views:

4339

answers:

10

I would like to compare two collections (in C#), but I'm not sure of the best way to implement this efficiently.

I've read the other thread about Enumerable.SequenceEqual, but it's not exactly what I'm looking for.

In my case, two collections would be equal if they both contain the same items (no matter the order).

Example:

collection1 = {1, 2, 3, 4};
collection2 = {2, 4, 1, 3};

collection1 == collection2; // true

What I usually do is to loop through each item of one collection and see if it exists in the other collection, then loop through each item of the other collection and see if it exists in the first collection. (I start by comparing the lengths).

if (collection1.Count != collection2.Count)
    return false; // the collections are not equal

foreach (Item item in collection1)
{
    if (!collection2.Contains(item))
        return false; // the collections are not equal
}

foreach (Item item in collection2)
{
    if (!collection1.Contains(item))
        return false; // the collections are not equal
}

return true; // the collections are equal

However, this is not entirely correct, and it's probably not the most efficient way to do compare two collections for equality.

An example I can think of that would be wrong is:

collection1 = {1, 2, 3, 3, 4}
collection2 = {1, 2, 2, 3, 4}

Which would be equal with my implementation. Should I just count the number of times each item is found and make sure the counts are equal in both collections?


The examples are in some sort of C# (let's call it pseudo-C#), but give your answer in whatever language you wish, it does not matter.

Note: I used integers in the examples for simplicity, but I want to be able to use reference-type objects too (they do not behave correctly as keys because only the reference of the object is compared, not the content).

+6  A: 

Create a Dictionary "dict" and then for each member in the first collection, do dict[member]++;

Then, loop over the second collection in the same way, but for each member do dict[member]--.

At the end, loop over all of the members in the dictionary:

    private bool SetEqual (List<int> left, List<int> right) {

        if (left.Count != right.Count)
            return false;

        Dictionary<int, int> dict = new Dictionary<int, int>();

        foreach (int member in left) {
            if (dict.ContainsKey(member) == false)
                dict[member] = 1;
            else
                dict[member]++;
        }

        foreach (int member in right) {
            if (dict.ContainsKey(member) == false)
                return false;
            else
                dict[member]--;
        }

        foreach (KeyValuePair<int, int> kvp in dict) {
            if (kvp.Value != 0)
                return false;
        }

        return true;

    }

Edit: As far as I can tell this is on the same order as the most efficient algorithm. This algorithm is O(N), assuming that the Dictionary uses O(1) lookups.

Daniel Jennings
This is almost what I want. However, I'd like to be able to do this even if I am not using integers. I'd like to use reference objects, but they do not behave properly as keys in dictionaries.
GoodEnough
Mono, your question is moot if your Items are not comparable. If they cannot be used as keys in Dictionary, there is no solution available.
skolima
I think Mono meant the keys are not sortable. But Daniel's solution is clearly intended to be implemented with a hashtable, not a tree, and will work as long as there's an equivalence test and a hash function.
erickson
Upvoted of course for the help, but not accepted since it's missing an important point (which I cover in my answer).
GoodEnough
+1  A: 

erickson is almost right: since you want to match on counts of duplicates, you want a Bag. In Java, this looks something like:

(new HashBag(collection1)).equals(new HashBag(collection2))

I'm sure C# has a built-in Set implementation. I would use that first; if performance is a problem, you could always use a different Set implementation, but use the same Set interface.

James A. Rosen
+5  A: 

A simple and fairly efficient solution is to sort both collections and then compare them for equality:

bool equal = collection1.OrderBy(i => i).SequenceEqual(
                 collection2.OrderBy(i => i);

This algorithm is O(N*logN), while your solution above is O(N^2).

If the collections have certain properties, you may be able to implement a faster solution. For example, if both of your collections are hash sets, they cannot contain duplicates. Also, checking whether a hash set contains some element is very fast.In that case, an algorithm similar to yours would likely be fastest.

I gave a O(N) answer earlier.
Daniel Jennings
I have to give you high marks on the cleanest implementation. Thanks!
Michael La Voie
You just have to add a using System.Linq; first to make it work
Junior Mayhé
if this code is within a loop and collection1 gets updated and collection2 remains untouched, notice even when both collections have the same object, debugger would show false for this "equal" variable.
Junior Mayhé
+8  A: 

This is my (heavily influenced by D.Jennings) generic implementation of the comparison method (in C#):

/// <summary>
/// Represents a service used to compare two collections for equality.
/// </summary>
/// <typeparam name="T">The type of the items in the collections.</typeparam>
public class CollectionComparer<T>
{
    /// <summary>
    /// Compares the content of two collections for equality.
    /// </summary>
    /// <param name="foo">The first collection.</param>
    /// <param name="bar">The second collection.</param>
    /// <returns>True if both collections have the same content, false otherwise.</returns>
    public bool Execute(ICollection<T> foo, ICollection<T> bar)
    {
        // Declare a dictionary to count the occurence of the items in the collection
        Dictionary<T, int> itemCounts = new Dictionary<T,int>();

        // Increase the count for each occurence of the item in the first collection
        foreach (T item in foo)
        {
            if (itemCounts.ContainsKey(item))
            {
                itemCounts[item]++;
            }
            else
            {
                itemCounts[item] = 1;
            }
        }

        // Wrap the keys in a searchable list
        List<T> keys = new List<T>(itemCounts.Keys);

        // Decrease the count for each occurence of the item in the second collection
        foreach (T item in bar)
        {
            // Try to find a key for the item
            // The keys of a dictionary are compared by reference, so we have to
            // find the original key that is equivalent to the "item"
            // You may want to override ".Equals" to define what it means for
            // two "T" objects to be equal
            T key = keys.Find(
                delegate(T listKey)
                {
                    return listKey.Equals(item);
                });

            // Check if a key was found
            if(key != null)
            {
                itemCounts[key]--;
            }
            else
            {
                // There was no occurence of this item in the first collection, thus the collections are not equal
                return false;
            }
        }

        // The count of each item should be 0 if the contents of the collections are equal
        foreach (int value in itemCounts.Values)
        {
            if (value != 0)
            {
                return false;
            }
        }

        // The collections are equal
        return true;
    }
}
GoodEnough
Sounds good! Glad my answer helped.
Daniel Jennings
Nice job, but Note: 1. In contrast to Daniel Jennings solution, This is not O(N) but rather O(N^2), because of the find function inside the foreach loop on the bar collection; 2. You can generalize the method to accept IEnumerable<T> instead of ICollection<T> with no further modification to the code
ohadsc
+1  A: 

You could use a Hashset. Look at the SetEquals method.

Joel Gauvreau
of course, using a HashSet assumes no duplicates but if so HashSet is the best way to go
Mark Cidade
A: 

There are many solutions to this problem. If you don't care about duplicates, you don't have to sort both. First make sure that they have the same number of items. After that sort one of the collections. Then binsearch each item from the second collection in the sorted collection. If you don't find a given item stop and return false. The complexity of this: - sorting the first collection: N*Log(N) - searching each item from second into the first: N*LOG(N) so you end up with 2*N*LOG(N) assuming that they match and you look up everything. This is similar to the complexity of sorting both. Also this gives you the benefit to stop earlier if there's a difference. However, keep in mind that if both are sorted before you step into this comparison and you try sorting by use something like a qsort, the sorting will be more expensive. There are optimizations for this. Another alternative, which is great for small collections where you know the range of the elements is to use a bitmask index. This will give you a O(n) performance. Another alternative is to use a hash and look it up. For small collections it is usually a lot better to do the sorting or the bitmask index. Hashtable have the disadvantage of worse locality so keep that in mind. Again, that's only if you don't care about duplicates. If you want to account for duplicates go with sorting both.

+2  A: 

EDIT: I realized as soon as I posed that this really only works for sets -- it will not properly deal with collections that have duplicate items. For example { 1, 1, 2 } and { 2, 2, 1 } will be considered equal from this algorithm's perspective. If your collections are sets (or their equality can be measured that way), however, I hope you find the below useful.

The solution I use is:

return c1.Count == c2.Count && c1.Intersect(c2).Count() == c1.Count;

Linq does the dictionary thing under the covers, so this is also O(N). (Note, it's O(1) if the collections aren't the same size).

I did a sanity check using the "SetEqual" method suggested by Daniel, the OrderBy/SequenceEquals method suggested by Igor, and my suggestion. The results are below, showing O(N*LogN) for Igor and O(N) for mine and Daniel's.

I think the simplicity of the Linq intersect code makes it the preferable solution.

__Test Latency(ms)__
N, SetEquals, OrderBy, Intersect    
1024, 0, 0, 0    
2048, 0, 0, 0    
4096, 31.2468, 0, 0    
8192, 62.4936, 0, 0    
16384, 156.234, 15.6234, 0    
32768, 312.468, 15.6234, 46.8702    
65536, 640.5594, 46.8702, 31.2468    
131072, 1312.3656, 93.7404, 203.1042    
262144, 3765.2394, 187.4808, 187.4808    
524288, 5718.1644, 374.9616, 406.2084    
1048576, 11420.7054, 734.2998, 718.6764    
2097152, 35090.1564, 1515.4698, 1484.223
The only issue with this code is that it only works when comparing value types or comparing the pointers to reference types. I could have two different instances of the same object in the collections, so I need to be able to specify how to compare each. Can you pass a comparison delegate to the intersect method?
GoodEnough
Sure, you can pass a comparer delegate. But, note the above limitation regarding sets that I added, which puts a significant limit on its applicability.
A: 

In the case of no repeats and no order, the following EqualityComparer can be used to allow collections as dictionary keys:

[Serializable]
public class SetComparer<T> : IEqualityComparer<IEnumerable<T>> 
where T:IComparable<T>
{
    public bool Equals(IEnumerable<T> first, IEnumerable<T> second)
    {
        if (first == second)
            return true;
        if ((first == null) || (second == null))
            return false;
        return first.ToHashSet().SetEquals(second);
    }

    public int GetHashCode(IEnumerable<T> enumerable)
    {
        int hash = 17;

        foreach (T val in enumerable.OrderBy(x => x))
            hash = hash * 23 + val.GetHashCode();

        return hash;
    }
}

Here is the ToHashSet() implementation I used. The hash code algorithm comes from Effective Java (by way of Jon Skeet).

ohadsc
A: 

A duplicate post of sorts, but check out my solution for comparing collections. It's pretty simple:

http://robertbouillon.com/2010/04/29/comparing-collections-in-net/

This will perform an equality comparison regardless of order:

var list1 = new[] { "Bill", "Bob", "Sally" };
var list2 = new[] { "Bob", "Bill", "Sally" };
bool isequal = list1.Compare(list2).IsSame;

This will check to see if items were added / removed:

var list1 = new[] { "Billy", "Bob" };
var list2 = new[] { "Bob", "Sally" };
var diff = list1.Compare(list2);
var onlyinlist1 = diff.Removed; //Billy
var onlyinlist2 = diff.Added;   //Sally
var inbothlists = diff.Equal;   //Bob

This will see what items in the dictionary changed:

var original = new Dictionary<int, string>() { { 1, "a" }, { 2, "b" } };
var changed = new Dictionary<int, string>() { { 1, "aaa" }, { 2, "b" } };
var diff = original.Compare(changed, (x, y) => x.Value == y.Value, (x, y) => x.Value == y.Value);
foreach (var item in diff.Different)
  Console.Write("{0} changed to {1}", item.Key.Value, item.Value.Value);
//Will output: a changed to aaa

(original post here: ht tp:// stackoverflow.com /questions/43500/is-there-a-built-in-method-to-compare-collections-in-c/2740552#2740552)

+5  A: 

It turns out Microsoft already has this covered in its testing framework: CollectionAssert.AreEquivalent

Remarks

Two collections are equivalent if they have the same elements in the same quantity, but in any order. Elements are equal if their values are equal, not if they refer to the same object.

Using reflector, I modified the code behind AreEquivalent() to create a corresponding equality comparer. It is more complete than existing answers, since it takes nulls into account, implements IEqualityComparer and has some efficiency and edge case checks. plus, it's Microsoft :)

public class CollectionComparer<T> : IEqualityComparer<IEnumerable<T>>
{
    public bool Equals(IEnumerable<T> first, IEnumerable<T> second)
    {
        if ((first == null) != (second == null))
            return false;

        if (!object.ReferenceEquals(first, second) && (first != null))
        {
            if (first.Count() != second.Count())
                return false;

            if ((first.Count() != 0) && HaveMismatchedElement(first, second))
                return false;
        }

        return true;
    }

    private static bool HaveMismatchedElement(IEnumerable<T> first, 
                                             IEnumerable<T> second)
    {
        int firstCount;
        int secondCount;

        var firstElementCounts = GetElementCounts(first, out firstCount);
        var secondElementCounts = GetElementCounts(second, out secondCount);

        if (firstCount != secondCount)
            return true;

        foreach (var kvp in firstElementCounts)
        {
            firstCount = kvp.Value;
            secondElementCounts.TryGetValue(kvp.Key, out secondCount);

            if (firstCount != secondCount)
                return true;
        }

        return false;
    }

    private static Dictionary<T, int> GetElementCounts(IEnumerable<T> enumerable, 
                                                       out int nullCount)
    {
        var dictionary = new Dictionary<T, int>();
        nullCount = 0;

        foreach (T element in enumerable)
        {
            if (element == null)
            {
                nullCount++;
            }
            else
            {
                int num;
                dictionary.TryGetValue(element, out num);
                num++;
                dictionary[element] = num;
            }
        }

        return dictionary;
    }

    public int GetHashCode(IEnumerable<T> enumerable)
    {
        int hash = 17;

        foreach (T val in enumerable.OrderBy(x => x))
            hash = hash * 23 + val.GetHashCode();

        return hash;
    }
}
ohadsc
Thanks for the answer, I didn't know Microsoft had it covered. For the matter, I actually use a variant of the answer on top which allows me to define how elements are compared for equality and checks for nulls.
GoodEnough
No problem. It's also easy to add an IEqualityComparer<T> parameter to the above implementation in order to support a custom equality definition for T - In GetElementCounts(), simply use the Dictionary ctor that accepts IEqualityComparer<T>.
ohadsc