如何在Matplotlib中为分组子图绘制侧边弧形标注括号

完整实现代码 #

import matplotlib.pyplot as plt
import numpy as np
from matplotlib.path import Path
from matplotlib.patches import PathPatch

fig, axs = plt.subplots(8, 1, figsize=(8,9), sharex=True)
for i, ax in enumerate(axs.flat):
    ax.plot([1,2])
    ax.set_ylabel('Label ' + str(i))
axs[7].set_xlabel('The common x Label')

plt.annotate('Group Label (0,1,2,3)', xy=(.02, 0.7), rotation=90, xycoords='figure fraction',
            horizontalalignment='center', verticalalignment='center', fontsize=14, color='k')
plt.annotate('Group Label (4,5,6,7)', xy=(.02, 0.3), rotation=90, xycoords='figure fraction',
            horizontalalignment='center', verticalalignment='center', fontsize=14, color='k')

# 生成括号原始坐标
npoints = 100
td = np.linspace(np.pi*3/4, np.pi*5/4, npoints)
xd = np.cos(td)
yd = np.sin(td)

# 定义坐标变换参数，适配侧边空白区域
x_scale = 0.068
x_offset = 0.108
# 第一个括号对应0-3子图分组，中心y坐标0.7
y_scale_1 = 0.283
y_offset_1 = 0.7
trans_x1 = xd * x_scale + x_offset
trans_y1 = yd * y_scale_1 + y_offset_1
# 第二个括号对应4-7子图分组，中心y坐标0.3
y_scale_2 = 0.283
y_offset_2 = 0.3
trans_x2 = xd * x_scale + x_offset
trans_y2 = yd * y_scale_2 + y_offset_2

# 构造路径对象
def create_bracket_path(x, y):
    verts = np.column_stack([x, y])
    codes = [Path.MOVETO] + [Path.LINETO]*(len(verts)-1)
    return Path(verts, codes)

path1 = create_bracket_path(trans_x1, trans_y1)
path2 = create_bracket_path(trans_x2, trans_y2)

# 添加括号到画布
patch1 = PathPatch(path1, facecolor='none', edgecolor='red', linewidth=2, transform=fig.transFigure)
patch2 = PathPatch(path2, facecolor='none', edgecolor='red', linewidth=2, transform=fig.transFigure)
fig.patches.extend([patch1, patch2])

plt.show()

核心逻辑说明 #

• 直接复用了你已经实现的弧形点生成逻辑，不需要调整原始参数
• 通过线性缩放和平移把标准弧形映射到侧边空白区域，缩放系数可以根据实际显示效果灵活调整
• 用Path封装点序列生成路径对象，再通过PathPatch添加到画布，指定transform=fig.transFigure使用全局画布分数坐标，适配跨子图的显示需求
• 可以通过修改PathPatch的edgecolor、linewidth参数调整括号的样式

内容的提问来源于stack exchange，提问作者Pedro