Java Stream groupingBy() 操作

本文主要是介绍Java Stream groupingBy() 操作，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

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对Stream的List<T>等更多操作进行操可参考：https://blog.csdn.net/u011663149/article/details/86743930

前言： 

       groupingBy() 提供与SQL的GROUP BY子句类似的功能，只有Java Stream API才有。为了使用它，我们需要指定一个用于执行分组的属性。我们通过提供功能接口的实现来实现这一点。通常通过传递lambda表达式。

Example1

//根据字符串长度
List<String> strings = List.of("a", "bb", "cc", "ddd"); 
Map<Integer, List<String>> result = strings.stream().collect(groupingBy(String::length)); 
System.out.println(result); // {1=[a], 2=[bb, cc], 3=[ddd]}

Example2

//分组输出指定的Map集合
List<String> strings = List.of("a", "bb", "cc", "ddd");
TreeMap<Integer, List<String>> result = strings.stream().collect(groupingBy(String::length, TreeMap::new, toList()));
System.out.println(result); // {1=[a], 2=[bb, cc], 3=[ddd]}

//同理上面
List<String> strings = List.of("a", "bb", "cc", "ddd");
Map<Integer, TreeSet<String>> result = strings.stream().collect(groupingBy(String::length, toCollection(TreeSet::new)));
System.out.println(result); // {1=[a], 2=[bb, cc], 3=[ddd]}

Example3

// counting() 计数收集器
List<String> strings = List.of("a", "bb", "cc", "ddd");
Map<Integer, Long> result = strings.stream().collect(groupingBy(String::length, counting()));
System.out.println(result); // {1=1, 2=2, 3=1}

Example4

//join连接结果数据
List<String> strings = List.of("a", "bb", "cc", "ddd");
Map<Integer, String> result = strings.stream().collect(groupingBy(String::length, joining(",", "[", "]")));
System.out.println(result); // {1=[a], 2=[bb,cc], 3=[ddd]}

Example5 

//分组过滤操作 filtering
List<String> strings = List.of("a", "bb", "cc", "ddd");
Map<Integer, List<String>> result = strings.stream().collect(groupingBy(String::length, filtering(s -> !s.contains("c"), toList())));
System.out.println(result); // {1=[a], 2=[bb], 3=[ddd]}

Example6

 *averagingInt() *averagingLong() *averagingDouble() 
//分组取 平均数
List<String> strings = List.of("a", "bb", "cc", "ddd");
Map<Integer, Double> result = strings.stream().collect(groupingBy(String::length, averagingInt(String::hashCode)));
System.out.println(result); // {1=97.0, 2=3152.0, 3=99300.0}*summingInt() *summingLong() *summingDouble()
//分组求和
List<String> strings = List.of("a", "bb", "cc", "ddd");
Map<Integer, Integer> result = strings.stream().collect(groupingBy(String::length, summingInt(String::hashCode)));
System.out.println(result); // {1=97, 2=6304, 3=99300}*summarizingInt() *summarizingLong() *summarizingDouble() 
List<String> strings = List.of("a", "bb", "cc", "ddd");
Map<Integer, IntSummaryStatistics> result = strings.stream().collect(groupingBy(String::length, summarizingInt(String::hashCode)));
System.out.println(result);//返回结果：
{1=IntSummaryStatistics{count=1, sum=97, min=97, average=97.000000, max=97}, 2=IntSummaryStatistics{count=2, sum=6304, min=3136, average=3152.000000, max=3168}, 3=IntSummaryStatistics{count=1, sum=99300, min=99300, average=99300.000000, max=99300}
}

Example7

// group reducing  分组换算
List<String> strings = List.of("a", "bb", "cc", "ddd");
Map<Integer, List<Character>> result = strings.stream().map(toStringList()).collect(groupingBy(List::size, reducing(List.of(), (l1, l2) -> Stream.concat(l1.stream(), l2.stream()).collect(Collectors.toList()))));
System.out.println(result); // {1=[a], 2=[b, b, c, c], 3=[d, d, d]}

Example8

//分组 使用Collectors 获取max、min
List<String> strings = List.of("a", "bb", "cc", "ddd");
Map<Integer, Optional<String>> result = strings.stream().collect(groupingBy(String::length, Collectors.maxBy(Comparator.comparing(String::toUpperCase))));
System.out.println(result); // {1=Optional[a], 2=Optional[cc], 3=Optional[ddd]}

Example9

//分组获得长度大于1的字符串 作为一个新的TreeSet
var result = strings.stream().collect(groupingBy(String::length,mapping(String::toUpperCase,filtering(s -> s.length() > 1,toCollection(TreeSet::new)))));
//result
{1=[], 2=[BB, CC], 3=[DDD]}

Example10

//字符串列表，按照它们的匹配长度对它们进行分组仅保留具有非零长度的不同元素,最终格式化
var result = strings.stream().collect(groupingBy(String::length,mapping(toStringList(),flatMapping(s -> s.stream().distinct(),filtering(s -> s.length() > 0,mapping(String::toUpperCase,reducing("", (s, s2) -> s + s2)))))));
//result 
{1=A, 2=BC, 3=D}

 

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