削减绘制艺术家的部分

Question

假设我画了一些像这样的背景图片：

fig, ax = plt.subplots()
imdata = np.random.randn(10, 10)
im = ax.imshow(imdata, extent=(0, 1, 0, 1), aspect='auto',
               cmap='coolwarm', interpolation='nearest')

现在我添加了许多矩形，如：

rect = matplotlib.patches.Rectangle((0.3,0.3),0.4,0.4)
ax.add_artist(rect)

现在我想从先前添加的矩形中剪切出其他几个矩形，以便再次显示基础图像。通过剪切，我的意思是指定这样一个＆＃34;删除矩形＆＃34;将从先前绘制的矩形中剪切出部分。因此，如果它们重叠，则只会切掉重叠的部分。 ＆＃34;删除矩形＆＃34;不要与上面的矩形占据的空间相交，可见区域不会发生任何事情。

我怎样才能做到这一点？

Answer 1

您可以使用路径构建矩形。要定位矩形，可以翻译和转换路径的顶点。然后，使用倒置顶点将从路径中切出的事实，在外部矩形中创建孔。

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


fig, ax = plt.subplots()
imdata = np.random.randn(10, 10)


# create rectangle, coordinates are ignored
rec = Rectangle((0,0),1,1).get_path()

#the big rectangle
r0 = rec.vertices+0.5
# r1 and r2 are the rectangles to cut out of r0
r1 = 0.6+rec.vertices[::-1]*0.35
r2 = 1+rec.vertices[::-1]*0.35

path = Path(vertices=np.concatenate([r0, r1, r2]),
               codes=np.concatenate([rec.codes]*3)) 

im = ax.imshow(imdata, extent=(0, 2, 0, 2), aspect='equal',
               cmap='coolwarm', interpolation='nearest')


patch = PathPatch(path, facecolor='w')
ax.add_patch(patch)

plt.tight_layout()
plt.show()

<小时/> 或者，一种解决方案，可以更容易地指定矩形的坐标：

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


fig, ax = plt.subplots()
imdata = np.random.randn(10, 10)


def create_rec(x0, y0, width, height):
    rec_patch = Rectangle((x0, y0),width, height)
    rec_path = rec_patch.get_path()
    rec_path = rec_patch.get_patch_transform().transform_path(rec_path) 
    return rec_path.vertices, rec_path.codes

#the big rectangle
r0,c = create_rec(0.3, 0.6, 1, 1.2)
# r1 and r2 are the rectangles to cut out of r0
r1,c = create_rec(0.4, 0.7, 0.3, 0.4)
r2,c = create_rec(0.8, 1, 0.4, 0.5)

path = Path(vertices=np.concatenate([r0, r1[::-1], r2[::-1]]),
               codes=np.concatenate([c]*3)) 

im = ax.imshow(imdata, extent=(0, 2, 0, 2), aspect='equal',
               cmap='coolwarm', interpolation='nearest')


patch = PathPatch(path, facecolor='w')
ax.add_patch(patch)

plt.tight_layout()
plt.show()

<小时/> 为了解释矩形部分位于原始矩形之外的情况，以下（基于第二个解决方案）可能有所帮助：

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


fig, ax = plt.subplots()
imdata = np.random.randn(10, 10)


def create_rec(x0, y0, width, height):
    rec_patch = Rectangle((x0, y0),width, height)
    rec_path = rec_patch.get_path()
    rec_path = rec_patch.get_patch_transform().transform_path(rec_path) 
    return rec_path.vertices, rec_path.codes

#the big rectangle
r0,c = create_rec(0.3, 0.6, 1, 1.2)
# r1 and r2 are the rectangles to cut out of r0
r1,c = create_rec(0.2, 0.5, 0.3, 0.4)
r2,c = create_rec(0.8, 1, 0.4, 0.5)

path = Path(vertices=np.concatenate([r0, r1[::-1], r2[::-1]]),
               codes=np.concatenate([c]*3)) 

im = ax.imshow(imdata, extent=(0, 2, 0, 2), aspect='equal',
               cmap='coolwarm', interpolation='nearest')

patho = Path(vertices=r0,codes=c)
patcho = PathPatch(patho, facecolor='none', edgecolor="none")
ax.add_patch(patcho)
patch = PathPatch(path, facecolor='w', clip_path=patcho, edgecolor="none")
ax.add_patch(patch)

plt.show()