map方法compute/computeIfAbsent/computeIfPresent/merge/putIfAbsent/forEach/getOrDefault

拼搏现实的明天。 2022-05-19 07:11 232阅读 0赞

/**
             *  forEach(BiConsumer<? super K, ? super V> action)
             *  循环key,value
             */
            @Test
            public void testHashMapForEach(){
                Map<String,String> map = new HashMap<String, String>();
                map.put("a","aaa");
                map.put("b","bbb");
                map.put("c","ccc");
                map.put("d","ddd");
    
                map.forEach((k,v)->{
                    System.out.println(k+"  "+v);
                });
                /**
                 * a  aaa
                 * b  bbb
                 * c  ccc
                 * d  ddd
                 */
    
            }
    
            /**
             *  V getOrDefault(Object key, V defaultValue)
             *  如果根据key得到value为null并且不包含这个key，则返回默认设置的值
             */
            @Test
            public void testHashMapGetOrDefault(){
                Map<String,String> map = new HashMap<String, String>();
                map.put("a","aaa");
                map.put("b",null);
                map.put("c","ccc");
                map.put("d","ddd");
    
                String orDefault1 = map.getOrDefault("a", "eeee");
                String orDefault2 = map.getOrDefault("b", "eeee");
                String orDefault3 = map.getOrDefault("m", "eeee");
                System.out.println(orDefault1+"  "+orDefault2+"   "+orDefault3);
                /**
                 * aaa  null   eeee
                 */
    
            }
    
    
            /**
             * compute(K key, BiFunction<? super K, ? super V, ? extends V> remappingFunction) {
             *         Objects.requireNonNull(remappingFunction);
             *         V oldValue = get(key);
             *         V newValue = remappingFunction.apply(key, oldValue);
             *
             *   只要新value不为null就会插入新key或者覆盖原来value,如果新value为null,就会把这个键值对直接删除
             */
            @Test
            public void testHashMapCompute(){
                List<Student> studentList = Arrays.asList(
                        new Student("apple",null,10),
                        //new Student("orange","女",30),
                        //  new Student("pink",null,60),
                        new Student("pink","男",60),
                        new Student("black","男",60),
                        new Student("orange",null,150)
                );
                Map<String,String> map = new HashMap<String, String>();
                map.put("pink",null);//原来为空，测试现在不为空
                map.put("orange","女");//原来不为空，测试现在为空
                map.put("apple",null);//原来为空，测试现在为空
                studentList.stream().forEachOrdered(student->map.compute(student.getName(),(k,v)->{
                    return student.getSex();}));
                System.out.println(map);
                // {pink=男, black=男}
            }
    
            /**
             * computeIfAbsent(K key, Function<? super K, ? extends V> mappingFunction) {
             *         Objects.requireNonNull(mappingFunction);
             *         返回值就是Function里面的V，所以computeIfAbsent方法执行后还可以继续做一系列的操作,
             *         但是value已经放到集合中了，所以只有引用类型才可以继续操作，String不行
             *
             * 只要是原来的value为null,就会插入或者更新，如果不为null，不变
             * 如果新的value为null,则不做操作,原来的数据不会删除
             *
             */
            @Test
            public void testHashMapComputeIfAbsent(){
                List<Student> studentList = Arrays.asList(
                        new Student("apple",null,10),
                        //new Student("orange","女",30),
                        //  new Student("pink",null,60),
                        new Student("pink","男",60),
                        new Student("black","男",60),
                        new Student("orange",null,150)
                );
                Map<String,String> map = new HashMap<String, String>();
                map.put("pink",null);//原来为空，测试现在不为空
                map.put("orange","女");//原来不为空，测试现在不为空
                studentList.stream().forEachOrdered(student->map.computeIfAbsent(student.getName(),k->k).toUpperCase());
                //  studentList.stream().forEachOrdered(student->map.computeIfAbsent(student.getName(),k->null));
                System.out.println(map);
                // {
       {orange=女, apple=apple, pink=pink, black=black}   toUpperCase没有达到效果
            }
    
            @Test
            public void testHashMapComputeIfAbsent2(){
                List<Student> studentList = Arrays.asList(
                        new Student("apple",null,10),
                        new Student("pink","男",60),
                        new Student("black","男",60),
                        new Student("black","女",80)
                );
                Map<String,List<Student>> map = new HashMap<>();
                studentList.stream().forEachOrdered(
                        student->map.computeIfAbsent(student.getName(),k->new ArrayList<Student>()).add(student)
                );
                System.out.println(map);
                // {
                // apple=[Student{name='apple', sex='null', money=10.0}],
                // pink=[Student{name='pink', sex='男', money=60.0}],
                // black=[Student{name='black', sex='男', money=60.0}, Student{name='black', sex='女', money=80.0}]
                // }   add达到效果了
            }
    
    
            /**
             * computeIfPresent(K key,BiFunction<? super K, ? super V, ? extends V> remappingFunction)
             *
             * 不会插入新key-value，
             * 根据key获取value,
             * 如果oldValue不为null, newValue为null则删除，newValue不为null则更新，
             * 如果oldValue为null，当不存在这个键值对处理，不做任何操作
             */
            @Test
            public void testHashMapComputeIfPresent(){
                List<Student> studentList = Arrays.asList(
                        new Student("apple","女",10),
                        new Student("pink","男",60),
                        new Student("black","男",60),
                        new Student("red","女",60),
                        new Student("orange",null,150)
                );
                Map<String,String> map = new HashMap<String, String>();
                map.put("pink",null);//原来为空，测试现在不为空
                map.put("orange","女");//原来不为空，测试现在为空
                map.put("apple","男");//原来不为空，测试现在不为空
                studentList.stream().forEachOrdered(student->map.computeIfPresent(student.getName(),(k,v)->{
                    return student.getSex();}));
                System.out.println(map);
                // {apple=女, pink=null}
            }
    
            /**
             *  V merge(K key, V value, BiFunction<? super V, ? super V, ? extends V> remappingFunction)
             *  根据key获取value为null,则用新value，
             *  若不为null, oldvalue与newValue会执行这个函数remappingFunction，赋值给当前key
             */
            @Test
            public void testHashMapMerge(){
                Map<String,String> map = new HashMap<String, String>();
                map.put("a","aaa");
                map.put("b",null);
    
                map.merge("a","hahaha",String::concat);
                map.merge("b","hahaha",String::concat);
                System.out.println(map);
                //{a=aaahahaha, b=hahaha}
            }
    
    
            /**
             * default V putIfAbsent(K key, V value)
             *
             * 如果获取的value为null,则用赋予新的value（插入或者更新）
             */
            @Test
            public void testHashMapPutIfAbsent(){
                Map<String,String> map = new HashMap<String, String>();
                map.put("a","aaa");
                map.put("b",null);
    
                map.putIfAbsent("a","ddd");
                map.putIfAbsent("b","ddd");
                map.putIfAbsent("c","ddd");
                System.out.println(map);
                //{a=aaa, b=ddd, c=ddd}
            }

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