在三维坐标中查找球体中的每个点

Question

我正在尝试解决这个算法。给定一个半径和一个点。找到三维坐标系中以给定点为中心的半径球体中的每个点，并将它们存储在列表中。

Answer 1

您可以使用下面的numpy来完成此操作。

请注意，这里的代码将为您提供相对于以您选择的点为中心的球体的坐标，半径由您选择。您需要确保设置了下面的输入尺寸"dim"，以便球体首先完全包含在该体积中。它也只适用于积极的迹象。如果你的点有任何负坐标，请使用该坐标的正坐标，然后在输出中翻转该轴坐标的符号。

import numpy as np
dim = 15
# get 3 arrays representing indicies along each axis
xx, yy, zz = np.ogrid[:dim, :dim, :dim]
# set you center point and radius you want
center = [7, 7, 7]
radius = 3
# create 3d array with values that are the distance from the
# center squared
d2 = (xx-center[0])**2 + (yy-center[1])**2 + (zz-center[2])**2
# create a logical true/false array based on whether the values in d2
# above are less than radius squared
#
# so this is what you want - all the values within "radius" of the center
# are now set to True
mask = d2 <= radius**2
# calculate distance squared and compare to radius squared to avoid having to use
# slow sqrt()
# now you want to get the indicies from the mask array where the value of the
# array is True.  numpy.nonzero does that, and gives you 3 numpy 1d arrays of
# indicies along each axis
s, t, u = np.nonzero(mask)
# finally, to get what you want, which is all those indicies in a list, zip them together:
coords = list(zip(s, t, u))
print(coords)
>>>
[(2, 5, 6),
(3, 4, 5),
(3, 4, 6),
(3, 4, 7),
(3, 5, 5),
(3, 5, 6),
(3, 5, 7),
(3, 6, 5),
(3, 6, 6),
(3, 6, 7),
(4, 3, 6),
(4, 4, 5),
(4, 4, 6),
(4, 4, 7),
(4, 5, 4),
(4, 5, 5),
(4, 5, 6),
(4, 5, 7),
(4, 5, 8),
(4, 6, 5),
(4, 6, 6),
(4, 6, 7),
(4, 7, 6),
(5, 4, 5),
(5, 4, 6),
(5, 4, 7),
(5, 5, 5),
(5, 5, 6),
(5, 5, 7),
(5, 6, 5),
(5, 6, 6),
(5, 6, 7),
(6, 5, 6)]