Python 如何删除剪裁圆的对角线的外部部分

发布于03月08日

在这answer本书中，它展示了如何使用transdata用对角线剪裁一个圆.代码和图像如下:

 import matplotlib.pyplot as plt

# Create the circle with radius 6
circle = plt.Circle((0, 0), 6, color='r', fill=False)

# Set up the plot (reuse the previous grid settings)
plt.figure(figsize=(8, 8))
plt.xlim(0, 10)
plt.ylim(0, 10)
plt.grid()

# Add the circle to the plot
ax = plt.gca()
ax.add_patch(circle)

# Draw a diagonal line
plt.plot([0, 7], [7, 0], color='b', linestyle='--')

# Set aspect ratio to ensure square grid cells
ax.set_aspect("equal")

polygon = plt.Polygon([[0, 0], [7, 0], [0, 7]], transform=ax.transData)

# Clip the circle using the diagonal line.
circle.set_clip_path(polygon)

# Show the plot
plt.title("Circle Centered at (0,0) Clipped by Diagonal Line")
plt.show()

我想要相同的图像，但与外对角线的圆圈移动的一部分.所以剩下的就是截断的圆弧了.

推荐答案

您可以使三角形稍微大一些，适当地设置其样式，然后将相应的面片添加到绘图中，而不是绘制特殊的对角线.然后，您可以添加圆，使用多边形对其进行剪裁，最后也使用圆对多边形进行剪裁:

import matplotlib.pyplot as plt

# Create the circle with radius 6
circle = plt.Circle((0, 0), 6, color='r', fill=False)

# Set up the plot (reuse the previous grid settings)
plt.figure(figsize=(8, 8))
plt.xlim(0, 10)
plt.ylim(0, 10)
plt.grid()

# Add the circle to the plot
ax = plt.gca()


# Set aspect ratio to ensure square grid cells
ax.set_aspect("equal")

# Make a polygon looking like a diagonal line on the visible part of the plot
polygon = plt.Polygon([[-1, -1], [7, 0], [0, 7]], transform=ax.transData, edgecolor='b', facecolor='w', linestyle='--')

# Add the polygon and circle to the plot
ax.add_patch(polygon)
ax.add_patch(circle)

# Clip the circle using the polygon appearing as a "diagonal line"
circle.set_clip_path(polygon)

# Clip the polygon ("exterior parts of the diagonal line") using the circle
polygon.set_clip_path(circle)

# Show the plot
plt.title("Circle Centered at (0,0) Clipped by Diagonal Line")
plt.show()

The output: