C# アプリでできるだけ速く数値を処理しようとしています。を使用しThread.Sleep()て、処理と乱数をシミュレートします。私は3つの異なるテクニックを使用しています。
これは私が使用したテストコードです:
using System;
using System.Collections.Concurrent;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using System.Threading;
using System.Threading.Tasks;
namespace Test
{
internal class Program
{
private static void Main()
{
var data = new int[500000];
var random = new Random();
for (int i = 0; i < 500000; i++)
{
data[i] = random.Next();
}
var partialTimes = new Dictionary<int, double>();
var iterations = 5;
for (int i = 1; i < iterations + 1; i++)
{
Console.Write($"ProcessData3 {i}\t");
StartProcessing(data, partialTimes, ProcessData3);
GC.Collect();
}
Console.WriteLine();
Console.WriteLine("Press Enter to Exit");
Console.ReadLine();
}
private static void StartProcessing(int[] data, Dictionary<int, double> partialTimes, Action<int[], Dictionary<int, double>> processData)
{
var stopwatch = Stopwatch.StartNew();
try
{
processData?.Invoke(data, partialTimes);
stopwatch.Stop();
Console.WriteLine($"{stopwatch.Elapsed.ToString(@"mm\:ss\:fffffff")} total = {partialTimes.Sum(s => s.Value)} max = {partialTimes.Values.Max()}");
}
finally
{
partialTimes.Clear();
}
}
private static void ProcessData1(int[] data, Dictionary<int, double> partialTimes)
{
Parallel.ForEach(data, number =>
{
var partialStopwatch = Stopwatch.StartNew();
Thread.Sleep(1);
partialStopwatch.Stop();
lock (partialTimes)
{
partialTimes[number] = partialStopwatch.Elapsed.TotalMilliseconds;
}
});
}
private static void ProcessData3(int[] data, Dictionary<int, double> partialTimes)
{
// Partition the entire source array.
var rangePartitioner = Partitioner.Create(0, data.Length);
// Loop over the partitions in parallel.
Parallel.ForEach(rangePartitioner, (range, loopState) =>
{
// Loop over each range element without a delegate invocation.
for (int i = range.Item1; i < range.Item2; i++)
{
var number = data[i];
var partialStopwatch = Stopwatch.StartNew();
Thread.Sleep(1);
partialStopwatch.Stop();
lock (partialTimes)
{
partialTimes[number] = partialStopwatch.Elapsed.TotalMilliseconds;
}
}
});
}
private static void ProcessData2(int[] data, Dictionary<int, double> partialTimes)
{
var tasks = new Task[data.Count()];
for (int i = 0; i < data.Count(); i++)
{
var number = data[i];
tasks[i] = Task.Factory.StartNew(() =>
{
var partialStopwatch = Stopwatch.StartNew();
Thread.Sleep(1);
partialStopwatch.Stop();
lock (partialTimes)
{
partialTimes[number] = partialStopwatch.Elapsed.TotalMilliseconds;
}
});
}
Task.WaitAll(tasks);
}
}
}
テクニックごとに、プログラムを再起動します。そして、次の結果を取得
しThread.Sleep( 1 )ます。
ProcessData1 1 00:56:1796688 total = 801335,282599955 max = 16,8783
ProcessData1 2 00:23:5390014 total = 816167,642100022 max = 14,5913
ProcessData1 3 00:14:7090566 total = 827589,675899998 max = 13,2617
ProcessData1 4 00:10:8929177 total = 829296,528300007 max = 15,0175
ProcessData1 5 00:10:6333310 total = 839282,123200008 max = 29,2738
ProcessData2 1 00:37:8084153 total = 824507,174200022 max = 112,071
ProcessData2 2 00:16:3762096 total = 849272,47810001 max = 77,1514
ProcessData2 3 00:12:9177717 total = 854012,353100029 max = 67,5684
ProcessData2 4 00:10:4798701 total = 857396,642899983 max = 92,9408
ProcessData2 5 00:09:2206146 total = 870966,655499989 max = 51,8945
ProcessData3 1 01:13:6814541 total = 803581,718699918 max = 25,6815
ProcessData3 2 01:07:9809277 total = 814069,532899922 max = 26,0671
ProcessData3 3 01:07:9857984 total = 814148,329399928 max = 21,3116
ProcessData3 4 01:07:4812183 total = 808042,695499966 max = 16,8601
ProcessData3 5 01:07:2954614 total = 805895,325499903 max = 23,8517
各関数内で一緒に費やさ
totalれた時間の合計であり、各関数の最大時間です。Parallel.ForEach()
max
最初のループがなぜこんなに遅いのですか? 他の試行がこれほど迅速に処理されるのはどうしてですか? 最初の試行でより高速な並列処理を実現するにはどうすればよいですか?
編集：
だから私も結果を持って試してみましたThread.Sleep( 10 )
:
ProcessData1 1 02:50:2845698 total = 5109831,95429994 max = 12,0612
ProcessData1 2 00:56:3361645 total = 5125884,05919954 max = 12,7666
ProcessData1 3 00:53:4911541 total = 5131105,15209993 max = 12,7486
ProcessData1 4 00:49:5665628 total = 5144654,75829992 max = 13,2678
ProcessData1 5 00:46:0218194 total = 5152955,19509996 max = 13,702
ProcessData2 1 01:21:7207557 total = 5121889,31579983 max = 73,8152
ProcessData2 2 00:39:6660074 total = 5175557,68889969 max = 59,369
ProcessData2 3 00:31:9036416 total = 5193819,89889973 max = 56,2895
ProcessData2 4 00:27:4616803 total = 5207168,56969977 max = 65,5495
ProcessData2 5 00:24:4270755 total = 5222567,9044998 max = 65,368
ProcessData3 1 02:44:9985645 total = 5110117,19019997 max = 11,7172
ProcessData3 2 02:25:6533128 total = 5237779,27010012 max = 26,3171
ProcessData3 3 02:22:2771259 total = 5116123,45259975 max = 12,0581
ProcessData3 4 02:22:1678911 total = 5112574,93779995 max = 11,5334
ProcessData3 5 02:21:9418178 total = 5104980,07120004 max = 11,5583
したがって、最初のループは他のループよりもはるかに多くの秒を要します..