A way of dispatching action based on type in Python.

dispatch_on_type.py

"""
This demonstrates how to use singledispatch for dispatching on type

We want to draw each based on the type. Typically, we can use if else statement to check the type
then call the function that does draw the shape, however, we will keep having multiple if elif statement.

This is just another pythonic way of doing it
"""

from functools import singledispatch


class Shape:
    def __init__(self, solid):
        self.solid = solid


class Circle(Shape):
    def __init__(self, center, radius, *args, **kwargs):
        super().__init__(*args, **kwargs)
        self.center = center
        self.radius = radius


class Parallelogram(Shape):
    def __init__(self, pa, pb, pc, *args, **kwargs):
        super().__init__(*args, **kwargs)
        self.pa = pa
        self.pb = pb
        self.pc = pc


class Triangle(Shape):
    def __init__(self, pa, pb, pc, *args, **kwargs):
        super().__init__(*args, **kwargs)
        self.pa = pa
        self.pb = pb
        self.pc = pc


@singledispatch
def draw(shape):
    raise TypeError("Don't know how to draw {!r}".format(shape))


@draw.register(Circle)
def draw_circle(shape):
    print("\u25CF" if shape.solid else "\u25A1")


@draw.register(Parallelogram)
def draw_parallelogram(shape):
    print("\u25B0" if shape.solid else "\u25B1")


@draw.register(Triangle)
def draw_triangle(shape):
    print("\u25B2" if shape.solid else "\u25B3")


def main():
    shapes = [Circle(center=(0, 0), radius=5, solid=False),
              Parallelogram(pa=(0, 0), pb=(2, 0), pc=(1, 1), solid=False),
              Triangle(pa=(0, 0), pb=(1, 2), pc=(2, 0), solid=True)
              ]
    for shape in shapes:
        draw(shape)

if __name__ == '__main__':
    main()