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我有以下示例数据框：

No  category    problem_definition_stopwords
175 2521       ['coffee', 'maker', 'brewing', 'properly', '2', '420', '420', '420']
211 1438       ['galley', 'work', 'table', 'stuck']
912 2698       ['cloth', 'stuck']
572 2521       ['stuck', 'coffee']

'problem_definition_stopwords' 字段已经被标记化，去除了停用词。

我想从“problem_definition_stopwords”字段创建 n-gram。具体来说，我想从我的数据中提取 n-gram 并找到具有最高点互信息 (PMI) 的那些。

本质上，我想找到同时出现的单词比我偶然期望的要多得多。

我尝试了以下代码：

import nltk
from nltk.collocations import *

bigram_measures = nltk.collocations.BigramAssocMeasures()
trigram_measures = nltk.collocations.TrigramAssocMeasures()

# errored out here 
finder = BigramCollocationFinder.from_words(nltk.corpus.genesis.words(df['problem_definition_stopwords']))

# only bigrams that appear 3+ times
finder.apply_freq_filter(3) 

# return the 10 n-grams with the highest PMI
finder.nbest(bigram_measures.pmi, 10)

我收到的错误是在第三块代码上... TypeError: join() argument must be str or bytes, not 'list'

编辑：DataFrame 的一种更便携的格式：

>>> df.columns
Index(['No', 'category', 'problem_definition_stopwords'], dtype='object')
>>> df.to_dict()
{'No': {0: 175, 1: 211, 2: 912, 3: 572}, 'category': {0: 2521, 1: 1438, 2: 2698, 3: 2521}, 'problem_definition_stopwords': {0: ['coffee', 'maker', 'brewing', 'properly', '2', '420', '420', '420'], 1: ['galley', 'work', 'table', 'stuck'], 2: ['cloth', 'stuck'], 3: ['stuck', 'coffee']}}
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1 回答 1

2

看起来你没有from_words以正确的方式使用呼叫，看着help(nltk.corpus.genesis.words)

Help on method words in module nltk.corpus.reader.plaintext:

words(fileids=None) method of nltk.corpus.reader.plaintext.PlaintextCorpusReader instance
    :return: the given file(s) as a list of words
        and punctuation symbols.
    :rtype: list(str)
(END)

这是你要找的吗？由于您已经将文档表示为字符串列表，根据我的经验，这与 NLTK 配合得很好，我认为您可以使用以下from_documents方法：

finder = BigramCollocationFinder.from_documents(
    df['problem_definition_stopwords']
)

# only bigrams that appear 3+ times
# Note, I limited this to 1 since the corpus you provided
# is very small and it'll be tough to find repeat ngrams
finder.apply_freq_filter(1) 

# return the 10 n-grams with the highest PMI
finder.nbest(bigram_measures.pmi, 10) 

[('brewing', 'properly'), ('galley', 'work'), ('maker', 'brewing'), ('properly', '2'), ('work', 'table'), ('coffee', 'maker'), ('2', '420'), ('cloth', 'stuck'), ('table', 'stuck'), ('420', '420')]
于 2018-11-30T15:47:52.360 回答