java - 在 Hadoop 中，如果你想将每个键值对的值保存到一个数组中，为什么你添加的所有元素都是相同的？

Question

我正在尝试存储 Map 函数获取的键值对中的值并进一步使用它们。给定以下输入:

Hello hadoop goodbye hadoop
Hello world goodbye world
Hello thinker goodbye thinker

如下代码:

注意 - map 是简单的 WordCount 示例

public class Inception extends Configured implements Tool{

public Path workingPath;

 public static class Map extends Mapper<LongWritable, Text, Text, IntWritable> {
    private final static IntWritable one = new IntWritable(1);
    private Text word = new Text();

     // initialising the arrays that contain the values and the keys
    public ArrayList<LongWritable> keyBuff = new ArrayList<LongWritable>();
    public ArrayList<Text> valueBuff = new ArrayList<Text>();


    public void map(LongWritable key, Text value, Context context) throws IOException, InterruptedException {
        String line = value.toString();
        StringTokenizer tokenizer = new StringTokenizer(line);

        while (tokenizer.hasMoreTokens()) {
            word.set(tokenizer.nextToken());
            context.write(word, one);
            System.out.println(word + " / " + one);
        }
    }   

    public void innerMap(LongWritable key, Text value, Context context) throws IOException, InterruptedException {

            // adding the value to the bufferr
        valueBuff.add(value);
        System.out.println("ArrayList addValue -> " + value);
        for (Text v : valueBuff){
            System.out.println("ArrayList containedValue -> " + value);
        }

        keyBuff.add(key);

        }   

    public void run(Context context) throws IOException, InterruptedException {
        setup(context);

        // going over the key-value pairs and storing them into the arrays
        while(context.nextKeyValue()){
            innerMap(context.getCurrentKey(), context.getCurrentValue(), context);
        }


        Iterator itrv = valueBuff.iterator();
        Iterator itrk = keyBuff.iterator();
        while(itrv.hasNext()){
            LongWritable nextk = (LongWritable) itrk.next();
            Text nextv = (Text) itrv.next();
            System.out.println("Value iterator -> " + nextv);
            System.out.println("Key iterator -> " + nextk);

            // iterating over the values and running the map on them.

            map(nextk, nextv, context);
        }

        cleanup(context);
    }
 }

 public int run(String[] args) throws Exception { ... }

 public static void main (..) { ... }

好的，现在日志输出:

标准输出日志

ArrayList addValue -> Hello hadoop goodbye hadoop
ArrayList containedValue -> Hello hadoop goodbye hadoop
ArrayList addValue -> Hello world goodbye world
ArrayList containedValue -> Hello world goodbye world
ArrayList containedValue -> Hello world goodbye world
ArrayList addValue -> Hello thinker goodbye thinker
ArrayList containedValue -> Hello thinker goodbye thinker
ArrayList containedValue -> Hello thinker goodbye thinker
ArrayList containedValue -> Hello thinker goodbye thinker
Value iterator -> Hello thinker goodbye thinker
Key iterator -> 84
Hello / 1
thinker / 1
goodbye / 1
thinker / 1
Value iterator -> Hello thinker goodbye thinker
Key iterator -> 84
Hello / 1
thinker / 1
goodbye / 1
thinker / 1
Value iterator -> Hello thinker goodbye thinker
Key iterator -> 84
Hello / 1
thinker / 1
goodbye / 1
thinker / 1

所以您会注意到，每次我向 ArrayList valueBuff 添加一个新值时，列表中的所有值都会被覆盖。有谁知道为什么会这样，为什么数组中的值没有正确添加？

Answer 1

TextInputFormat使用 LineRecordReader .当 Context#nextKeyValue 被调用时，LineRecordReader#nextKeyValue 被调用。

在 LineRecordReader 中，每次调用 nextKeyValue 方法时都使用相同的键和值对象，只是更改了它们的内容。如果要保留键和值数据，则必须在用户代码中创建对象的副本。

这对于优化是有意义的，如果为每条记录创建一个新的键和值对象，那么系统很容易 OOM。