切割件

Question

假設我已經得出了一些背景圖片是這樣的：

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') 

現在我加入了許多類似的矩形：

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

現在我希望削減其他幾個矩形不在之前添加的矩形中，因此底層圖像會再次顯示。通過剪切，我的意思是說，指定這樣的「刪除矩形」將從先前繪製的矩形中剪除部分。所以如果它們重疊，只有重疊的部分會被切掉。如果「刪除矩形」不與上述矩形佔據的空間相交，則可見區域不會發生任何變化。

我該如何做到這一點？

Answer 1

您可以使用路徑來構造矩形。爲了定位矩形，路徑的頂點可以被轉換和轉換。然後，使用倒轉頂點從路徑中切出的事實，在外部矩形中創建一個孔。

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() 

---

或者，溶液，其可以更容易地指定矩形的座標：

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() 

---

爲了解釋的情況下的矩形部分是外原始的矩形，以下（基於第二種解決方案）可能會有所幫助：

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()