在Map中迭代两次减少

Question

我写了一个Reducer作业，其中我的键和值是复合的。我需要在值中迭代两次，因此尝试缓存值，但重复相同的值。请帮帮我。

以下是我的减速机课程。

this->f == af_system

输入文件：

 public static class Reducerclass  extends Reducer<Text,Text,Text,Text> {
            DateFormat dateFormat = new SimpleDateFormat("yyyy-MM-dd hh:mm:ss a");

            private MultipleOutputs<Text, Text> multipleOutputs;

            @Override
            public void setup(Context context){
                multipleOutputs = new MultipleOutputs<Text, Text>(context);
            }
            public void reduce(Text rkey, Iterable<Text> rvalue, Context context) throws IOException, InterruptedException {             
                ArrayList<Text> ArrayList  = new ArrayList<Text>();
                Iterator<Text> iterator = rvalue.iterator();

                while (iterator.hasNext()) {
                    Text writable = iterator.next();
                    System.out.println("first iteration: " + writable);
                    ArrayList.add(new Text(writable));
context.write(new Text(rkey + ", "),new Text(writable + "--> first iteration"));
                }

                 int size = ArrayList.size();
                    for (int i = 0; i < size; ++i) {
                        System.out.println("second iteration: " + ArrayList.get(i));
context.write(new Text(rkey + ", "),new Text(ArrayList.get(i) + "--> Second iteration--->" + "Array Size -->" + size));
                    }


            }


        }

预期产出：

1509075052824 13.0619798 80.1468367
1509075112825 13.07537311 80.19612851
1509073985114 13.0507832 80.25069245
1509075072824 12.91690859 80.06168244

当前输出：

first iteration: 1509075052824 13.0619798 80.1468367
first iteration: 1509075112825 13.07537311 80.19612851
first iteration: 1509073985114 13.0507832 80.25069245
first iteration: 1509075072824 12.91690859 80.06168244

second iteration: 1509075052824 13.0619798 80.1468367
second iteration: 1509075112825 13.07537311 80.19612851
second iteration: 1509073985114 13.0507832 80.25069245
second iteration: 1509075072824 12.91690859 80.06168244

提前致谢！