in ndimage or other libraries. I know that mayavi2 and vtk have a lot of tools to deal with iso-surfaces, but my aim here is not visualisation -- I want to perform calculations on those iso-surface values, not display them. Plus, a lot of the iso-surface methods of vtk seem to involve polygons and the like, and what I need is just a 2D array of positions for each iso-surface value.

推荐答案

仅使用numpy，您可以使用argsort，sort，take和适当的数组操作来获得良好的解决方案.下面的函数使用加权平均值来计算等值面:

Using only numpy you can get a good solution using argsort, sort, take and the proper array manipulation. The function below uses a weighted average to compute the iso-surface:

def calc_iso_surface(my_array, my_value, zs, interp_order=6, power_parameter=0.5):
    if interp_order < 1: interp_order = 1
    from numpy import argsort, take, clip, zeros
    dist = (my_array - my_value)**2
    arg = argsort(dist,axis=2)
    dist.sort(axis=2)
    w_total = 0.
    z = zeros(my_array.shape[:2], dtype=float)
    for i in xrange(int(interp_order)):
        zi = take(zs, arg[:,:,i])
        valuei = dist[:,:,i]
        wi = 1/valuei
        clip(wi, 0, 1.e6, out=wi) # avoiding overflows
        w_total += wi**power_parameter
        z += zi*wi**power_parameter
    z /= w_total
    return z

此解决方案无法处理与my_value对应的多个z的情况.以下代码给出了构建下面的等值面的应用示例:

This solution does not handle situations where there is more than one z corresponding to my_value. An application example to build the iso-surfaces below is given in the following code:

from numpy import meshgrid, sin, cos, pi, linspace
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
dx = 100; dy =  50; dz = 25
nx = 200; ny = 100; nz = 100
xs = linspace(0,dx,nx)
ys = linspace(0,dy,ny)
zs = linspace(0,dz,nz)
X,Y,Z = meshgrid( xs, ys, zs, dtype=float)
my_array = sin(0.3*pi+0.4*pi*X/dx)*sin(0.3*pi+0.4*pi*Y/dy)*(Z/dz)

fig = plt.figure()
ax = fig.gca(projection='3d')

z = calc_iso_surface( my_array, my_value=0.1, zs=zs, interp_order=6 )
ax.plot_surface(X[:,:,0], Y[:,:,0], z, cstride=4, rstride=4, color='g')

z = calc_iso_surface( my_array, my_value=0.2, zs=zs, interp_order=6 )
ax.plot_surface(X[:,:,0], Y[:,:,0], z, cstride=4, rstride=4, color='y')

z = calc_iso_surface( my_array, my_value=0.3, zs=zs, interp_order=6 )
ax.plot_surface(X[:,:,0], Y[:,:,0], z, cstride=4, rstride=4, color='b')

plt.ion()
plt.show()

您也可以使用不同的插值功能.参见下面的示例，该示例取两个最接近的zs的平均值:

You can also play with different interpolation functions. See below one example that takes the average of the two closest zs:

def calc_iso_surface_2(my_array, my_value, zs):
    '''Takes the average of the two closest zs
    '''
    from numpy import argsort, take
    dist = (my_array - my_value)**2
    arg = argsort(dist,axis=2)
    z0 = take(zs, arg[:,:,0])
    z1 = take(zs, arg[:,:,1])
    z = (z0+z1)/2
    return z

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