I have an ArrayList<String>, and I want to remove repeated strings from it. How can I do this?
If you don't want duplicates in a Collection, you should consider why you're using a Collection that allows duplicates. The easiest way to remove repeated elements is to add the contents to a Set (which will not allow duplicates) and then add the Set back to the ArrayList:
Set<String> set = new HashSet<>(yourList);
yourList.clear();
yourList.addAll(set);
Of course, this destroys the ordering of the elements in the ArrayList.
Although converting the ArrayList to a HashSet effectively removes duplicates, if you need to preserve insertion order, I'd rather suggest you to use this variant
// list is some List of Strings
Set<String> s = new LinkedHashSet<>(list);
Then, if you need to get back a List reference, you can use again the conversion constructor.
In Java 8:
List<String> deduped = list.stream().distinct().collect(Collectors.toList());
Please note that the hashCode-equals contract for list members should be respected for the filtering to work properly.
Suppose we have a list of String like:
List<String> strList = new ArrayList<>(5);
// insert up to five items to list.
Then we can remove duplicate elements in multiple ways.
Prior to Java 8
List<String> deDupStringList = new ArrayList<>(new HashSet<>(strList));
Note: If we want to maintain the insertion order then we need to use LinkedHashSet in place of HashSet
Using Guava
List<String> deDupStringList2 = Lists.newArrayList(Sets.newHashSet(strList));
Using Java 8
List<String> deDupStringList3 = strList.stream().distinct().collect(Collectors.toList());
Note: In case we want to collect the result in a specific list implementation e.g. LinkedList then we can modify the above example as:
List<String> deDupStringList3 = strList.stream().distinct()
.collect(Collectors.toCollection(LinkedList::new));
We can use parallelStream also in the above code but it may not give expected performace benefits. Check this question for more.
If you don't want duplicates, use a Set instead of a List. To convert a List to a Set you can use the following code:
// list is some List of Strings
Set<String> s = new HashSet<String>(list);
If really necessary you can use the same construction to convert a Set back into a List.
Java 8 streams provide a very simple way to remove duplicate elements from a list. Using the distinct method.
If we have a list of cities and we want to remove duplicates from that list it can be done in a single line -
List<String> cityList = new ArrayList<>();
cityList.add("Delhi");
cityList.add("Mumbai");
cityList.add("Bangalore");
cityList.add("Chennai");
cityList.add("Kolkata");
cityList.add("Mumbai");
cityList = cityList.stream().distinct().collect(Collectors.toList());
How to remove duplicate elements from an arraylist
You can also do it this way, and preserve order:
// delete duplicates (if any) from 'myArrayList'
myArrayList = new ArrayList<String>(new LinkedHashSet<String>(myArrayList));
Here's a way that doesn't affect your list ordering:
ArrayList l1 = new ArrayList();
ArrayList l2 = new ArrayList();
Iterator iterator = l1.iterator();
while (iterator.hasNext()) {
YourClass o = (YourClass) iterator.next();
if(!l2.contains(o)) l2.add(o);
}
l1 is the original list, and l2 is the list without repeated items
(Make sure YourClass has the equals method according to what you want to stand for equality)
this can solve the problem:
private List<SomeClass> clearListFromDuplicateFirstName(List<SomeClass> list1) {
Map<String, SomeClass> cleanMap = new LinkedHashMap<String, SomeClass>();
for (int i = 0; i < list1.size(); i++) {
cleanMap.put(list1.get(i).getFirstName(), list1.get(i));
}
List<SomeClass> list = new ArrayList<SomeClass>(cleanMap.values());
return list;
}
It is possible to remove duplicates from arraylist without using HashSet or one more arraylist.
Try this code..
ArrayList<String> lst = new ArrayList<String>();
lst.add("ABC");
lst.add("ABC");
lst.add("ABCD");
lst.add("ABCD");
lst.add("ABCE");
System.out.println("Duplicates List "+lst);
Object[] st = lst.toArray();
for (Object s : st) {
if (lst.indexOf(s) != lst.lastIndexOf(s)) {
lst.remove(lst.lastIndexOf(s));
}
}
System.out.println("Distinct List "+lst);
Output is
Duplicates List [ABC, ABC, ABCD, ABCD, ABCE]
Distinct List [ABC, ABCD, ABCE]
There is also ImmutableSet from Guava as an option (here is the documentation):
ImmutableSet.copyOf(list);
Probably a bit overkill, but I enjoy this kind of isolated problem. :)
This code uses a temporary Set (for the uniqueness check) but removes elements directly inside the original list. Since element removal inside an ArrayList can induce a huge amount of array copying, the remove(int)-method is avoided.
public static <T> void removeDuplicates(ArrayList<T> list) {
int size = list.size();
int out = 0;
{
final Set<T> encountered = new HashSet<T>();
for (int in = 0; in < size; in++) {
final T t = list.get(in);
final boolean first = encountered.add(t);
if (first) {
list.set(out++, t);
}
}
}
while (out < size) {
list.remove(--size);
}
}
While we're at it, here's a version for LinkedList (a lot nicer!):
public static <T> void removeDuplicates(LinkedList<T> list) {
final Set<T> encountered = new HashSet<T>();
for (Iterator<T> iter = list.iterator(); iter.hasNext(); ) {
final T t = iter.next();
final boolean first = encountered.add(t);
if (!first) {
iter.remove();
}
}
}
Use the marker interface to present a unified solution for List:
public static <T> void removeDuplicates(List<T> list) {
if (list instanceof RandomAccess) {
// use first version here
} else {
// use other version here
}
}
EDIT: I guess the generics-stuff doesn't really add any value here.. Oh well. :)
public static void main(String[] args){
ArrayList<Object> al = new ArrayList<Object>();
al.add("abc");
al.add('a');
al.add('b');
al.add('a');
al.add("abc");
al.add(10.3);
al.add('c');
al.add(10);
al.add("abc");
al.add(10);
System.out.println("Before Duplicate Remove:"+al);
for(int i=0;i<al.size();i++){
for(int j=i+1;j<al.size();j++){
if(al.get(i).equals(al.get(j))){
al.remove(j);
j--;
}
}
}
System.out.println("After Removing duplicate:"+al);
}
If you're willing to use a third-party library, you can use the method distinct() in Eclipse Collections (formerly GS Collections).
ListIterable<Integer> integers = FastList.newListWith(1, 3, 1, 2, 2, 1);
Assert.assertEquals(
FastList.newListWith(1, 3, 2),
integers.distinct());
The advantage of using distinct() instead of converting to a Set and then back to a List is that distinct() preserves the order of the original List, retaining the first occurrence of each element. It's implemented by using both a Set and a List.
MutableSet<T> seenSoFar = UnifiedSet.newSet();
int size = list.size();
for (int i = 0; i < size; i++)
{
T item = list.get(i);
if (seenSoFar.add(item))
{
targetCollection.add(item);
}
}
return targetCollection;
If you cannot convert your original List into an Eclipse Collections type, you can use ListAdapter to get the same API.
MutableList<Integer> distinct = ListAdapter.adapt(integers).distinct();
Note: I am a committer for Eclipse Collections.
If you are using model type List< T>/ArrayList< T> . Hope,it's help you.
Here is my code without using any other data structure like set or hashmap
for (int i = 0; i < Models.size(); i++){
for (int j = i + 1; j < Models.size(); j++) {
if (Models.get(i).getName().equals(Models.get(j).getName())) {
Models.remove(j);
j--;
}
}
}
If you want to preserve your Order then it is best to use LinkedHashSet.
Because if you want to pass this List to an Insert Query by Iterating it, the order would be preserved.
Try this
LinkedHashSet link=new LinkedHashSet();
List listOfValues=new ArrayList();
listOfValues.add(link);
This conversion will be very helpful when you want to return a List but not a Set.
This three lines of code can remove the duplicated element from ArrayList or any collection.
List<Entity> entities = repository.findByUserId(userId);
Set<Entity> s = new LinkedHashSet<Entity>(entities);
entities.clear();
entities.addAll(s);
for(int a=0;a<myArray.size();a++){
for(int b=a+1;b<myArray.size();b++){
if(myArray.get(a).equalsIgnoreCase(myArray.get(b))){
myArray.remove(b);
dups++;
b--;
}
}
}
When you are filling the ArrayList, use a condition for each element. For example:
ArrayList< Integer > al = new ArrayList< Integer >();
// fill 1
for ( int i = 0; i <= 5; i++ )
if ( !al.contains( i ) )
al.add( i );
// fill 2
for (int i = 0; i <= 10; i++ )
if ( !al.contains( i ) )
al.add( i );
for( Integer i: al )
{
System.out.print( i + " ");
}
We will get an array {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Code:
List<String> duplicatList = new ArrayList<String>();
duplicatList = Arrays.asList("AA","BB","CC","DD","DD","EE","AA","FF");
//above AA and DD are duplicate
Set<String> uniqueList = new HashSet<String>(duplicatList);
duplicatList = new ArrayList<String>(uniqueList); //let GC will doing free memory
System.out.println("Removed Duplicate : "+duplicatList);
Note: Definitely, there will be memory overhead.
ArrayList<String> city=new ArrayList<String>();
city.add("rajkot");
city.add("gondal");
city.add("rajkot");
city.add("gova");
city.add("baroda");
city.add("morbi");
city.add("gova");
HashSet<String> hashSet = new HashSet<String>();
hashSet.addAll(city);
city.clear();
city.addAll(hashSet);
Toast.makeText(getActivity(),"" + city.toString(),Toast.LENGTH_SHORT).show();
you can use nested loop in follow :
ArrayList<Class1> l1 = new ArrayList<Class1>();
ArrayList<Class1> l2 = new ArrayList<Class1>();
Iterator iterator1 = l1.iterator();
boolean repeated = false;
while (iterator1.hasNext())
{
Class1 c1 = (Class1) iterator1.next();
for (Class1 _c: l2) {
if(_c.getId() == c1.getId())
repeated = true;
}
if(!repeated)
l2.add(c1);
}
LinkedHashSet will do the trick.
String[] arr2 = {"5","1","2","3","3","4","1","2"};
Set<String> set = new LinkedHashSet<String>(Arrays.asList(arr2));
for(String s1 : set)
System.out.println(s1);
System.out.println( "------------------------" );
String[] arr3 = set.toArray(new String[0]);
for(int i = 0; i < arr3.length; i++)
System.out.println(arr3[i].toString());
//output: 5,1,2,3,4
List<String> result = new ArrayList<String>();
Set<String> set = new LinkedHashSet<String>();
String s = "ravi is a good!boy. But ravi is very nasty fellow.";
StringTokenizer st = new StringTokenizer(s, " ,. ,!");
while (st.hasMoreTokens()) {
result.add(st.nextToken());
}
System.out.println(result);
set.addAll(result);
result.clear();
result.addAll(set);
System.out.println(result);
output:
[ravi, is, a, good, boy, But, ravi, is, very, nasty, fellow]
[ravi, is, a, good, boy, But, very, nasty, fellow]
This is used for your Custom Objects list
public List<Contact> removeDuplicates(List<Contact> list) {
// Set set1 = new LinkedHashSet(list);
Set set = new TreeSet(new Comparator() {
#Override
public int compare(Object o1, Object o2) {
if (((Contact) o1).getId().equalsIgnoreCase(((Contact) o2).getId()) /*&&
((Contact)o1).getName().equalsIgnoreCase(((Contact)o2).getName())*/) {
return 0;
}
return 1;
}
});
set.addAll(list);
final List newList = new ArrayList(set);
return newList;
}
As said before, you should use a class implementing the Set interface instead of List to be sure of the unicity of elements. If you have to keep the order of elements, the SortedSet interface can then be used; the TreeSet class implements that interface.
import java.util.*;
class RemoveDupFrmString
{
public static void main(String[] args)
{
String s="appsc";
Set<Character> unique = new LinkedHashSet<Character> ();
for(char c : s.toCharArray()) {
System.out.println(unique.add(c));
}
for(char dis:unique){
System.out.println(dis);
}
}
}
public Set<Object> findDuplicates(List<Object> list) {
Set<Object> items = new HashSet<Object>();
Set<Object> duplicates = new HashSet<Object>();
for (Object item : list) {
if (items.contains(item)) {
duplicates.add(item);
} else {
items.add(item);
}
}
return duplicates;
}
ArrayList<String> list = new ArrayList<String>();
HashSet<String> unique = new LinkedHashSet<String>();
HashSet<String> dup = new LinkedHashSet<String>();
boolean b = false;
list.add("Hello");
list.add("Hello");
list.add("how");
list.add("are");
list.add("u");
list.add("u");
for(Iterator iterator= list.iterator();iterator.hasNext();)
{
String value = (String)iterator.next();
System.out.println(value);
if(b==unique.add(value))
dup.add(value);
else
unique.add(value);
}
System.out.println(unique);
System.out.println(dup);
If you want to remove duplicates from ArrayList means find the below logic,
public static Object[] removeDuplicate(Object[] inputArray)
{
long startTime = System.nanoTime();
int totalSize = inputArray.length;
Object[] resultArray = new Object[totalSize];
int newSize = 0;
for(int i=0; i<totalSize; i++)
{
Object value = inputArray[i];
if(value == null)
{
continue;
}
for(int j=i+1; j<totalSize; j++)
{
if(value.equals(inputArray[j]))
{
inputArray[j] = null;
}
}
resultArray[newSize++] = value;
}
long endTime = System.nanoTime()-startTime;
System.out.println("Total Time-B:"+endTime);
return resultArray;
}
Related
I need to check if all Strings from ArrayList are present in another ArrayList. I can use containsAll but this is not what I want to achieve. Let's me show you this on example:
assertThat(firstArray).containsAll(secondArray);
This code will check if all items from one array is in another one. But I need to check that every single item from one array is contained in any place in the second array.
List<String> firstArray = new ArrayList<>;
List<String> secondArray = new ArrayList<>;
firstArray.add("Bari 1908")
firstArray.add("Sheffield United")
firstArray.add("Crystal Palace")
secondArray.add("Bari")
secondArray.add("Sheffield U")
secondArray.add("C Palace")
So I want to check if first item from secondArray is in firstArray(true) than that second(true) and third(false). I wrote the code which is doing this job but it's quite complicated and I would like to know if there is any simpler way to achieve this goal (maybe with using hamcrest matchers or something like that)
ArrayList<String> notMatchedTeam = new ArrayList<>();
for (int i = 0; i < secondArray.size(); i++) {
String team = secondArray.get(i);
boolean teamMatched = false;
for (int j = 0; j < firstArray.size(); j++) {
teamMatched = firstArray.get(j).contains(team);
if (teamMatched) {
break;
}
}
if (!teamMatched) {
notMatchedTeam.add(team);
}
}
You can do something like this
List<String> firstArray = new ArrayList<>();
List<String> secondArray = new ArrayList<>();
firstArray.add("Bari 1908");
firstArray.add("Sheffield United");
firstArray.add("Crystal Palace");
secondArray.add("Bari");
secondArray.add("Sheffield U");
secondArray.add("C Palace");
Set<String> firstSet= firstArray
.stream()
.collect(Collectors.toSet());
long count= secondArray.stream().filter(x->firstSet.contains(x)).count();
///
Map<String, Boolean> result =
secondArray.stream().collect(Collectors.toMap(s->s, firstSet::contains));
If count >0, then there are some items in second array which are not there in first.
result contains the string with its status.
Thanks
If you have space concerns like you have millions of words in one file and need to check entry of second file in first then use trie. From first make trie and check every entry of second in first.
Situation:
In your question you said that you wanted to return for each element if it exists or not, and in your actual code you are only returning a list of matching elements.
Solution:
You need to return a list of Boolean results instead, this is the code you need:
public static List<Boolean> whichElementsFound(List<String> firstList, List<String> secondList){
ArrayList<Boolean> resultList = new ArrayList<>();
for (int i = 0; i < secondList.size(); i++) {
String team = secondList.get(i);
resultList.add(firstList.contains(team));
}
return resultList;
}
Demo:
This is a working Demo using this method, returning respectively a List<Boolean> to reflects which element from the first list are found in the second.
Edit:
If you want to return the list of elements that were not found, use the following code:
public static List<String> whichElementsAreNotFound(List<String> firstList, List<String> secondList){
ArrayList<String> resultList = new ArrayList<>();
for (int i = 0; i < secondList.size(); i++) {
String team = secondList.get(i);
if(!firstList.contains(team)){
resultList.add(team);
}
}
return resultList;
}
This is the Demo updated.
I would like to compare String arrays to a list with market objects.
I implemented the code like that:
private List<Data> addMarketData(List<Data> list) {
String[] SEE = new String[]{"Albania", "Bosnia and Herzegovina", "Bulgaria", "Croatia", "Macedonia FYR", "Moldavia", "Montenegro", "Romania", "Serbia", "Slovenia" };
List<String> seeList = Arrays.asList(SEE);
String[] CEE = new String[]{"Czech Republic", "Hungary", "Poland", "Slovakia"};
List<String> ceeList = Arrays.asList(CEE);
for (int i = 0; i < list.size(); i++) {
for (int j = 0; j < seeList.size(); j++) {
if(list.get(i).getPropertyCountry().equals(seeList.get(j).toString())) {
list.get(i).setMarket("SEE");
}
}
for (int k = 0; k < ceeList.size(); k++) {
if(list.get(i).getPropertyCountry().equals(ceeList.get(k).toString())) {
list.get(i).setMarket("CEE");
}
}
}
return list;
}
However, I believe that this code produces more overhead than it really should. Especially the for loops. Could I just use one loop?
Therefore, how to make this piece of code much faster?
I appreciate your answer!
Move all the data into a Set<String>:
String[] SEE = ...
Set<String> setSEE = new HashSet<>(Arrays.asList(SEE));
String[] CEE = ...
Set<String> setCEE = new HashSet<>(Arrays.asList(CEE));
for (Data data : list) {
if (setSEE.contains(data.getPropertyCountry()) {
data.setMarket("SEE");
} else if (setCEE.contains(data.getPropertyCountry()) {
data.setMarket("CEE");
}
}
This won't generate the overhead you may think. Also, it is faster than your current O(N^2) approach.
Another idea is to move the data of these arrays into a Map<String, String> as proposed by #Narmer, but in this case you should define a value when the country is not found as key in the map.
Since Java 7, you can use diamond operator. For Java 5 and 6, you have to specify the whole generics usage:
Set<String> setSEE = new HashSet<String>(Arrays.asList(SEE));
//...
Set<String> setCEE = new HashSet<String>(Arrays.asList(CEE));
Define seeList and ceeList as HashSets and then use its contains() method. HashSet's contains() has constant time complexity.
Set<String> seeSet = new HashSet<>();
Collections.addAll(seeSet, SEE);
Set<String> ceeSet = new HashSet<>();
Collections.addAll(ceeSet, CEE);
And then:
for (int i = 0; i < list.size(); i++) {
if (seeSet.contains(list.get(i).getPropertyCountry()) {
list.get(i).setMarket("SEE");
}
if (ceeSet.contains(list.get(i).getPropertyCountry()) {
list.get(i).setMarket("CEE");
}
}
You could use a Map instead of list.
private static final Map<String, String> markets = new HashMap<String,String>(){{
put("Albania", "SEE");
put("Bosnia and Herzegovina", "SEE");
...
put("Hungary", "CEE");
...
}}
Then consult it
for(Data data: list){
data.setMarket(markets.get(data.getPropertyCountry()));
}
EDIT
As per the comments, the above is the optimal situation. You should check that data.getPropertyCountry() is not null (if permitted) and that the value returned by the list ins't null either:
for(Data data: list){
if(data.getPropertyCountry()!=null){
String market = markets.get(data.getPropertyCountry());
data.setMarket(market==null?"default market":market);
}
else data.setMarket("default value"); //if needed
}
Or using the beatiful Java 8 stream interface:
for(Data data: list.stream().filter(p -> p.getPropertyCountry() != null).collect(Collectors.toList())){
String market = markets.get(data.getPropertyCountry());
data.setMarket(market==null?"default market":market);
}
Well you can simply use two HashSet<String> collections to store the name of the countries in. A HashSet<String> performs lookups in approximately O(1) time per item, this O(n) for the entire array. Or you could use one HashMap<String,String> to perform lookups resulting in "SEE" or "CEE".
Example
Map<String,String> lut = new HashMap<String,String>();
for(String s : new String[]{"Albania", "Bosnia and Herzegovina", "Bulgaria", "Croatia", "Macedonia FYR", "Moldavia", "Montenegro", "Romania", "Serbia", "Slovenia" }) {
lut.put(s,"SEE");
}
for(String s : new String[]{"Czech Republic", "Hungary", "Poland", "Slovakia"}) {
lut.put(s,"CEE");
}
for (Data data : list) {
data.setMarket(lut.get(data.getPropertyCountry()));
}
The generation of the HashMap<String,String> (and putting data into it) should only be executed once (at startup). This will increase performance with a factor equal to the number of elements you put into the HashMap<String,String> (in this case 14).
I have one Arraylist of String and I have added Some Duplicate Value in that. and i just wanna remove that Duplicate value So how to remove it.
Here Example I got one Idea.
List<String> list = new ArrayList<String>();
list.add("Krishna");
list.add("Krishna");
list.add("Kishan");
list.add("Krishn");
list.add("Aryan");
list.add("Harm");
System.out.println("List"+list);
for (int i = 1; i < list.size(); i++) {
String a1 = list.get(i);
String a2 = list.get(i-1);
if (a1.equals(a2)) {
list.remove(a1);
}
}
System.out.println("List after short"+list);
But is there any Sufficient way remove that Duplicate form list. with out using For loop ?
And ya i can do it by using HashSet or some other way but using array list only.
would like to have your suggestion for that. thank you for your answer in advance.
You can create a LinkedHashSet from the list. The LinkedHashSet will contain each element only once, and in the same order as the List. Then create a new List from this LinkedHashSet. So effectively, it's a one-liner:
list = new ArrayList<String>(new LinkedHashSet<String>(list))
Any approach that involves List#contains or List#remove will probably decrease the asymptotic running time from O(n) (as in the above example) to O(n^2).
EDIT For the requirement mentioned in the comment: If you want to remove duplicate elements, but consider the Strings as equal ignoring the case, then you could do something like this:
Set<String> toRetain = new TreeSet<String>(String.CASE_INSENSITIVE_ORDER);
toRetain.addAll(list);
Set<String> set = new LinkedHashSet<String>(list);
set.retainAll(new LinkedHashSet<String>(toRetain));
list = new ArrayList<String>(set);
It will have a running time of O(n*logn), which is still better than many other options. Note that this looks a little bit more complicated than it might have to be: I assumed that the order of the elements in the list may not be changed. If the order of the elements in the list does not matter, you can simply do
Set<String> set = new TreeSet<String>(String.CASE_INSENSITIVE_ORDER);
set.addAll(list);
list = new ArrayList<String>(set);
if you want to use only arraylist then I am worried there is no better way which will create a huge performance benefit. But by only using arraylist i would check before adding into the list like following
void addToList(String s){
if(!yourList.contains(s))
yourList.add(s);
}
In this cases using a Set is suitable.
You can make use of Google Guava utilities, as shown below
list = ImmutableSet.copyOf(list).asList();
This is probably the most efficient way of eliminating the duplicates from the list and interestingly, it preserves the iteration order as well.
UPDATE
But, in case, you don't want to involve Guava then duplicates can be removed as shown below.
ArrayList<String> list = new ArrayList<String>();
list.add("Krishna");
list.add("Krishna");
list.add("Kishan");
list.add("Krishn");
list.add("Aryan");
list.add("Harm");
System.out.println("List"+list);
HashSet hs = new HashSet();
hs.addAll(list);
list.clear();
list.addAll(hs);
But, of course, this will destroys the iteration order of the elements in the ArrayList.
Shishir
Java 8 stream function
You could use the distinct function like above to get the distinct elements of the list,
stringList.stream().distinct();
From the documentation,
Returns a stream consisting of the distinct elements (according to Object.equals(Object)) of this stream.
Another way, if you do not wish to use the equals method is by using the collect function like this,
stringList.stream()
.collect(Collectors.toCollection(() ->
new TreeSet<String>((p1, p2) -> p1.compareTo(p2))
));
From the documentation,
Performs a mutable reduction operation on the elements of this stream using a Collector.
Hope that helps.
Simple function for removing duplicates from list
private void removeDuplicates(List<?> list)
{
int count = list.size();
for (int i = 0; i < count; i++)
{
for (int j = i + 1; j < count; j++)
{
if (list.get(i).equals(list.get(j)))
{
list.remove(j--);
count--;
}
}
}
}
Example:
Input: [1, 2, 2, 3, 1, 3, 3, 2, 3, 1, 2, 3, 3, 4, 4, 4, 1]
Output: [1, 2, 3, 4]
List<String> list = new ArrayList<String>();
list.add("Krishna");
list.add("Krishna");
list.add("Kishan");
list.add("Krishn");
list.add("Aryan");
list.add("Harm");
HashSet<String> hs=new HashSet<>(list);
System.out.println("=========With Duplicate Element========");
System.out.println(list);
System.out.println("=========Removed Duplicate Element========");
System.out.println(hs);
I don't think the list = new ArrayList<String>(new LinkedHashSet<String>(list)) is not the best way , since we are using the LinkedHashset(We could use directly LinkedHashset instead of ArrayList),
Solution:
import java.util.ArrayList;
public class Arrays extends ArrayList{
#Override
public boolean add(Object e) {
if(!contains(e)){
return super.add(e);
}else{
return false;
}
}
public static void main(String[] args) {
Arrays element=new Arrays();
element.add(1);
element.add(2);
element.add(2);
element.add(3);
System.out.println(element);
}
}
Output:
[1, 2, 3]
Here I am extending the ArrayList , as I am using the it with some changes by overriding the add method.
public List<Contact> removeDuplicates(List<Contact> list) {
// Set set1 = new LinkedHashSet(list);
Set set = new TreeSet(new Comparator() {
#Override
public int compare(Object o1, Object o2) {
if(((Contact)o1).getId().equalsIgnoreCase(((Contact)2).getId()) ) {
return 0;
}
return 1;
}
});
set.addAll(list);
final List newList = new ArrayList(set);
return newList;
}
This will be the best way
List<String> list = new ArrayList<String>();
list.add("Krishna");
list.add("Krishna");
list.add("Kishan");
list.add("Krishn");
list.add("Aryan");
list.add("Harm");
Set<String> set=new HashSet<>(list);
It is better to use HastSet
1-a) A HashSet holds a set of objects, but in a way that it allows you to easily and quickly determine whether an object is already in the set or not. It does so by internally managing an array and storing the object using an index which is calculated from the hashcode of the object. Take a look here
1-b) HashSet is an unordered collection containing unique elements. It has the standard collection operations Add, Remove, Contains, but since it uses a hash-based implementation, these operation are O(1). (As opposed to List for example, which is O(n) for Contains and Remove.) HashSet also provides standard set operations such as union, intersection, and symmetric difference.Take a look here
2) There are different implementations of Sets. Some make insertion and lookup operations super fast by hashing elements. However that means that the order in which the elements were added is lost. Other implementations preserve the added order at the cost of slower running times.
The HashSet class in C# goes for the first approach, thus not preserving the order of elements. It is much faster than a regular List. Some basic benchmarks showed that HashSet is decently faster when dealing with primary types (int, double, bool, etc.). It is a lot faster when working with class objects. So that point is that HashSet is fast.
The only catch of HashSet is that there is no access by indices. To access elements you can either use an enumerator or use the built-in function to convert the HashSet into a List and iterate through that.Take a look here
Without a loop, No! Since ArrayList is indexed by order rather than by key, you can not found the target element without iterate the whole list.
A good practice of programming is to choose proper data structure to suit your scenario. So if Set suits your scenario the most, the discussion of implementing it with List and trying to find the fastest way of using an improper data structure makes no sense.
public static void main(String[] args) {
#SuppressWarnings("serial")
List<Object> lst = new ArrayList<Object>() {
#Override
public boolean add(Object e) {
if(!contains(e))
return super.add(e);
else
return false;
}
};
lst.add("ABC");
lst.add("ABC");
lst.add("ABCD");
lst.add("ABCD");
lst.add("ABCE");
System.out.println(lst);
}
This is the better way
list = list.stream().distinct().collect(Collectors.toList());
This could be one of the solutions using Java8 Stream API. Hope this helps.
public void removeDuplicates() {
ArrayList<Object> al = new ArrayList<Object>();
al.add("java");
al.add('a');
al.add('b');
al.add('a');
al.add("java");
al.add(10.3);
al.add('c');
al.add(14);
al.add("java");
al.add(12);
System.out.println("Before Remove Duplicate elements:" + al);
for (int i = 0; i < al.size(); i++) {
for (int j = i + 1; j < al.size(); j++) {
if (al.get(i).equals(al.get(j))) {
al.remove(j);
j--;
}
}
}
System.out.println("After Removing duplicate elements:" + al);
}
Before Remove Duplicate elements:
[java, a, b, a, java, 10.3, c, 14, java, 12]
After Removing duplicate elements:
[java, a, b, 10.3, c, 14, 12]
Using java 8:
public static <T> List<T> removeDuplicates(List<T> list) {
return list.stream().collect(Collectors.toSet()).stream().collect(Collectors.toList());
}
In case you just need to remove the duplicates using only ArrayList, no other Collection classes, then:-
//list is the original arraylist containing the duplicates as well
List<String> uniqueList = new ArrayList<String>();
for(int i=0;i<list.size();i++) {
if(!uniqueList.contains(list.get(i)))
uniqueList.add(list.get(i));
}
Hope this helps!
private static void removeDuplicates(List<Integer> list)
{
Collections.sort(list);
int count = list.size();
for (int i = 0; i < count; i++)
{
if(i+1<count && list.get(i)==list.get(i+1)){
list.remove(i);
i--;
count--;
}
}
}
public static List<String> removeDuplicateElements(List<String> array){
List<String> temp = new ArrayList<String>();
List<Integer> count = new ArrayList<Integer>();
for (int i=0; i<array.size()-2; i++){
for (int j=i+1;j<array.size()-1;j++)
{
if (array.get(i).compareTo(array.get(j))==0) {
count.add(i);
int kk = i;
}
}
}
for (int i = count.size()+1;i>0;i--) {
array.remove(i);
}
return array;
}
}
I have a generic arraylist of an object here I want to remove certain elements, The problem is when I iterate the list with for loop, I can't do a simple sequence of remove()'s because the elements are shifted after each removal.
Thanks
Use Iterator to remove element
Like
Iterator itr = list.iterator();
String strElement = "";
while (itr.hasNext()) {
strElement = (String) itr.next();
if (strElement.equals("2")) {
itr.remove();
}
}
See here
You can iterate the list this way ...
public void clean(List<Kopek> kopeks) {
for(Kopek kopek : kopeks) {
if (kopek.isDirty())
kopeks.remove(kopek);
}
}
Which is equiv to ...
public void clean1(List<Kopek> kopeks) {
Iterator<Kopek> kopekIter = kopeks.iterator();
while (kopekIter.hasNext()) {
Kopek kopek = kopekIter.next();
if (kopek.isDirty())
kopeks.remove(kopek);
}
}
Don't do this ... (due to the reason you have already observed.)
public void clean(List<Kopek> kopeks) {
for(int i=0; i<kopeks.size(); i++) {
Kopek kopek = kopeks.get(i);
if (kopek.isDirty())
kopeks.remove(i);
}
}
However, I believe removal by index rather than by object is more efficient. Removal by object is not efficient because the list is in most cases not a hashed list.
kopeks.remove(kopek);
vs
kopeks.remove(i);
To achieve positional remove, by treating a moving target appropriately ...
public void clean(List<Kopek> kopeks) {
int i=0;
while(i<kopeks.size()) {
Kopek kopek = kopeks.get(i);
if (kopek.isDirty()) // no need to increment.
kopeks.remove(i);
else
i++;
}
}
If you have the objects that you want to remove from your ArrayList<T> you can use :
mArrayList.remove(object);
or you can use an Iterator to remove your objects:
while(iterator.hasNext()){
if(iterator.next() == some condition for removal){
iterator.remove();
}
}
You could iterate backwards and remove as you go through the ArrayList. This has the advantage of subsequent elements not needing to shift and is easier to program than moving forwards.
List<String> arr = new ArrayList<String>();
ListIterator<String> li = arr.listIterator(arr.size());
// Iterate in reverse.
while(li.hasPrevious()) {
String str=li.previous();
if(str.equals("A"))
{
li.remove();
}
}
Create a separate ArrayList of Index of the data to be removed from the original ArrayList, then remove those elements by looping over it with for loop.
ArrayList<Myobj> arr = new ArrayList<Myobj>();
for (Myobj o : arr){
arr.remove(arr.indexOf(o));
}
without using iterators also solves the issue.. All i wanted to do is get the index which are to be deleted and sort it in decending order then remove it from the list.
check the code below
Arraylist<obj> addlist = getlist();
List<Integer> indices = new ArrayList<Integer>();
for(int i=0; i<addlist.size() ;i++){
if(addlist.get(i).getDelete()){
indices.add(i);
}
}
Collections.sort(indices, Collections.reverseOrder());
for (int i : indices)
addlist.remove(i);
Is it possible to use a single
Vector vector1 = new Vector();
Vector vector2 = new Vector();
Map map = new HashMap();
for(String key:vector1&&String value:vector2)
{
map.put(key,value);
}
or something similar.I am trying to poulate a map with these two vectors.I tried the above for each statement but it gave me a syntax error.
Any help.?
This can't work.
this:
for(String key:vector1){
}
is a shortcut for this:
for(Iterator<String> it = vector1.iterator();it.hasNext();){
String key = it.next();
}
And it works with one iterator only.
Here's some code that gets close to what you want:
// don't use vector, it's ancient and unofficially deprecated
final List<String> keyList = new ArrayList<String>();
final List<String> valueList = new ArrayList<String>();
//...
// add keys and values here
final Map<String, String> map = new HashMap<String, String>();
for(
Iterator<String> it1 = keyList.iterator(), it2 = valueList.iterator();
// this works even if the lists have different sizes
it1.hasNext() && it2.hasNext();){
map.put(it1.next(), it2.next());
}
I also took the liberty to address some flaws in your code:
Don't use Raw Types in new code (use generics)
Don't use Vector, Hashtable or Enumeration, use ArrayList<E>, HashMap<K,V>, Iterator<E> (respectively) instead. Many questions here cover this, including this one
Why not just use the old style?
for (int i = 0; i < vector1.size() && i < vector2.size(); i++) {
map.put(vector1.get(i), vector2.get(i));
}
Its not allowed You can however use this.
Vector<String> vector1 = new Vector();
Vector<String> vector2 = new Vector();
Map <String,String> map = new HashMap<>();
for (int index = 0 ; index < vector1.size() ; index ++ ) //vector1 & vector are of same length
{
map.put(vector1.elementAt(index), vector2.elementAt(index));
}
No, there's nothing like that. You'll have to use two loops.
There is no such construct but you could do something like :
Iterator i1 = vector1.iterator();
Iterator i2 = vector2.iterator();
while (i1.hasNext() ) {
map.put(i1.next(),i2.next());
}
Assuming the same size for both vectors.
No, that doesn't work. You can get that effect like this:
Iterator<String> i1 = v1.iterator();
Iterator<String> i2 = v2.iterator();
while (i1.hasNext()) {
map.put(i1.next(), i2.next();
}
assert !i2.hasNext();
There's no parallel for statement, but you can use Iterators:
Vector vector1 = new Vector();
// ...populate vector1...
Vector vector2 = new Vector();
// ...populate vector2...
Map map = new HashMap();
Iterator it1, it2;
it1 = vector1.iterator();
it2 = vector2.iterator();
while (it1.hasNext() && it2.hasNext())
{
map.put(it1.next(), it2.next());
}
for-each loop behind the scene is converted to typical old fashion iterator so
for(String item : items){
...
}
Are converted by the compiler into
Iterator iItems = items.iterator()
while(iItems.hasNext()){
String item = iItems.next();
...
}
So you could do sth like that:
Iterator iVector2 = vector2.iterator();
for(String item1: vector1){
String item2 = iVector2.next();
....
}