标签：图片 inf bubuko png 参考 hunk 生成 http cos
算法 1
设半径为$R$。
$x = r \ast cos(\theta)$
$y = r \ast sin(\theta)$
其中 $0\leqslant r \leqslant R$，$t$为0-1均匀分布产生的随机数，$r = sqrt(t) \ast R$，$\theta = 2\pi \ast t, t \sim U(0, 1)$ 证明：url
1 import numpy as np 2 import random 3 import math 4 import matplotlib.pyplot as plt 5 import matplotlib as mpl 6 mpl.rcParams[‘agg.path.chunksize‘] = 10000 # the default is 0 7 N = int(10000) 8 x1 = 1.5 9 x2 = 0.5 10 radius = 2 11 a = 2 * math.pi * np.array([random.random() for _ in range(N)]) 12 r = np.array([random.random() for _ in range(N)]) 13 x = radius*np.sqrt(r)*np.cos(a) + x1; 14 y = radius*np.sqrt(r)*np.sin(a) + x2; 15 plt.scatter(x, y, s=1) 16 plt.show()
下面的算法是错误的
原因在于对$R$求开方，导致$r = sqrt(R)*t$平方后不在满足均匀分布。
1 import numpy as np 2 import random 3 import math 4 import matplotlib.pyplot as plt 5 import matplotlib as mpl 6 mpl.rcParams[‘agg.path.chunksize‘] = 10000 # the default is 0 7 N = int(10000) 8 x1 = 1.5 9 x2 = 0.5 10 radius = 2 11 a = 2 * math.pi * np.array([random.random() for _ in range(N)]) 12 r = np.array([random.random() for _ in range(N)]) 13 #x = radius*np.sqrt(r)*np.cos(a) + x1; 14 #y = radius*np.sqrt(r)*np.sin(a) + x2; 15 x = np.sqrt(radius)*r*np.cos(a) + x1; 16 y = np.sqrt(radius)*r*np.sin(a) + x2; 17 plt.scatter(x, y, s=1) 18 plt.show()
参考链接：JustDoIT
标签：图片 inf bubuko png 参考 hunk 生成 http cos
原文地址：https://www.cnblogs.com/Atanisi/p/8849215.html
