java8 How to use Stream to check whether an attribute of an List object is duplicated

Directory uses Stream to check whether an attribute of List object has repeated exercise 1. Some uses of stream under exercise 1. Five duplicate removal methods of list Method 1. Use java8 new feature stream to remove List duplicate method 2. Double for loop duplicate removal method 3. set set to judge duplicate without disturbing the order Method 4. Judge to assign to another list set after traversal Method 5. set and list transform duplicate removal

Use Stream to check whether an attribute of an List object is duplicated

In the development of Java8, it is often necessary to judge whether there are duplicate values in an attribute for List object collection. The results can be obtained conveniently by Stream stream processing.

Exercise 1 Some Uses of stream

Test sample Java code

  @Test
    public void t2() {
        List<User> list = new ArrayList<>();
        User user1 = new User("zhangsan", "beijing", 30);
        User user2 = new User("zhangsan", "beijing", 40);
        User user3 = new User("lisi", "shanghai", 35);
        User user4 = new User("lisi", "shanghai", 28);
        User user5 = new User("lisim", "shanghai", 32);
        list.add(user1); list.add(user2);list.add(user3); list.add(user4);list.add(user5);
        System.out.println(" Original data: "+list);
        // Determine whether the name is duplicated , Practice using java8 Adj. stream Method 
        // Method 1. distinct,  Compare sizes directly, only knowing if there are duplicates 
        List<String> collect1 = list.stream().map(User::getName).distinct().collect(Collectors.toList());
        System.out.println(collect1.size()!=list.size()?" Method 1- Duplicate name ":" No repetition ");
        // Method 2. User name count 
        Map<Object, Long> collect2 = list.stream().collect(
                Collectors.groupingBy( User::getName , Collectors.counting()  )   );
        System.out.println(" Name duplicate count: "+collect2);
        // Screen out duplicate names 
        List<Object> collect3 = collect2.keySet().stream().
                filter(key -> collect2.get(key) > 1).collect(Collectors.toList());
        // You can know which names have duplicates 
        System.out.println(" Method 2- Duplicate names   :  "+collect3);
        // Method 3 Retain a count for duplicate names 
        List<Map<String, Long>> collect4 = collect2.keySet().stream().
                filter(key -> collect2.get(key) > 1).map(key -> {
            Map<String, Long> map = new HashMap<>();
            map.put((String) key, collect2.get(key));
            return map;
        }).collect(Collectors.toList());
        System.out.println(" Method 3- Duplicate names and counts: "+collect4);
    }

Run the results, which is convenient to verify whether it is what it needs.

Original data: [User (name=zhangsan, address=beijing, age=30), User (name=zhangsan, address=beijing, address=zhangsan, address=User, age=35), User (name=lisi, address=shanghai, age=28), ES60name=lisim, address=age=32)]
Method 1-Duplicate name
Name duplicate count: {lisi=2, zhangsan=2, lisim=1}
Method 2-Duplicate names: [lisi, zhangsan]
Method 3-Duplicate names and counts: [{lisi=2}, {zhangsan=2}]

Five Ways to Deduplicate list

Interview is often asked how list, 1 is oral, do not need to reflect the code, at this time, thinking 1 must be clear, you can list the methods of centralized de-heavy, to show you the list data structure, and related methods to master, reflect your java basic learning is solid or not

Next, I will show five methods one by one

Create a new list array:

List list = new ArrayList(); 
list.add(26); 
list.add(39); 
list.add(39); 
list.add(39); 
list.add(39); 
list.add(5); 
list.add(40); 
list.add(39); 
list.add(25); 
System.out.println(list); 

Method 1: Use stream, a new feature of java8, to deduplicate List

List newList = list.stream().distinct().collect(Collectors.toList()); 
System.out.println( " java8 New features stream Weight removal : " +newList); 
list.add(39); 

Method 2: Double for cycle deduplication

for (int i = 0; i < list.size(); i++) { 
for (int j = 0; j < list.size(); j++) { 
if(i!=j&&list.get(i)==list.get(j)) { 
list.remove(list.get(j)); 
} 
} 
} 

After studying the above method, it is really a little bit of a problem, and the optimized method is put below (it is not recommended to use, the speed is too slow)

        for (int i = 0; i < list.size(); i++) {
            for (int j = 0; j < list.size(); ) {
//                    System.out.println(i+"-"+list.get(i)+"-"+j+" ! ! ! ! "+list.get(j));
                if (i != j && list.get(i) == list.get(j)) {
//                    System.out.println(j+":"+list.get(j));
                    list.remove(j);
                } else {
                    j++;
                }
            }
        }
System.out.println( "Double for Cyclic weight removal : " +list); 
list.add(39); 

Method 3: set set judges duplication removal without disturbing the order

Set set1 = new HashSet(); 
List newList1 = new ArrayList(); 
for (Integer integer : list) { 
if(set1.add(integer)) { 
newList1.add(integer); 
} 
} 
System.out.println( " set Set judgment deduplication : " +list); 
list.add(39); 

Method 4: After traversing, decide to assign to another list set

List newList2 = new ArrayList(); 
for (Integer integer : list) { 
if(!newList2.contains(integer)){ 
newList2.add(integer); 
} 
} 
System.out.println( "Assign new list Weight removal : " +newList2); 
list.add(39); 

Method 5: set and list conversion deduplication

Set set2 = new HashSet(); 
List newList3 = new ArrayList(); 
set2.addAll(list); 
newList3.addAll(set2); 
System.out.println( " set And list Conversion de-duplication : " +newList3);