检查一个列表是否包含来自另一个列表的元素

Question

I have two lists with different objects in them.

List<Object1> list1;
List<Object2> list2;

I want to check if element from list1 exists in list2, based on specific attribute (Object1 and Object2 have (among others), one mutual attribute (with type Long), named attributeSame).

right now, I do it like this:

boolean found = false;
for(Object1 object1 : list1){
   for(Object2 object2: list2){
       if(object1.getAttributeSame() == object2.getAttributeSame()){
           found = true;
           //also do something
       }
    }
    if(!found){
        //do something
    }
    found = false;
}

But I think there is a better and faster way to do this :) Can someone propose it?

Thanks!

Answer 1

If you just need to test basic equality, this can be done with the basic JDK without modifying the input lists in the one line

!Collections.disjoint(list1, list2);

If you need to test a specific property, that's harder. I would recommend, by default,

list1.stream()
   .map(Object1::getProperty)
   .anyMatch(
     list2.stream()
       .map(Object2::getProperty)
       .collect(toSet())
       ::contains)

...which collects the distinct values in list2 and tests each value in list1 for presence.

Answer 2

You can use Apache Commons CollectionUtils :

if(CollectionUtils.containsAny(list1,list2)) {  
    // do whatever you want
} else { 
    // do other thing 
}  

This assumes that you have properly overloaded the equals functionality for your custom objects.

Answer 3

为了缩短纳伦德拉的逻辑，你可以使用这个：

boolean var = lis1.stream().anyMatch(element -> list2.contains(element));

Answer 4

There is one method of Collection named retainAll but having some side effects for you reference

Retains only the elements in this list that are contained in the specified collection (optional operation). In other words, removes from this list all of its elements that are not contained in the specified collection.

true if this list changed as a result of the call

Its like

boolean b = list1.retainAll(list2);

Answer 5

Loius answer is correct, I just want to add an example:

listOne.add("A");
listOne.add("B");
listOne.add("C");

listTwo.add("D");
listTwo.add("E");
listTwo.add("F");      

boolean noElementsInCommon = Collections.disjoint(listOne, listTwo); // true

Answer 6

to make it faster, you can add a break; that way the loop will stop if found is set to true:

boolean found = false;
for(Object1 object1 : list1){
   for(Object2 object2: list2){
       if(object1.getAttributeSame() == object2.getAttributeSame()){
           found = true;
           //also do something  
           break;
       }
    }
    if(!found){
        //do something
    }
    found = false;
}

If you would have maps in stead of lists with as keys the attributeSame, you could check faster for a value in one map if there is a corresponding value in the second map or not.

Answer 7

According to the JavaDoc for the .contains(Object obj) :

Returns true if this list contains the specified element. More formally, returns true if and only if this list contains at least one element e such that (o==null ? e==null : o.equals(e)).

So if you override your .equals() method for your given object, you should be able to do: if(list1.contains(object2))...

If the elements will be unique (ie. have different attributes) you could override the .equals() and .hashcode() and store everything inHashSets . This will allow you to check if one contains another element in constant time.

Answer 8

faster way will require additional space .

For example:

1. put all items in one list into a HashSet ( you have to implement the hash function by yourself to use object.getAttributeSame() )

2. Go through the other list and check if any item is in the HashSet.

In this way each object is visited at most once. and HashSet is fast enough to check or insert any object in O(1).

Answer 9

Can you define the type of data you hold ? is it big data ? is it sorted ? I think that you need to consider different efficiency approaches depending on the data.

For example, if your data is big and unsorted you could try and iterate the two lists together by index and store each list attribute in another list helper. then you could cross check by the current attributes in the helper lists.

good luck

edited : and I wouldn't recommend overloading equals. its dangerous and probably against your object oop meaning.

Answer 10

org.springframework.util.CollectionUtils

boolean containsAny(java.util.Collection<?> source, java.util.Collection<?> candidates)

Return true if any element in 'candidates' is contained in 'source'; otherwise returns false

Answer 11

使用java 8 ，我们可以像下面那样检查一个列表是否包含其他列表的任何元素

boolean var = lis1.stream().filter(element -> list2.contains(element)).findFirst().isPresent();

Answer 12

I have two lists with different objects in them.

List<Object1> list1;
List<Object2> list2;

I want to check if element from list1 exists in list2, based on specific attribute (Object1 and Object2 have (among others), one mutual attribute (with type Long), named attributeSame).

right now, I do it like this:

boolean found = false;
for(Object1 object1 : list1){
   for(Object2 object2: list2){
       if(object1.getAttributeSame() == object2.getAttributeSame()){
           found = true;
           //also do something
       }
    }
    if(!found){
        //do something
    }
    found = false;
}

But I think there is a better and faster way to do this :) Can someone propose it?

Thanks!