Java : Create new ArrayLists based on some condition from existing ArrayList

Question

I am new to java .

I have 2 ArrayLists of Strings

List<String> a= [2,14]
List<String> b= [2,3,4,5]

I want two new ArrayLists

1) List has the value which is in b but not in a

  List<String> c= [3,4,5]

2) List has the value a but not in b

  List<String> d=[14]

I tried:

    List<String> c = new ArrayList<String>(b);
    c.removeAll(a);
   System.out.println("c::::::::::::::::::::::::::::::"+c);  // 2,3,4,5

which is not removing the values of List a

Complete Code

   public static void updatePartyType(List<String> oldPartyKeys, List<String> newPartyKeys, String customerCode) {
   System.out.println("oldPartyKeys--->"+oldPartyKeys);// 2,14
   System.out.println("newPartyKeys--->"+newPartyKeys); // 2,3,4,5

   System.out.println("oldPartyKeys class --->"+oldPartyKeys.getClass());// class java.util.ArrayList

   List<String> newlySelectedPartyKeys = new ArrayList<String>(newPartyKeys);
   newlySelectedPartyKeys.removeAll(oldPartyKeys);
                    System.out.println("newlySelectedPartyKeys::::::::::::::::::::::::::::"+newlySelectedPartyKeys);

Is the ordering important? These are set-based operations more than list-based ones. — Jon Skeet
– Jon Skeet, Commented Jan 1, 2014 at 10:06
Please edit the question to be a short but complete program - you've only given pseudo-code for the list initialization at the moment. — Jon Skeet
– Jon Skeet, Commented Jan 1, 2014 at 10:29
The code you've been trying should work which means the problem is elsewhere. You can go to my answer and see how it's done using your provided example using the same method you've tried. — Reut Sharabani
– Reut Sharabani, Commented Jan 1, 2014 at 10:38
That's still not complete code. Complete code is code which we can copy into a text editor, compile and run. — Jon Skeet
– Jon Skeet, Commented Jan 1, 2014 at 11:20

Jon Skeet · Accepted Answer · 2014-01-01 10:27:18Z

6

You're really proposing set operations more than list operations - in which case you'd be better off using a HashSet than an ArrayList. Then you could use Collection<E>.removeAll:

Set<String> a = ...;
Set<String> b = ...;

Set<String> c = new HashSet<String>(b);
c.removeAll(a);

Set<String> d = new HashSet<String>(a);
d.removeAll(b);

(This will work for ArrayList as well as HashSet - I've only changed to using sets because it's a more appropriate type when you want set-based operations.)

Or better, use Guava's Sets.difference method:

Set<String> a = ...;
Set<String> b = ...;

Set<String> c = Sets.difference(b, a);
Set<String> d = Sets.difference(a, b);

This will create views on the differences - so changes to the original sets will be reflected in the views. You can effectively take a snapshot of a view by creating a new HashSet:

Set<String> snapshot = new HashSet<String>(c);

edited Jan 1, 2014 at 10:27

answered Jan 1, 2014 at 10:10

Jon Skeet

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Comments

Mureinik · Accepted Answer · 2014-01-01 10:19:29Z

3

This can be done by using the removeAll method:

List<String> c = new ArrayList<>(b);
c.removeAll(a);

List<String> d = new ArrayList<>(a);
d.removeAll(b);

edited Jan 1, 2014 at 10:19

answered Jan 1, 2014 at 10:09

Mureinik

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5 Comments

Reut Sharabani Over a year ago

@monda I've added an example

Jon Skeet Over a year ago

@monda: "not working" doesn't give us any significant information. What happened when you tried it? It should be fine.

Reut Sharabani Over a year ago

@monda, I've added an example of a static "main" function in my code. You do know how to run a "main" function? Simple drop my code in, import "List" and "ArrayList" (your IDE should mark this for you really...) and follow the comments in the code.

monda Over a year ago

@Mureinik :Updated my question

Dawood ibn Kareem Over a year ago

@monda Maybe you only have Java 6. This answer requires Java 7. Try putting String inside each pair of <>.

Reut Sharabani · Accepted Answer · 2014-01-01 10:40:22Z

take a look at the addAll() and removeAll() in ArrayList

now you need b\a which is b.removeAll(a) and a\b which is a.removeAll(b)

A working example (this is for those who are new to Java, so it's verbose):

public static void main(String[] args) {
    // first we want to create the lists

    // create list a
    List<String> a = new ArrayList<String>();
    // add members to a
    a.add("2");
    a.add("14");
    // create list b
    List<String> b = new ArrayList<String>();
    // add members to b
    b.add("2");
    b.add("3");
    b.add("4");
    b.add("5");

    // create a list in which we store the "filtered" list - duplicated from
    // a
    List<String> aMinusB = new ArrayList<>(a);
    // "filter" using "removeAll" and giving the list b as the argument
    aMinusB.removeAll(b);
    System.out.println("A minus b:");
    // this is short for
    // "iterate over the entire list, naming the currently iterated node s"
    for (String s : aMinusB) {
        System.out.println(s);
    }

    // duplicate list b in the same manner as above
    List<String> bMinusA = new ArrayList<>(b);
    // "filter" using "removeAll" and giving the list a as the argument in
    // the same manner as above
    bMinusA.removeAll(a);
    System.out.println("B minus a:");
    // this is short for
    // "iterate over the entire list, naming the currently iterated node s"
    // in the same manner as above
    for (String s : bMinusA) {
        System.out.println(s);
    }

}

Juvanis · Accepted Answer · 2014-01-01 10:10:31Z

2

Convert your lists to Set instances. Then you can easily find the difference of sets by your own implementation or using a 3rd party library like Google Guava which has Sets.difference(), for example.

edited Jan 1, 2014 at 10:10

user180100

answered Jan 1, 2014 at 10:08

Juvanis

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4 Comments

user180100 Over a year ago

Beware that Sets.difference returns a view, not a new Set

Jon Skeet Over a year ago

@RC.: It returns a new Set instance - it's a view, but it's still a new set (SetView implements Set). It's not an independent set, but you could create one easily enough.

user180100 Over a year ago

@JonSkeet I know but you can run into issues if you use the view directly (like adding something to it)

Jon Skeet Over a year ago

@RC.: My point is that claiming it's "not a new set" is misleading IMO. I've now mentioned that it's a view in my answer though.

Maroun · Accepted Answer · 2014-01-01 10:08:37Z

1

addAll elements, then removeAll elements that appears in a.

The time complexity is O(n).

answered Jan 1, 2014 at 10:08

Maroun

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