一段代碼x次，以及如何循環取平均值

這裏收集的結果的平均值是代碼：

class PingPongVariousLengths { 

    static public void main(String[] args) { 

     MPI.Init(args); 
     int myrank = MPI.COMM_WORLD.Rank(); 
     int tag = 99; 
     int maxlen = 104857600; //200 megabytes  104857600 characters * 2 bytes per character = 209715200 bytes total, or 200 megabytes 
     int minlen = 1; // 2 bytes 
     char [] sendbuff = new char [maxlen]; 
     char [] recvbuff = new char [maxlen]; 
     long speedKbps; 
     long speedMbps; 


     if (myrank == 0) { 
      for (int len = minlen; len <= maxlen; len *= 2) { //len=*2 doubles the ping size each time 
       long startTime = System.nanoTime();     
       MPI.COMM_WORLD.Send(sendbuff, 0, len, MPI.CHAR, 1, tag); 
       MPI.COMM_WORLD.Recv(recvbuff, 0, len, MPI.CHAR, 1, tag); 
       long endTime = System.nanoTime(); 
       long duration = endTime - startTime; 
       long durationseconds = (duration * 10-9); // Converts nanoseconds to seconds 
       System.out.println("Time for the ping to be sent and recived of " + (len*2) + " bytes is " + durationseconds + " seconds"); // multiples by 2 becuase 1 character is 2 bytes 
       //double transferRate = ((len*2.0)/durationseconds) ; //amount of data in bytes transferred in 1 second. Currently returning 0 for every result 
       //System.out.println("transferRate: " + transferRate + " bytes per second"); 
       double transferRateMb = ((len*524288.0)/durationseconds) ; //amount of data in megabytes transferred in 1 second. 
       System.out.println("transferRate (megabytes) : " + transferRateMb + " megabytes per second"); 
       } 
     } else if (myrank == 1) { 
      for (int len = minlen; len <= maxlen; len *= 2) { 
       MPI.COMM_WORLD.Recv(recvbuff, 0, len, MPI.CHAR, 0, tag); 
       MPI.COMM_WORLD.Send(recvbuff, 0, len, MPI.CHAR, 0, tag); 
      } 
     } 


     MPI.Finalize(); 
    } 
}

我試圖循環它說10或20倍，將代碼從

  for (int len = minlen; len <= maxlen; len *= 2) { //len=*2 doubles the ping size each time

到

   long durationseconds = (duration * 10-9); // Converts nanoseconds to seconds

我要運行20次，然後將結果duration是平均超過10或20

我怎樣才能最好的和最有效地去了解這個

來源

2013-04-13 user2065929

不知道，如果你實際上是在問這個...在一般平均可以通過添加所有時長來計算，然後通過循環次數劃分的？... 1/N *總和（持續時間） – gausss

「和然後計算平均值「你在說什麼？ –

上面編輯，希望它更有意義，對第一個令人困惑的問題抱歉 – user2065929

如果我理解你正確地，你需要每個長度增量的持續時間的平均值。爲了獲得持續時間的平均秒數，您需要在平均循環之外移動長度循環。請參閱下面的代碼片段。

long durationseconds; 
int MAX_LOOPS = 20; 

for (int len = minlen; len <= maxlen; len *= 2) { 
     if (myrank == 0) { 
       durationseconds = 0; 
       for (int i = 0; i < MAX_LOOPS; i++) { 
         long startTime = System.nanoTime();   
         MPI.COMM_WORLD.Send(sendbuff, 0, len, MPI.CHAR, 1, tag); 
         MPI.COMM_WORLD.Recv(recvbuff, 0, len, MPI.CHAR, 1, tag); 
         long endTime = System.nanoTime(); 
         long duration = endTime - startTime; 
         durationseconds = durationseconds + (duration * 10-9); 
       } 
       durationseconds = durationseconds/MAX_LOOPS; 
       System.out.println("Average time for the ping to be sent and recived of " + (len*2) + " bytes is " + durationseconds + " seconds"); 
       double transferRateMb = ((len*524288.0)/durationseconds); 
       System.out.println("average transferRate (megabytes) : " + transferRateMb + " megabytes per second"); 
     } else if (myrank == 1) { 
       MPI.COMM_WORLD.Recv(recvbuff, 0, len, MPI.CHAR, 0, tag); 
       MPI.COMM_WORLD.Send(recvbuff, 0, len, MPI.CHAR, 0, tag); 
     } 
}

來源

2013-04-13 23:10:08

這是完美的，正是我想解釋的:)謝謝 – user2065929

可愛的，唯一的區別是我和30分鐘前發佈的是，你實際上他的代碼位於循環中間並更改了一些變量名稱。這太棒了，你們倆:) –

long totalDuration = 0; 
for (int i = 0; i<nTimes;i++){ 
... //Code N stuff here 
totalDuration += duration; 
} 
long avgDuration = totalTime /nTimes;

來源

2013-04-13 22:51:21

一段代碼x次，以及如何循環取平均值

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