来自圆柱坐标的 Numpy 掩码答案

【问题标题】：Numpy mask from cylinder coordinates来自圆柱坐标的 Numpy 掩码
【发布时间】：2019-01-31 00:55:45
【问题描述】：

我生成了圆柱体的坐标。它的两个面连接已经给定的两个任意点。

是否可以使用标准 Python 库从坐标构建填充圆柱体的 3D numpy 掩码？创建 2D 蒙版似乎很简单，但我在 3D 方面遇到了一些困难。

这里是生成圆柱体的代码，取自here和here：

import scipy
import scipy.linalg
import numpy as np
import nibabel as nib
import matplotlib
matplotlib.use('TkAgg')
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

# defining mask
shape = (100, 100, 100)
image = np.zeros(shape=shape)

# set radius and centres values
r = 3
start = [30, 45, 60]
end = [40, 58, 70]
p1 = np.array(start)
p2 = np.array(end)

# # calculate p2-p1 distance
# dx = p2[0] - p1[0]
# dy = p2[1] - p1[1]
# dz = p2[2] - p1[2]
# dist = math.sqrt(dx**2 + dy**2 + dz**2)

# vector in direction of axis
v = p2 - p1
# find magnitude of vector
mag = scipy.linalg.norm(v)
# unit vector in direction of axis
v = v / mag
# make some vector not in the same direction as v
not_v = np.array([1, 0, 0])
if (v == not_v).all():
    not_v = np.array([0, 1, 0])
# make vector perpendicular to v
n1 = np.cross(v, not_v)
# normalize n1
n1 /= scipy.linalg.norm(n1)
# make unit vector perpendicular to v and n1
n2 = np.cross(v, n1)
#surface ranges over t from 0 to length of axis and 0 to 2*pi
t = np.linspace(0, mag, 100)
theta = np.linspace(0, 2 * np.pi, 100)
rsample = np.linspace(0, r, 2)

#use meshgrid to make 2d arrays
t, theta2 = np.meshgrid(t, theta)
rsample, theta = np.meshgrid(rsample, theta)

# generate coordinates for surface
# "Tube"
X, Y, Z = [p1[i] + v[i] * t + r * np.sin(theta2) * n1[i] + r * np.cos(theta2) * n2[i] for i in [0, 1, 2]]
# "Bottom"
X2, Y2, Z2 = [p1[i] + rsample[i] * np.sin(theta) * n1[i] + rsample[i] * np.cos(theta) * n2[i] for i in [0, 1, 2]]
# "Top"
X3, Y3, Z3 = [p1[i] + v[i] * mag + rsample[i] * np.sin(theta) * n1[i] + rsample[i] * np.cos(theta) * n2[i] for i in [0, 1, 2]]

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_surface(X, Y, Z)
ax.plot_surface(X2, Y2, Z2)
ax.plot_surface(X3, Y3, Z3)

plt.show()

我需要 3D numpy 掩码来选择 3D 图像圆柱体内的所有值。 mask和image的形状是一样的。

【问题讨论】：

您能否详细说明 2d 和 3d 蒙版的含义？
@Bazingaa 对于 2D 蒙版，我的意思是形状维度 = 2 的 numpy 数组。3D 蒙版形状维度 = 3 的 numpy 数组。
口罩应该怎么做？
@kevinkayaks 掩码应该从相同尺寸的图像中选择圆柱体内的所有像素。
图片的形状是什么？

标签： python numpy geometry mask

【解决方案1】：

最后，我遍历了管和面的坐标。我通过这个链接得到了坐标：3D points from Numpy meshgrid coordinates

tube = np.stack((X.ravel(), Y.ravel(), Z.ravel()), axis=1)
face1 = np.stack((X2.ravel(), Y2.ravel(), Z2.ravel()), axis=1)
face2 = np.stack((X3.ravel(), Y3.ravel(), Z3.ravel()), axis=1)

# filling numpy mask

for i in range(len(tube)):
    image[int(tube[i][0]), int(tube[i][1]), int(tube[i][2])] = 255

for j in range(len(face1)):
    image[int(face1[j][0]), int(face1[j][1]), int(face1[j][2])] = 255

for k in range(len(face2)):
    image[int(face2[k][0]), int(face2[k][1]), int(face2[k][2])] = 255

mask_new = nib.Nifti1Image(image.astype(np.float32), ctsurg_file.affine)

nib.save(mask_new, os.path.join(currdir, 'mask_cyl.nii.gz'))

【讨论】：