首先,我知道我的标题可以更好地表达,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？
编辑 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,我纯粹出于个人的好奇而寻找它.
Ani*_*Ani 32
使用LINQ:
int dotProduct = digits1.Zip(digits2, (d1, d2) => d1 * d2)
.Sum();
Zip将产生一个流序列,其中包含来自两个数组的相应元素的乘积,然后将其求和为整数Sum.
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指出的那样,Enumerable.Zip只是添加到.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();
Bru*_*oLM 10
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
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];
}
繁荣.
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
Intel Core 2 Quad 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)