这样一个有趣的问题。 一个点被选择的概率随着距离轴原点的增加而降低的基本原理在上面已经解释了多次。我们通过取U[0,1]的根来解释这一点。 下面是Python 3中正r的通解。

import numpy
import math
import matplotlib.pyplot as plt

def sq_point_in_circle(r):
    """
    Generate a random point in an r radius circle 
    centered around the start of the axis
    """

    t = 2*math.pi*numpy.random.uniform()
    R = (numpy.random.uniform(0,1) ** 0.5) * r

    return(R*math.cos(t), R*math.sin(t))

R = 200 # Radius
N = 1000 # Samples

points = numpy.array([sq_point_in_circle(R) for i in range(N)])
plt.scatter(points[:, 0], points[:,1])

朴素解不起作用的原因是它给了靠近圆中心的点更高的概率密度。换句话说，半径为r/2的圆被选中点的概率为r/2，但它的面积(点的数量)为*r^2/4。

因此，我们希望半径概率密度具有以下性质:

选择半径小于或等于给定r的概率必须与半径为r的圆的面积成正比(因为我们希望在点上有一个均匀的分布，面积越大意味着点越多)。

换句话说，我们希望在[0,r]之间选择半径的概率等于它在圆的总面积中所占的份额。圆的总面积是*R^2，半径为R的圆的面积是*R^2。因此，我们希望在[0,r]之间选择半径的概率为(pi*r^2)/(pi* r^2) = r^2/ r^2。

现在来算算:

The probability of choosing a radius between [0,r] is the integral of p(r) dr from 0 to r (that's just because we add all the probabilities of the smaller radii). Thus we want integral(p(r)dr) = r^2/R^2. We can clearly see that R^2 is a constant, so all we need to do is figure out which p(r), when integrated would give us something like r^2. The answer is clearly r * constant. integral(r * constant dr) = r^2/2 * constant. This has to be equal to r^2/R^2, therefore constant = 2/R^2. Thus you have the probability distribution p(r) = r * 2/R^2

Note: Another more intuitive way to think about the problem is to imagine that you are trying to give each circle of radius r a probability density equal to the proportion of the number of points it has on its circumference. Thus a circle which has radius r will have 2 * pi * r "points" on its circumference. The total number of points is pi * R^2. Thus you should give the circle r a probability equal to (2 * pi * r) / (pi * R^2) = 2 * r/R^2. This is much easier to understand and more intuitive, but it's not quite as mathematically sound.

我不知道这个问题是否还有新的答案，但我自己碰巧也遇到过同样的问题。我试着跟自己“讲道理”寻找解决办法，我找到了一个。这可能和一些人在这里提出的建议是一样的，但不管怎样，它是这样的:

in order for two elements of the circle's surface to be equal, assuming equal dr's, we must have dtheta1/dtheta2 = r2/r1. Writing expression of the probability for that element as P(r, theta) = P{ r1< r< r1 + dr, theta1< theta< theta + dtheta1} = f(r,theta)*dr*dtheta1, and setting the two probabilities (for r1 and r2) equal, we arrive to (assuming r and theta are independent) f(r1)/r1 = f(r2)/r2 = constant, which gives f(r) = c*r. And the rest, determining the constant c follows from the condition on f(r) being a PDF.

Java解决方案和分发示例(2000分)

public void getRandomPointInCircle() {
    double t = 2 * Math.PI * Math.random();
    double r = Math.sqrt(Math.random());
    double x = r * Math.cos(t);
    double y = r * Math.sin(t);
    System.out.println(x);
    System.out.println(y);
}

基于以前的解决方案https://stackoverflow.com/a/5838055/5224246从@sigfpe

我认为在这种情况下，使用极坐标是一种使问题复杂化的方法，如果你在一个边长为2R的正方形中随机选择点，然后选择点(x,y)使x^2+y^2<=R^2，这将会容易得多。

我仍然不确定确切的“(2/R2)×r”，但显而易见的是，在给定的单位“dr”中需要分配的点的数量，即r的增加将与R2成正比，而不是r。

check this way...number of points at some angle theta and between r (0.1r to 0.2r) i.e. fraction of the r and number of points between r (0.6r to 0.7r) would be equal if you use standard generation, since the difference is only 0.1r between two intervals. but since area covered between points (0.6r to 0.7r) will be much larger than area covered between 0.1r to 0.2r, the equal number of points will be sparsely spaced in larger area, this I assume you already know, So the function to generate the random points must not be linear but quadratic, (since number of points required to be distributed in given unit 'dr' i.e. increase in r will be proportional to r2 and not r), so in this case it will be inverse of quadratic, since the delta we have (0.1r) in both intervals must be square of some function so it can act as seed value for linear generation of points (since afterwords, this seed is used linearly in sin and cos function), so we know, dr must be quadratic value and to make this seed quadratic, we need to originate this values from square root of r not r itself, I hope this makes it little more clear.