在 matplotlib 中绘制带有中点箭头的圆形 fancyarrowpatch

Question

我一直在尝试突破 matplotlib 补丁的界限，并指示它在中点绘制一个带有方向箭头的圆形 FancyArrowPatch。这在我试图创建的网络表示中非常有用。

我使用 python 的编码时间还没有达到两位数，所以我不能说我对 matplotlib 的 patches.py 有清楚的了解，但我已经将解决方案缩小到两个可能的策略：

聪明的，可能是 pythonic 方式：创建自定义 arrowstyle class，这进一步需要修改 _get_arrow_wedge() 函数以包含中点坐标。这可能超出了我现在的可能性，或者
懒惰的方法：从引出的 FancyArrowPatch 中提取中点坐标，并在此类坐标上绘制所需的 arrowstyle。

当然，到目前为止我选择了懒惰的方式。我做了一些早期实验，使用 get_path() 和 get_path_in_displaycoord() 提取弯曲 FancyArrowPatch 的中点坐标，但我似乎无法预测精确的中点坐标。如果能提供一些帮助，我们将不胜感激。

到目前为止我的摆弄：

import matplotlib.pyplot as plt
from matplotlib.patches import FancyArrowPatch

n1 = (2,3)
n2 = (4,6)


# Try with multiple arc radius sizes, draw a separate plot each time
for rad in range(20):
    #setup figure
    figure = plt.figure()
    ax = plt.subplot(111)
    plt.annotate('rad:' + str(rad/25.),xy=(2,5))

    # create rounded fancyarrowpatch   
    t = FancyArrowPatch(posA=n1,posB=n2,
                        connectionstyle='arc3,rad=%s'%float(rad/25.),
                        arrowstyle='->',
                        shrinkA=0,
                        shrinkB=0,
                        mutation_scale=0.5)

    # extract vertices from get_path: points P#
    path = t.get_path().vertices.tolist()
    lab, px, py = ['P{0}'.format(i) for i in range(len(path))], [u[0] for u in path],[u[1] for u in path]
    for i in range(len(path)):
        plt.annotate(lab[i],xy=(px[i],py[i]))

    # extract vertices from get_path_in_displaycoord (but they are useless) : points G#
    newpath = t.get_path_in_displaycoord()
    a,b = newpath[0][0].vertices.tolist(), newpath[0][1].vertices.tolist()
    a.extend(b)
    glab, gx, gy = ['G{0}'.format(i) for i in range(len(a))], [u[0] for u in a],[u[1] for u in a]
    for i in range(len(a)):
        plt.annotate(glab[i],xy=(gx[i],gy[i]))    

    #point A: start
    x1, y1 = n1
    plt.annotate('A',xy=(x1,y1))    

    #point B:end
    x2, y2 = n2
    plt.annotate('B',xy=(x2,y2))

    #point M: the 'midpoint' as defined by class Arc3, specifically its connect() function
    x12, y12 = (x1 + x2) / 2., (y1 + y2) / 2.
    dx, dy = x2 - x1, y2 - y1
    cx, cy = x12 + (rad/100.) * dy, y12 - (rad/100.) * dx    
    plt.annotate('M',xy=(cx,cy))

    #point O : midpoint between M and P1, the second vertex from get_path   
    mx,my = (cx + px[1])/2., (cy + py[1])/2.
    plt.annotate('O',xy=(mx,my))

    ax.add_patch(t)
    plt.scatter([x1,cx,x2,mx,gx].extend(px),[y1,cy,y2,my,gy].extend(py))


plt.show()

编辑：采纳@cphlewis 的建议：我试图重建贝塞尔曲线：

def bezcurv(start,control,end,tau):
    ans = []
    for t in tau:
        B = [(1-t)**2 * start[i] + 2*(1-t)*t*end[i] + (t**2)*control[i] for i in range(len(start))]
        ans.append(tuple(B))
    return ans

因此我将生成的线添加到原始图中：

tau = [time/100. for time in range(101)]
bezsim = bezcurv(n1,n2,(cx,cy),tau)
simx,simy = [b[0] for b in bezsim], [b[1] for b in bezsim]

下面的绿线是（应该是？）重建的贝塞尔曲线，但显然不是。

Answer 1

经过一番挣扎，我说服自己要解决这个问题，我必须离开 FancyArrowPatch 套件并从头开始创建一些东西。这是一个可行的解决方案，远非实现任何完美主义精神，但令我满意：

import matplotlib.pyplot as plt
import numpy as np
from numpy.random import seed, randint

# Build function that connects two points with a curved line, 
# and an arrow on the middle of it

seed(1679)

narrow = 3
rad_one = 50
numpoints = 3

random_points = list(randint(1,20,[numpoints,4]))
rpoints = [[(a,b),(c,d)] for a,b,c,d in random_points]

def curvline(start,end,rad,t=100,arrows=1,push=0.8):
    #Compute midpoint
    rad = rad/100.    
    x1, y1 = start
    x2, y2 = end
    y12 = (y1 + y2) / 2
    dy = (y2 - y1)
    cy = y12 + (rad) * dy
    #Prepare line
    tau = np.linspace(0,1,t)
    xsupport = np.linspace(x1,x2,t)
    ysupport = [(1-i)**2 * y1 + 2*(1-i)*i*cy + (i**2)*y2 for i in tau]
    #Create arrow data    
    arset = list(np.linspace(0,1,arrows+2))
    c = zip([xsupport[int(t*a*push)] for a in arset[1:-1]],
                      [ysupport[int(t*a*push)] for a in arset[1:-1]])
    dt = zip([xsupport[int(t*a*push)+1]-xsupport[int(t*a*push)] for a in arset[1:-1]],
                      [ysupport[int(t*a*push)+1]-ysupport[int(t*a*push)] for a in arset[1:-1]])
    arrowpath = zip(c,dt)
    return xsupport, ysupport, arrowpath

def plotcurv(start,end,rad,t=100,arrows=1,arwidth=.25):
    x, y, c = curvline(start,end,rad,t,arrows)
    plt.plot(x,y,'k-')
    for d,dt in c:
        plt.arrow(d[0],d[1],dt[0],dt[1], shape='full', lw=0, 
                  length_includes_head=False, head_width=arwidth)
    return c

#Create figure
figure = plt.figure()
ax = plt.subplot(111)
for n1,n2 in rpoints:    
    #First line
    plotcurv(n1,n2,rad_one,200,narrow,0.5)
    #Second line
    plotcurv(n2,n1,rad_one,200,narrow,0.5)
ax.set_xlim(0,20)
ax.set_ylim(0,20)
plt.show

我用三个随机点测试了它，来回绘制线条。其中给出了下图：

该函数允许用户设置多个所需的箭头，并将它们均匀地放置在绘制的贝塞尔曲线上，确保表示正确的方向。然而，因为贝塞尔曲线不完全是 'arc'，我试探性地推动箭头的开始，使它们看起来更居中。对此解决方案的任何改进将不胜感激。

在 matplotlib 中绘制带有中点箭头的圆形 fancyarrowpatch

Draw rounded fancyarrowpatch with midpoint arrow in matplotlib

python

patch

matplotlib