Map Reduce - 如何在单个作业中对多个属性进行分组和聚合

时间:2018-06-08 05:40:08

标签: java hadoop mapreduce grouping aggregation

我目前正在努力解决MapReduce问题。 我有以下数据集:

1,John,Computer
2,Anne,Computer
3,John,Mobile
4,Julia,Mobile
5,Jack,Mobile
6,Jack,TV
7,John,Computer
8,Jack,TV
9,Jack,TV
10,Anne,Mobile
11,Anne,Computer
12,Julia,Mobile

现在我想将MapReduce应用于分组和 聚合在这个数据集上,以便输出 并不只显示哪个人买了什么东西, 而且产品是什么,这个人最有责任。

所以输出应该如下:

John 3 Computer
Anne 3 Mobile
Jack 4 TV
Julia 2 Mobile

我当前实现的mapper和reducer 看起来像这样,完美地返回了多少订单 然而,由个人提出,我真的很无能为力 获得所需的输出。

static class CountMatchesMapper extends Mapper<Object,Text,Text,IntWritable> {
    @Override
    protected void map(Object key, Text value, Context ctx) throws IOException, InterruptedException {
        String row = value.toString();
        String[] row_part = row.split(",");


            try{
                ctx.write(new Text(row_part[1]), new IntWritable(1));

            catch (IOException e) {
            }
            catch (InterruptedException e) {
            }

        }

    }
}


static class CountMatchesReducer extends Reducer<Text,IntWritable,Text,IntWritable> {
    @Override
    protected void reduce(Text key, Iterable<IntWritable> values, Context ctx) throws IOException, InterruptedException {
        int i = 0;
        for (IntWritable value : values) i += value.get();
        try{
            ctx.write(key, new IntWritable(i));
        }
        catch (IOException e) {
        }
        catch (InterruptedException e) {
        }
    }
}

我真的很感激任何有效的解决方案和帮助。

提前致谢!

1 个答案:

答案 0 :(得分:1)

如果我理解你想要什么,我认为第二个输出行应该是:

Anne 3 Computer

基于输入。 Anne共购买了3款产品:2台电脑和1台手机。

我这里有一个非常基本和简单的方法,它没有考虑边缘情况等,但可以给你一些方向:

    static class CountMatchesMapper extends Mapper<LongWritable, Text, Text, Text> {
    private Text outputKey = new Text();
    private Text outputValue = new Text();

    @Override
    protected void map(LongWritable key, Text value, Context ctx) throws IOException, InterruptedException {
        String row = value.toString();
        String[] row_part = row.split(",");
        outputKey.set(row_part[1]);
        outputValue.set(row_part[2]);
        ctx.write(outputKey, outputValue);
    }
}

static class CountMatchesReducer extends Reducer<Text, Text, Text, NullWritable> {
    private Text output = new Text();

    @Override
    protected void reduce(Text key, Iterable<Text> values, Context ctx) throws IOException, InterruptedException {
        HashMap<String, Integer> productCounts = new HashMap();

        int totalProductsBought = 0;
        for (Text value : values) {
            String productBought = value.toString();
            int count = 0;
            if (productCounts.containsKey(productBought)) {
                count = productCounts.get(productBought);
            }
            productCounts.put(productBought, count + 1);
            totalProductsBought += 1;
        }

        String topProduct = getTopProductForPerson(productCounts);
        output.set(key.toString() + " " + totalProductsBought + " " + topProduct);
        ctx.write(output, NullWritable.get());
    }

    private String getTopProductForPerson(Map<String, Integer> productCounts) {
        String topProduct = "";
        int maxCount = 0;
        for (Map.Entry<String, Integer> productCount : productCounts.entrySet()) {
            if (productCount.getValue() > maxCount) {
                maxCount = productCount.getValue();
                topProduct = productCount.getKey();
            }
        }
        return topProduct;
    }
}

以上将给出您描述的输出。

如果你想要一个适当的解决方案,那么你可能需要一个复合键和自定义GroupComparator。这样您就可以添加Combiner并使其更加高效。但是,上述方法适用于普通情况。