Generating a filled polygon inside a numpy array

2
votes

I'm looking for a way to 'draw' a filled polygon into a numpy array based upon a set of polygon vertices. I'd prefer to use as few external libraries as possible.

For example: I have a 20x20 numpy array and I'd like the region bounded by points (3,12), (8,18), (13,14), (11,6) and (4,6) to be filled with 1.0 while the rest of the array contains 0.0

2
pythonarraysnumpypolygon
Simon, see if this is a duplicate of your question (considering that the answer allows for an arbitrary polygon): stackoverflow.com/questions/12638790/… – Warren Weckesser
@Will - not actually drawing or GUI at all. Just filling an array with ones and zeros. – Christian
Here's one approach that uses matplotlib (specifically the part where I generate the mask) – ali_m
Take a look at skimage.draw.polygon(). – Jaime

2 Answers

6
votes

The following solution requires only numpy. It works for vertices (defined in clockwise order in [Row, Column] coordinate system) for convex polygons. Concave polygons will work, but will end up cutting off the protruding points.

import numpy as np

def check(p1, p2, base_array):
    """
    Uses the line defined by p1 and p2 to check array of 
    input indices against interpolated value

    Returns boolean array, with True inside and False outside of shape
    """
    idxs = np.indices(base_array.shape) # Create 3D array of indices

    p1 = p1.astype(float)
    p2 = p2.astype(float)

    # Calculate max column idx for each row idx based on interpolated line between two points
    max_col_idx = (idxs[0] - p1[0]) / (p2[0] - p1[0]) * (p2[1] - p1[1]) +  p1[1]    
    sign = np.sign(p2[0] - p1[0])
    return idxs[1] * sign <= max_col_idx * sign

def create_polygon(shape, vertices):
    """
    Creates np.array with dimensions defined by shape
    Fills polygon defined by vertices with ones, all other values zero"""
    base_array = np.zeros(shape, dtype=float)  # Initialize your array of zeros

    fill = np.ones(base_array.shape) * True  # Initialize boolean array defining shape fill

    # Create check array for each edge segment, combine into fill array
    for k in range(vertices.shape[0]):
        fill = np.all([fill, check(vertices[k-1], vertices[k], base_array)], axis=0)

    # Set all values inside polygon to one
    base_array[fill] = 1

    return base_array


# (Row, Col) Vertices of Polygon (Defined Clockwise)
vertices = np.array([
    [5,12],
    [8,18],
    [13,14],
    [11,6],
    [4,6],
])

polygon_array = create_polygon([20,20], vertices)

# This section prints numbers at each vertex for visual check, just comment out 
# to print an array of only zeros and ones
for n, vertex in enumerate(vertices):
    polygon_array[vertex[0],vertex[1]] = 10*(n+1)

# Simple routine to print the final array
for row in polygon_array.tolist():
    for c in row:
        print '{:4.1f}'.format(c),
    print ''
3
votes

I found one bug in @schoolie response and stackoverflow do not allow add multi line code snippets in comments the problem is when p1[0] == p2[0]. My suggestion of update in function check: 

if p1[0] == p2[0]:
    max_col_idx = (idxs[0] - p1[0]) * idxs.shape[1]
    sign = np.sign(p2[1] - p1[1])
else:
    max_col_idx = (idxs[0] - p1[0]) / (p2[0] - p1[0]) * (p2[1] - p1[1]) + p1[1]
    sign = np.sign(p2[0] - p1[0])