remove list from another list in dart

Convert to sets and take the difference, then convert back to a list (with the caveat that neither duplicates nor the ordering of elements will be preserved):

void main() {  var lst1 = ["t1" , "t2" , "t3" , "t4"];  var lst2 = ["t2" , "t4" , "t5"];  var set1 = Set.from(lst1);  var set2 = Set.from(lst2);  print(List.from(set1.difference(set2)));}

Output

[t1, t3]

Try it online

The method using sets suggested by Ray Toal is probably the fastest, but if your first list contains duplicates that you want to keep, sets will completely destroy them.

Instead, you could use a simple list filtering method.

void main() {  var lst1 = ["t1" , "t2" , "t2", "t3" , "t3", "t4"];  // introduce duplicates  var lst2 = ["t2" , "t4" , "t5"];  var set1 = Set.from(lst1);  var set2 = Set.from(lst2);  print(List.from(set1.difference(set2)));  // Output : [t1, t3]  var filtered_lst = List.from(lst1.where(    (value) => !lst2.contains(value)));  print(filtered_lst);  // Output: [t1, t3, t3]}

If there is duplicates in both lists and you actually want to subtract list item per item, you could use the remove method (warning: this will actually remove items from your first list, so you might need to create a copy first).

void main() {  var lst1 = ["t1" , "t2" , "t2", "t3" , "t3", "t4"];  // introduce duplicates  var lst2 = ["t2" , "t4" , "t5"];  for (var elem in lst2) {    lst1.remove(elem);  }  print(lst1);  // Output : [t1, t2, t3, t3]  // only one occurrence of "t2" was removed.}

If you want to do this in a non-destructive way and preserve the order, you can create an Iterable extension for this.

void main() {  final lst1 = ["t1" , "t2" , "t3" , "t4"];  final lst2 = ["t2" , "t4" , "t5"];  final diff = lst1.whereNotIn(lst2).toList();  print(diff);}extension WhereNotInExt<T> on Iterable<T> {  Iterable<T> whereNotIn(Iterable<T> reject) {    final rejectSet = reject.toSet();    return where((el) => !rejectSet.contains(el));  }}

Try it out here: https://dartpad.dev/88c5c949b6ac8c2d0812e15ce10e40ce?null_safety=true

Some benefits to this approach:

• Doesn't mutate the original list
• Preserves the original order
• Works on Iterable too
• It uses a set to keep track of the elements to reject, so it doesn't slow down a lot if both lists are large