python - 主题分布：在python中做LDA后如何查看哪个文档属于哪个主题

Question

我能够从 gensim 运行 LDA 代码，并获得前 10 个主题及其各自的关键字。

现在我想更进一步，通过查看它们聚集到每个主题中的文档来了解 LDA 算法的准确性。这在gensim LDA中可能吗？

基本上我想做这样的事情，但是在 python 中并使用 gensim。

LDA 与 topicmodels，我如何查看不同文档属于哪些主题？

Answer 1

使用主题的概率，您可以尝试设置一些阈值并将其用作聚类基线，但我相信有比这种“hacky”方法更好的聚类方法。

from gensim import corpora, models, similarities
from itertools import chain

""" DEMO """
documents = ["Human machine interface for lab abc computer applications",
             "A survey of user opinion of computer system response time",
             "The EPS user interface management system",
             "System and human system engineering testing of EPS",
             "Relation of user perceived response time to error measurement",
             "The generation of random binary unordered trees",
             "The intersection graph of paths in trees",
             "Graph minors IV Widths of trees and well quasi ordering",
             "Graph minors A survey"]

# remove common words and tokenize
stoplist = set('for a of the and to in'.split())
texts = [[word for word in document.lower().split() if word not in stoplist]
         for document in documents]

# remove words that appear only once
all_tokens = sum(texts, [])
tokens_once = set(word for word in set(all_tokens) if all_tokens.count(word) == 1)
texts = [[word for word in text if word not in tokens_once] for text in texts]

# Create Dictionary.
id2word = corpora.Dictionary(texts)
# Creates the Bag of Word corpus.
mm = [id2word.doc2bow(text) for text in texts]

# Trains the LDA models.
lda = models.ldamodel.LdaModel(corpus=mm, id2word=id2word, num_topics=3, \
                               update_every=1, chunksize=10000, passes=1)

# Prints the topics.
for top in lda.print_topics():
  print top
print

# Assigns the topics to the documents in corpus
lda_corpus = lda[mm]

# Find the threshold, let's set the threshold to be 1/#clusters,
# To prove that the threshold is sane, we average the sum of all probabilities:
scores = list(chain(*[[score for topic_id,score in topic] \
                      for topic in [doc for doc in lda_corpus]]))
threshold = sum(scores)/len(scores)
print threshold
print

cluster1 = [j for i,j in zip(lda_corpus,documents) if i[0][1] > threshold]
cluster2 = [j for i,j in zip(lda_corpus,documents) if i[1][1] > threshold]
cluster3 = [j for i,j in zip(lda_corpus,documents) if i[2][1] > threshold]

print cluster1
print cluster2
print cluster3

[out]：

0.131*trees + 0.121*graph + 0.119*system + 0.115*user + 0.098*survey + 0.082*interface + 0.080*eps + 0.064*minors + 0.056*response + 0.056*computer
0.171*time + 0.171*user + 0.170*response + 0.082*survey + 0.080*computer + 0.079*system + 0.050*trees + 0.042*graph + 0.040*minors + 0.040*human
0.155*system + 0.150*human + 0.110*graph + 0.107*minors + 0.094*trees + 0.090*eps + 0.088*computer + 0.087*interface + 0.040*survey + 0.028*user

0.333333333333

['The EPS user interface management system', 'The generation of random binary unordered trees', 'The intersection graph of paths in trees', 'Graph minors A survey']
['A survey of user opinion of computer system response time', 'Relation of user perceived response time to error measurement']
['Human machine interface for lab abc computer applications', 'System and human system engineering testing of EPS', 'Graph minors IV Widths of trees and well quasi ordering']

---

只是为了更清楚：

# Find the threshold, let's set the threshold to be 1/#clusters,
# To prove that the threshold is sane, we average the sum of all probabilities:
scores = []
for doc in lda_corpus
    for topic in doc:
        for topic_id, score in topic:
            scores.append(score)
threshold = sum(scores)/len(scores)

上面的代码是所有文档的所有单词和所有主题的得分之和。然后通过分数的数量对总和进行归一化。

Answer 2

如果你想使用的技巧

cluster1 = [j for i,j in zip(lda_corpus,documents) if i[0][1] > threshold]
cluster2 = [j for i,j in zip(lda_corpus,documents) if i[1][1] > threshold]
cluster3 = [j for i,j in zip(lda_corpus,documents) if i[2][1] > threshold]

在 alvas 的上一个答案中，确保在 LdaModel 中设置 minimum_probability=0

gensim.models.ldamodel.LdaModel(corpus,
            num_topics=num_topics, id2word = dictionary,
            passes=2, minimum_probability=0)

否则，lda_corpus 和文档的维度可能不一致，因为 gensim 会抑制概率低于 minimum_probability 的任何语料库。

将文档分组为主题的另一种方法是根据最大概率分配主题

    lda_corpus = [max(prob,key=lambda y:y[1])
                    for prob in lda[mm] ]
    playlists = [[] for i in xrange(topic_num])]
    for i, x in enumerate(lda_corpus):
        playlists[x[0]].append(documents[i])

注意lda[mm]粗略地说是列表列表或二维矩阵。行数是文档数，列数是主题数。例如，每个矩阵元素都是形式的元组(3,0.82)。这里 3 是指主题索引，0.82 是该主题的相应概率。默认情况下，minimum_probability=0.01任何概率小于 0.01 的元组在lda[mm]. 如果您使用最大概率的分组方法，则可以将其设置为 1/#topics。

Answer 3

lda_corpus[i][j] 的形式为 [(0,t1),(0,t2)...,(0,t10),....(n,t10)] 其中第一项表示文档索引和第二项表示该特定文档中主题的概率。