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Java 基于mahout用户的协同过滤推荐中intro.csv的FileNotFountException_Java_Maven_Csv_Mahout_Mahout Recommender - Fatal编程技术网

Java 基于mahout用户的协同过滤推荐中intro.csv的FileNotFountException

Java 基于mahout用户的协同过滤推荐中intro.csv的FileNotFountException,java,maven,csv,mahout,mahout-recommender,Java,Maven,Csv,Mahout,Mahout Recommender,我创建了一个maven项目,并在pom文件中添加了以下依赖项 <dependency> <groupId>org.apache.mahout</groupId> <artifactId>mahout-mr</artifactId> <version>0.11.0</version> </dependency> 我在mahout中使用了这个例子,

我创建了一个maven项目,并在pom文件中添加了以下依赖项

<dependency>
        <groupId>org.apache.mahout</groupId>
        <artifactId>mahout-mr</artifactId>
        <version>0.11.0</version>
    </dependency>
我在mahout中使用了这个例子,并添加了RecommenderIntro 类,并将intro.csv文件复制到src/main/resource文件夹中

错误发生在:

DataModel model = new FileDataModel (new File("resources\intro.csv"));

Exception in thread "main" java.io.FileNotFoundException: resources/intro.csv
    at org.apache.mahout.cf.taste.impl.model.file.FileDataModel.<init>(FileDataModel.java:180)
    at org.apache.mahout.cf.taste.impl.model.file.FileDataModel.<init>(FileDataModel.java:167)
    at org.apache.mahout.cf.taste.impl.model.file.FileDataModel.<init>(FileDataModel.java:147)
    at recstest.UserBased.main(UserBased.java:19)

我还将该文件复制到main src\main\java并进行了检查,仍然出现相同的错误。原因可能是什么?

使用此选项从类路径加载资源:

    URL url = getClass().getResource("/intro.csv");
    Path path = Paths.get(url.toURI());
    File file = path.toFile();

您只需在类上使用getResource即可获得所需的资源

DataModel model = new FileDataModel(new File(this.getClass().getResource("/intro.csv").toURI()));
当然,您需要以某种方式处理可能的URISyntaxException和IOException