stream使用案例
1.1 查找所有的偶数并求和
public static void p1() {List<Integer> numbers = Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9, 10);int sum = numbers.stream().filter(num -> num % 2 == 0).mapToInt(Integer::intValue).sum() ;System.err.printf("result: %s%n", sum) ;}
1.2 查找并打印长度大于 5 的字符串个数
public static void p2() {List<String> strings = Arrays.asList("apple", "banana", "grape","watermelon", "kiwi", "orange");Long count = strings.stream().filter(str -> str.length() > 5).count() ;System.err.printf("result: %s%n", count) ;}
1.3 处理每一个元素最后返回新集合
public static void p3() {List<Integer> numbers = Arrays.asList(1, 2, 3, 4, 5);List<Integer> squares = numbers.stream().map(num -> num * num).collect(Collectors.toList()) ;System.err.printf("result: %s%n", squares) ;}
将每一个元素进行平方操作，最后返回一个新的集合。输出结果：
result: [1, 4, 9, 16, 25]1.4 找出整数列表中的最大元素
public static void p4() {List<Integer> numbers = Arrays.asList(10, 5, 25, 15, 30);int max = numbers.stream().mapToInt(Integer::intValue).max().getAsInt();System.err.printf("result: %s%n", max) ;}
1.5 将列表中的所有字符串连接成一个字符串
public static void p5() {List<String> fruits = Arrays.asList("apple", "banana", "cherry","coconut", "apple");String concat =fruits.stream().collect(Collectors.joining()) ;System.err.printf("result: %s%n", concat) ;}
1.6 转成大写再排序
public static void p6() {List<String> fruits = Arrays.asList("apple", "Banana", "Grape", "orange", "kiwi");List<String> sortedUppercase = fruits.stream().map(String::toUpperCase).sorted().collect(Collectors.toList());System.err.printf("result: %s%n", sortedUppercase) ;}
先将字符串转为大写，然后在进行排序，输出结果：
result: [APPLE, BANANA, GRAPE, KIWI, ORANGE]1.7 计算double类型平均值
public static void p7() {List<Double> doubles = Arrays.asList(1.0, 2.0, 3.0, 4.0, 5.0);double average = doubles.stream().mapToDouble(Double::doubleValue).average().getAsDouble() ;System.err.printf("result: %s%n", average) ;}
1.8 删除重复元素
public static void p8() {List<String> words = Arrays.asList("apple", "banana", "apple", "orange", "banana", "kiwi");List<String> uniqueWords = words.stream().distinct().collect(Collectors.toList()) ;System.err.printf("result: %s%n", uniqueWords) ;}
1.9 检查所有元素是否符合条件
public static void p9() {List<Integer> numbers = Arrays.asList(2, 4, 6, 8, 10);boolean allEven = numbers.stream().allMatch(n -> n%2 == 0) ;System.err.printf("result: %s%n", allEven) ;}
1.10 检查集合中是否包含特定元素
public static void p10() {List<Integer> numbers = Arrays.asList(2, 4, 6, 8, 10);boolean exists = numbers.stream().anyMatch(n -> n.equals(8)) ;System.err.printf("result: %s%n", exists) ;}
1.11 查找流中最长的字符串
public static void p11() {List<String> fruits = Arrays.asList("apple", "banana", "cherry", "coconut", "apple") ;int max = fruits.stream().mapToInt(String::length).max().getAsInt() ;System.err.printf("result: %s%n", max) ;}
1.12 从流中删除null值
public static void p12() {List<String> fruits = Arrays.asList("apple", "banana", "cherry", null,"coconut", "apple");List<String> nonNullValues = fruits.stream().filter(Objects::nonNull).collect(Collectors.toList()) ;System.err.printf("result: %s%n", nonNullValues) ;}
过滤为null的值，输出结果：
result: [apple, banana, cherry, coconut, apple]1.13 分组并查找最大值
public static void p13() {List<Employee> employees =new ArrayList<>() ;employees.add(new Employee("Alice","HR",50000.0)) ;employees.add(new Employee("Bob","IT",60000.0)) ;employees.add(new Employee("Charlie","Finance",55000.0)) ;employees.add(new Employee("David","IT",70000.0)) ;employees.add(new Employee("Eva", "HR", 45000.0)) ;employees.add(new Employee("Frank","Finance",58000.0));Map<String, Optional<Employee>> highestSalaryPerDept = employees.stream().collect(Collectors.groupingBy(Employee::getDepartment,Collectors.maxBy(Comparator.comparingDouble(Employee::getSalary))));highestSalaryPerDept.forEach((key, value) -> {System.err.printf("部门: %s, \t最高薪: %s%n", key, value.get()) ;});}
输出结果：
2.14 查找列表中第二小的元素
public static void p14() {List<Integer> numbers = Arrays.asList(2, 4, 6, 8, 10) ;Optional<Integer> secondSmallest = numbers.stream().distinct().sorted().skip(1).findFirst() ;System.err.printf("result: %s%n", secondSmallest) ;}
1.15 查找两个列表的交集
public static void p15() {List<Integer> list1 = Arrays.asList(1, 2, 3, 4, 5) ;List<Integer> list2 = Arrays.asList(4, 5, 6, 7,8) ;List<Integer> intersection = list1.stream().filter(list2::contains).collect(Collectors.toList()) ;System.err.printf("result: %s%n", intersection) ;}
1.16 并行处理提升性能
使用并行流可以通过BaseStream.parallel或Collection#parallelStream操作，如下计算1亿个数的求和
public static void main(String[] args) {double[] arr = IntStream.range(0, 100_000_000).mapToDouble(i -> new Random().nextDouble() * 100000).toArray() ;computeSumOfSquareRoots(arr);}public static void computeSumOfSquareRoots(double[] arr) {double serialSum = computeSerialSum(DoubleStream.of(arr));System.out.println("Serial Sum: " + serialSum);double parallelSum = computeParallelSum(DoubleStream.of(arr));System.out.println("Parallel Sum: " + parallelSum);}public static double computeSerialSum(DoubleStream stream) {long startTime = System.currentTimeMillis();double sum = stream.reduce(0.0D, (l, r) -> l + r) ;long endTime = System.currentTimeMillis();System.out.println("Serial Computation Time: " + (endTime - startTime) + " ms");return sum;}public static double computeParallelSum(DoubleStream stream) {long startTime = System.currentTimeMillis();double sum = stream.parallel().reduce(0, (l, r) -> l + r) ;long endTime = System.currentTimeMillis();System.out.println("Parallel Computation Time: " + (endTime - startTime) + " ms");return sum;}
运行结果
SerialComputation Time: 73 msSerialSum: 5.000114159154823E12ParallelComputation Time: 38 msParallelSum: 5.000114159151367E12