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

Answer 1

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);
  ...

Answer 2

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

Answer 3

This is a slightly difficult problem, which I think reduces to: "Test if two lists are permutations of each other."

I believe the solutions provided by others only indicate whether the 2 lists contain the same unique elements. This is a necessary but insufficient test, for example {1, 1, 2, 3} is not a permutation of {3, 3, 1, 2} although their counts are equal and they contain the same distinct elements.

I believe this should work though, although it's not the most efficient:

static bool ArePermutations<T>(IList<T> list1, IList<T> list2)
{
   if(list1.Count != list2.Count)
         return false;

   var l1 = list1.ToLookup(t => t);
   var l2 = list2.ToLookup(t => t);

   return l1.Count == l2.Count 
       && l1.All(group => l2.Contains(group.Key) && l2[group.Key].Count() == group.Count()); 
}