如何将.txt文件转换为Hadoop的序列文件格式

Question

为了在Hadoop中有效地利用map-reduce作业,我需要以hadoop的序列文件格式存储数据.但是,目前数据只是平面.txt格式.任何人都建议我可以将.txt文件转换为序列文件？

Answer 1

因此,最简单的答案只是一个具有SequenceFile输出的"身份"作业.

在java中看起来像这样:

    public static void main(String[] args) throws IOException,
        InterruptedException, ClassNotFoundException {

    Configuration conf = new Configuration();
    Job job = new Job(conf);
    job.setJobName("Convert Text");
    job.setJarByClass(Mapper.class);

    job.setMapperClass(Mapper.class);
    job.setReducerClass(Reducer.class);

    // increase if you need sorting or a special number of files
    job.setNumReduceTasks(0);

    job.setOutputKeyClass(LongWritable.class);
    job.setOutputValueClass(Text.class);

    job.setOutputFormatClass(SequenceFileOutputFormat.class);
    job.setInputFormatClass(TextInputFormat.class);

    TextInputFormat.addInputPath(job, new Path("/lol"));
    SequenceFileOutputFormat.setOutputPath(job, new Path("/lolz"));

    // submit and wait for completion
    job.waitForCompletion(true);
   }

Answer 2

import java.io.IOException;
import java.net.URI;

import org.apache.hadoop.conf.Configuration;
import org.apache.hadoop.fs.FileSystem;
import org.apache.hadoop.fs.Path;
import org.apache.hadoop.io.IOUtils;
import org.apache.hadoop.io.IntWritable;
import org.apache.hadoop.io.SequenceFile;
import org.apache.hadoop.io.Text;

//White, Tom (2012-05-10). Hadoop: The Definitive Guide (Kindle Locations 5375-5384). OReilly Media - A. Kindle Edition. 

public class SequenceFileWriteDemo { 

    private static final String[] DATA = { "One, two, buckle my shoe", "Three, four, shut the door", "Five, six, pick up sticks", "Seven, eight, lay them straight", "Nine, ten, a big fat hen" };

    public static void main( String[] args) throws IOException { 
        String uri = args[ 0];
        Configuration conf = new Configuration();
        FileSystem fs = FileSystem.get(URI.create( uri), conf);
        Path path = new Path( uri);
        IntWritable key = new IntWritable();
        Text value = new Text();
        SequenceFile.Writer writer = null;
        try { 
            writer = SequenceFile.createWriter( fs, conf, path, key.getClass(), value.getClass());
            for (int i = 0; i < 100; i ++) { 
                key.set( 100 - i);
                value.set( DATA[ i % DATA.length]);
                System.out.printf("[% s]\t% s\t% s\n", writer.getLength(), key, value); 
                writer.append( key, value); } 
        } finally 
        { IOUtils.closeStream( writer); 
        } 
    } 
}

Answer 3

这取决于TXT文件的格式.每条记录是一行吗？如果是这样,您可以简单地使用TextInputFormat,为每行创建一条记录.在您的映射器中,您可以解析该行并以您选择的方式使用它.

如果每条记录不是一行,则可能需要编写自己的InputFormat实现.有关详细信息,请查看本教程.