如何在圆形分布中生成随机点

Question

I am wondering how i could generate random numbers that appear in a circular distribution.

I am able to generate random points in a rectangular distribution such that the points are generated within the square of (0 <= x < 1000, 0 <= y < 1000):

How would i go upon to generate the points within a circle such that:

(x−500)^2 + (y−500)^2 < 250000 ?

Answer 1

import random
import math

# radius of the circle
circle_r = 10
# center of the circle (x, y)
circle_x = 5
circle_y = 7

# random angle
alpha = 2 * math.pi * random.random()
# random radius
r = circle_r * math.sqrt(random.random())
# calculating coordinates
x = r * math.cos(alpha) + circle_x
y = r * math.sin(alpha) + circle_y

print("Random point", (x, y))

In your example circle_x is 500 as circle_y is. circle_r is 500.

Another version of calculating radius to get uniformly distributed points, based on this answer

u = random.random() + random.random()
r = circle_r * (2 - u if u > 1 else u)

Answer 2

FIRST ANSWER: An easy solution would be to do a check to see if the result satisfies your equation before proceeding.

Generate x, y (there are ways to randomize into a select range)

Check if ((x−500)^2 + (y−500)^2 < 250000) is true if not, regenerate.

The only downside would be inefficiency.

SECOND ANSWER:

OR, you could do something similar to riemann sums like for approximating integrals. Approximate your circle by dividing it up into many rectangles. (the more rectangles, the more accurate), and use your rectangle algorithm for each rectangle within your circle.

Answer 3

What you need is to sample from (polar form):

r, theta = [math.sqrt(random.randint(0,500))*math.sqrt(500), 2*math.pi*random.random()]

You can then transform r and theta back to cartesian coordinates x and y via

x = 500 + r * math.cos(theta) 
y = 500 + r * math.sin(theta)

Related (although not Python), but gives the idea.

Answer 4

You can use below the code and if want to learn more https://programming.guide/random-point-within-circle.html

import random
import math
circle_x = 500
circle_y = 500
a = random.randint(0,500) * 2 * math.pi
r = 1 * math.sqrt(random.randint(0,500))
x = r * math.cos(a) + circle_x
y = r * math.sin(a) + circle_y

Answer 5

here's an example hope could help someone :).

randProba = lambda a: a/sum(a)
npoints = 5000 # points to chose from
r = 1 # radius of the circle

plt.figure(figsize=(5,5))
t = np.linspace(0, 2*np.pi, npoints, endpoint=False)
x = r * np.cos(t)
y = r * np.sin(t)
plt.scatter(x, y, c='0.8')

n = 2 # number of points to chose
t = np.linspace(0, 2*np.pi, npoints, endpoint=False)[np.random.choice(range(npoints), n, replace=False, p=randProba(np.random.random(npoints)))]
x = r * np.cos(t)
y = r * np.sin(t)

plt.scatter(x, y)

Answer 6

I would use polar coordinates:

r_squared, theta = [random.randint(0,250000), 2*math.pi*random.random()]

Then r is always less than or equal to the radius, and theta always between 0 and 2*pi radians.

Since r is not at the origin, you will always convert it to a vector centered at 500, 500, if I understand correctly

x = 500 + math.sqrt(r_squared)*math.cos(theta) y = 500 + math.sqrt(r_squared)*math.sin(theta)

Choose r_squared randomly because of this

Answer 7

您可以使用拒绝采样，在覆盖圆的(2r)×(2r)方格内生成一个随机点，重复直到在圆内得到一个点。