Question

我在matplotlib中绘制一个多边形。我输入了所有坐标点。在某些点之间我想要'圆形'或'径向'边缘而不是直线（比如图中的点1和2。这可能吗？如果不是最有效的绘制方法？

example drawing

编辑： Rutger的解决方案很有效。

enter image description here

Answer 1

您可以通过从路径制作多边形来使用弧。

正常的方格：

import matplotlib.path as mpath
import matplotlib.patches as patches

verts = [(0,0),
         (1,0),
         (1,1),
         (0,1),
         (0,0)]

codes = [mpath.Path.MOVETO] + (len(x)-1)*[mpath.Path.LINETO]
square_verts = mpath.Path(verts, codes)

fig, ax = plt.subplots(subplot_kw={'aspect': 1.0, 'xlim': [-0.2,1.2], 'ylim': [-0.2,1.2]})

square = patches.PathPatch(square_verts, facecolor='orange', lw=2)
ax.add_patch(square)

enter image description here

圆角正方形可以用：

verts = [(0.2, 0.0),
         (0.8, 0.0), # start of the lower right corner
         (1.0, 0.0), # intermediate point (as if it wasn't rounded)
         (1.0, 0.2), # end point of the lower right corner
         (1.0, 0.8), # move to the next point etc.
         (1.0, 1.0),
         (0.8, 1.0),
         (0.2, 1.0),
         (0.0, 1.0),
         (0.0, 0.8),
         (0.0, 0.2),
         (0.0, 0.0),
         (0.2, 0.0)]

codes = [mpath.Path.MOVETO,
         mpath.Path.LINETO,
         mpath.Path.CURVE3,
         mpath.Path.CURVE3,
         mpath.Path.LINETO,
         mpath.Path.CURVE3,
         mpath.Path.CURVE3,
         mpath.Path.LINETO,
         mpath.Path.CURVE3,
         mpath.Path.CURVE3,
         mpath.Path.LINETO,
         mpath.Path.CURVE3,
         mpath.Path.CURVE3]


rounded_verts = mpath.Path(verts, codes)

fig, ax = plt.subplots(subplot_kw={'aspect': 1.0, 'xlim': [-0.2,1.2], 'ylim': [-0.2,1.2]})

rounded_verts = patches.PathPatch(rounded_verts, facecolor='orange', lw=2)
ax.add_patch(rounded_verts)

enter image description here

对于您的示例，您需要指定一个中间点，该点使用Point1中的x-coordinate和Point2中的y-coordinate。

matplotlib路径教程提供了如何制作路径的详细说明： http://matplotlib.org/users/path_tutorial.html

Answer 2

此类适用于一般多边形。只需指定顶点和角的填充/半径。

import numpy as np
from matplotlib import patches, path, pyplot as plt


class RoundedPolygon(patches.PathPatch):
    def __init__(self, xy, pad, **kwargs):
        p = path.Path(*self.__round(xy=xy, pad=pad))
        super().__init__(path=p, **kwargs)

    def __round(self, xy, pad):
        n = len(xy)

        for i in range(0, n):

            x0, x1, x2 = np.atleast_1d(xy[i - 1], xy[i], xy[(i + 1) % n])

            d01, d12 = x1 - x0, x2 - x1
            d01, d12 = d01 / np.linalg.norm(d01), d12 / np.linalg.norm(d12)

            x00 = x0 + pad * d01
            x01 = x1 - pad * d01
            x10 = x1 + pad * d12
            x11 = x2 - pad * d12

            if i == 0:
                verts = [x00, x01, x1, x10]
            else:
                verts += [x01, x1, x10]
        codes = [path.Path.MOVETO] + n*[path.Path.LINETO, path.Path.CURVE3, path.Path.CURVE3]

        return np.atleast_1d(verts, codes)


# Test
xy = np.array([(0, 0), (0.25, 0), (0.5, -0.25), (0.75, 0),
               (1, 0), (1, 0.25), (1.25, 0.5), (1, 0.75),
               (1, 1), (0.75, 1), (0.5, 1.25), (0.25, 1),
               (0, 1), (0, 0.75), (-0.25, 0.5), (0, 0.25)])
rp = RoundedPolygon(xy=xy, pad=0.1, facecolor='red', edgecolor='magenta', lw=3)

fig, ax = plt.subplots()
ax.add_patch(rp)

ax.set_aspect(1)
ax.axis('off')
ax.set_xlim(-1, 2)
ax.set_ylim(-1, 2)
plt.savefig('star.png')

matplotlib - 多边形边缘的半径 - 是否可能？

2 个答案: