的FancyArrowPatch类需要一个路径作为参数,所以我想你可以使用它。
1)对于每个线段,创建一个matplotlib.path.Path
实例。
2)使用路径实例绘制箭头。
import numpy as np
import matplotlib.pyplot as plt; plt.ion()
from matplotlib.patches import FancyArrowPatch, PathPatch
from matplotlib.path import Path
def create_path(x,y):
vertices = zip(x,y)
codes = [Path.MOVETO] + (len(vertices)-1) * [Path.CURVE3]
return Path(vertices, codes)
X = np.linspace(0,4*np.pi,10000)
Y = np.sin(X)
fig, ax = plt.subplots(1,1)
ax.plot(X,Y,color='blue')
shift = 0.1
seg_size = 300
i = 0
while i +seg_size < len(X):
x = X[i:i+seg_size]
y = Y[i:i+seg_size]+shift
path = create_path(x,y)
# for testing path
# patch = PathPatch(path, facecolor='none', lw=2)
# ax.add_patch(patch)
arrow = FancyArrowPatch(path=path, color='r')
ax.add_artist(arrow)
i += seg_size*2
不幸的是,这并不工作,因为传递给FancyArrowPatch不能有超过2段的路径(没有记录,但有一个支票ensure_quadratic_bezier
)。
所以你必须作弊。在下面我使用每段的最后2点绘制箭头。
import numpy as np
import matplotlib.pyplot as plt; plt.ion()
from matplotlib.patches import FancyArrowPatch
X = np.linspace(0,4*np.pi,10000)
Y = np.sin(X)
fig, ax = plt.subplots(1,1)
ax.plot(X,Y,color='blue')
shift = 0.1
seg_size = 300
i = 0
while i +seg_size < len(X):
x = X[i:i+seg_size]
y = Y[i:i+seg_size]+shift
ax.plot(x, y, 'k')
posA, posB = zip(x[-2:], y[-2:])
edge_width = 2.
arrowstyle = "fancy,head_length={},head_width={},tail_width={}".format(2*edge_width, 3*edge_width, edge_width)
arrow = FancyArrowPatch(posA=posA, posB=posB, arrowstyle=arrowstyle, color='k')
ax.add_artist(arrow)
i += seg_size*2
''numpy''或''pyplot''?你给出的链接是后者,它不允许将曲线变成箭头,只能是直线。我可以“在曲线末尾作弊并放置注释 – Yotam
如果我们取曲线的终点并将其用作pyplot注释的起点(箭头),它会起作用吗? –
是的,我做了类似的事情该点不能太远,因为曲线并不总是接近于线性的 – Yotam