我正在尝试绘制图片,并且绘制了一个矩形,然后我想绘制一个弧形元素,但是此元素必须精确,并且它只是矩形外部的一个圆的一部分形状。因此,我尝试使用Arc补丁创建相同的内容,但是形状不匹配。
因此,我想知道是否可以绘制圆形,但仅保留矩形之外的部分?更具体地说,我想在下图中丢弃/隐藏/消除“蓝色箭头”部分,并保留“红色箭头”部分,该部分位于矩形外(如弧形)。有什么办法吗?
这是我的代码:
from matplotlib.patches import Circle, Rectangle, Arc, Ellipse
def plot_pic(ax=None, color='black', lw=2, scale = 15):
# get the current ax if ax is None
if ax is None:
ax = plt.gca()
# Plot the rectangle
rec = Rectangle((-(7.32 * scale / 2+ 5.5 * scale +11 * scale),0), width = (5.5 * scale * 2 + 11 * scale * 2 + 7.32 * scale), height = 16.5 * scale, linewidth = lw, color = color, fill = False)
testCircle = Circle((0, 11 * scale), radius = 9.15 * scale, color = color, lw = lw, fill = False)
# List of elements to be plotted
pic_elements = [rec, testCircle]
# Add the elements onto the axes
for element in pic_elements:
ax.add_patch(element)
return ax
之后,运行以下命令:
plt.figure(figsize=(16, 22))
plt.xlim(-600,600)
plt.ylim(-100,1700)
plot_pic()
plt.show()
非常感谢您的帮助。
答案 0 :(得分:1)
如果仅是按照您说的去做,则可以将矩形的面色设置为white
,将圆的zorder
设置为0
,以便在后面进行绘制:
def plot_pic(ax=None, color='black', lw=2, scale = 15):
# get the current ax if ax is None
if ax is None:
ax = plt.gca()
# Plot the rectangle
rec = Rectangle((-(7.32 * scale / 2+ 5.5 * scale +11 * scale),0), width = (5.5 * scale * 2 + 11 * scale * 2 + 7.32 * scale), height = 16.5 * scale, linewidth = lw, color = color, fc='white')
testCircle = Circle((0, 11 * scale), radius = 9.15 * scale, color = color, lw = lw, fill = False, zorder=0)
# List of elements to be plotted
pic_elements = [rec, testCircle]
# Add the elements onto the axes
for element in pic_elements:
ax.add_patch(element)
return ax