Draw an Arc as Polygon using Start, End, Center and Radius using Python Matplotlib

Question

I have used the below code with only inputs start_point and end_point:

import matplotlib.pyplot as plt
from matplotlib.path import Path
import matplotlib.patches as patches

start_point = (25, 50)
end_point = (50, 25)
center = (25, 25)
radius = (25
# We need to some how automatically calculate this midpoint
mid_point = (45, 45)
verts = [start_point, mid_point, end_point]
codes = [Path.MOVETO, Path.CURVE3, Path.CURVE3]#, Path.CLOSEPOLY]

path = Path(verts, codes)
shape = patches.PathPatch(path, facecolor='none', lw=0.75)
plt.gca().add_patch(shape)
plt.axis('scaled')

I get the following output

I need the following output (I created the below output using MS Paint)

[Explanation: Converted the arc into set of points joined as straight lines. I need those set of points as list]

I do not understand the question. If you already have created the desired output, what do you need help with? — ImportanceOfBeingErnest, May 15 '18 at 10:55
I created the desired output using MS Paint (I have edited the question with that information). I need the output using `matplotlib` in `python` — Kartheek Palepu, May 15 '18 at 11:05

ImportanceOfBeingErnest · Accepted Answer · 2018-05-20T12:28:16.160

To obtain the coordinates along a Path that has been defined via curves (CURVE3 or CURVE4) you may use the Path's .to_polygons() method. This will give a N x 2 shaped array of coordinates.

poly, = path.to_polygons()
plt.plot(*zip(*poly), marker="o", ls="None")

There is no way to manipulate the number of points created by .to_polygon. If that is a requirement, you may instead create your own Bezier curve from the given points as shown below. Here we choose 32 points along the path.

import matplotlib.pyplot as plt
import numpy as np
from scipy.special import binom

bernstein = lambda n, k, t: binom(n,k)* t**k * (1.-t)**(n-k)

def bezier(points, num=200):
    N = len(points)
    t = np.linspace(0, 1, num=num)
    curve = np.zeros((num, 2))
    for i in range(N):
        curve += np.outer(bernstein(N - 1, i, t), points[i])
    return curve

start_point = (25, 50)
end_point = (50, 25)
mid_point = (45, 45)

nodes1 = np.array([start_point, mid_point, end_point])
curve1 = bezier(nodes1, num=32)

plt.plot(curve1[:,0], curve1[:,1], color="red", ls="", marker="o", ms=3)
plt.axis('scaled')

plt.show()

Thanks for the quick answer. Can we decide on the number of coordinates to be formed on the arc? Is there some parameter for that? — Kartheek Palepu, May 16 '18 at 05:53
Unfortunately that does not seem to be possible. But instead you may of course create your own polygon using a Bezier curve. I updated the answer. — ImportanceOfBeingErnest, May 20 '18 at 12:29
Wonderful. This is exactly what I need. Cheers @ImportanceOfBeingErnest — Kartheek Palepu, Jun 07 '18 at 06:07

Draw an Arc as Polygon using Start, End, Center and Radius using Python Matplotlib

1 Answers1