Question

我的印象是，在词典中查找项目而不是列表更快，以下代码似乎暗示：

词典：66滴答

列表：32个刻度

我假设我已经搞砸了？

static void Main(string[] args)
    {
        // Speed test.
        Dictionary<string, int> d = new Dictionary<string, int>()
        {
            {"P1I1-1MS    P2I1-1MS    3I-1MS    4I-1MS", 2},
            {"P1I2-1MS    P2I1-1MS    3I-1MS    4I-1MS", 1},
            {"P1I3-1MS    P2I1-1MS    3I-1MS    4I-1MS", 0},
            {"P1I4-1MS    P2I1-1MS    3I-1MS    4I-1MS", -1},
            {"P1I5-1MS    P2I1-1MS    3I-1MS    4I-1MS", 0},
            {"P1I1-1MS    P2I2-1MS    3I-1MS    4I-1MS", 0},
            {"P1I2-1MS    P2I2-1MS    3I-1MS    4I-1MS", 0},
            {"P1I3-1MS    P2I2-1MS    3I-1MS    4I-1MS", 0},
            {"P1I4-1MS    P2I2-1MS    3I-1MS    4I-1MS", 0},
            {"P1I5-1MS    P2I2-1MS    3I-1MS    4I-1MS", 0},
            {"P1I1-1MS    P2I3-1MS    3I-1MS    4I-1MS", 0},
            {"P1I2-1MS    P2I3-1MS    3I-1MS    4I-1MS", 0},
            {"P1I3-1MS    P2I3-1MS    3I-1MS    4I-1MS", 0},
            {"P1I4-1MS    P2I3-1MS    3I-1MS    4I-1MS", 0},
            {"P1I5-1MS    P2I3-1MS    3I-1MS    4I-1MS", 0},
            {"P1I1-1MS    P2I4-1MS    3I-1MS    4I-1MS", 2} 
        };

        List<string> l = new List<string>();
            l.Add("P1I1-1MS    P2I1-1MS    3I-1MS    4I-1MS");
            l.Add("P1I2-1MS    P2I1-1MS    3I-1MS    4I-1MS");
            l.Add("P1I3-1MS    P2I1-1MS    3I-1MS    4I-1MS");
            l.Add("P1I4-1MS    P2I1-1MS    3I-1MS    4I-1MS");
            l.Add("P1I5-1MS    P2I1-1MS    3I-1MS    4I-1MS");
            l.Add("P1I1-1MS    P2I2-1MS    3I-1MS    4I-1MS");
            l.Add("P1I2-1MS    P2I2-1MS    3I-1MS    4I-1MS");
            l.Add("P1I3-1MS    P2I2-1MS    3I-1MS    4I-1MS");
            l.Add("P1I4-1MS    P2I2-1MS    3I-1MS    4I-1MS");
            l.Add("P1I5-1MS    P2I2-1MS    3I-1MS    4I-1MS");
            l.Add("P1I1-1MS    P2I3-1MS    3I-1MS    4I-1MS");
            l.Add("P1I2-1MS    P2I3-1MS    3I-1MS    4I-1MS");
            l.Add("P1I3-1MS    P2I3-1MS    3I-1MS    4I-1MS");
            l.Add("P1I4-1MS    P2I3-1MS    3I-1MS    4I-1MS");
            l.Add("P1I5-1MS    P2I3-1MS    3I-1MS    4I-1MS");
            l.Add("P1I1-1MS    P2I4-1MS    3I-1MS    4I-1MS");


        Stopwatch sw = new Stopwatch();

        string temp = "P1I1-1MS    P2I4-1MS    3I-1MS    4I-1MS";

        bool inDictionary = false;

        sw.Start();
        if (d.ContainsKey(temp))
        {
            sw.Stop();
            inDictionary = true;
        }
        else sw.Reset();

        Console.WriteLine(sw.ElapsedTicks.ToString());
        Console.WriteLine(inDictionary.ToString());


        bool inList = false;

        sw.Reset();
        sw.Start();
        if (l.Contains(temp))
        {
            sw.Stop();
            inList = true;
        }
        else sw.Reset();

        Console.WriteLine(sw.ElapsedTicks.ToString());
        Console.WriteLine(inList.ToString());

        Console.ReadLine();
    }

修改

对Matthew Watson代码的修改。

Answer 1

这是一个合适的测试：

Dictionary<string, int> d = new Dictionary<string, int>()
{
    {"P1I1-1MS    P2I1-1MS    3I-1MS    4I-1MS", 2},
    {"P1I2-1MS    P2I1-1MS    3I-1MS    4I-1MS", 1},
    {"P1I3-1MS    P2I1-1MS    3I-1MS    4I-1MS", 0},
    {"P1I4-1MS    P2I1-1MS    3I-1MS    4I-1MS", -1},
    {"P1I5-1MS    P2I1-1MS    3I-1MS    4I-1MS", 0},
    {"P1I1-1MS    P2I2-1MS    3I-1MS    4I-1MS", 0},
    {"P1I2-1MS    P2I2-1MS    3I-1MS    4I-1MS", 0},
    {"P1I3-1MS    P2I2-1MS    3I-1MS    4I-1MS", 0},
    {"P1I4-1MS    P2I2-1MS    3I-1MS    4I-1MS", 0},
    {"P1I5-1MS    P2I2-1MS    3I-1MS    4I-1MS", 0},
    {"P1I1-1MS    P2I3-1MS    3I-1MS    4I-1MS", 0},
    {"P1I2-1MS    P2I3-1MS    3I-1MS    4I-1MS", 0},
    {"P1I3-1MS    P2I3-1MS    3I-1MS    4I-1MS", 0},
    {"P1I4-1MS    P2I3-1MS    3I-1MS    4I-1MS", 0},
    {"P1I5-1MS    P2I3-1MS    3I-1MS    4I-1MS", 0},
    {"P1I1-1MS    P2I4-1MS    3I-1MS    4I-1MS", 2} 
};

List<string> l = new List<string>
{
    "P1I1-1MS    P2I1-1MS    3I-1MS    4I-1MS", 
    "P1I2-1MS    P2I1-1MS    3I-1MS    4I-1MS", 
    "P1I3-1MS    P2I1-1MS    3I-1MS    4I-1MS", 
    "P1I4-1MS    P2I1-1MS    3I-1MS    4I-1MS", 
    "P1I5-1MS    P2I1-1MS    3I-1MS    4I-1MS",
    "P1I1-1MS    P2I2-1MS    3I-1MS    4I-1MS", 
    "P1I2-1MS    P2I2-1MS    3I-1MS    4I-1MS",
    "P1I3-1MS    P2I2-1MS    3I-1MS    4I-1MS",
    "P1I4-1MS    P2I2-1MS    3I-1MS    4I-1MS",
    "P1I5-1MS    P2I2-1MS    3I-1MS    4I-1MS", 
    "P1I1-1MS    P2I3-1MS    3I-1MS    4I-1MS", 
    "P1I2-1MS    P2I3-1MS    3I-1MS    4I-1MS",
    "P1I3-1MS    P2I3-1MS    3I-1MS    4I-1MS", 
    "P1I4-1MS    P2I3-1MS    3I-1MS    4I-1MS",
    "P1I5-1MS    P2I3-1MS    3I-1MS    4I-1MS", 
    "P1I1-1MS    P2I4-1MS    3I-1MS    4I-1MS"
};

int trials = 4;
int iters  = 1000000;

Stopwatch sw = new Stopwatch();

string target = "P1I1-1MS    P2I4-1MS    3I-1MS    4I-1MS";

for (int trial = 0; trial < trials; ++trial)
{
    sw.Restart();

    for (int i = 0; i < iters; ++i)
        d.ContainsKey(target);

    sw.Stop();
    Console.WriteLine("Dictionary took " + sw.Elapsed);
    sw.Restart();

    for (int i = 0; i < iters; ++i)
        l.Contains(target);

    sw.Stop();
    Console.WriteLine("List took " + sw.Elapsed);
}

运行此OUTSIDE任何调试器的RELEASE版本。

我的结果：

Dictionary took 00:00:00.0587588
List took 00:00:00.2018361
Dictionary took 00:00:00.0578586
List took 00:00:00.2003057
Dictionary took 00:00:00.0611053
List took 00:00:00.2033325
Dictionary took 00:00:00.0583175
List took 00:00:00.2056591

字典显然更快，即使条目很少。

使用更多条目，与列表相比，词典会更快。

使用Dictionary的查找时间是O（1），而List是O（N）。这当然会对N的大值产生巨大的差异。

Answer 2

字典上）;它意味着有一个大小X，在这个大小X上，Dictionary开始比List快。例如。如果词典/列表包含10000个或更多项目，则词典总是更快（如果更少要存储的项目，比如5，列表可以更快）。 Dictionary还有另一个问题：它需要 GetHashCode（）方法;如果GetHashCode（）实现不好，Dictionary可能会非常慢。

Answer 3

这是由于law of large numbers。简而言之

根据法律规定，从大的平均结果中获得试验次数应接近预期值，并且趋于一致随着更多的试验进行而越来越近。

另一个约束是Big-O notation ，实际上在小规模上是无用的。例如，对于给定的 n ，可以说O（1）~O（N）~O（n！）小于某个小数。

运行一个好的实验需要满足一些非常严格的条件，例如：

确保算法具有可比性
算法中有大量迭代
在同一硬件上运行
使用max optimization
不应附加调试器或性能分析工具
进行多次实验并计算平均值+标准差
...

用于查找项目的字典与列表

3 个答案: