SciPy创建2D多边形蒙版

Question

我需要使用标准的Python包创建一个numpy 2D数组,它表示多边形的二进制掩码.

• 输入:多边形顶点,图像尺寸
• 输出:多边形的二进制掩码(numpy 2D数组)

(更大的上下文:我想使用scipy.ndimage.morphology.distance_transform_edt获取此多边形的距离变换.)

谁能告诉我怎么做？

Answer 1

答案结果很简单:

import numpy
from PIL import Image, ImageDraw

# polygon = [(x1,y1),(x2,y2),...] or [x1,y1,x2,y2,...]
# width = ?
# height = ?

img = Image.new('L', (width, height), 0)
ImageDraw.Draw(img).polygon(polygon, outline=1, fill=1)
mask = numpy.array(img)

Answer 2

作为@Anil答案的一个更直接的替代品,matplotlib matplotlib.nxutils.points_inside_poly可以用来快速栅格化任意多边形.例如

import numpy as np
from matplotlib.nxutils import points_inside_poly

nx, ny = 10, 10
poly_verts = [(1,1), (5,1), (5,9),(3,2),(1,1)]

# Create vertex coordinates for each grid cell...
# (<0,0> is at the top left of the grid in this system)
x, y = np.meshgrid(np.arange(nx), np.arange(ny))
x, y = x.flatten(), y.flatten()

points = np.vstack((x,y)).T

grid = points_inside_poly(points, poly_verts)
grid = grid.reshape((ny,nx))

print grid

产生(布尔numpy数组):

[[False False False False False False False False False False]
 [False  True  True  True  True False False False False False]
 [False False False  True  True False False False False False]
 [False False False False  True False False False False False]
 [False False False False  True False False False False False]
 [False False False False  True False False False False False]
 [False False False False False False False False False False]
 [False False False False False False False False False False]
 [False False False False False False False False False False]
 [False False False False False False False False False False]]

你应该能够grid很好地传递给任何scipy.ndimage.morphology函数.

Answer 3

Joe的评论更新.自评论发布以来,Matplotlib API已发生变化,现在您需要使用子模块提供的方法matplotlib.path.

工作代码如下.

import numpy as np
from matplotlib.path import Path

nx, ny = 10, 10
poly_verts = [(1,1), (5,1), (5,9),(3,2),(1,1)]

# Create vertex coordinates for each grid cell...
# (<0,0> is at the top left of the grid in this system)
x, y = np.meshgrid(np.arange(nx), np.arange(ny))
x, y = x.flatten(), y.flatten()

points = np.vstack((x,y)).T

path = Path(poly_verts)
grid = path.contains_points(points)
grid = grid.reshape((ny,nx))

print grid

Answer 4

作为@Yusuke N.答案的一个轻微替代方案，请考虑使用matplotlib.path，它与 one by 一样有效from PIL import Image, ImageDraw（无需安装Pillow，无需考虑integer或float。对我有用吗？）

工作代码如下：

import pylab as plt
import numpy as np
from matplotlib.path import Path

width, height=2000, 2000

polygon=[(0.1*width, 0.1*height), (0.15*width, 0.7*height), (0.8*width, 0.75*height), (0.72*width, 0.15*height)]
poly_path=Path(polygon)

x, y = np.mgrid[:height, :width]
coors=np.hstack((x.reshape(-1, 1), y.reshape(-1,1))) # coors.shape is (4000000,2)

mask = poly_path.contains_points(coors)
plt.imshow(mask.reshape(height, width))
plt.show()

结果图像如下，其中暗区为False，亮区为True。