MapReduce编程模型

Posted 2021-11-25 pipiyan

# 文本前期处理
strl_ist = str.replace(‘\\n‘, ‘‘).lower().split(‘ ‘)
count_dict = {}
# 如果字典里有该单词则加 1，否则添加入字典
for str in strl_ist:
if str in count_dict.keys():
    count_dict[str] = count_dict[str] + 1
    else:
        count_dict[str] = 1

建一个hash表，将文本中的每个词都放在这个hash表里面，如果这个词第一次放入，就新建一个kry，Value对，key是这个词，Value是1；如果已经有这个词，那么给Value+1。

# Ｍapeduce 计算框架会将这些<word，1>收集起来，将相同的word放在一起，形成<word,<1,1,1,1,1,1…>>这样的<key,value集合>数据，然后将其输入给reduce函数

public class WordCount {

  public static class TokenizerMapper
       extends Mapper<Object, Text, Text, IntWritable>{

    private final static IntWritable one = new IntWritable(1);
    private Text word = new Text();

    public void map(Object key, Text value, Context context
                    ) throws IOException, InterruptedException {
      StringTokenizer itr = new StringTokenizer(value.toString());
      while (itr.hasMoreTokens()) {
        word.set(itr.nextToken());
        context.write(word, one);
      }
    }
  }

  public static class IntSumReducer
       extends Reducer<Text,IntWritable,Text,IntWritable> {
    private IntWritable result = new IntWritable();

    public void reduce(Text key, Iterable<IntWritable> values,
                       Context context
                       ) throws IOException, InterruptedException {
      int sum = 0;
      for (IntWritable val : values) {
        sum += val.get();
      }
      result.set(sum);
      context.write(key, result);
    }
  }
}

reduce函数的计算过程：将这个集合中的1求和，再将单词（word）和这个和（sum）组成一个<key,Value>，也就是<word,sum>输出。