Question

我想在C＃中为双精度实现一个vector类，需要覆盖Equals和GetHashCode，这样我就可以将我的vector类用作Dictionary中的键或使用HashSets。由于我需要对相等性有一定的容忍度，我知道没有办法实现传递的Equals方法和相应的GetHashCode方法。

我在类似的帖子上偶然发现了答案：https://stackoverflow.com/a/580972/5333340

我想知道，有没有办法在C＃中更改HashSet / Dictionaries的查找行为，这样它不仅可以检查一个桶，还可以检查多个桶？

或者是否有一些类对C＃有这种行为？

Answer 1

由于HashSet不提供自定义存储区搜索行为的方法，因此我编写了一个执行多桶搜索的自定义类。包括现实生活用途的一个例子：三维矢量类。

// Implementing this interface introduces the concept of neighbouring buckets.
public interface IHasNeighbourConcept
{
    int[] GetSeveralHashCodes();
    // The returned int[] must at least contain the return value of GetHashCode.
}

// Custom HashSet-like class that can search in several buckets.
public class NeighbourSearchHashSet<T> where T : IHasNeighbourConcept
{
    // Internal data storage.
    private Dictionary<int, List<T>> buckets;

    // Constructor.
    public NeighbourSearchHashSet()
    {
        buckets = new Dictionary<int, List<T>>();
    }

    // Classic implementation utilizing GetHashCode.
    public bool Add(T elem)
    {
        int hash = elem.GetHashCode();

        if(!buckets.ContainsKey(hash))
        {
            buckets[hash] = new List<T>();
            buckets[hash].Add(elem);
            return true;
        }

        foreach(T t in buckets[hash])
        {
            if(elem.Equals(t))
                return false;
        }

        buckets[hash].Add(elem);
        return true;
    }

    /// Nonclassic implementation utilizing GetSeveralHashCodes.
    public bool Contains(T elem)
    {
        int[] hashes = elem.GetSeveralHashCodes();

        foreach(int h in hashes)
            foreach(T t in buckets[h])
                if(elem.Equals(t))
                    return true;
        return false;
    }


}


// A 3-dimensional vector class. Since its Equals method is not transitive,
// there can be vectors that are considered equal but have different HashCodes.
// So the Contains method of HashSet<Vector> does not work as expected.
public class Vector : IHasNeighbourConcept
{
    private double[] coords;
    private static double TOL = 1E-10;
    // Tolerance for considering two doubles as equal

    public Vector(double x, double y, double z)
    {
        if(double.IsNaN(x) || double.IsInfinity(x) ||
           double.IsNaN(y) || double.IsInfinity(y) ||
           double.IsNaN(z) || double.IsInfinity(z))
            throw new NotFiniteNumberException("All input must be finite!");

        coords = new double[] { x, y, z };
    }

    // Two vectors are equal iff the distance of each
    // corresponding component pair is significantly small.
    public override bool Equals(object obj)
    {
        if(!(obj is Vector))
            throw new ArgumentException("Input argument is not a Vector!");

        Vector other = obj as Vector;

        bool retval = true;
        for(int i = 0; i < 2; i++)
            retval = retval && (Math.Abs(coords[i] - other.coords[i]) < TOL);

        return retval;

    }

    // The set of all Vectors with the same HashCode
    // is a cube with side length TOL.
    // Two Vectors considered equal may have different
    // HashCodes, but the x, y, z intermediate values
    // differ by at most 1.
    public override int GetHashCode()
    {
        int x =(int) Math.Truncate(coords[0] / TOL);
        int y =(int) Math.Truncate(coords[1] / TOL);
        int z =(int) Math.Truncate(coords[2] / TOL);
        return x + 3*y + 5*z; // The purpose of the factors is to make
                              // permuting the coordinates result
                              // in different HashCodes.
    }

    // Gets the HashCode of the given Vector as well as the 26
    // HashCodes of the surrounding cubes.
    public int[] GetSeveralHashCodes()
    {
        int[] hashes = new int[27];
        int x =(int) Math.Truncate(coords[0] / TOL);
        int y =(int) Math.Truncate(coords[1] / TOL);
        int z =(int) Math.Truncate(coords[2] / TOL);

        for(int i = -1; i <= 1; i++)
            for(int j = -1; j <= 1; j++)
                for(int k = -1; k <= 1; k++)
                    hashes[(i+1)+3*(j+1)+9*(k+1)] = (x+i) + 3*(y+j) + 5*(z+k);
        return hashes;
    }
}

编辑：

上述实现扩展了HashSet的概念，即使没有传递Equals方法，集合的Contains方法也能正常工作。这是有效的，因为对于Contains，我们不需要知道我们所寻找的元素所处的确切等价类。

但是，对于词典来说，它是不同的。我们确实需要获得正确的等价类（即hashCode），否则我们会获得不同的图像。因此，不同的HashCodes必然导致元素不相等。

C＃覆盖Dictionary / HashSet中

1 个答案: