我找到了this answer,这似乎与这个问题有些相关,但是我想知道是否可以一个接一个地生成坐标而没有额外的〜22%( 1 - π/ 4)将每个点与圆的半径进行比较的损失(通过计算圆的中心与该点之间的距离)。在网格中生成坐标在一个圆内的坐标
到目前为止,我在Python中有以下功能。我知道Gauss' circle problem号码坐标我会结束,但我想要一个一个地产生这些点。
from typing import Iterable
from math import sqrt, floor
def circCoord(sigma: float =1.0, centroid: tuple =(0, 0)) -> Iterable[tuple]:
r""" Generate all coords within $3\vec{\sigma}$ of the centroid """
# The number of least iterations is given by Gauss' circle problem:
# http://mathworld.wolfram.com/GausssCircleProblem.html
maxiterations = 1 + 4 * floor(3 * sigma) + 4 * sum(\
floor(sqrt(9 * sigma**2 - i**2)) for i in range(1, floor(3 * sigma) + 1)
)
for it in range(maxiterations):
# `yield` points in image about `centroid` over which we loop
我试图做的是遍历只有那些(在上面的功能在centroid
)趴在3 *西格玛像素的像素。
我写了下面的示例脚本,证明下面的解决方案是准确的。
#! /usr/bin/env python3
# -*- coding: utf-8 -*-
import matplotlib.pyplot as plt
from matplotlib.patches import Ellipse
import numpy as np
import argparse
from typing import List, Tuple
from math import sqrt
def collect(x: int, y: int, sigma: float =3.0) -> List[Tuple[int, int]]:
""" create a small collection of points in a neighborhood of some point
"""
neighborhood = []
X = int(sigma)
for i in range(-X, X + 1):
Y = int(pow(sigma * sigma - i * i, 1/2))
for j in range(-Y, Y + 1):
neighborhood.append((x + i, y + j))
return neighborhood
def plotter(sigma: float =3.0) -> None:
""" Plot a binary image """
arr = np.zeros([sigma * 2 + 1] * 2)
points = collect(int(sigma), int(sigma), sigma)
# flip pixel value if it lies inside (or on) the circle
for p in points:
arr[p] = 1
# plot ellipse on top of boxes to show their centroids lie inside
circ = Ellipse(\
xy=(int(sigma), int(sigma)),
width=2 * sigma,
height=2 * sigma,
angle=0.0
)
fig = plt.figure(0)
ax = fig.add_subplot(111, aspect='equal')
ax.add_artist(circ)
circ.set_clip_box(ax.bbox)
circ.set_alpha(0.2)
circ.set_facecolor((1, 1, 1))
ax.set_xlim(-0.5, 2 * sigma + 0.5)
ax.set_ylim(-0.5, 2 * sigma + 0.5)
plt.scatter(*zip(*points), marker='.', color='white')
# now plot the array that's been created
plt.imshow(-arr, interpolation='none', cmap='gray')
#plt.colorbar()
plt.show()
if __name__ == '__main__':
parser = argparse.ArgumentParser()
parser.add_argument('-s', '--sigma', type=int, \
help='Circle about which to collect points'
)
args = parser.parse_args()
plotter(args.sigma)
而且输出
./circleCheck.py -s 4
是:
这看起来更好,因为它不是浪费。 – bjd2385
我已经在上面添加了一个小测试脚本,这也使得一个很好的视觉'确认'情节。 – bjd2385