5

最近，我使用 numpy 在 Python 上实现了 LDA 主题模型的 Gibbs 采样，参考了网站上的一些代码。在 Gibbs 采样的每次迭代中，我们删除一个（当前）单词，根据从 LDA 模型推断的后验条件概率分布为该单词采样一个新主题，并更新单词主题计数，如下所示：

for m, doc in enumerate(docs): #m: doc id
  for n, t in enumerate(doc): #n: id of word inside document, t: id of the word globally
    # discount counts for word t with associated topic z
    z = z_m_n[m][n]
    n_m_z[m][z] -= 1
    n_z_t[z, t] -= 1 
    n_z[z] -= 1
    n_m[m] -= 1

    # sample new topic for multinomial                
    p_z_left = (n_z_t[:, t] + beta) / (n_z + V * beta)
    p_z_right = (n_m_z[m] + alpha) / ( n_m[m] + alpha * K)
    p_z = p_z_left * p_z_right
    p_z /= numpy.sum(p_z)
    new_z = numpy.random.multinomial(1, p_z).argmax() 

    # set z as the new topic and increment counts
    z_m_n[m][n] = new_z
    n_m_z[m][new_z] += 1
    n_z_t[new_z, t] += 1
    n_z[new_z] += 1
    n_m[m] += 1

在上面的代码中，我们使用多项式 scipy 函数对一个新的（单个）z 进行采样。

现在，我想实现本文的联合情感主题模型。现在，我需要以下结构来跟踪所需的计数：

3D matrix containing # occurrences for a word for each topic, for each sentiment
3D matrix containing # occurrences for a topic, for each sentiment, for each document
2D matrix containing # occurrences for a topic, for each sentiment
2D matrix containing # occurrences for a sentiment for each document

现在问题来了：在这个 Gibbs 采样器中，对于文档中看到的每个单词，现在都从条件后验中采样了一个新主题和一个情感标签（论文的第 4 页等式 5）。我现在如何在 Python 中“对这两个值进行采样”？

提前致谢...

4

1 回答 1

2

试试这个。从主题和情感标签的联合分布中采样仅意味着整个 T x S 矩阵的总和应为 1。

docs=[[0,1],[0,0],[1,0,1]]
D=len(docs)
z_d_n=[[0 for _ in xrange(len(d))] for d in docs]
l_d_n=[[0 for _ in xrange(len(d))] for d in docs]

V=2
T=2
S=2
n_m_j_k=numpy.zeros( (V,T,S) )
n_j_k_d=numpy.zeros( (T,S,D) )
n_j_k=numpy.zeros( (T,S) )
n_k_d=numpy.zeros( (S,D) )
n_d=numpy.zeros( (D) )

beta=.1
alpha=.1
gamma=.1

for d, doc in enumerate(docs): #d: doc id
    for n, m in enumerate(doc): #i: index of the word inside document, m: id of the word in the vocabulary
        # j is the topic
        j = z_d_n[d][n]
        # k is the sentiment
        k = l_d_n[d][n]
        n_m_j_k[m][j][k] += 1
        n_j_k_d[j][k][d] += 1
        n_j_k[j][k] += 1
        n_k_d[k][d] += 1
        n_d[d] += 1 

for d, doc in enumerate(docs): #d: doc id
    for n, m in enumerate(doc): #i: index of the word inside document, m: id of the word in the vocabulary
        # j is the topic
        j = z_d_n[d][n]
        # k is the sentiment
        k = l_d_n[d][n]
        n_m_j_k[m][j][k] -= 1
        n_j_k_d[j][k][d] -= 1
        n_j_k[j][k] -= 1
        n_k_d[k][d] -= 1
        n_d[d] -= 1 

        # sample a new topic and sentiment label jointly
        # T is the number of topics
        # S is the number of sentiments
        p_left = (n_m_j_k[m] + beta) / (n_j_k + V * beta) # T x S array
        p_mid = (n_j_k_d[:,:,d] + alpha) / numpy.tile(n_k_d[:,d] + T * alpha, (T,1) )
        p_right = numpy.tile(n_k_d[:,d] + gamma,(T,1)) /  numpy.tile(n_d[d] + S * gamma,(T,S))
        p = p_left * p_mid * p_right
        p /= numpy.sum(p)
        new_jk = numpy.random.multinomial(1, numpy.reshape(p, (T*S) )).argmax()
        j=new_jk/T
        k=new_jk%T

        z_d_n[d][n]=j
        l_d_n[d][n]=k
        n_m_j_k[m][j][k] += 1
        n_j_k[j][k] += 1
        n_k_d[k][d] += 1
        n_d[d] += 1
于 2012-07-10T02:26:38.243 回答