将 Voronoi 图渲染为 numpy 数组

Question

I'd like to generate Voronoi regions, based on a list of centers and an image size.

I'm tryed the next code, based on https://rosettacode.org/wiki/Voronoi_diagram

def generate_voronoi_diagram(width, height, centers_x, centers_y):
    image = Image.new("RGB", (width, height))
    putpixel = image.putpixel
    imgx, imgy = image.size
    num_cells=len(centers_x)
    nx = centers_x
    ny = centers_y
    nr,ng,nb=[],[],[]
    for i in range (num_cells):
        nr.append(randint(0, 255));ng.append(randint(0, 255));nb.append(randint(0, 255));

    for y in range(imgy):
        for x in range(imgx):
            dmin = math.hypot(imgx-1, imgy-1)
            j = -1
            for i in range(num_cells):
                d = math.hypot(nx[i]-x, ny[i]-y)
                if d < dmin:
                    dmin = d
                    j = i
            putpixel((x, y), (nr[j], ng[j], nb[j]))
    image.save("VoronoiDiagram.png", "PNG")
    image.show()

I have the desired output:

But it takes too much to generate the output.

I also tried https://stackoverflow.com/a/20678647 It is fast, but I didn't find the way to translate it to numpy array of img_width X img_height. Mostly, because I don't know how to give image size parameter to scipy Voronoi class .

Is there any faster way to have this output? No centers or polygon edges are needed

Thanks in advance

Edited 2018-12-11: Using @tel "Fast Solution"

The code execution is faster, it seems that the centers have been transformed. Probably this method is adding a margin to the image

Answer 1

Fast solution

Here's how you can convert the output of the fast solution based on scipy.spatial.Voronoi that you linked to into a Numpy array of arbitrary width and height. Given the set of regions, vertices that you get as output from the voronoi_finite_polygons_2d function in the linked code, here's a helper function that will convert that output to an array:

import numpy as np
import matplotlib.pyplot as plt
from matplotlib.backends.backend_agg import FigureCanvasAgg as FigureCanvas

def vorarr(regions, vertices, width, height, dpi=100):
    fig = plt.Figure(figsize=(width/dpi, height/dpi), dpi=dpi)
    canvas = FigureCanvas(fig)
    ax = fig.add_axes([0,0,1,1])

    # colorize
    for region in regions:
        polygon = vertices[region]
        ax.fill(*zip(*polygon), alpha=0.4)

    ax.plot(points[:,0], points[:,1], 'ko')
    ax.set_xlim(vor.min_bound[0] - 0.1, vor.max_bound[0] + 0.1)
    ax.set_ylim(vor.min_bound[1] - 0.1, vor.max_bound[1] + 0.1)

    canvas.draw()
    return np.frombuffer(canvas.tostring_rgb(), dtype='uint8').reshape(height, width, 3)

Testing it out

Here's a complete example of vorarr in action:

from scipy.spatial import Voronoi

# get random points
np.random.seed(1234)
points = np.random.rand(15, 2)

# compute Voronoi tesselation
vor = Voronoi(points)

# voronoi_finite_polygons_2d function from https://stackoverflow.com/a/20678647/425458
regions, vertices = voronoi_finite_polygons_2d(vor)

# convert plotting data to numpy array
arr = vorarr(regions, vertices, width=1000, height=1000)

# plot the numpy array
plt.imshow(arr)

Output:

As you can see, the resulting Numpy array does indeed have a shape of (1000, 1000) , as specified in the call to vorarr .

If you want to fix up your existing code

Here's how you could alter your current code to work with/return a Numpy array:

import math
import matplotlib.pyplot as plt
import numpy as np

def generate_voronoi_diagram(width, height, centers_x, centers_y):
    arr = np.zeros((width, height, 3), dtype=int)
    imgx,imgy = width, height
    num_cells=len(centers_x)

    nx = centers_x
    ny = centers_y

    randcolors = np.random.randint(0, 255, size=(num_cells, 3))

    for y in range(imgy):
        for x in range(imgx):
            dmin = math.hypot(imgx-1, imgy-1)
            j = -1
            for i in range(num_cells):
                d = math.hypot(nx[i]-x, ny[i]-y)
                if d < dmin:
                    dmin = d
                    j = i
            arr[x, y, :] = randcolors[j]

    plt.imshow(arr.transpose(1, 0, 2))
    plt.scatter(cx, cy, c='w', edgecolors='k')
    plt.show()
    return arr

Example usage:

np.random.seed(1234)

width = 500
cx = np.random.rand(15)*width

height = 300
cy = np.random.rand(15)*height

arr = generate_voronoi_diagram(width, height, cx, cy)

Example output:

Answer 2

A fast solution without using matplotlib is also possible. Your solution is slow because you're iterating over all pixels, which incurs a lot of overhead in Python. A simple solution to this is to compute all distances in a single numpy operation and assigning all colors in another single operation.

def generate_voronoi_diagram_fast(width, height, centers_x, centers_y):
    # Create grid containing all pixel locations in image
    x, y = np.meshgrid(np.arange(width), np.arange(height))

    # Find squared distance of each pixel location from each center: the (i, j, k)th
    # entry in this array is the squared distance from pixel (i, j) to the kth center.
    squared_dist = (x[:, :, np.newaxis] - centers_x[np.newaxis, np.newaxis, :]) ** 2 + \
                   (y[:, :, np.newaxis] - centers_y[np.newaxis, np.newaxis, :]) ** 2
    
    # Find closest center to each pixel location
    indices = np.argmin(squared_dist, axis=2)  # Array containing index of closest center

    # Convert the previous 2D array to a 3D array where the extra dimension is a one-hot
    # encoding of the index
    one_hot_indices = indices[:, :, np.newaxis, np.newaxis] == np.arange(centers_x.size)[np.newaxis, np.newaxis, :, np.newaxis]

    # Create a random color for each center
    colors = np.random.randint(0, 255, (centers_x.size, 3))

    # Return an image where each pixel has a color chosen from `colors` by its
    # closest center
    return (one_hot_indices * colors[np.newaxis, np.newaxis, :, :]).sum(axis=2)

Running this function on my machine obtains a ~10x speedup relative to the original iterative solution (not taking plotting and saving the result to disk into account). I'm sure there are still a lot of other tweaks which could further accelerate my solution.