优雅的将一个对象的集合转化成另一个对象的集合

List<OrderDetail> orderDetailList = orderDetailService.listOrderDetails(); List<CartDTO> cartDTOList = orderDetailList.stream()                 .map(e -> new CartDTO(e.getProductId(), e.getProductQuantity()))                 .collect(Collectors.toList()); 

交集 (list1 + list2)

List<T> intersect = list1.stream()                          .filter(list2::contains)                          .collect(Collectors.toList()); 

差集

//(list1 - list2) List<String> reduce1 = list1.stream().filter(item -> !list2.contains(item)).collect(toList());  //(list2 - list1) List<String> reduce2 = list2.stream().filter(item -> !list1.contains(item)).collect(toList()); 

并集

//使用并行流  List<String> listAll = list1.parallelStream().collect(toList()); List<String> listAll2 = list2.parallelStream().collect(toList()); listAll.addAll(listAll2); 

去重并集

List<String> listAllDistinct = listAll.stream() .distinct().collect(toList()); 

从List中过滤出一个元素

User match = users.stream() .filter((user) -> user.getId() == 1).findAny().get(); 

Map集合转 List

List<Person> list = map.entrySet().stream().sorted(Comparator.comparing(e -> e.getKey()))         .map(e -> new Person(e.getKey(), e.getValue())).collect(Collectors.toList());          List<Person> list = map.entrySet().stream().sorted(Comparator.comparing(Map.Entry::getValue)).map(e -> new Person(e.getKey(), e.getValue())).collect(Collectors.toList());  List<Person> list = map.entrySet().stream().sorted(Map.Entry.comparingByKey()).map(e -> new Person(e.getKey(), e.getValue())).collect(Collectors.toList()); 

Collectors toList

streamArr.collect(Collectors.toList()); List<Integer> collectList = Stream.of(1, 2, 3, 4)         .collect(Collectors.toList()); System.out.println("collectList: " + collectList); // 打印结果 collectList: [1, 2, 3, 4] 

Collectors toMap

map value 为对象 student Map<Integer, Student> map = list.stream().collect(Collectors.toMap(Student::getId, student -> student)); // 遍历打印结果 map.forEach((key, value) -> {     System.out.println("key: " + key + " value: " + value); }); map value 为对象中的属性 Map<Integer, String> map = list.stream().collect(Collectors.toMap(Student::getId, Student::getName)); map.forEach((key, value) -> {     System.out.println("key: " + key + " value: " + value); }); 

List集合转 Map

/*使用Collectors.toMap形式*/ Map result = peopleList.stream().collect(Collectors.toMap(p -> p.name, p -> p.age, (k1, k2) -> k1)); //其中Collectors.toMap方法的第三个参数为键值重复处理策略，如果不传入第三个参数，当有相同的键时，会抛出一个IlleageStateException。 //或者 Map<Integer, String> result1 = list.stream().collect(Collectors.toMap(Hosting::getId, Hosting::getName)); //List<People> -> Map<String,Object> List<People> peopleList = new ArrayList<>(); peopleList.add(new People("test1", "111")); peopleList.add(new People("test2", "222")); Map result = peopleList.stream().collect(HashMap::new,(map,p)->map.put(p.name,p.age),Map::putAll); 

List 转 Map<Integer,Apple>

/** * List<Apple> -> Map<Integer,Apple> * 需要注意的是： * toMap 如果集合对象有重复的key，会报错Duplicate key .... * apple1,apple12的id都为1。 * 可以用 (k1,k2)->k1 来设置，如果有重复的key,则保留key1,舍弃key2 */ Map<Integer, Apple> appleMap = appleList.stream().collect(Collectors.toMap(Apple::getId, a -> a,(k1, k2) -> k1)); 

List 转 List<Map<String,Object>>

List<Map<String,Object>> personToMap = peopleList.stream().map((p) -> {     Map<String, Object> map = new HashMap<>();     map.put("name", p.name);     map.put("age", p.age);     return map; }).collect(Collectors.toList()); //或者 List<Map<String,Object>> personToMap = peopleList.stream().collect(ArrayList::new, (list, p) -> {    Map<String, Object> map = new HashMap<>();     map.put("name", p.name);     map.put("age", p.age);     list.add(map); }, List::addAll); 

字典查询和数据转换 toMap时，如果value为null,会报空指针异常
解决办法一：

Map<String, List<Dict>> resultMaps = Arrays.stream(dictTypes) .collect(Collectors.toMap(i -> i, i -> Optional.ofNullable(dictMap.get(i)).orElse(new ArrayList<>()), (k1, k2) -> k2)); 

解决办法二：

Map<String, List<Dict>> resultMaps = Arrays.stream(dictTypes) .filter(i -> dictMap.get(i) != null).collect(Collectors.toMap(i -> i, dictMap::get, (k1, k2) -> k2)); 

解决办法三：

Map<String, String> memberMap = list.stream().collect(HashMap::new, (m,v)->     m.put(v.getId(), v.getImgPath()),HashMap::putAll); System.out.println(memberMap); 

解决办法四：

Map<String, String> memberMap = new HashMap<>(); list.forEach((answer) -> memberMap.put(answer.getId(), answer.getImgPath())); System.out.println(memberMap);  Map<String, String> memberMap = new HashMap<>(); for (Member member : list) {     memberMap.put(member.getId(), member.getImgPath()); } 

假设有一个User实体类，有方法getId(),getName(),getAge()等方法，现在想要将User类型的流收集到一个Map中，示例如下：

Stream<User> userStream = Stream.of(new User(0, "张三", 18), new User(1, "张四", 19), new User(2, "张五", 19), new User(3, "老张", 50));  Map<Integer, User> userMap = userSteam.collect(Collectors.toMap(User::getId, item -> item)); 

假设要得到按年龄分组的Map<Integer,List>,可以按这样写：

Map<Integer, List<User>> ageMap = userStream.collect(Collectors.toMap(User::getAge, Collections::singletonList, (a, b) -> {             List<User> resultList = new ArrayList<>(a);             resultList.addAll(b);             return resultList;         }));  Map<Integer, String> map = persons     .stream()     .collect(Collectors.toMap(         p -> p.age,         p -> p.name,         (name1, name2) -> name1 + ";" + name2));  System.out.println(map); // {18=Max, 23=Peter;Pamela, 12=David} 

Map 转 另一个Map

//示例1 Map<String, List<String>> 转 Map<String,User> Map<String,List<String>> map = new HashMap<>(); map.put("java", Arrays.asList("1.7", "1.8")); map.entrySet().stream();  @Getter @Setter @AllArgsConstructor public static class User{     private List<String> versions; }  Map<String, User> collect = map.entrySet().stream()                 .collect(Collectors.toMap(                         item -> item.getKey(),                         item -> new User(item.getValue())));  //示例2 Map<String,Integer> 转 Map<String,Double> Map<String, Integer> pointsByName = new HashMap<>(); Map<String, Integer> maxPointsByName = new HashMap<>();  Map<String, Double> gradesByName = pointsByName.entrySet().stream()         .map(entry -> new AbstractMap.SimpleImmutableEntry<>(                 entry.getKey(), ((double) entry.getValue() /                         maxPointsByName.get(entry.getKey())) * 100d))         .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue)); 

List<String> 转String

//java8 String.join 方式  List<String> webs = Arrays.asList("voidcc.com", "voidmvn.com", "voidtool.com"); //webs 必须是List<String> String allwebs = String.join(",", webs); System.out.println(allwebs);  //stream  List<String> webs = Arrays.asList("voidcc.com", "voidmvn.com", "voidtool.com"); String allwebs = webs.stream().collect(Collectors.joining(",")); System.out.println(allwebs);  

Collectors toSet

Set<String> result = Stream.of("aa", "bb", "cc", "aa").collect(HashSet::new, HashSet::add, HashSet::addAll); //Collectors类中已经预定义好了toList，toSet，toMap，toCollection等方便使用的方法，所以以上代码还可以简化如下： Set<String> result2 = Stream.of("aa", "bb", "cc", "aa").collect(Collectors.toSet());  Set<Integer> collectSet = Stream.of(1, 2, 3, 4).collect(Collectors.toSet()); System.out.println("collectSet: " + collectSet); // 打印结果 collectSet: [1, 2, 3, 4]  Stack stack1 = stream.collect(Collectors.toCollection(Stack::new)); // collect toString String str = stream.collect(Collectors.joining()).toString(); 

排序

//按照自然顺序进行排序 如果要自定义排序sorted 传入自定义的 Comparator list.stream()     .sorted()     .filter((s) -> s.startsWith("a"))     .forEach(System.out::println);  //对象排序比较 请重写对象的equals()和hashCode()方法 list.sorted((a, b) -> b.compareTo(a))  Collections.sort(names, (a, b) -> b.compareTo(a)); 

比较

Comparator<Person> comparator = (p1, p2) -> p1.firstName.compareTo(p2.firstName);  Person p1 = new Person("John", "Doe"); Person p2 = new Person("Alice", "Wonderland");  comparator.compare(p1, p2);             // > 0 comparator.reversed().compare(p1, p2);  // < 0 

Collectors groupingBy 分组

Map<Integer, List<User>> ageMap2 = userStream .collect(Collectors.groupingBy(User::getAge)); 

对集合按照多个属性分组
将多个字段拼接成一个新字段，然后再使用groupBy分组

Map<String, List<EntryDeliveryDetailywk>> detailmap = details.stream() .collect(Collectors.groupingBy(this::fetchGroupKey));  private String fetchGroupKey(EntryDeliveryDetailywk detail){         return detail.getSkuId().toString()          + detail.getItemsName()          + detail.getWarehouseId().toString()            + detail.getSupplierId().toString();     } 

groupingBy 分组后操作
//Collectors中还提供了一些对分组后的元素进行downStream处理的方法：
//counting方法返回所收集元素的总数；
//summing方法会对元素求和；
//maxBy和minBy会接受一个比较器，求最大值，最小值；
//mapping函数会应用到downstream结果上，并需要和其他函数配合使用；

Map<Integer, Long> sexCount = userStream.collect(Collectors.groupingBy(User::getSex,Collectors.counting()));  Map<Integer, Integer> ageCount = userStream.collect(Collectors.groupingBy(User::getSex,Collectors.summingInt(User::getAge)));  Map<Integer, Optional<User>> ageMax =  userStream.collect(Collectors.groupingBy(User::getSex,Collectors.maxBy(Comparator.comparing(User::getAge))));  Map<Integer, List<String>> nameMap =  userStream.collect(Collectors.groupingBy(User::getSex,Collectors.mapping(User::getName,Collectors.toList()))); 

groupingBy 根据年龄来分组：

Map<Integer, List> peopleByAge = peoples.stream() .filter(p -> p.age > 12).collect(Collectors.groupingBy(p -> p.age, Collectors.toList())); 

groupingBy 根据年龄分组，年龄对应的键值List存储的为Person的姓名:

Map<Integer, List> peopleByAge = people.stream() .collect(Collectors.groupingBy(p -> p.age, Collectors.mapping((Person p) -> p.name, Collectors.toList()))); //mapping即为对各组进行投影操作，和Stream的map方法基本一致。 

groupingBy 根据姓名分组，获取每个姓名下人的年龄总和:

Map sumAgeByName = peoples.stream().collect(Collectors.groupingBy(p -> p.name, Collectors.reducing(0, (Person p) -> p.age, Integer::sum))); /* 或者使用summingInt方法 */ sumAgeByName = peoples.stream().collect(Collectors.groupingBy(p -> p.name, Collectors.summingInt((Person p) -> p.age))); 

groupingBy Boolean分组:

Map<Boolean, List<Integer>> collectGroup = Stream.of(1, 2, 3, 4)             .collect(Collectors.groupingBy(it -> it > 3)); System.out.println("collectGroup : " + collectGroup); // 打印结果 // collectGroup : {false=[1, 2, 3], true=[4]} 

groupingBy 按年龄分组

Map<Integer, List<Person>> personsByAge = persons.stream().collect(Collectors.groupingBy(p -> p.age)); personsByAge.forEach((age, p) -> System.out.format("age %s: %s\n", age, p)); // age 18: [Max] // age 23: [Peter, Pamela] // age 12: [David] 

Map.merge() 类似于分组之后sum

 Map<String, Integer> studentScoreMap2 = new HashMap<>();         studentScoreList.forEach(studentScore -> studentScoreMap2.merge(           studentScore.getStuName(),           studentScore.getScore(),           Integer::sum)); 

Collectors partitioningBy
Collectors中还提供了partitioningBy方法，接受一个Predicate函数，该函数返回boolean值，用于将内容分为两组。假设User实体中包含性别信息getSex()，可以按如下写法将userStream按性别分组：

Map<Boolean, List<User>> sexMap = userStream .collect(Collectors.partitioningBy(item -> item.getSex() > 0)); 

可以看到Java8的分组功能相当强大，当然你还可以完成更复杂的功能。另外Collectors中还存在一个类似groupingBy的方法：partitioningBy，它们的区别是partitioningBy为键值为Boolean类型的groupingBy，这种情况下它比groupingBy更有效率。
partitioningBy 将数字的Stream分解成奇数集合和偶数集合。

Map<Boolean, List<Integer>> collectParti = Stream.of(1, 2, 3, 4)             .collect(Collectors.partitioningBy(it -> it % 2 == 0)); System.out.println("collectParti : " + collectParti); // 打印结果 // collectParti : {false=[1, 3], true=[2, 4]} 

Collectors joining
Collectors.joining 收集Stream中的值，该方法可以方便地将Stream得到一个字符串。joining函数接受三个参数，分别表示允（用以分隔元素）、前缀和后缀:

String names = peoples.stream().map(p->p.name).collect(Collectors.joining(","))  String strJoin = Stream.of("1", "2", "3", "4")         .collect(Collectors.joining(",", "[", "]")); System.out.println("strJoin: " + strJoin); // 打印结果 // strJoin: [1,2,3,4]  //字符串连接 String phrase = persons     .stream()     .filter(p -> p.age >= 18)     .map(p -> p.name)     .collect(Collectors.joining(" and ", "In Germany ", " are of legal age.")); System.out.println(phrase); // In Germany Max and Peter and Pamela are of legal age. 

组合 Collectors:

Map<Boolean, Long> partiCount = Stream.of(1, 2, 3, 4)         .collect(Collectors.partitioningBy(it -> it.intValue() % 2 == 0,                 Collectors.counting())); System.out.println("partiCount: " + partiCount); // 打印结果 // partiCount: {false=2, true=2} 

Collectors分别提供了求平均值averaging、总数couting、最小值minBy、最大值maxBy、求和suming等操作。但是假如你希望将流中结果聚合为一个总和、平均值、最大值、最小值，那么Collectors.summarizing(Int/Long/Double)就是为你准备的，它可以一次行获取前面的所有结果，其返回值为(Int/Long/Double)SummaryStatistics。

DoubleSummaryStatistics dss = people.collect(Collectors.summarizingDouble((Person p)->p.age)); double average=dss.getAverage(); double max=dss.getMax(); double min=dss.getMin(); double sum=dss.getSum(); double count=dss.getCount();  IntSummaryStatistics ageSummary = persons         .stream()         .collect(Collectors.summarizingInt(p -> p.age));  System.out.println(ageSummary); // IntSummaryStatistics{count=4, sum=76, min=12, average=19.000000, max=23} 

使用collect可以将Stream转换成值。maxBy和minBy允许用户按照某个特定的顺序生成一个值。
averagingDouble:求平均值，Stream的元素类型为double
averagingInt:求平均值，Stream的元素类型为int
averagingLong:求平均值，Stream的元素类型为long
counting:Stream的元素个数
maxBy:在指定条件下的，Stream的最大元素
minBy:在指定条件下的，Stream的最小元素
reducing: reduce操作
summarizingDouble:统计Stream的数据(double)状态，其中包括count，min，max，sum和平均。
summarizingInt:统计Stream的数据(int)状态，其中包括count，min，max，sum和平均。
summarizingLong:统计Stream的数据(long)状态，其中包括count，min，max，sum和平均。
summingDouble:求和，Stream的元素类型为double
summingInt:求和，Stream的元素类型为int
summingLong:求和，Stream的元素类型为long

Optional<Integer> collectMaxBy = Stream.of(1, 2, 3, 4)             .collect(Collectors.maxBy(Comparator.comparingInt(o -> o))); System.out.println("collectMaxBy:" + collectMaxBy.get()); // 打印结果 // collectMaxBy:4 

Collectors averagingInt
计算集合的平均年龄

Double averageAge = persons     .stream()     .collect(Collectors.averagingInt(p -> p.age));  System.out.println(averageAge);     // 19.0 

自定义 Collector

Collector<Person, StringJoiner, String> personNameCollector =     Collector.of(         () -> new StringJoiner(" | "),          // supplier         (j, p) -> j.add(p.name.toUpperCase()),  // accumulator         (j1, j2) -> j1.merge(j2),               // combiner         StringJoiner::toString);                // finisher  String names = persons     .stream()     .collect(personNameCollector);  System.out.println(names);  // MAX | PETER | PAMELA | DAVID 

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