这应该给你给予你所需要的：

from itertools import product
import numpy as np

def splitcubes(K, d):
    coords = [np.linspace(-1.0 , 1.0, num=K + 1) for i in range(d)]
    grid = np.stack(np.meshgrid(*coords)).T

    ks = list(range(1, K))
    for slices in product(*([[slice(b,e) for b,e in zip([None] + ks, [k+1 for k in ks] + [None])]]*d)):
        yield grid[slices]

def cubesets(K, d):
    if (K & (K - 1)) or K < 2:
        raise ValueError('K must be a positive power of 2. K: %s' % K)

    return [set(tuple(p.tolist()) for p in c.reshape(-1, d)) for c in splitcubes(K, d)]

二维案例演示

下面是2D案例的一个小演示：

import matplotlib.pyplot as plt

def assemblecube(c, spread=.03):
    c = np.array(list(c))
    c = c[np.lexsort(c.T[::-1])]

    d = int(np.log2(c.size))
    for i in range(d):
        c[2**i:2**i + 2] = c[2**i + 1:2**i - 1:-1]

    # get the point farthest from the origin
    sp = c[np.argmax((c**2).sum(axis=1)**.5)]
    # shift all points a small distance towards that farthest point
    c += sp * .1 #np.copysign(np.ones(sp.size)*spread, sp)

    # create several different orderings of the same points so that matplotlib will draw a closed shape
    return [(np.roll(c, i, axis=1) - (np.roll(c, i, axis=1)[0] - c[0])[None,:]).T for i in range(d)]

fig = plt.figure(figsize=(6,6))
ax = fig.gca()

for i,c in enumerate(cubesets(4, 2)):
    for cdata in assemblecube(c):
        p = ax.plot(*cdata, c='C%d' % (i % 9))

ax.set_aspect('equal', 'box')
fig.show()

输出量：

为了可视化的目的，立方体已经稍微分开了（这样它们就不会重叠和互相覆盖）。

3D案例演示

对于3D的情况也是一样的：

import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

fig = plt.figure(figsize=(6,6))
ax = fig.add_subplot(111, projection='3d')

for i,c in enumerate(cubesets(2,3)):
    for cdata in assemblecube(c, spread=.05):
        ax.plot(*cdata, c=('C%d' % (i % 9)))

plt.gcf().gca().set_aspect('equal', 'box')
plt.show()

输出量：

K=4演示

下面是与上面相同的2D和3D演示的输出，但是使用K=4：