您需要SentimentCoreAnnotations.SentimentAnnotatedTree.class
从带注释的句子中获取树对象。RNNCoreAnnotations
然后,您可以通过班级获得预测。我在下面编写了以下独立的演示代码,展示了如何获取 CoreNLP 情绪预测的每个标签的分数。
import java.util.Arrays;
import java.util.List;
import java.util.Properties;
import org.ejml.simple.SimpleMatrix;
import edu.stanford.nlp.ling.CoreAnnotations;
import edu.stanford.nlp.neural.rnn.RNNCoreAnnotations;
import edu.stanford.nlp.pipeline.Annotation;
import edu.stanford.nlp.pipeline.StanfordCoreNLP;
import edu.stanford.nlp.sentiment.SentimentCoreAnnotations;
import edu.stanford.nlp.trees.Tree;
import edu.stanford.nlp.util.CoreMap;
public class DemoSentiment {
public static void main(String[] args) {
final List<String> texts = Arrays.asList("I am happy.", "This is a neutral sentence.", "I am very angry.");
final Properties props = new Properties();
props.setProperty("annotators", "tokenize, ssplit, parse, sentiment");
final StanfordCoreNLP pipeline = new StanfordCoreNLP(props);
for (String text : texts) {
final Annotation doc = new Annotation(text);
pipeline.annotate(doc);
for (CoreMap sentence : doc.get(CoreAnnotations.SentencesAnnotation.class)) {
final Tree tree = sentence.get(SentimentCoreAnnotations.SentimentAnnotatedTree.class);
final SimpleMatrix sm = RNNCoreAnnotations.getPredictions(tree);
final String sentiment = sentence.get(SentimentCoreAnnotations.SentimentClass.class);
System.out.println("sentence: "+sentence);
System.out.println("sentiment: "+sentiment);
System.out.println("matrix: "+sm);
}
}
}
}
输出将与以下内容相似(一些浮点舍入错误或更新的模型可能会改变分数)。
对于第一句I am happy.
,您可以看到情绪是,并且,在将矩阵解释为有序列表时,Positive
返回矩阵中的最大值是,在第四个位置。0.618
第二个句子This is a neutral sentence.
在中间的得分最高,在0.952
,因此是Neutral
情绪。
最后一句话有相应的Negative
情感,最高分0.652
在第二位。
sentence: I am happy.
sentiment: Positive
matrix: Type = dense , numRows = 5 , numCols = 1
0.016
0.037
0.132
0.618
0.196
sentence: This is a neutral sentence.
sentiment: Neutral
matrix: Type = dense , numRows = 5 , numCols = 1
0.001
0.007
0.952
0.039
0.001
sentence: I am very angry.
sentiment: Negative
matrix: Type = dense , numRows = 5 , numCols = 1
0.166
0.652
0.142
0.028
0.012