Matplotlib:旋转贴片

Question

我想在matplotlib中旋转一个Rectangle,但是当我应用转换时,矩形不再显示:

rect = mpl.patches.Rectangle((0.0120,0),0.1,1000)
t = mpl.transforms.Affine2D().rotate_deg(45)
rect.set_transform(t)

这是一个已知的错误还是我犯了错误？

Answer 1

提供的代码中的补丁很难说明发生了什么,所以我已经清楚地演示了我从matplotlib示例中得到的结论:

import numpy as np
import matplotlib.pyplot as plt
import matplotlib.patches as patches
import matplotlib as mpl

fig = plt.figure()
ax = fig.add_subplot(111)

r1 = patches.Rectangle((0,0), 20, 40, color="blue", alpha=0.50)
r2 = patches.Rectangle((0,0), 20, 40, color="red",  alpha=0.50)

t2 = mpl.transforms.Affine2D().rotate_deg(-45) + ax.transData
r2.set_transform(t2)

ax.add_patch(r1)
ax.add_patch(r2)

plt.xlim(-20, 60)
plt.ylim(-20, 60)

plt.grid(True)

plt.show()

Answer 2

显然,补丁上的变换是几个变换的组合,用于处理缩放和边界框.将变换添加到现有的绘图变换似乎可以提供更符合您期望的内容.虽然它看起来仍然存在偏差.

import numpy as np
import matplotlib.pyplot as plt
import matplotlib.patches as patches
import matplotlib as mpl
fig = plt.figure()
ax = fig.add_subplot(111)

rect = patches.Rectangle((0.0120,0),0.1,1000)

t_start = ax.transData
t = mpl.transforms.Affine2D().rotate_deg(-45)
t_end = t_start + t

rect.set_transform(t_end)

print repr(t_start)
print repr(t_end)
ax.add_patch(rect)

plt.show()