如何在Matplotlib中仅显示圆形位于x、y0-9范围的区域

问题描述 #

现有如下Matplotlib代码，绘制了网格、散点和两个同心圆，但需要仅保留圆形中x、y范围为0到9的区域着色，其余区域为白色背景：

import matplotlib.pyplot as plt
from matplotlib.patches import Circle
import numpy as np
plt.yticks(np.arange(-10, 10.01, 1))
plt.xticks(np.arange(-10, 10.01, 1))
plt.xlim(-10,9)
plt.ylim(-10,9)
plt.gca().invert_yaxis()
# Set aspect ratio to be equal
plt.gca().set_aspect('equal', adjustable='box')
plt.grid()   
np.random.seed(40)
square = np.empty((10, 10), dtype=np.int_)
for x in np.arange(0, 10, 1):
    for y in np.arange(0, 10, 1):
        plt.scatter(x, y, color="blue", s=2, zorder=2, clip_on=False) 
for x in np.arange(0, 10, 1):
    for y in np.arange(0, 10, 1):
        value = np.random.randint(-3, 4)
        square[int(x), int(y)] = value

r1 = 3
circle1 = Circle((0, 0), r1,  color="blue", alpha=0.5, ec='k', lw=1)
r2 = 6
circle2 = Circle((0, 0), r2,  color="blue", alpha=0.5, ec='k', lw=1)
plt.gca().add_patch(circle1)
plt.gca().add_patch(circle2)

解决方案 #

核心思路是通过**裁剪（clipping）**限制圆形的显示范围，只保留与x∈[0,9]、y∈[0,9]矩形区域的交集部分。以下提供两种实现方式：

方法1：使用边界框裁剪（简洁高效） #

直接给圆形patch设置裁剪边界框，只保留矩形范围内的圆形部分：

import matplotlib.pyplot as plt
from matplotlib.patches import Circle
from matplotlib.transforms import Bbox
import numpy as np

plt.yticks(np.arange(-10, 10.01, 1))
plt.xticks(np.arange(-10, 10.01, 1))
plt.xlim(-10,9)
plt.ylim(-10,9)
plt.gca().invert_yaxis()
plt.gca().set_aspect('equal', adjustable='box')
plt.grid()   

np.random.seed(40)
square = np.empty((10, 10), dtype=np.int_)
for x in np.arange(0, 10, 1):
    for y in np.arange(0, 10, 1):
        plt.scatter(x, y, color="blue", s=2, zorder=2, clip_on=False) 
for x in np.arange(0, 10, 1):
    for y in np.arange(0, 10, 1):
        value = np.random.randint(-3, 4)
        square[int(x), int(y)] = value

# 创建0-9范围的裁剪边界框
clip_bbox = Bbox.from_extents(0, 0, 9, 9)

# 添加第一个带裁剪的圆
r1 = 3
circle1 = Circle((0, 0), r1, color="blue", alpha=0.5, ec='k', lw=1)
circle1.set_clip_bbox(clip_bbox)
plt.gca().add_patch(circle1)

# 添加第二个带裁剪的圆
r2 = 6
circle2 = Circle((0, 0), r2, color="blue", alpha=0.5, ec='k', lw=1)
circle2.set_clip_bbox(clip_bbox)
plt.gca().add_patch(circle2)

plt.show()

方法2：使用复合路径裁剪（灵活支持复杂形状） #

如果需要更复杂的裁剪形状（比如非矩形），可以通过路径交集创建复合裁剪区域：

import matplotlib.pyplot as plt
from matplotlib.patches import Circle, Rectangle
from matplotlib.path import Path
import numpy as np

plt.yticks(np.arange(-10, 10.01, 1))
plt.xticks(np.arange(-10, 10.01, 1))
plt.xlim(-10,9)
plt.ylim(-10,9)
plt.gca().invert_yaxis()
plt.gca().set_aspect('equal', adjustable='box')
plt.grid()   

np.random.seed(40)
square = np.empty((10, 10), dtype=np.int_)
for x in np.arange(0, 10, 1):
    for y in np.arange(0, 10, 1):
        plt.scatter(x, y, color="blue", s=2, zorder=2, clip_on=False) 
for x in np.arange(0, 10, 1):
    for y in np.arange(0, 10, 1):
        value = np.random.randint(-3, 4)
        square[int(x), int(y)] = value

# 获取0-9矩形的路径（转换为轴坐标）
rect = Rectangle((0, 0), 9, 9)
rect_path = rect.get_path().transformed(plt.gca().transData.inverted())

# 处理第一个圆
r1 = 3
circle1 = Circle((0, 0), r1, color="blue", alpha=0.5, ec='k', lw=1)
# 获取圆形路径并转换坐标
circle_path = circle1.get_path().transformed(circle1.get_transform())
# 创建圆形与矩形的交集路径
intersection_path = Path.make_compound_path(circle_path, rect_path)
# 设置裁剪路径
circle1.set_clip_path(intersection_path, plt.gca().transData)
plt.gca().add_patch(circle1)

# 处理第二个圆
r2 = 6
circle2 = Circle((0, 0), r2, color="blue", alpha=0.5, ec='k', lw=1)
circle_path2 = circle2.get_path().transformed(circle2.get_transform())
intersection_path2 = Path.make_compound_path(circle_path2, rect_path)
circle2.set_clip_path(intersection_path2, plt.gca().transData)
plt.gca().add_patch(circle2)

plt.show()

内容的提问来源于stack exchange，提问作者Simd