联盟与添加范围在列表中，哪个是最快和良好的性能?

Question

哪个是你认为下面两个代码之间最好的，还是你有另一种更有效的替代方案

第一

List<int> g = new List<int>();
g.AddRange(listof1);
g.AddRange(listof2);
return g.GroupBy(a=>a)           
.Select(root=>root.FirstOrDefault())
.ToList();

第二

var rootIds = listof1.
Union(listof2)
.Select(rootId => rootId).ToList();
return rootIds;

Answer 1

1. 没关系

2. 让我们看看:)

实验

候选人

static int M1(List<int> l1, List<int> l2)
{
List<int> g = new List<int>();
g.AddRange(l1);
g.AddRange(l2);
var x = g.GroupBy(a => a)
.Select(root => root.FirstOrDefault())
.ToList();
return x.Count();
}
static int M2(List<int> l1, List<int> l2)
{
var x = l1
.Union(l2)
.Select(rootId => rootId).ToList();
return x.Count();
}
static int M3(List<int> l1, List<int> l2)
{
var g = new HashSet<int>(l1);
g.UnionWith(l2);
return g.Count();
}

结果

100.00
0.057       0.038       0.017           res: 200,200,200
0.040       0.015       0.011           res: 200,200,200
0.055       0.023       0.009           res: 200,200,200
0.062       0.036       0.022           res: 200,200,200
0.055       0.023       0.027           res: 200,200,200
1,000,000.00
1788.493    513.422     292.815         res: 1999068,1999068,1999068
1795.416    591.142     453.076         res: 1999068,1999068,1999068
2540.099    529.801     303.729         res: 1999068,1999068,1999068
1797.251    527.849     428.672         res: 1999068,1999068,1999068
2008.561    527.680     501.997         res: 1999068,1999068,1999068
10,000,000.00
34458.838   6764.472    3679.255        res: 19906255,19906255,19906255
32756.011   8507.293    3875.931        res: 19906255,19906255,19906255

测试代码

public class Program
{
static Random r = new Random(1);
public static async Task Main(string[] args)
{
Try(100);
Try(1000000);
Try(10000000);
Console.ReadLine();
}
public static void Try(int i)
{
var l1 = Enumerable.Range(0, i).Select(i => r.Next(0, int.MaxValue)).ToList();
var l2 = Enumerable.Range(0, i).Select(i => r.Next(0, int.MaxValue)).ToList();
NapkinQualityCompareMeasureIt(i, 5, fs: new List<Func<int>> { () => M1(l1, l2), () => M2(l1, l2), () => M3(l1, l2) });
}
public static void NapkinQualityCompareMeasureIt<T>(int count, int howManyTimes, List<Func<T>> fs)
{
//This is good only for detecting significant differences
Console.WriteLine(count.ToString("N"));
for (int i = 0; i < howManyTimes; i++)
{
var metrics = fs.Select(f =>
{
var sw = Stopwatch.StartNew();
var r = f();
return (Elapsed: sw.Elapsed, Result: r);
});
var res = string.Join(",", metrics.Select(m=>m.Result));
var times = string.Join("", metrics.Select(m => $"{m.Elapsed.TotalMilliseconds.ToString("F3").PadRight(12)}"));
Console.WriteLine($"{times}tres: {res}");
}
}
}

Answer 2

如果你想要一组唯一的值，HashSet<int>值得考虑：

var g = new HashSet<int>(listof1);
g.UnionWith(listof2);

使用HashSet的优点是它不可能包含重复值 - 使使用g的代码更易于阅读和推理。