hadoop map任务怎么一次读取一整块数据而非一次读取一行数据

发布网友 发布时间：2022-04-14 05:10

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共2个回答

懂视网 时间：2022-04-14 09:32

代码里面按行读取：

 @Override
 protected void map(LongWritable key, Text value, Context context) throws IOException, InterruptedException {
  FileSystem fs = FileSystem.get(context.getConfiguration());
  for (FileStatus status : fs.listStatus(new Path(value.toString()))) {
  // process file
  }
 }

都不能满足需求，还是自己实现一个 OneMapOneDirectoryInputFormat 吧，也很简单：

import java.io.IOException;
import java.util.*;
import org.apache.commons.logging.Log;
import org.apache.commons.logging.LogFactory;
import org.apache.hadoop.fs.FileStatus;
import org.apache.hadoop.fs.Path;
import org.apache.hadoop.mapreduce.InputSplit;
import org.apache.hadoop.mapreduce.JobContext;
import org.apache.hadoop.mapreduce.lib.input.CombineFileInputFormat;
import org.apache.hadoop.mapreduce.lib.input.CombineFileSplit;
/**
 * 一个map处理一个目录的数据
 */
public abstract class OneMapOneDirectoryInputFormat extends CombineFileInputFormat {
 private static final Log LOG = LogFactory.getLog(OneMapOneDirectoryInputFormat.class);
 @Override
 protected boolean isSplitable(JobContext context, Path file) {
 return false;
 }
 @Override
 public List getSplits(JobContext job) throws IOException {
 // get all the files in input path
 List stats = listStatus(job);
 List splits = new ArrayList();
 if (stats.size() == 0) {
  return splits;
 }
 LOG.info("fileNums=" + stats.size());
 Map> map = new HashMap>();
 for (FileStatus stat : stats) {
  String directory = stat.getPath().getParent().toString();
  if (map.containsKey(directory)) {
  map.get(directory).add(stat);
  } else {
  List fileList = new ArrayList();
  fileList.add(stat);
  map.put(directory, fileList);
  }
 }
 // 设置inputSplit
 long currentLen = 0;
 List pathLst = new ArrayList();
 List offsetLst = new ArrayList();
 List lengthLst = new ArrayList();
 Iterator itr = map.keySet().iterator();
 while (itr.hasNext()) {
  String dir = itr.next();
  List fileList = map.get(dir);
  for (int i = 0; i < fileList.size(); i++) {
  FileStatus stat = fileList.get(i);
  pathLst.add(stat.getPath());
  offsetLst.add(0L);
  lengthLst.add(stat.getLen());
  currentLen += stat.getLen();
  }
  Path[] pathArray = new Path[pathLst.size()];
  CombineFileSplit thissplit = new CombineFileSplit(pathLst.toArray(pathArray),
   getLongArray(offsetLst), getLongArray(lengthLst), new String[0]);
  LOG.info("combineFileSplit(" + splits.size() + ") fileNum(" + pathLst.size()
   + ") length(" + currentLen + ")");
  for (int i = 0; i < pathArray.length; i++) {
  LOG.info(" -> path[" + i + "]=" + pathArray[i].toString());
  }
  splits.add(thissplit);
  pathLst.clear();
  offsetLst.clear();
  lengthLst.clear();
  currentLen = 0;
 }
 return splits;
 }
 private long[] getLongArray(List lst) {
 long[] rst = new long[lst.size()];
 for (int i = 0; i < lst.size(); i++) {
  rst[i] = lst.get(i);
 }
 return rst;
 }
}

这个InputFormat的具体使用方法就不说了。其实与“一个Hadoop程序的优化过程 – 根据文件实际大小实现CombineFileInputFormat”中的MultiFileInputFormat比较类似。

原文地址：Hadoop : 一个目录下的数据只由一个map处理, 感谢原作者分享。

热心网友 时间：2022-04-14 06:40