How can I delete duplicates in a Dart List? list.distinct()?

Question

How do I delete duplicates from a list without fooling around with a set? Is there something like list.distinct()? or list.unique()?

void main() {
  print("Hello, World!");

  List<String> list = ['abc',"abc",'def'];
  list.forEach((f) => print("this is list $f"));

  Set<String> set = new Set<String>.from(list);
  print("this is #0 ${list[0]}");
  set.forEach((f) => print("set: $f"));

  List<String> l2= new List<String>.from(set);
  l2.forEach((f) => print("This is new $f"));
}

Hello, World!
this is list abc
this is list abc
this is list def
this is #0 abc
set: abc
set: def
This is new abc
This is new def

---

Set seems to be way faster!! But it loses the order of the items :/

Answer 1

Use toSet and then toList

  var ids = [1, 4, 4, 4, 5, 6, 6];
  var distinctIds = ids.toSet().toList();

Result: [1, 4, 5, 6]

Or with spread operators:

var distinctIds = [...{...ids}];

Answer 2

I didn't find any of the provided answers very helpful. Here is what I generally do:

final ids = Set();
myList.retainWhere((x) => ids.add(x.id));

Of course you can use any attribute which uniquely identifies your objects. It doesn't have to be an id field.

Benefits over other approaches:

• Preserves the original order of the list
• Works for rich objects not just primitives/hashable types
• Doesn't have to copy the entire list to a set and back to a list

Update 09/12/21
You can also declare an extension method once for lists:

extension Unique<E, Id> on List<E> {
  List<E> unique([Id Function(E element)? id, bool inplace = true]) {
    final ids = Set();
    var list = inplace ? this : List<E>.from(this);
    list.retainWhere((x) => ids.add(id != null ? id(x) : x as Id));
    return list;
  }
}

This extension method does the same as my original answer. Usage:

// Use a lambda to map an object to its unique identifier.
myRichObjectList.unique((x) => x.id);
// Don't use a lambda for primitive/hashable types.
hashableValueList.unique();

Answer 3

Set works okay, but it doesn't preserve the order. Here's another way using LinkedHashSet:

import "dart:collection";

void main() {
  List<String> arr = ["a", "a", "b", "c", "b", "d"];
  List<String> result = LinkedHashSet<String>.from(arr).toList();
  print(result); // => ["a", "b", "c", "d"]
}

https://api.dart.dev/stable/2.4.0/dart-collection/LinkedHashSet/LinkedHashSet.from.html

Answer 4

Try the following:

List<String> duplicates = ["a", "c", "a"];

duplicates = duplicates.toSet().toList();

Check this code on Dartpad.

Answer 5

If you want to keep ordering or are dealing with more complex objects than primitive types, store seen ids to the Set and filter away those ones that are already in the set.

final list = ['a', 'a', 'b'];
final seen = Set<String>();
final unique = list.where((str) => seen.add(str)).toList();

print(unique); // => ['a', 'b']

Answer 6

//This easy way works fine

List<String> myArray = [];
myArray = ['x', 'w', 'x', 'y', 'o', 'x', 'y', 'y', 'r', 'a'];

myArray = myArray.toSet().toList();

print(myArray);

// result => myArray =['x','w','y','o','r', 'a']

Answer 7

Remove duplicates from a list of objects:

class Stock {
  String? documentID; //key
  Make? make;
  Model? model;
  String? year;

  Stock({
    this.documentID,
    this.make,
    this.model,
    this.year,
  });
}

List of stock, from where we want to remove duplicate stocks

List<Stock> stockList = [stock1, stock2, stock3];

Remove duplicates

final ids = stockList.map((e) => e.documentID).toSet();
stockList.retainWhere((x) => ids.remove(x.documentID));

Answer 8

I am adding this to atreeon's answer. For anyone that want use this with Object:

class MyObject{
  int id;

  MyObject(this.id);


  @override
  bool operator ==(Object other) {
    return other != null && other is MyObject && hashCode == other.hashCode;
  }


  @override
  int get hashCode => id;
}

main(){
   List<MyObject> list = [MyObject(1),MyObject(2),MyObject(1)];

   // The new list will be [MyObject(1),MyObject(2)]
   List<MyObject> newList = list.toSet().toList();
}

Answer 9

Using Dart 2.3+, you can use the spread operators to do this:

final ids = [1, 4, 4, 4, 5, 6, 6];
final distinctIds = [...{...ids}];

Whether this is more or less readable than ids.toSet().toList() I'll let the reader decide :)

Answer 10

For distinct list of objects you can use Equatable package.

Example:

// ignore: must_be_immutable
class User extends Equatable {
  int id;
  String name;

  User({this.id, this.name});

  @override
  List<Object> get props => [id];
}

List<User> items = [
  User(
    id: 1,
    name: "Omid",
  ),
  User(
    id: 2,
    name: "Raha",
  ),
  User(
    id: 1,
    name: "Omid",
  ),
  User(
    id: 2,
    name: "Raha",
  ),
];

print(items.toSet().toList());

Output:

[User(1), User(2)]

Answer 11

Here it is, a working solution:

var sampleList = ['1', '2', '3', '3', '4', '4'];
//print('original: $sampleList');
sampleList = Set.of(sampleList).toList();
//print('processed: $sampleList');

Output:

original: [1, 2, 3, 3, 4, 4]
processed: [1, 2, 3, 4]

Answer 12

Using the fast_immutable_collections package:

[1, 2, 3, 2].distinct();

Or

[1, 2, 3, 2].removeDuplicates().toList();

Note: While distinct() returns a new list, removeDuplicates() does it lazily by returning an Iterable. This means it is much more efficient when you are doing some extra processing. For example, suppose you have a list with a million items, and you want to remove duplicates and get the first five:

// This will process five items:
List<String> newList = list.removeDuplicates().take(5).toList();

// This will process a million items:
List<String> newList = list.distinct().sublist(0, 5);

// This will also process a million items:
List<String> newList = [...{...list}].sublist(0, 5);

Both methods also accept a by parameter. For example:

// Returns ["a", "yk", "xyz"]
["a", "yk", "xyz", "b", "xm"].removeDuplicates(by: (item) => item.length);

If you don't want to include a package into your project but needs the lazy code, here it is a simplified removeDuplicates():

Iterable<T> removeDuplicates<T>(Iterable<T> iterable) sync* {
  Set<T> items = {};
  for (T item in iterable) {
    if (!items.contains(item)) yield item;
    items.add(item);
  }
}

Note: I am one of the authors of the fast_immutable_collections package.

Answer 13

void uniqifyList(List<Dynamic> list) {
  for (int i = 0; i < list.length; i++) {
    Dynamic o = list[i];
    int index;
    // Remove duplicates
    do {
      index = list.indexOf(o, i+1);
      if (index != -1) {
        list.removeRange(index, 1);
      }
    } while (index != -1);
  }
}

void main() {
  List<String> list = ['abc', "abc", 'def'];
  print('$list');
  uniqifyList(list);
  print('$list');
}

Gives output:

[abc, abc, def]
[abc, def]

Answer 14

As for me, one of the best practices is sort the array, and then deduplicate it. The idea is stolen from low-level languages. So, first make the sort by your own, and then deduplicate equal values that are going after each other.

// Easy example
void dedup<T>(List<T> list, {removeLast: true}) {
  int shift = removeLast ? 1 : 0;
  T compareItem;
  for (int i = list.length - 1; i >= 0; i--) {
    if (compareItem == (compareItem = list[i])) {
      list.removeAt(i + shift);
    }
  }
}

// Harder example
void dedupBy<T, I>(List<T> list, I Function(T) compare, {removeLast: true}) {
  int shift = removeLast ? 1 : 0;
  I compareItem;
  for (int i = list.length - 1; i >= 0; i--) {
    if (compareItem == (compareItem = compare(list[i]))) {
      list.removeAt(i + shift);
    }
  }
}


void main() {
  List<List<int>> list = [[1], [1], [2, 1], [2, 2]];
  print('$list');
  dedupBy(list, (innerList) => innerList[0]);
  print('$list');

  print('\n removeLast: false');

  List<List<int>> list2 = [[1], [1], [2, 1], [2, 2]];
  print('$list2');
  dedupBy(list2, (innerList) => innerList[0], removeLast: false);
  print('$list2');
}

Output:

[[1], [1], [2, 1], [2, 2]]
[[1], [2, 1]]

removeLast: false
[[1], [1], [2, 1], [2, 2]]
[[1], [2, 2]]

Answer 15

This is another way...

final reducedList = [];

list.reduce((value, element) {
    if (value != element) 
        reducedList.add(value);
    return element;
});

reducedList.add(list.last);

print(reducedList);

Answer 16

Mind this extension

extension IterableExtension<T> on Iterable<T> {
  List<T> distinct<U>({required U Function(T t) by}) {
    final unique = <U, T>{};

    for (final item in this) {
      unique.putIfAbsent(by(item), () => item);
    }

    return unique.values.toList();
  }
}

class Item {
  Item(this.name, this.id);

  final String name;
  final int id;

  @override
  String toString() {
    return 'Item{name: $name, id: $id}';
  }
}

void main() {
  final list = [1, 1, 1, 2, 3, 4, 4, 5, 6, 6, 7, 7];
  print(list.distinct(by: (item) => item));
  // [1, 2, 3, 4, 5, 6, 7]

  final items = [Item('foo', 1), Item('bar', 2), Item('foo', 3), Item('bar', 4), Item('foo', 4), Item('bar', 3)];
  print(items.distinct(by: (item) => item.name));
  // [Item{name: foo, id: 1}, Item{name: bar, id: 2}]

  print(items.distinct(by: (item) => item.id));
  // [Item{name: foo, id: 1}, Item{name: bar, id: 2}, Item{name: foo, id: 3}, Item{name: bar, id: 4}]
}

Answer 17

It works for me.

var list = [
 {"id": 1, "name": "Joshua"},
 {"id": 2, "name": "Joshua"},
 {"id": 3, "name": "Shinta"},
 {"id": 4, "name": "Shinta"},
 {"id": 5, "name": "Zaidan"}
];
list.removeWhere((element) => element.name == element.name.codeUnitAt(1));
list.sort((a, b) => a.name.compareTo(b.name));

Output:

[{"id": 1, "name": "Joshua"}, 
{"id": 3, "name": "Shinta"}, 
{"id": 5, "name": "Zaidan"}]

Answer 18

List<Model> bigList = [];
List<ModelNew> newList = [];  

for (var element in bigList) {
      var list = newList.where((i) => i.type == element.type).toList();
      if(list.isEmpty){
       newList.add(element);
      }
    }

Answer 19

Create method to remove duplicates from Array and return Array of unique elements.

class Utilities {
  static List<String> uniqueArray(List<String> arr) {
    List<String> newArr = [];
    for (var obj in arr) {
      if (newArr.contains(obj)) {
        continue;
      }
      newArr.add(obj);
    }
    return newArr;
  }
}

Answer 20

You can use the following way:

void main(List <String> args){
    List<int> nums = [1, 2, 2, 2, 3, 4, 5, 5];
    List<int> nums2 = nums.toSet().toList();
}

NOTE: This will not work if the items in the list are objects of class and have the same attributes. So, to solve this, you can use the following way:

void main() {
  List<Medicine> objets = [Medicine("Paracetamol"),Medicine("Paracetamol"), Medicine("Benylin")];
  
  List <String> atributs = [];
  objets.forEach((element){
    atributs.add(element.name);
  });
  
  List<String> noDuplicates = atributs.toSet().toList();
  print(noDuplicates);
}

class Medicine{
  final String name;
  Medicine(this.name);
}

Answer 21

Remove duplicate data from any type of map or list.

 var list = [
     {"id": 1, "name": "piyush"},
     {"id": 2, "name": "jay"},
     {"id": 3, "name": "premal"},
     {"id": 4, "name": "piyush"},
     {"id": 5, "name": "nishant"}
    ];
    
    final ids = Set();
    randomProgramsList.retainWhere((x) => ids.add(x["name"]));

Answer 22

This is my solution

    List<T> removeDuplicates<T>(List<T> list, IsEqual isEqual) {
      List<T> output = [];
      for(var i = 0; i < list.length; i++) {
        bool found = false;
        for(var j = 0; j < output.length; j++) {
          if (isEqual(list[i], output[j])) {
            found = true;
          }
        }
        if (found) {
          output.add(list[i]);
        }
      }

      return output;
    }

Use it like this:

  var theList = removeDuplicates(myOriginalList, (item1, item2) => item1.documentID == item2.documentID);

or...

  var theList = removeDuplicates(myOriginalList, (item1, item2) => item1.equals(item2));

or...

Answer 23

I have a library called Reactive-Dart that contains many composable operators for terminating and non-terminating sequences. For your scenario it would look something like this:

final newList = [];
Observable
   .fromList(['abc', 'abc', 'def'])
   .distinct()
   .observe((next) => newList.add(next), () => print(newList));

Yielding:

[abc, def]

I should add that there are other libraries out there with similar features. Check around on GitHub and I'm sure you'll find something suitable.