我有一个关于并行for循环的问题。我有以下代码:
public static void MultiplicateArray(double[] array, double factor)
{
for (int i = 0; i < array.Length; i++)
{
array[i] = array[i] * factor;
}
}
public static void MultiplicateArray(double[] arrayToChange, double[] multiplication)
{
for (int i = 0; i < arrayToChange.Length; i++)
{
arrayToChange[i] = arrayToChange[i] * multiplication[i];
}
}
public static void MultiplicateArray(double[] arrayToChange, double[,] multiArray, int dimension)
{
for (int i = 0; i < arrayToChange.Length; i++)
{
arrayToChange[i] = arrayToChange[i] * multiArray[i, dimension];
}
}
现在我尝试添加并行功能:
public static void MultiplicateArray(double[] array, double factor)
{
Parallel.For(0, array.Length, i =>
{
array[i] = array[i] * factor;
});
}
public static void MultiplicateArray(double[] arrayToChange, double[] multiplication)
{
Parallel.For(0, arrayToChange.Length, i =>
{
arrayToChange[i] = arrayToChange[i] * multiplication[i];
});
}
public static void MultiplicateArray(double[] arrayToChange, double[,] multiArray, int dimension)
{
Parallel.For(0, arrayToChange.Length, i =>
{
arrayToChange[i] = arrayToChange[i] * multiArray[i, dimension];
});
}
问题是,我想节省时间,而不是浪费时间。使用标准for循环,它计算大约2分钟,但使用并行for循环需要3分钟。为什么呢?
答案 0 :(得分:58)
Parallel.For()
可以通过并行化代码来提高性能,但它也有开销(线程之间的同步,在每次迭代时调用委托)。因为在你的代码中,每次迭代都非常简短(基本上只是一些CPU指令),这种开销会变得很突出。
因此,我认为使用Parallel.For()
不适合您。相反,如果您手动并行化代码(在这种情况下非常简单),您可能会看到性能提高。
为了验证这一点,我进行了一些测量:我在一个包含200 000 000个项目的数组上运行MultiplicateArray()
的不同实现(我使用的代码如下)。在我的机器上,串行版本一直需要0.21秒,而Parallel.For()
通常需要大约0.45秒,但有时它会飙升到8-9秒!
首先,我会尝试改进常见情况,稍后我会看到这些尖峰。我们希望通过 N CPU处理数组,因此我们将它分成 N 大小相等的部分并分别处理每个部分。结果? 0.35秒这仍然比串行版本更糟糕。但for
循环遍历数组中的每个项目是最优化的构造之一。我们不能帮助编译器吗?提取计算循环的边界可能会有所帮助。事实证明它确实:0.18秒。这比串行版更好,但不是很多。而且,有趣的是,在我的4核机器上没有将并行度从4改为2(没有HyperThreading)并没有改变结果:仍然是0.18秒。这让我得出结论,CPU不是这里的瓶颈,内存带宽是。
现在,回到峰值:我的自定义并行化没有它们,但是Parallel.For()
呢,为什么? Parallel.For()
确实使用范围分区,这意味着每个线程都处理它自己的数组部分。但是,如果一个线程提前完成,它将尝试帮助处理尚未完成的另一个线程的范围。如果发生这种情况,您将获得大量的虚假共享,这可能会大大减慢代码的速度。我自己的强制虚假分享测试似乎表明这确实是问题所在。强制Parallel.For()
的并行度似乎有助于尖峰。
当然,所有这些测量都是针对我计算机上的硬件而定的,并且对您来说会有所不同,所以您应该自己进行测量。
我使用的代码:
static void Main()
{
double[] array = new double[200 * 1000 * 1000];
for (int i = 0; i < array.Length; i++)
array[i] = 1;
for (int i = 0; i < 5; i++)
{
Stopwatch sw = Stopwatch.StartNew();
Serial(array, 2);
Console.WriteLine("Serial: {0:f2} s", sw.Elapsed.TotalSeconds);
sw = Stopwatch.StartNew();
ParallelFor(array, 2);
Console.WriteLine("Parallel.For: {0:f2} s", sw.Elapsed.TotalSeconds);
sw = Stopwatch.StartNew();
ParallelForDegreeOfParallelism(array, 2);
Console.WriteLine("Parallel.For (degree of parallelism): {0:f2} s", sw.Elapsed.TotalSeconds);
sw = Stopwatch.StartNew();
CustomParallel(array, 2);
Console.WriteLine("Custom parallel: {0:f2} s", sw.Elapsed.TotalSeconds);
sw = Stopwatch.StartNew();
CustomParallelExtractedMax(array, 2);
Console.WriteLine("Custom parallel (extracted max): {0:f2} s", sw.Elapsed.TotalSeconds);
sw = Stopwatch.StartNew();
CustomParallelExtractedMaxHalfParallelism(array, 2);
Console.WriteLine("Custom parallel (extracted max, half parallelism): {0:f2} s", sw.Elapsed.TotalSeconds);
sw = Stopwatch.StartNew();
CustomParallelFalseSharing(array, 2);
Console.WriteLine("Custom parallel (false sharing): {0:f2} s", sw.Elapsed.TotalSeconds);
}
}
static void Serial(double[] array, double factor)
{
for (int i = 0; i < array.Length; i++)
{
array[i] = array[i] * factor;
}
}
static void ParallelFor(double[] array, double factor)
{
Parallel.For(
0, array.Length, i => { array[i] = array[i] * factor; });
}
static void ParallelForDegreeOfParallelism(double[] array, double factor)
{
Parallel.For(
0, array.Length, new ParallelOptions { MaxDegreeOfParallelism = Environment.ProcessorCount },
i => { array[i] = array[i] * factor; });
}
static void CustomParallel(double[] array, double factor)
{
var degreeOfParallelism = Environment.ProcessorCount;
var tasks = new Task[degreeOfParallelism];
for (int taskNumber = 0; taskNumber < degreeOfParallelism; taskNumber++)
{
// capturing taskNumber in lambda wouldn't work correctly
int taskNumberCopy = taskNumber;
tasks[taskNumber] = Task.Factory.StartNew(
() =>
{
for (int i = array.Length * taskNumberCopy / degreeOfParallelism;
i < array.Length * (taskNumberCopy + 1) / degreeOfParallelism;
i++)
{
array[i] = array[i] * factor;
}
});
}
Task.WaitAll(tasks);
}
static void CustomParallelExtractedMax(double[] array, double factor)
{
var degreeOfParallelism = Environment.ProcessorCount;
var tasks = new Task[degreeOfParallelism];
for (int taskNumber = 0; taskNumber < degreeOfParallelism; taskNumber++)
{
// capturing taskNumber in lambda wouldn't work correctly
int taskNumberCopy = taskNumber;
tasks[taskNumber] = Task.Factory.StartNew(
() =>
{
var max = array.Length * (taskNumberCopy + 1) / degreeOfParallelism;
for (int i = array.Length * taskNumberCopy / degreeOfParallelism;
i < max;
i++)
{
array[i] = array[i] * factor;
}
});
}
Task.WaitAll(tasks);
}
static void CustomParallelExtractedMaxHalfParallelism(double[] array, double factor)
{
var degreeOfParallelism = Environment.ProcessorCount / 2;
var tasks = new Task[degreeOfParallelism];
for (int taskNumber = 0; taskNumber < degreeOfParallelism; taskNumber++)
{
// capturing taskNumber in lambda wouldn't work correctly
int taskNumberCopy = taskNumber;
tasks[taskNumber] = Task.Factory.StartNew(
() =>
{
var max = array.Length * (taskNumberCopy + 1) / degreeOfParallelism;
for (int i = array.Length * taskNumberCopy / degreeOfParallelism;
i < max;
i++)
{
array[i] = array[i] * factor;
}
});
}
Task.WaitAll(tasks);
}
static void CustomParallelFalseSharing(double[] array, double factor)
{
var degreeOfParallelism = Environment.ProcessorCount;
var tasks = new Task[degreeOfParallelism];
int i = -1;
for (int taskNumber = 0; taskNumber < degreeOfParallelism; taskNumber++)
{
tasks[taskNumber] = Task.Factory.StartNew(
() =>
{
int j = Interlocked.Increment(ref i);
while (j < array.Length)
{
array[j] = array[j] * factor;
j = Interlocked.Increment(ref i);
}
});
}
Task.WaitAll(tasks);
}
示例输出:
Serial: 0,20 s
Parallel.For: 0,50 s
Parallel.For (degree of parallelism): 8,90 s
Custom parallel: 0,33 s
Custom parallel (extracted max): 0,18 s
Custom parallel (extracted max, half parallelism): 0,18 s
Custom parallel (false sharing): 7,53 s
Serial: 0,21 s
Parallel.For: 0,52 s
Parallel.For (degree of parallelism): 0,36 s
Custom parallel: 0,31 s
Custom parallel (extracted max): 0,18 s
Custom parallel (extracted max, half parallelism): 0,19 s
Custom parallel (false sharing): 7,59 s
Serial: 0,21 s
Parallel.For: 11,21 s
Parallel.For (degree of parallelism): 0,36 s
Custom parallel: 0,32 s
Custom parallel (extracted max): 0,18 s
Custom parallel (extracted max, half parallelism): 0,18 s
Custom parallel (false sharing): 7,76 s
Serial: 0,21 s
Parallel.For: 0,46 s
Parallel.For (degree of parallelism): 0,35 s
Custom parallel: 0,31 s
Custom parallel (extracted max): 0,18 s
Custom parallel (extracted max, half parallelism): 0,18 s
Custom parallel (false sharing): 7,58 s
Serial: 0,21 s
Parallel.For: 0,45 s
Parallel.For (degree of parallelism): 0,40 s
Custom parallel: 0,38 s
Custom parallel (extracted max): 0,18 s
Custom parallel (extracted max, half parallelism): 0,18 s
Custom parallel (false sharing): 7,58 s
答案 1 :(得分:17)
Svick已经提供了一个很好的答案,但我想强调一点,关键是不要手动并行化你的代码&#34;而不是使用Parallel.For()
,而是必须处理更大的数据块。
这仍然可以使用Parallel.For()
完成,如下所示:
static void My(double[] array, double factor)
{
int degreeOfParallelism = Environment.ProcessorCount;
Parallel.For(0, degreeOfParallelism, workerId =>
{
var max = array.Length * (workerId + 1) / degreeOfParallelism;
for (int i = array.Length * workerId / degreeOfParallelism; i < max; i++)
array[i] = array[i] * factor;
});
}
与svicks CustomParallelExtractedMax()
做同样的事情,但更短,更简单,并且(在我的机器上)表现得更快:
Serial: 3,94 s
Parallel.For: 9,28 s
Parallel.For (degree of parallelism): 9,58 s
Custom parallel: 2,05 s
Custom parallel (extracted max): 1,19 s
Custom parallel (extracted max, half parallelism): 1,49 s
Custom parallel (false sharing): 27,88 s
My: 0,95 s
顺便说一下,所有其他答案中缺少的关键字是粒度。
答案 2 :(得分:6)
Parallel.For涉及更复杂的内存管理。结果可能会因cpu规格而异,例如#cores,L1&amp;二级缓存......
请看一下这篇有趣的文章:
答案 3 :(得分:6)
请参阅Custom Partitioners for PLINQ and TPL:
在
For
循环中,循环体作为委托提供给方法。调用该委托的成本与虚拟方法调用大致相同。在某些情况下,并行循环的主体可能足够小,以致每次循环迭代的委托调用的成本变得很大。在这种情况下,您可以使用Create
重载之一在源元素上创建IEnumerable<T>
范围分区。然后,您可以将此范围集合传递给ForEach
方法,该方法的主体由常规for循环组成。这种方法的好处是委托调用成本每个范围只发生一次,而不是每个元素发生一次。
在循环体中,您正在执行单个乘法,并且委托调用的开销将非常明显。
试试这个:
public static void MultiplicateArray(double[] array, double factor)
{
var rangePartitioner = Partitioner.Create(0, array.Length);
Parallel.ForEach(rangePartitioner, range =>
{
for (int i = range.Item1; i < range.Item2; i++)
{
array[i] = array[i] * factor;
}
});
}
另请参阅:Parallel.ForEach
documentation和Partitioner.Create
documentation。
答案 4 :(得分:-6)
来自http://msdn.microsoft.com/en-us/library/system.threading.tasks.parallel.aspx和http://msdn.microsoft.com/en-us/library/dd537608.aspx
你没有创建执行你的for的三个线程/进程,但是for的迭代尝试并行执行,所以即使只有一个你正在使用多个线程/进程。
这意味着可以同时执行与index = 0和index = 1的交互。
很可能你强迫使用过多的线程/进程,并且创建/执行它们的开销大于速度增加。
尝试使用三个正常但在三个不同的线程/进程中,如果你的系统是多核的(至少3倍)它应该不到一分钟