为什么HashSet< Point>比HashSet< string>慢多少? [英] Why is HashSet<Point> so much slower than HashSet<string>?

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问题描述

我想存储一些像素位置而不允许重复,所以首先想到的是 HashSet< Point> 或类似的类。但是,与 HashSet< string> 之类的东西相比,这似乎非常慢。



例如,以下代码:

  HashSet< Point> points = new HashSet< Point>(); 
使用(位图img =新位图(1000,1000))
{
for(int x = 0; x< img.Width; x ++)
{
表示(int y = 0; y< img.Height; y ++)
{
points.Add(new Point(x,y));
}
}
}

大约需要22.5秒。 / p>

以下代码(出于明显的原因,这不是一个好选择)仅需1.6秒:

  HashSet< string> points = new HashSet< string>(); 
使用(位图img =新位图(1000,1000))
{
for(int x = 0; x< img.Width; x ++)
{
表示(int y = 0; y< img.Height; y ++)
{
points.Add(x +, + y);
}
}
}

所以,我的问题是:




  • 是否有原因?我检查了此答案,但是22.5秒远比该答案中显示的数字还要长。

  • 是否有更好的方法来存储点而不重复?


解决方案

有Point结构引起的两个性能问题。将 Console.WriteLine(GC.CollectionCount(0)); 添加到测试代码中时,您可以看到一些内容。您会看到Point测试需要大约3720个集合,而字符串测试只需要大约18个集合。不是免费的。当您看到一个值类型引发了如此多的集合时,您需要得出结论呃,哦,太多拳击。



问题在于 HashSet< T> 需要 IEqualityComparer< T> 才能完成工作。由于您没有提供,因此需要回退到 EqualityComparer.Default< T>()返回的值。该方法可以很好地处理字符串,它实现了IEquatable。但是不是Point的一种,它是从.NET 1.0衍生出来的一种类型,从未受到泛型的热爱。它所能做的就是使用Object方法。



另一个问题是Point.GetHashCode()在此测试中做得不好,有太多的冲突,所以它非常重锤Object.Equals()。 String具有出色的GetHashCode实现。



您可以通过为HashSet提供良好的比较器来解决这两个问题。像这样:

  class PointComparer:IEqualityComparer< Point> {
public bool Equals(Point x,y点){
return x.X == y.X&& x.Y == y.Y;
}

public int GetHashCode(Point obj){
//适用于实际位图的完美哈希,其宽度/高度从不> = 65536
return(obj .Y<< 16)^ obj.X;
}
}

并使用它:

  HashSet< Point> list = new HashSet< Point>(new PointComparer()); 

现在它的速度快了150倍左右,很容易击败了字符串测试。


I wanted to store some pixels locations without allowing duplicates, so the first thing comes to mind is HashSet<Point> or similar classes. However this seems to be very slow compared to something like HashSet<string>.

For example, this code:

HashSet<Point> points = new HashSet<Point>();
using (Bitmap img = new Bitmap(1000, 1000))
{
    for (int x = 0; x < img.Width; x++)
    {
        for (int y = 0; y < img.Height; y++)
        {
            points.Add(new Point(x, y));
        }
    }
}

takes about 22.5 seconds.

While the following code (which is not a good choice for obvious reasons) takes only 1.6 seconds:

HashSet<string> points = new HashSet<string>();
using (Bitmap img = new Bitmap(1000, 1000))
{
    for (int x = 0; x < img.Width; x++)
    {
        for (int y = 0; y < img.Height; y++)
        {
            points.Add(x + "," + y);
        }
    }
}

So, my questions are:

  • Is there a reason for that? I checked this answer, but 22.5 sec is way more than the numbers shown in that answer.
  • Is there a better way to store points without duplicates?

解决方案

There are two perf problems induced by the Point struct. Something you can see when you add Console.WriteLine(GC.CollectionCount(0)); to the test code. You'll see that the Point test requires ~3720 collections but the string test only needs ~18 collections. Not for free. When you see a value type induce so many collections then you need to conclude "uh-oh, too much boxing".

At issue is that HashSet<T> needs an IEqualityComparer<T> to get its job done. Since you did not provide one, it needs to fall back to one returned by EqualityComparer.Default<T>(). That method can do a good job for string, it implements IEquatable. But not for Point, it is a type that harks from .NET 1.0 and never got the generics love. All it can do is use the Object methods.

The other issue is that Point.GetHashCode() does not do a stellar job in this test, too many collisions, so it hammers Object.Equals() pretty heavily. String has an excellent GetHashCode implementation.

You can solve both problems by providing the HashSet with a good comparer. Like this one:

class PointComparer : IEqualityComparer<Point> {
    public bool Equals(Point x, Point y) {
        return x.X == y.X && x.Y == y.Y;
    }

    public int GetHashCode(Point obj) {
        // Perfect hash for practical bitmaps, their width/height is never >= 65536
        return (obj.Y << 16) ^ obj.X;
    }
}

And use it:

HashSet<Point> list = new HashSet<Point>(new PointComparer());

And it is now about 150 times faster, easily beating the string test.

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