Question

我正在尝试生成所有可能的句子组合。作为变量我有两个字符串，一个字符串将成为主题，例如health，其中一个将是一个对象，例如fruit，但是我会有一个List<String>值与一个“头”字相关联，因此与上面提到的两个组件保持一致，它们将与列表[improve, change, alter, modify]相关联。我想生成这些句子的所有可能组合，并将每个句子添加到List<Sentences>，例如：

Sentence example_sentence = new Sentence(verb, object, subject);
sentences.add(example_sentence);

现在，发生这种情况的更大功能如下所示：

public Sentence dataPreprocessing(String raw_subject, String raw_object, String raw_verb, List<Sentence> sentences) throws IOException {
    WordNet wordnet = new WordNet();
    String verb = wordnet.getStem(raw_verb);
    String object = wordnet.getStem(raw_object);        
    String subject = wordnet.getStem(raw_subject);
    List<String> verb_hypernym_container = new ArrayList<>();       
    verb_hypernym_container = wordnet.getHypernyms(verb, POS.VERB);
    //wordnet.getHypernyms(this.object, POS.NOUN);  
    //wordnet.getHypernyms(this.subject, POS.NOUN); 
    Sentence return_sentence = new Sentence( verb, object, subject );
    return return_sentence;
}

如何才能最有效地实现产生所有可能句子的目标？

Answer 1

由于您有固定数量的列表，最简单的方法是使用嵌套循环：

List<Sentence> sentences = new ArrayList<>();

for(String verb_hypernym : wordnet.getHypernyms(verb, POS.VERB))
    for(String object_hypernym : wordnet.getHypernyms(object, POS.NOUN))
        for(String subject_hypernym : wordnet.getHypernyms(subject, POS.NOUN))
            sentences.add(new Sentence(verb_hypernym, object_hypernym, subject_hypernym));

return sentences;

或者，为避免更频繁地调用getHypernyms：

List<String> verb_hypernyms = wordnet.getHypernyms(verb, POS.VERB);
List<String> object_hypernyms = wordnet.getHypernyms(object, POS.NOUN);
List<String> subject_hypernyms = wordnet.getHypernyms(subject, POS.NOUN);


for(String verb_hypernym : verb_hypernyms)
    for(String object_hypernym : object_hypernyms)
        for(String subject_hypernym : subject_hypernyms)
            sentences.add(new Sentence(verb_hypernym, object_hypernym, subject_hypernym));

return sentences;

Answer 2

一旦你有一个名词和动词列表，你可以使用流来返回一个句子列表。这也使您有机会删除任何重复项，排序或您需要对流做的任何其他事情。

List<Sentence> sentences = subjectList.stream()
        .flatMap(object -> verbList.stream()
            .flatMap(verb -> objectList.stream()
                .map(subject -> new Sentence(object, verb, subject))))
        .distinct()
        .collect(Collectors.toList());

生成并输出所有可能的组合，每个唯一项</string>只有三个List <string>

2 个答案: