Bruno Teixeira Bruno Teixeira - 2 months ago 13
C# Question

Compare two List<T> objects for equality, ignoring order

Yet another list-comparing question.

List<MyType> list1;
List<MyType> list2;

I need to check that they both have the same elements, regardless of their position within the list. Each MyType object may appear multiple times on a list. Is there a built-in function that checks this? What if I guarantee that each element appears only once in a list?

EDIT: Guys thanks for the answers but I forgot to add something, the number of occurrences of each element should be the same on both lists.


If you want them to be really equal (i.e. the same items and the same number of each item), I think that the simplest solution is to sort before comparing:

Enumerable.SequenceEqual(list1.OrderBy(t => t), list2.OrderBy(t => t))


Here is a solution that performs a bit better (about ten times faster), and only requires IEquatable, not IComparable:

public static bool ScrambledEquals<T>(IEnumerable<T> list1, IEnumerable<T> list2) {
  var cnt = new Dictionary<T, int>();
  foreach (T s in list1) {
    if (cnt.ContainsKey(s)) {
    } else {
      cnt.Add(s, 1);
  foreach (T s in list2) {
    if (cnt.ContainsKey(s)) {
    } else {
      return false;
  return cnt.Values.All(c => c == 0);

Edit 2:

To handle any data type as key (for example nullable types as Frank Tzanabetis pointed out), you can make a version that takes a comparer for the dictionary:

public static bool ScrambledEquals<T>(IEnumerable<T> list1, IEnumerable<T> list2, IEqualityComparer<T> comparer) {
  var cnt = new Dictionary<T, int>(comparer);