def draw_dashed_line(surf, color, start_pos, end_pos, width=1, dash_length=10):
    x1, y1 = start_pos
    x2, y2 = end_pos
    dl = dash_length

    if (x1 == x2):
        ycoords = [y for y in range(y1, y2, dl if y1 < y2 else -dl)]
        xcoords = [x1] * len(ycoords)
    elif (y1 == y2):
        xcoords = [x for x in range(x1, x2, dl if x1 < x2 else -dl)]
        ycoords = [y1] * len(xcoords)
        a = abs(x2 - x1)
        b = abs(y2 - y1)
        c = round(math.sqrt(a**2 + b**2))
        dx = dl * a / c
        dy = dl * b / c

        xcoords = [x for x in numpy.arange(x1, x2, dx if x1 < x2 else -dx)]
        ycoords = [y for y in numpy.arange(y1, y2, dy if y1 < y2 else -dy)]

    next_coords = list(zip(xcoords[1::2], ycoords[1::2]))
    last_coords = list(zip(xcoords[0::2], ycoords[0::2]))
    for (x1, y1), (x2, y2) in zip(next_coords, last_coords):
        start = (round(x1), round(y1))
        end = (round(x2), round(y2))
        pygame.draw.line(surf, color, start, end, width)

This function takes two coordinates and draws a colored dashed line from the first to the second coordinate. Line a = abs(x2 - x1) to line dy = dl * b / c calculates the amount x and y change for dl (dash_length). Because dx and dy are floats, I had to use numpy.arange (built-in range() doesn't allow floating-point).

Here's an example:

draw_dashed_line(screen, RED, (0, 0), (800, 600), dash_length=5)


Documentation for pygame.draw.line: http://www.pygame.org/docs/ref/draw.html#pygame.draw.line


2 Answers 2


Instead of storing all the values you're trying to draw, it might be more efficient to just get the slope of the drawn line, and compute dashes at draw time.

Considering that there's a lot of mirrored code here, consider using a point class to handle both the x and the y values at the same time!

class Point:
    # constructed using a normal tupple
    def __init__(self, point_t = (0,0)):
        self.x = float(point_t[0])
        self.y = float(point_t[1])
    # define all useful operators
    def __add__(self, other):
        return Point((self.x + other.x, self.y + other.y))
    def __sub__(self, other):
        return Point((self.x - other.x, self.y - other.y))
    def __mul__(self, scalar):
        return Point((self.x*scalar, self.y*scalar))
    def __div__(self, scalar):
        return Point((self.x/scalar, self.y/scalar))
    def __len__(self):
        return int(math.sqrt(self.x**2 + self.y**2))
    # get back values in original tuple format
    def get(self):
        return (self.x, self.y)

def draw_dashed_line(surf, color, start_pos, end_pos, width=1, dash_length=10):
    origin = Point(start_pos)
    target = Point(end_pos)
    displacement = target - origin
    length = len(displacement)
    slope = displacement/length

    for index in range(0, length/dash_length, 2):
        start = origin + (slope *    index    * dash_length)
        end   = origin + (slope * (index + 1) * dash_length)
        pygame.draw.line(surf, color, start.get(), end.get(), width)

You could make this function way shorter, by using all of NumPy's features ...

import pygame as pg
import numpy  as np

# -------------------------------- #

def draw_line_dashed(surface, color, start_pos, end_pos, width = 1, dash_length = 10, exclude_corners = True):

    # convert tuples to numpy arrays
    start_pos = np.array(start_pos)
    end_pos   = np.array(end_pos)

    # get euclidian distance between start_pos and end_pos
    length = np.linalg.norm(end_pos - start_pos)

    # get amount of pieces that line will be split up in (half of it are amount of dashes)
    dash_amount = int(length / dash_length)

    # x-y-value-pairs of where dashes start (and on next, will end)
    dash_knots = np.array([np.linspace(start_pos[i], end_pos[i], dash_amount) for i in range(2)]).transpose()

    return [pg.draw.line(surface, color, tuple(dash_knots[n]), tuple(dash_knots[n+1]), width)
            for n in range(int(exclude_corners), dash_amount - int(exclude_corners), 2)]

# -------------------------------- #
  • 2
    \$\begingroup\$ Welcome to Code Review! You have presented an alternative solution, but haven't reviewed the code. Please explain your reasoning (how your solution works and why it is better than the original) so that the author and other readers can learn from your thought process. \$\endgroup\$
    – Vogel612
    Commented Sep 2, 2020 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.