Java 8 Streams如何避免使用map或set进行过滤? [英] Java 8 Streams how to avoid filtering with map or set?

问题描述

我不断遇到需要/认为我需要通过Map或Set保存状态的解决方案. 例如创建一个方法，该方法返回在输入中找到的重复项

I keep running into solutions where I want/think i need to save state via either a Map or a Set. e.g. create a method that returns duplicates found in the input

// non streams solution
public int[] getDuplicates(int[] input){
  Set<Integer> allSet = new HashSet<Integer>();
  Set<Integer> duplicates = new HashSet<Integer>();

  int[] dups = new int[input.length];
  int j = 0;
  for (Integer i : input) {
    if (!allSet.add(i)) {
      if(duplicates.add(i)) {
        dups[j++] = i;
      }
    }
  }
  return Arrays.copyOfRange(dups, 0, j);
}

不幸的是，我的Java 8 Streams解决方案使用的是HashSet进行过滤.我了解这不是适当的"，因为它取决于状态. 没有国家提出建议或硬性规定吗?运行并行流时仅是一个问题吗?有人可以建议一种不使用HashSet的方法吗?

My Java 8 Streams solution, unfortunately I am using a HashSet for filtering. I understand this is not "proper" as it depends on state. Is no state a suggestion or a hard-fast rule? Is it only an issue when running a parallel stream? Can someone suggest a way not to use HashSet here?

public static int[] getDuplicatesStreamsToArray(int[] input) {
  Set<Integer> allSet = new HashSet<>();
  int[] dups = Arrays.stream(input)
      .sequential()                   // prevents parallel processing
      .unordered()                    // speed up distinct operation
      .boxed()                        // int to Integer
      .filter(n -> !allSet.add(n))    // passes dups, but uses STATE
      .distinct()                     // uses internal Set of dups
      .mapToInt(i -> i)               // Integer back to int
      .toArray();
  return dups;
}

推荐答案

这是怎么回事:

基本上，创建类型为Map<Integer,Long>的频率计数，并返回value大于1的那些keys.

Basically, creates a frequency count of type Map<Integer,Long> and returns those keys where the value is greater than 1.

    public static int[] getDuplicatesStreamsToArray(int[] input) {

      int[] dups = Arrays.stream(input).boxed().collect(
            Collectors.groupingBy(Function.identity(),
                  Collectors.counting())).entrySet().stream().filter(
                        e -> e.getValue() > 1).mapToInt(
                              e -> e.getKey()).toArray();
      return dups;
   }

我误解了你以前想做什么.

I misunderstood what you were trying to do earlier.

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