如何计算流中的唯一单词? [英] How to count unique words in a stream?

问题描述

有没有办法用 Flink Streaming 计算流中唯一单词的数量?结果将是不断增加的数字流.

Is there a way to count the number of unique words in a stream with Flink Streaming? The results would be a stream of number which keeps increasing.

推荐答案

您可以通过存储您已经看过的所有单词来解决问题.有了这些知识，您可以过滤掉所有重复的单词.然后可以通过具有并行性 1 的映射运算符计算其余部分.下面的代码片段正是这样做的.

You can solve the problem by storing all words which you've already seen. Having this knowledge you can filter out all duplicate words. The rest can then be counted by a map operator with parallelism 1. The following code snippet does exactly that.

val env = StreamExecutionEnvironment.getExecutionEnvironment

val inputStream = env.fromElements("foo", "bar", "foobar", "bar", "barfoo", "foobar", "foo", "fo")

// filter words out which we have already seen
val uniqueWords = inputStream.keyBy(x => x).filterWithState{
  (word, seenWordsState: Option[Set[String]]) => seenWordsState match {
    case None => (true, Some(HashSet(word)))
    case Some(seenWords) => (!seenWords.contains(word), Some(seenWords + word))
  }
}

// count the number of incoming (first seen) words
val numberUniqueWords = uniqueWords.keyBy(x => 0).mapWithState{
  (word, counterState: Option[Int]) =>
    counterState match {
      case None => (1, Some(1))
      case Some(counter) => (counter + 1, Some(counter + 1))
    }
}.setParallelism(1)

numberUniqueWords.print();

env.execute()

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