首先，我知道我的头衔可以更好地制定，但我的数学课程已经不复存在了，我不记得正确的单词了。
我需要做这样的事情（伪c＃）
int[] digits1 = new int[10]{0,1,2,3,4,5,6,7,8,9};
int[] digits2 = new int[10]{0,1,2,3,4,5,6,7,8,9};
int result = digits1*digits2
这将是每个数组的元素[i]的乘积之和。
这显然不起作用。 对任何更好的头衔或解决方案的任何建议？
修改 澄清：我知道我可以循环它们并进行数学计算。 基本上我认为有一个更好的方法来做到这一点，我纯粹是出于个人的好奇而寻找它。
答案 0 :(得分：32)
使用LINQ：
int dotProduct = digits1.Zip(digits2, (d1, d2) => d1 * d2)
.Sum();
Zip将生成一个流序列，其中包含来自两个数组的相应元素的乘积，然后将其加总为Sum的整数。
请注意，这不会像在长度不等的数组时那样失败，因此您可能需要验证输入：
//null checks here
if(digits1.Length != digits2.Length)
throw new ArgumentException("...");
编辑：
正如Jeff M所指出的那样，Enumerable.Zip只被添加到.NET 4.0中的框架中。在.NET 3.5中，您可以这样做（这个想法仅对暴露快速索引器的集合有效）：
int dotProduct = Enumerable.Range(0, digits1.Length)
.Sum(i => digits1[i] * digits2[i]);
//from Jeff M's comment:
int dotProduct = digits1.Select((n, i) => n * digits2[i])
.Sum();
答案 1 :(得分：10)
使用LINQ的解决方案
int[] digits1 = new int[10]{0,1,2,3,4,5,6,7,8,9};
int[] digits2 = new int[10]{0,1,2,3,4,5,6,7,8,9};
int result1 = digits1.Zip(digits2, (x, y) => x * y).Sum();
int result2 = digits1.Select((x, y) => x * digits2.ElementAt(y)).Sum();
int result3 = digits1.Select((n, i) => n * digits2[i]).Sum();
// Ani answer
int result4 = Enumerable.Range(0, digits1.Length)
.Sum(i => digits1[i] * digits2[i]);
性能测试 100000次迭代：
Queries
Fn: Result 1 Ticks 135306
Fn: Result 2 Ticks 2470614
Fn: Result 3 Ticks 130034
Fn: Result 4 Ticks 123374
-------------
Fastest
Fn: Result 4 Ticks 123374
Fn: Result 3 Ticks 130034
Fn: Result 1 Ticks 135306
Fn: Result 2 Ticks 2470614
答案 2 :(得分：8)
很简单，做一个循环。
int sum = 0;
for(int i = 0; i < digits1.length && i < digits2.length; i++)
{
sum += digits1[i] * digits2[i];
}
动臂。
答案 3 :(得分：7)
我写了一个测试台来比较这些方法在我的机器上的时间。
规格：
Windows 7专业版64位
英特尔酷睿2四核Q9550 @ 2.83GHz
4x1GiB Corsair Dominator DDR2 1066（PC2-8500）
using System;
using System.Linq;
namespace Testbench
{
class Program
{
static void Main(string[] args)
{
var digits1 = Enumerable.Range(0, 500).ToArray();
var digits2 = digits1.ToArray(); // create a copy
Test("Regular Loop", () =>
{
int result = 0;
for (int i = 0; i < digits1.Length; i++)
{
result += digits1[i] * digits2[i];
}
return result;
});
// using LINQ
Test("Enumerable \"Loop\"", () => Enumerable.Range(0, digits1.Length).Sum(i => digits1[i] * digits2[i]));
Test("Using Zip", () => digits1.Zip(digits2, (x, y) => x * y).Sum());
Test("Using Indexed Select", () => digits1.Select((n, i) => n * digits2[i]).Sum());
Test("Using Indexed Select with ElementAt", () => digits1.Select((n, i) => n * digits2.ElementAt(i)).Sum());
// using PLINQ
Test("Parallel Enumerable \"Loop\"", () => ParallelEnumerable.Range(0, digits1.Length).Sum(i => digits1[i] * digits2[i]));
Test("Using Parallel Zip", () => digits1.AsParallel().Zip(digits2.AsParallel(), (x, y) => x * y).Sum());
Test("Using Parallel Indexed Select", () => digits1.AsParallel().Select((n, i) => n * digits2[i]).Sum());
Test("Using Parallel Indexed Select with ElementAt", () => digits1.AsParallel().Select((n, i) => n * digits2.ElementAt(i)).Sum());
Console.Write("Press any key to continue . . . ");
Console.ReadKey(true);
Console.WriteLine();
}
static void Test<T>(string testName, Func<T> test, int iterations = 1000000)
{
Console.WriteLine(testName);
Console.WriteLine("Iterations: {0}", iterations);
var results = Enumerable.Repeat(0, iterations).Select(i => new System.Diagnostics.Stopwatch()).ToList();
var timer = System.Diagnostics.Stopwatch.StartNew();
for (int i = 0; i < results.Count; i++)
{
results[i].Start();
test();
results[i].Stop();
}
timer.Stop();
Console.WriteLine("Time(ms): {0,3}/{1,10}/{2,8} ({3,10})", results.Min(t => t.ElapsedMilliseconds), results.Average(t => t.ElapsedMilliseconds), results.Max(t => t.ElapsedMilliseconds), timer.ElapsedMilliseconds);
Console.WriteLine("Ticks: {0,3}/{1,10}/{2,8} ({3,10})", results.Min(t => t.ElapsedTicks), results.Average(t => t.ElapsedTicks), results.Max(t => t.ElapsedTicks), timer.ElapsedTicks);
Console.WriteLine();
}
}
}
32位目标：
Regular Loop Iterations: 1000000 Time(ms): 0/ 0/ 0 ( 1172) Ticks: 3/ 3.101365/ 526 ( 3244251) Enumerable "Loop" Iterations: 1000000 Time(ms): 0/ 4.3E-05/ 25 ( 9054) Ticks: 24/ 24.93989/ 69441 ( 25052172) Using Zip Iterations: 1000000 Time(ms): 0/ 2.4E-05/ 16 ( 16282) Ticks: 41/ 44.941406/ 45395 ( 45052491) Using Indexed Select Iterations: 1000000 Time(ms): 0/ 5.3E-05/ 32 ( 13473) Ticks: 34/ 37.165088/ 89602 ( 37280177) Using Indexed Select with ElementAt Iterations: 1000000 Time(ms): 0/ 1.5E-05/ 6 ( 160215) Ticks: 405/443.154147/ 17821 ( 443306156) Parallel Enumerable "Loop" Iterations: 1000000 Time(ms): 0/ 0.000103/ 29 ( 17194) Ticks: 38/ 47.412312/ 81906 ( 47576133) Using Parallel Zip Iterations: 1000000 Time(ms): 0/ 9.4E-05/ 19 ( 21703) Ticks: 49/ 59.859005/ 53200 ( 60051081) Using Parallel Indexed Select Iterations: 1000000 Time(ms): 0/ 0.000114/ 27 ( 20579) Ticks: 45/ 56.758491/ 75455 ( 56943627) Using Parallel Indexed Select with ElementAt Iterations: 1000000 Time(ms): 0/ 8.1E-05/ 19 ( 61137) Ticks: 144/ 168.97909/ 53320 ( 169165086)
64位目标：
Regular Loop Iterations: 1000000 Time(ms): 0/ 0/ 0 ( 506) Ticks: 1/ 1.254137/ 1491 ( 1401969) Enumerable "Loop" Iterations: 1000000 Time(ms): 0/ 2.9E-05/ 15 ( 10118) Ticks: 27/ 27.850086/ 41954 ( 27995994) Using Zip Iterations: 1000000 Time(ms): 0/ 2.2E-05/ 13 ( 17089) Ticks: 45/ 47.132834/ 38506 ( 47284608) Using Indexed Select Iterations: 1000000 Time(ms): 0/ 3.1E-05/ 12 ( 14057) Ticks: 37/ 38.740923/ 33846 ( 38897274) Using Indexed Select with ElementAt Iterations: 1000000 Time(ms): 0/ 3.8E-05/ 29 ( 117412) Ticks: 304/324.711279/ 82726 ( 324872753) Parallel Enumerable "Loop" Iterations: 1000000 Time(ms): 0/ 9.9E-05/ 28 ( 24234) Ticks: 38/ 66.79389/ 77578 ( 67054956) Using Parallel Zip Iterations: 1000000 Time(ms): 0/ 0.000111/ 24 ( 30193) Ticks: 46/ 83.264037/ 69029 ( 83542711) Using Parallel Indexed Select Iterations: 1000000 Time(ms): 0/ 6.5E-05/ 20 ( 28417) Ticks: 45/ 78.337831/ 56354 ( 78628396) Using Parallel Indexed Select with ElementAt Iterations: 1000000 Time(ms): 0/ 9.2E-05/ 16 ( 65233) Ticks: 112/180.154663/ 44799 ( 180496754)
答案 4 :(得分：4)
int result = 0;
for(int i = 0; i < digits1.length; i++)
{
result += digits1[i] * digits2[i];
}
答案 5 :(得分：0)
更快甚至是展开循环
Test("Regular Loop", () =>
{
int result = 0;
for (int i = 0; i < digits1.Length; i++)
{
result += digits1[i] * digits2[i];
}
return result;
});
// This will fail if vectors are not a multiple of 4 in length.
Test("Unroll 4x", () =>
{
int result = 0;
for (int i = 0; i < digits1.Length; i+=4)
{
result += digits1[i] * digits2[i];
result += digits1[i+1] * digits2[i+1];
result += digits1[i+2] * digits2[i+2];
result += digits1[i+3] * digits2[i+3];
}
return result;
});
Test("Dynamic unroll", () =>
{
int result = 0;
int limit = (digits1.Length/8)*8;
int reminderLimit = digits1.Length;
if (digits1.Length >= 8)
{
for (int i = 0; i < limit; i+=8)
{
result += digits1[i] * digits2[i];
result += digits1[i+1] * digits2[i+1];
result += digits1[i+2] * digits2[i+2];
result += digits1[i+3] * digits2[i+3];
result += digits1[i+4] * digits2[i+4];
result += digits1[i+5] * digits2[i+5];
result += digits1[i+6] * digits2[i+6];
result += digits1[i+7] * digits2[i+7];
}
reminderLimit = digits1.Length % 8;
}
switch(reminderLimit)
{
case 7: {
result += digits1[limit] * digits2[limit];
result += digits1[limit+1] * digits2[limit+1];
result += digits1[limit+2] * digits2[limit+2];
result += digits1[limit+3] * digits2[limit+3];
result += digits1[limit+4] * digits2[limit+4];
result += digits1[limit+5] * digits2[limit+5];
result += digits1[limit+6] * digits2[limit+6];
break;
}
case 6: {
result += digits1[limit] * digits2[limit];
result += digits1[limit+1] * digits2[limit+1];
result += digits1[limit+2] * digits2[limit+2];
result += digits1[limit+3] * digits2[limit+3];
result += digits1[limit+4] * digits2[limit+4];
result += digits1[limit+5] * digits2[limit+5];
break;
}
case 5: {
result += digits1[limit] * digits2[limit];
result += digits1[limit+1] * digits2[limit+1];
result += digits1[limit+2] * digits2[limit+2];
result += digits1[limit+3] * digits2[limit+3];
result += digits1[limit+4] * digits2[limit+4];
break;
}
case 4: {
result += digits1[limit] * digits2[limit];
result += digits1[limit+1] * digits2[limit+1];
result += digits1[limit+2] * digits2[limit+2];
result += digits1[limit+3] * digits2[limit+3];
break;
}
case 3: {
result += digits1[limit] * digits2[limit];
result += digits1[limit+1] * digits2[limit+1];
result += digits1[limit+2] * digits2[limit+2];
break;
}
case 2: {
result += digits1[limit] * digits2[limit];
result += digits1[limit+1] * digits2[limit+1];
break;
}
case 1: {
result += digits1[limit] * digits2[limit];
break;
}
default :
{
break;
}
}
return result;
});
C＃代码的调试和发布模式之间的运行时间有很大的不同，这是在发布模式下运行的：
Regular Loop
Iterations: 1000000
Time(ms): 0/ 0/ 0 ( 596)
Ticks: 1/ 2,071213/ 455 ( 2154248)
Unroll 4x
Iterations: 1000000
Time(ms): 0/ 2E-06/ 1 ( 575)
Ticks: 1/ 1,984301/ 3876 ( 2076105)
Dynamic unroll
Iterations: 1000000
Time(ms): 0/ 0/ 0 ( 430)
Ticks: 1/ 1,4635/ 3228 ( 1554830)
调试模式：
Regular Loop
Iterations: 1000000
Time(ms): 0/ 1E-06/ 1 ( 1296)
Ticks: 4/ 4,529916/ 3907 ( 4678354)
Unroll 4x
Iterations: 1000000
Time(ms): 0/ 0/ 0 ( 871)
Ticks: 2/ 3,048466/ 701 ( 3145277)
Dynamic unroll
Iterations: 1000000
Time(ms): 0/ 0/ 0 ( 819)
Ticks: 2/ 2,858588/ 1398 ( 2957179)