我想根据中心列表和图像尺寸生成Voronoi区域。
我尝试了基于https://rosettacode.org/wiki/Voronoi_diagram的下一个代码
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()
我有想要的输出:
但是生成输出会花费太多时间。
我也尝试过https://stackoverflow.com/a/20678647 它很快,但是我没有找到将其转换为img_width X img_height的numpy数组的方法。通常是因为我不知道如何将图像大小参数赋予scipy Voronoi class。
有没有更快的方法来获得此输出?不需要中心或多边形边缘
预先感谢
编辑2018-12-11: 使用@tel“快速解决方案”
代码执行速度更快,似乎中心已经转换。可能是这种方法会在图片上添加边距
答案 0 :(得分:4)
这是将链接到的fast solution based on scipy.spatial.Voronoi
的输出转换为任意宽度和高度的Numpy数组的方法。给定您从链接代码中的regions, vertices
函数的输出中获得的voronoi_finite_polygons_2d
的集合,以下是一个辅助函数,该函数会将输出转换为数组:
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)
这是vorarr
的完整示例:
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)
输出:
如您所见,根据对(1000, 1000)
的调用所指定的,所得的Numpy数组确实具有vorarr
的形状。
以下是更改当前代码以使用/返回Numpy数组的方法:
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))
imgx,imgy = width, height
num_cells=len(centers_x)
nx = centers_x
ny = centers_y
nr = list(range(num_cells))
ng = nr
nb = nr
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, :] = (nr[j], ng[j], nb[j])
plt.imshow(arr.astype(int))
plt.show()
return arr
答案 1 :(得分:1)
也可以使用不使用matplotlib的快速解决方案。您的解决方案很慢,因为您要遍历所有像素,这会在Python中产生很多开销。一个简单的解决方案是在一个numpy操作中计算所有距离,并在另一个操作中分配所有颜色。
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)
相对于原始迭代解决方案,在我的机器上运行此功能可获得约10倍的加速(不考虑绘图并将结果保存到磁盘)。我确信还有很多其他调整可以进一步加快我的解决方案的速度。