设ρ(半径)和φ(方位角)是两个随机变量，对应于圆内任意一点的极坐标。如果这些点是均匀分布的，那么ρ和φ的分布函数是什么?

对于任意r: 0 < r < r，半径坐标ρ小于r的概率为

P[ρ < r] = P[点在半径r的圆内]= S1 / S0 =(r/ r)2

其中S1和S0分别是半径为r和r的圆的面积。 因此，CDF可表示为:

          0          if r<=0
  CDF =   (r/R)**2   if 0 < r <= R
          1          if r > R

和PDF格式:

PDF = d/dr(CDF) = 2 * (r/R**2) (0 < r <= R).

请注意，对于R=1随机变量根号(X)，其中X在[0,1]上是一致的，有这个确切的CDF(因为P[根号(X) < y] = P[X < y**2] = y**2对于0 < y <= 1)。

φ在0 ~ 2*π范围内分布明显均匀。现在你可以创建随机极坐标，并使用三角方程将其转换为笛卡尔坐标:

x = ρ * cos(φ)
y = ρ * sin(φ)

忍不住要发布R=1的python代码。

from matplotlib import pyplot as plt
import numpy as np

rho = np.sqrt(np.random.uniform(0, 1, 5000))
phi = np.random.uniform(0, 2*np.pi, 5000)

x = rho * np.cos(phi)
y = rho * np.sin(phi)

plt.scatter(x, y, s = 4)

你会得到

首先我们生成一个cdf[x]

一点到圆心的距离小于x的概率。假设圆的半径为R。

显然，如果x = 0，那么cdf[0] = 0

显然，如果x是R，则cdf[R] = 1

显然，如果x = r，则cdf[r] = (r^2)/(r^2)

这是因为圆上的每个“小区域”都有相同的被选中的概率，所以概率与问题区域成比例。距离圆心x的面积是r^2

所以cdf[x] = x^2/R^2因为两者相互抵消了

我们有cdf[x]=x^2/R^2其中x从0到R

我们解出x

R^2 cdf[x] = x^2

x = R Sqrt[ cdf[x] ]

现在我们可以用一个从0到1的随机数来替换cdf

x = R Sqrt[ RandomReal[{0,1}] ]

最后

r = R Sqrt[  RandomReal[{0,1}] ];
theta = 360 deg * RandomReal[{0,1}];
{r,theta}

我们得到极坐标 {0.601168 R, 311.915°}

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

程序员解决方案:

创建一个位图(布尔值的矩阵)。你想要多大就有多大。 在位图中画一个圆。 创建一个圆的点查找表。 在这个查找表中选择一个随机索引。

const int RADIUS = 64;
const int MATRIX_SIZE = RADIUS * 2;

bool matrix[MATRIX_SIZE][MATRIX_SIZE] = {0};

struct Point { int x; int y; };

Point lookupTable[MATRIX_SIZE * MATRIX_SIZE];

void init()
{
  int numberOfOnBits = 0;

  for (int x = 0 ; x < MATRIX_SIZE ; ++x)
  {
    for (int y = 0 ; y < MATRIX_SIZE ; ++y)
    {
      if (x * x + y * y < RADIUS * RADIUS) 
      {
        matrix[x][y] = true;

        loopUpTable[numberOfOnBits].x = x;
        loopUpTable[numberOfOnBits].y = y;

        ++numberOfOnBits;

      } // if
    } // for
  } // for
} // ()

Point choose()
{
  int randomIndex = randomInt(numberOfBits);

  return loopUpTable[randomIndex];
} // ()

位图仅用于解释逻辑。这是没有位图的代码:

const int RADIUS = 64;
const int MATRIX_SIZE = RADIUS * 2;

struct Point { int x; int y; };

Point lookupTable[MATRIX_SIZE * MATRIX_SIZE];

void init()
{
  int numberOfOnBits = 0;

  for (int x = 0 ; x < MATRIX_SIZE ; ++x)
  {
    for (int y = 0 ; y < MATRIX_SIZE ; ++y)
    {
      if (x * x + y * y < RADIUS * RADIUS) 
      {
        loopUpTable[numberOfOnBits].x = x;
        loopUpTable[numberOfOnBits].y = y;

        ++numberOfOnBits;
      } // if
    } // for
  } // for
} // ()

Point choose()
{
  int randomIndex = randomInt(numberOfBits);

  return loopUpTable[randomIndex];
} // ()

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.