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

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))

Edit:

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)) {
      cnt[s]++;
    } else {
      cnt.Add(s, 1);
    }
  }
  foreach (T s in list2) {
    if (cnt.ContainsKey(s)) {
      cnt[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);
  ...

If you don't care about the number of occurrences, I would approach it like this. Using hash sets will give you better performance than simple iteration.

var set1 = new HashSet<MyType>(list1);
var set2 = new HashSet<MyType>(list2);
return set1.SetEquals(set2);

This will require that you have overridden .GetHashCode() and implemented IEquatable<MyType> on MyType.

As written, this question is ambigous. The statement:

... they both have the same elements, regardless of their position within the list. Each MyType object may appear multiple times on a list.

does not indicate whether you want to ensure that the two lists have the same set of objects or the same distinct set.

If you want to ensure to collections have exactly the same set of members regardless of order, you can use:

// lists should have same count of items, and set difference must be empty
var areEquivalent = (list1.Count == list2.Count) && !list1.Except(list2).Any();

If you want to ensure two collections have the same distinct set of members (where duplicates in either are ignored), you can use:

// check that [(A-B) Union (B-A)] is empty
var areEquivalent = !list1.Except(list2).Union( list2.Except(list1) ).Any();

Using the set operations (Intersect, Union, Except) is more efficient than using methods like Contains. In my opinion, it also better expresses the expectations of your query.

EDIT: Now that you've clarified your question, I can say that you want to use the first form - since duplicates matter. Here's a simple example to demonstrate that you get the result you want:

var a = new[] {1, 2, 3, 4, 4, 3, 1, 1, 2};
var b = new[] { 4, 3, 2, 3, 1, 1, 1, 4, 2 };

// result below should be true, since the two sets are equivalent...
var areEquivalent = (a.Count() == b.Count()) && !a.Except(b).Any(); 

In addition to Guffa's answer, you could use this variant to have a more shorthanded notation.

public static bool ScrambledEquals<T>(this IEnumerable<T> list1, IEnumerable<T> list2)
{
  var deletedItems = list1.Except(list2).Any();
  var newItems = list2.Except(list1).Any();
  return !newItems && !deletedItems;          
}

Thinking this should do what you want:

list1.All(item => list2.Contains(item)) &&
list2.All(item => list1.Contains(item));

if you want it to be distinct, you could change it to:

list1.All(item => list2.Contains(item)) &&
list1.Distinct().Count() == list1.Count &&
list1.Count == list2.Count