CODEWITHSUNDEEP
flutterdart
Sheshank S.
Sheshank S.

Reputation: 3298

Get difference of lists flutter dart

I have two lists [1,2,3,4,5,6,7] and [3,5,6,7,9,10]. I want to get the difference of the first list and the second list.

The expected output would be [1,2,4] since those are the only elements in only list 1 and not list 2.

I'm using Flutter and Dart. I looked this up on the internet, I know it seems like a simple question but I couldn't find anything.

Should be irrelevant but I'm using publishing for iOS

I'd prefer an answer without just a foreach loop, im looking to see if there is a library for it.

Upvotes: 24

Views: 26168

Answers (9)

Logachu
Logachu

Reputation: 609

Some people may be searching for the symmetric difference rather than strict difference provided in the standard library. Here is an extension to Set which adds a symmetricDifference function:

   extension on Set {
      /// Creates a set that is the _symmetric difference_ of this set and
      /// another set. That is the set of all elements that are in this
      /// set _or_ the other set, but not in _both_ sets.
      Set<E> symmetricDifference<E>(Set<E> other) =>
          difference(other).union(other.difference(this)) as Set<E>;
    }
    
    void main() {
      final Set l = {1, 2, 3};
      final Set r = {2, 3, 4};
      assert(l.difference(r) == {1});
      assert(r.difference(l) == {4});
      assert(l.symmetricDifference(r) == {1, 4});
    }

Upvotes: 0

shtse8
shtse8

Reputation: 1365

Because list.contains() is an O(n) operation, other answers have terrible performance and don't consider duplicate elements in the lists.

Myers diff algorithm can help if you want to diff two lists.

Here is the dart implementation by @knaeckeKami.

https://github.com/knaeckeKami/diffutil.dart

Upvotes: 0

MOHAMMED MISHAL N.M
MOHAMMED MISHAL N.M

Reputation: 31

void main() {
    var listA = [1,2,3,4,5,6,7];
    var listB = [3,5,6,7,9,10];
    listB.map((e)=>listA.remove(e)).toList();
    print(listA);
}

Upvotes: 0

Mohammad Reza Mrg
Mohammad Reza Mrg

Reputation: 2033

Use the !.contains for nullable variable.

var a = [1,2,3,4,5,6,7];
var b = [3,5,6,7,9,10];
a.removeWhere((element) => b!.contains(element));

Upvotes: 0

Abhishek Ghimire
Abhishek Ghimire

Reputation: 2424

Simple & Recommended Way to do this is

var a = [1,2,3,4,5];
var b = [1,2];
a.removeWhere((element) => b.contains(element));
print(a); //[3, 4, 5]

Upvotes: 6

malik kurosaki
malik kurosaki

Reputation: 2042

if you need single line , just do like this one

print([1,2,3,4,5,6,7].where((e) => ![3,5,6,7,9,10].contains(e)).toList());

result is

[1, 2, 4]

Upvotes: 4

Pjotr Gainullin
Pjotr Gainullin

Reputation: 1057

Since you will be looking for unique elements you could use the difference method of the Set class (https://api.flutter.dev/flutter/dart-core/Set/difference.html) to do something like this:

List<int> first = [1,2,3,4,5,6,7];
List<int> second = [3,5,6,7,9,10];
List<int> difference = first.toSet().difference(second.toSet()).toList();
print(difference.toString());
// prints [1, 2, 4]

Upvotes: 33

csells
csells

Reputation: 2593

This seems to do the trick:

void main() {
  final l1 = [1, 2, 3];
  final l2 = [3, 4, 5];
  print(listDiff(l1, l2)); // [1, 2, 4, 5]
}

List<T> listDiff<T>(List<T> l1, List<T> l2) => (l1.toSet()..addAll(l2))
    .where((i) => !l1.contains(i) || !l2.contains(i))
    .toList();

Upvotes: 2

Karim Elghamry
Karim Elghamry

Reputation: 1421

you can do something like this:

List<double> first = [1,2,3,4,5,6,7];
List<double> second = [3,5,6,7,9,10];
List<double> output = [];

first.forEach((element) {
    if(!second.contains(element)){
    output.add(element);
}
});

//at this point, output list should have the answer

alternative answer:

List<double> first = [1,2,3,4,5,6,7];
List<double> second = [3,5,6,7,9,10];
List<double> output = first.where((element) => !second.contains(element));

note that for both cases, you need to loop over the larger list.

Upvotes: 24

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