Java8如何从一个Stream中过滤null值

package com.mkyong.java8; 
import java.util.List;
import java.util.stream.Collectors;
import java.util.stream.Stream;
 
public class Java8Examples { 
    public static void main(String[] args) { 
        Stream<String> language = Stream.of("java", "python", "node", null, "ruby", null, "PHP"); 
        List<String> result = language.collect(Collectors.toList()); 
        result.forEach(System.out::println); 
    }
}

package com.mkyong.java8; 
import java.util.List;
import java.util.stream.Collectors;
import java.util.stream.Stream;
 
public class Java8Examples { 
    public static void main(String[] args) { 
        Stream<String> language = Stream.of("java", "python", "node", null, "ruby", null, "php"); 
        //List<String> result = language.collect(Collectors.toList()); 
        List<String> result = language.filter(x -> x!=null).collect(Collectors.toList()); 
        result.forEach(System.out::println);  
    }
}

import java.util.List;
List<String> result = language.filter(Objects::nonNull).collect(Collectors.toList());

@Data 
@AllArgsConstructor    
public class User { 
    private Long id;      // id
    private Integer age;  // 年龄 
    private Byte gentle;  // 性别 
    private String name;  // 名字 
    private Integer rank; // 排名 
}
 
User user0 = new User(1L, 18, (byte) 0, "张三", 1); 
User user1 = new User(2L, 20, (byte) 1, "李四", 4); 
User user2 = new User(3L, 35, (byte) 0, "王五", 2); 
User user3 = new User(4L, 29, (byte) 1, "赵六", 3);

List<User> list = Arrays.asList(user0, user1, user2, user3);
Stream<User> stream = null;
stream = list.stream(); // 需要预判NPE
stream = Optional.of(list).orElseGet(Collections::emptyList).stream(); // 需要预判NPE
stream = Optional.ofNullable(list).orElseGet(Collections::emptyList).stream();
stream = Optional.ofNullable(list).orElseGet(Collections::emptyList).parallelStream(); // 并行处理流
stream = Stream.of(user0, user1, user2, user3).parallel(); // 直接构造
stream = Stream.of(Arrays.asList(user0, user1), Arrays.asList(user2, user3)).flatMap(Collection::stream); // flatMap合并

// 过滤出性别为0的user
List<User> userList = Optional.ofNullable(list).orElseGet(Collections::emptyList).stream().filter(user -> (byte) 0 == user.getGentle()).collect(Collectors.toList());
 
// 获取排名大于1的用户年龄set
Set<Integer> ageList = Optional.ofNullable(list).orElseGet(Collections::emptyList).stream().filter(user -> 1 < user.getRank()).map(User::getAge).collect(Collectors.toSet());
 
// 合计性别为0的user的年龄
Integer totalAge = Optional.ofNullable(userList).orElseGet(Collections::emptyList).stream().map(User::getAge).reduce(0, Integer::sum);
 
// 按排名倒序排列
List<User> sortedUserList = Optional.ofNullable(list).orElseGet(Collections::emptyList).stream().sorted(Comparator.comparing(User::getRank, Comparator.reverseOrder())).collect(Collectors.toList());
 
// 获取排名第2高的user
User rankUser = Optional.ofNullable(sortedUserList).orElseGet(Collections::emptyList).stream().skip(1).findFirst().get();
 
// 排名最高的user
User highestRankUser = Optional.ofNullable(list).orElseGet(Collections::emptyList).stream().max(Comparator.comparing(User::getRank)).get();
 
// 是否存在排名大于1的user
boolean flag = Optional.ofNullable(list).orElseGet(Collections::emptyList).stream().anyMatch(user -> user.getRank() > 1);
 
// 是否所有user排名都大于1
boolean flag = Optional.ofNullable(list).orElseGet(Collections::emptyList).stream().allMatch(user -> user.getRank() > 1);
 
// 是否所有user排名都不大于5
boolean flag = Optional.ofNullable(list).orElseGet(Collections::emptyList).stream().noneMatch(user -> user.getRank() > 5);
 
// 按唯一id分组
Map<Long, User> idUserMap = Optional.ofNullable(list).orElseGet(Collections::emptyList).stream().collect(Collectors.toMap(User::getId, Function.identity()));
 
// 按唯一id,名字分组
Map<Long, String> idNameMap = Optional.ofNullable(list).orElseGet(Collections::emptyList).stream().collect(Collectors.toMap(User::getId, User::getName));
 
// 按年龄,名字分组，相同年龄的后出现的被覆盖
Map<Integer, String> ageNameMap = Optional.ofNullable(list).orElseGet(Collections::emptyList).stream().collect(Collectors.toMap(User::getAge, User::getName, (a, b) -> a));
 
// 按性别分组
Map<Byte, List<User>> gentleUserMap = Optional.ofNullable(list).orElseGet(Collections::emptyList).stream().collect(Collectors.groupingBy(User::getGentle));
 
// 按排名是否大于3分组
Map<Boolean, List<User>> partitionUserMap = Optional.ofNullable(list).orElseGet(Collections::emptyList).stream().collect(Collectors.partitioningBy(user -> user.getRank() > 3));
 
// 按性别名字分组
Map<Byte, List<String>> gentleNameMap = Optional.ofNullable(list).orElseGet(Collections::emptyList).stream().collect(Collectors.groupingBy(User::getGentle, Collectors.mapping(User::getName, Collectors.toList())));
 
// 按性别年龄总和分组
Map<Byte, Integer> gentleTotalAgeMap = Optional.ofNullable(list).orElseGet(Collections::emptyList).stream().collect(Collectors.groupingBy(User::getGentle, Collectors.reducing(0, User::getAge, Integer::sum)));
 
// 迭代操作
Stream.iterate(0, i -> i + 1).limit(list.size()).forEach(i -> {
    System.out.println(list.get(i).getName());
});
 
// guava table转换
Table<Long, String, Integer> idNameRankTable = Optional.ofNullable(list).orElseGet(Collections::emptyList).stream().map(user -> ImmutableTable.of(user.getId(), user.getName(), user.getRank())).collect(HasHBasedTable::create, HashBasedTable::putAll, HashBasedTable::putAll);

// stream只能被terminal一次，下面是错误示范
Stream<User> stream = Optional.ofNullable(list).orElseGet(Collections::emptyList).stream();
stream.collect(Collectors.toMap(User::getId, Function.identity()));
stream.collect(Collectors.toMap(User::getId, User::getName)); // java.lang.IllegalStateException: stream has already been operated upon or closed 
 
// ssc-common的com.meicloud.mcu.common.util.StreamUtil简单封装了一些流操作，欢迎试用 
// 参考资料：https://www.ibm.com/developerworks/cn/java/j-lo-java8streamapi/index.html