带孔的 Matplotlib 补丁

Question

以下代码有效。问题是我不知道它为什么会起作用。该代码从中心绘制一个带有三角形切口的圆形补丁（使用 PathPatch）。我的猜测是内三角形被切掉了，因为它是顺时针绘制的，而外圆是逆时针绘制的。如果方向不反转，三角形就不会被切断。我在文档中没有找到有关所采用规则的任何内容。那么为什么会这样呢？

from matplotlib import pyplot
from matplotlib.path import Path
from matplotlib.patches import PathPatch
import numpy

#
# draw a triangle within a circle using PathPatch
#
phi = numpy.linspace(0, 2*numpy.pi, 100)
circle = 4 * numpy.exp(1j * phi)
circleV = [[p.real, p.imag] for p in circle]

phi = numpy.linspace(0, 2*numpy.pi, 4)
triangle = 2 * numpy.exp(1j * phi)
triangleV = [[p.real, p.imag] for p in triangle]

circleC = [Path.LINETO for p in circleV]
circleC[0] = Path.MOVETO
triangleC = [Path.LINETO for p in triangleV]
triangleC[0] = Path.MOVETO

vertices = []
vertices.extend(circleV)
vertices.extend(triangleV[::-1])

codes = []
codes.extend(circleC)
codes.extend(triangleC)

path = Path(vertices, codes)
patch = PathPatch(path, facecolor="#aa6677")

fig, ax = pyplot.subplots()
ax.add_patch(patch)
ax.set_xlim([-5, 5])
ax.set_ylim([-5, 5])
ax.set_aspect(1.0)
pyplot.show()

现在作为一个更复杂的示例，从主圆切出五边形，并且从与五边形部分相交的主圆也切出小圆。如果小圆圈是顺时针绘制的，它们在与五边形相交的地方被填充，在不与五边形相交的地方不被填充。这种方式符合上面的规则。但是如果逆时针绘制，则完全填充，不符合上述规则。

from matplotlib import pyplot
from matplotlib.path import Path
from matplotlib.patches import PathPatch
import numpy

clockwise_inner_circles = True

#
# draw a pentagon within a circle using PathPatch
#
phi = numpy.linspace(0, 2*numpy.pi, 100)
circle = 4 * numpy.exp(1j * phi)
circleV = [[p.real, p.imag] for p in circle]

phi = numpy.linspace(0, 2*numpy.pi, 6)
triangle = 2 * numpy.exp(1j * phi)
triangleV = [[p.real, p.imag] for p in triangle]

circleC = [Path.LINETO for p in circleV]
circleC[0] = Path.MOVETO
triangleC = [Path.LINETO for p in triangleV]
triangleC[0] = Path.MOVETO

vertices = []
vertices.extend(circleV)
vertices.extend(triangleV[::-1])

codes = []
codes.extend(circleC)
codes.extend(triangleC)

#
# draw circles in a circular pattern
#
phi = numpy.linspace(0, 2*numpy.pi, 100)
for theta in 2*numpy.pi*numpy.arange(5)/5:
    circle = 2*numpy.exp(1j*theta) + 0.5*numpy.exp(1j*phi)
    circleV = [[p.real, p.imag] for p in circle]
    circleC = [Path.LINETO for p in circleV]
    circleC[0] = Path.MOVETO
    if clockwise_inner_circles:
        vertices.extend(circleV[::-1])
    else:
        vertices.extend(circleV[::1])
    codes.extend(circleC)

path = Path(vertices, codes)
patch = PathPatch(path, facecolor="#aa6677")

fig, ax = pyplot.subplots()
ax.add_patch(patch)
ax.set_xlim([-5, 5])
ax.set_ylim([-5, 5])
ax.set_aspect(1.0)
pyplot.show()

Answer 1

这不是为什么此代码有效的答案，但现在https://matplotlib.org/gallery/shapes_and_collections/donut.html 有一个示例代码。 这似乎遵循 SVG 实现。

Answer 2

基于您的第一个示例，我添加了更多的 cmets，这应该很清楚。 PathPath 的关键是获取 vertices 和 codes。有关顶点和代码的更多详细信息可以找到here。

虽然这是一个很老的问题，但我希望我在这里的回答对需要它的人有用。

from matplotlib import pyplot
from matplotlib.path import Path
from matplotlib.patches import PathPatch
import numpy

# ##########################################
# draw a triangle within a circle using PathPatch
# ##########################################
phi = numpy.linspace(0, 2*numpy.pi, 100)
circle = 4 * numpy.exp(1j * phi)

## generate circle vertices --> "circleV"
circleV = [[p.real, p.imag] for p in circle]

phi = numpy.linspace(0, 2*numpy.pi, 4)
triangle = 2 * numpy.exp(1j * phi)
## generate triangle vertices --> "triangleV"
triangleV = [[p.real, p.imag] for p in triangle]

# generate codes for patch, "C" means "codes"
# codes for circle
circleC = [Path.LINETO for p in circleV]
circleC[0] = Path.MOVETO
# codes for triangle
triangleC = [Path.LINETO for p in triangleV]
triangleC[0] = Path.MOVETO

# combine vertices
vertices = []
vertices.extend(circleV)
vertices.extend(triangleV[::-1])
# combine codes
codes = []
codes.extend(circleC)
codes.extend(triangleC)

# create Path object from vertices and codes
path = Path(vertices, codes)
# create patch from path
patch = PathPatch(path, facecolor="#aa6677")

# plot fig and add patch
fig, ax = pyplot.subplots()
ax.add_patch(patch)
ax.set_xlim([-5, 5])
ax.set_ylim([-5, 5])
ax.set_aspect(1.0)