python - 来自python中文本的n-grams

Question

对我以前的post进行了更新，并做了一些更改：
说我有100条微博。
在这些推特中，我需要提取：1）食品名称，2）饮料名称。我还需要附上类型（饮料或食品）和一个id号（每个项目有一个唯一的id）为每个提取。
我已经有了名字、类型和ID号的词典：

lexicon = {
'dr pepper': {'type': 'drink', 'id': 'd_123'},
'coca cola': {'type': 'drink', 'id': 'd_234'},
'cola': {'type': 'drink', 'id': 'd_345'},
'banana': {'type': 'food', 'id': 'f_456'},
'banana split': {'type': 'food', 'id': 'f_567'},
'cream': {'type': 'food', 'id': 'f_678'},
'ice cream': {'type': 'food', 'id': 'f_789'}}

推特示例：
经过“tweetç1”的各种处理，我有以下句子：

sentences = [
'dr pepper is better than coca cola and suits banana split with ice cream', 
'coca cola and banana is not a good combo']

我请求的输出（可以是列表以外的其他类型）：

["tweet_id_1",
 [[["dr pepper"], ["drink", "d_124"]],
  [["coca cola"], ["drink", "d_234"]],
  [["banana split"], ["food", "f_567"]],
  [["ice cream"], ["food", "f_789"]]],

 "tweet_id_1",,
 [[["coca cola"], ["drink", "d_234"]],
  [["banana"], ["food", "f_456"]]]]

重要的是，输出不应在ngrams中提取unigrams（n>1）：

["tweet_id_1",
 [[["dr pepper"], ["drink", "d_124"]],
  [["coca cola"], ["drink", "d_234"]],
  [["cola"], ["drink", "d_345"]],
  [["banana split"], ["food", "f_567"]],
  [["banana"], ["food", "f_456"]],
  [["ice cream"], ["food", "f_789"]],
  [["cream"], ["food", "f_678"]]],

 "tweet_id_1",
 [[["coca cola"], ["drink", "d_234"]],
  [["cola"], ["drink", "d_345"]],
  [["banana"], ["food", "f_456"]]]]

理想情况下，我希望能够在提取之前在各种nltk过滤器中运行我的语句，比如lemmatize（）和pos_tag（），以获得如下输出。但是有了这个regexp解决方案，如果我这样做了，那么所有的单词都被拆分成unigram，或者它们将从字符串“coca cola”生成1个unigram和1个bigram，这将生成我不想要的输出（如上面的例子）。
理想输出（同样，输出类型不重要）：

["tweet_id_1",
 [[[("dr pepper", "NN")], ["drink", "d_124"]],
  [[("coca cola", "NN")], ["drink", "d_234"]],
  [[("banana split", "NN")], ["food", "f_567"]],
  [[("ice cream", "NN")], ["food", "f_789"]]],

 "tweet_id_1",
 [[[("coca cola", "NN")], ["drink", "d_234"]],
  [[("banana", "NN")], ["food", "f_456"]]]]

Answer 1

可能不是最有效的解决方案，但这肯定会让你开始-

sentences = [
'dr pepper is better than coca cola and suits banana split with ice cream', 
'coca cola and banana is not a good combo']

lexicon = {
'dr pepper': {'type': 'drink', 'id': 'd_123'},
'coca cola': {'type': 'drink', 'id': 'd_234'},
'cola': {'type': 'drink', 'id': 'd_345'},
'banana': {'type': 'food', 'id': 'f_456'},
'banana split': {'type': 'food', 'id': 'f_567'},
'cream': {'type': 'food', 'id': 'f_678'},
'ice cream': {'type': 'food', 'id': 'f_789'}}

lexicon_list = list(lexicon.keys())
lexicon_list.sort(key = lambda s: len(s.split()), reverse=True)

chunks = []

for sentence in sentences:
    for lex in lexicon_list:
        if lex in sentence:
                chunks.append({lex: list(lexicon[lex].values()) })
                sentence = sentence.replace(lex, '')

print(chunks)

输出

[{'dr pepper': ['drink', 'd_123']}, {'coca cola': ['drink', 'd_234']}, {'banana split': ['food', 'f_567']}, {'ice cream': ['food', 'f_789']}, {'coca cola': ['drink', 'd_234']}, {'banana': ['food', 'f_456']}]

解释
lexicon_list = list(lexicon.keys())获取需要搜索的短语列表，并按长度对它们进行排序（以便首先找到较大的块）
输出是一个 dict列表，其中每个dict都有 list值。