python - 如何使用python中的散点图创建一个在圆周上均匀分布点的圆

假设我有一个圆圈x**2 + y**2 = 20。
现在，我想在散点图中在圆的周长中绘制点数为n_dots的圆。所以我创建了如下代码：

n_dots = 200
x1 = np.random.uniform(-20, 20, n_dots//2)
y1_1 = (400 - x1**2)**.5
y1_2 = -(400 - x1**2)**.5
plt.figure(figsize=(8, 8))
plt.scatter(x1, y1_1, c = 'blue')
plt.scatter(x1, y1_2, c = 'blue')
plt.show()

但这表明圆点上的点并非均匀分布。输出为：

python - 如何使用python中的散点图创建一个在圆周上均匀分布点的圆-LMLPHP

那么，如何在散点图中创建带有点的圆，使所有点均匀分布在圆的周长中？

最佳答案

一个非常通用的答案也可以在2D模式下使用：

import numpy as np
import matplotlib.pyplot as plt


def u_sphere_pts(dim, N):
    """
    uniform  distribution points on hypersphere
    from uniform distribution in n-D (<-1, +1>) hypercube,
    clipped by unit 2 norm to get the points inside the insphere,
    normalize selected points to lie on surface of unit radius hypersphere
    """
    # uniform points in hypercube
    u_pts = np.random.uniform(low=-1.0, high=1.0, size=(dim, N))

    # n dimensional 2 norm squared
    norm2sq = (u_pts**2).sum(axis=0)

    # mask of points where 2 norm squared  < 1.0
    in_mask = np.less(norm2sq, np.ones(N))

    # use mask to select points, norms inside unit hypersphere
    in_pts = np.compress(in_mask, u_pts, axis=1)
    in_norm2 = np.sqrt(np.compress(in_mask, norm2sq))  # only sqrt selected

    # return normalized points, equivalently, projected to hypersphere surface
    return in_pts/in_norm2


# show some 2D "sphere" points
N = 1000
dim = 2
fig2, ax2 = plt.subplots()
ax2.scatter(*u_sphere_pts(dim, N))
ax2.set_aspect('equal')
plt.show()

# plot histogram of angles

pts = u_sphere_pts(dim, 1000000)
theta = np.arctan2(pts[0,:], pts[1,:])
num_bins = 360
fig1, ax1 = plt.subplots()
n, bins, patches = plt.hist(theta, num_bins, facecolor='blue', alpha=0.5)
plt.show()