在圆内的网格中生成坐标 [英] Generate coordinates in grid that lie within a circle
问题描述
我找到了
像这样的东西(对于原点处的圆)怎么办?
X = int(R)#R是半径对于范围内的x(-X,X + 1):Y = int(((R * R-x * x)** 0.5)#给定y给定x对于范围(-Y,Y + 1)中的y:产量(x,y)
当圆不以原点为中心时,可以很容易地适应一般情况.
I've found this answer, which seems to be somewhat related to this question, but I'm wondering if it's possible to generate the coordinates one by one without the additional ~22% (1 - pi / 4) loss of comparing each point to the radius of the circle (by computing the distance between the circle's center and that point).
So far I have the following function in Python. I know by Gauss' circle problem the number of coordinates I will end up with, but I'd like to generate those points one by one as well.
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
What I'm trying to do is iterate over only those pixels lying within 3 * sigma of a pixel (at centroid
in the above function).
I've since written the following example script that demonstrates that the solution below is accurate.
#! /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)
And the output for
./circleCheck.py -s 4
is:
What about simply something like this (for a circle at origin)?
X = int(R) # R is the radius
for x in range(-X,X+1):
Y = int((R*R-x*x)**0.5) # bound for y given x
for y in range(-Y,Y+1):
yield (x,y)
This can easily be adapted to the general case when the circle is not centred at the origin.
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