LDA说的比较利索参考：/a/1190000012215533
# -*- coding: utf-8 -*-'''Created on 5月16日@author: userp:输入的概率分布，离散情况采用元素为概率值的数组表示N:认为迭代N次马尔可夫链收敛Nlmax:马尔可夫链收敛后又取的服从p分布的样本数isMH:是否采用MH算法，默认为True'''from __future__ import divisionimport matplotlib.pyplot as pltimport numpy as npfrom array import arraydef mcmc(p,N=10000,Nlmax=10000,isMH=True):A = np.array([p for y in range(len(p))], dtype=np.float64) #第一步：构造转移概率矩阵X0 = np.random.randint(len(p))count = 0samplecount = 0L = array("d",[X0])l = array("d")while True:X = int(L[samplecount])#第二步：初始化x0cur = np.argmax(np.random.multinomial(1,A[X]))#第三步：采样候选样本count += 1if isMH:a = (p[cur]*A[cur][X])/(p[X]*A[X][cur])#第四步：计算是否满足马氏平稳条件alpha = min(a,1)else:alpha = p[cur]*A[cur][X]u = np.random.uniform(0,1) #第五步：生成阈值if u<alpha:#第六步：是否接受样本samplecount += 1L.append(cur)if count>N:l.append(cur)if len(l)>=Nlmax:breakelse:continueLa = np.frombuffer(L)la = np.frombuffer(l)return La,ladef count(q,n):L = array("d")l1 = array("d")l2 = array("d")for e in q:L.append(e)for e in range(n):l1.append(L.count(e))for e in l1:l2.append(e/sum(l1))return l1,l2if __name__ == '__main__': p = np.array([0.6,0.3,0.1]) a = mcmc(p)[1]l1 = ['state%d'% x for x in range(len(p))]plt.pie(count(a,len(p))[0],labels=l1,labeldistance=0.3,autopct='%1.2f%%')plt.title("sampling")plt.show()
