How do I get a coordinates list from an svgpathtools bezier curve?

Question

I have python code to create a bezier curve, from which I create a bezier path.

Here are my imports:

import from svgpathtools import Path, Line, CubicBezier

Here is my code:

    bezier_curve = CubicBezier(start_coordinate, control_point_1, control_point_2, end_coordinate)
    bezier_path = Path(bezier_curve)

I would like to create a list of coordinates that make up this curve, but none of the documentation I am reading gives a straightforward way to do that. bezier_curve and bezier_path only have parameters for the start point, end point, and control point.

What do you mean by "coordinates that make up this curve"? `bezier_curve.bpoints()` will return the tuple `(start_coordinate, control_point_1, control_point_2, end_coordinate)`, but it sounds like that's not what you're looking for. Bezier curves are polynomials, so for example, a cubic Bezier curve is completely determined by its four coefficients. For more info it may help to take a look at: https://en.wikipedia.org/wiki/B%C3%A9zier_curve#Cubic_B.C3.A9zier_curves — mathandy, Apr 26 '17 at 07:39

score 8 · Accepted Answer · answered Aug 22 '17 at 01:42

Seems like a pretty reasonable question. Surprised there's no answer. I had to do this myself recently, and the secret is point().

Here's how I got it done, using your boilerplate as a starting point:

from svgpathtools import Path, Line, CubicBezier

bezier_curve = CubicBezier(start=(300+100j), control1=(100+100j), control2=(200+200j), end=(200+300j))
bezier_path = Path(bezier_curve)

NUM_SAMPLES = 10

myPath = []
for i in range(NUM_SAMPLES):
    myPath.append(bezier_path.point(i/(NUM_SAMPLES-1)))

print(myPath)

Output:

[(300+100j), (243.8957475994513+103.56652949245542j), (206.72153635116598+113.71742112482853j), (185.1851851851852+129.62962962962962j), (175.99451303155004+150.480109739369j), (175.85733882030178+175.44581618655695j), (181.4814814814815+203.7037037037037j), (189.57475994513032+234.43072702331963j), (196.84499314128942+266.8038408779149j), (200+300j)]

score 2 · Answer 2 · answered Jun 08 '18 at 16:54

The answer given above worked very well for me. I only had to make a tiny modification to the code:

from svgpathtools import Path, Line, CubicBezier

bezier_curve = CubicBezier(start=(300+100j), control1=(100+100j), control2=(200+200j), end=(200+300j))
bezier_path = Path(bezier_curve)

NUM_SAMPLES = 10

myPath = []
for i in range(NUM_SAMPLES):
    myPath.append(bezier_path.point(i/(**float(NUM_SAMPLES)**-1)))

print(myPath)

Changing i/(NUM_SAMPLES -1) by i/(float(NUM_SAMPLES) -1) assures a correct behavior when the curve is parametrized from 0 to 1. Otherwise only an integer division is produced.

score 1 · Answer 3 · edited Oct 28 '21 at 23:31

1

#to demonstrate lines and cubics, improving readibility

from svgpathtools import Path, Line, CubicBezier

cubic = CubicBezier(300+100j, 100+100j, 200+200j, 200+300j)  # A cubic beginning at (300, 100) and ending at (200, 300)
line = Line(200+300j, 250+350j)  # A line beginning at (200, 300) and ending at (250, 350)

number_of_points = 10

cubic_points = []

for i in range(number_of_points):
    cubic_points.append(cubic.point(i/(NUM_SAMPLES-1)))

print('cubic points', path_points)

line_points = []

for i in range(number_of_points):
    line_points.append(line.point(i/(NUM_SAMPLES-1)))

print('line points', path_points)

edited Oct 28 '21 at 23:31

Robert

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answered Oct 28 '21 at 22:54

Carlos Denner

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As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Oct 28 '21 at 23:34

score 0 · Answer 4 · answered Oct 22 '22 at 22:43

I needed a more general form to extract multiple paths with different features. Its a recursive solution to work with path, list of path, list of segments and single segments simultaneously. And you can specify the sample_points per segment for curves, but a line stays 2 points to not add extra points without a need:

import svgpathtools.path

def svgpathtools_unpacker(obj, sample_points=10):
    path = []
    if isinstance(obj, (svgpathtools.path.Path, list)):
        for i in obj:
            path.extend(svgpathtools_unpacker(i, sample_points=sample_points))
    elif isinstance(obj, svgpathtools.path.Line):
        path.extend(obj.bpoints())
    elif isinstance(obj, (svgpathtools.path.CubicBezier, svgpathtools.path.QuadraticBezier)):
        path.extend(obj.points(np.linspace(0,1,sample_points)))
    else:
        print(type(obj))
    return np.array(path)

and how you can use it:

import matplotlib.pyplot as plt

bezier_curve = svgpathtools.path.CubicBezier(start=(300+100j), control1=(100+100j), control2=(200+200j), end=(200+300j))
bezier_quad = svgpathtools.path.QuadraticBezier(bezier_curve.end, control=(200+200j), end=(300+150j))
line = svgpathtools.path.Line(start=bezier_quad.end, end=bezier_curve.start)
bezier_path = svgpathtools.path.Path(bezier_curve, bezier_quad, line)

plt.figure()
for i in [10, 100]:
    xy = svgpathtools_unpacker(bezier_path, sample_points=i)
    plt.plot(xy.real, xy.imag, label=f'sample_points={i}')
plt.legend()

How do I get a coordinates list from an svgpathtools bezier curve?

4 Answers4