如何多次計數峯值？

Question

我打開一個閥門讓流體流動。這裏測量的壓力是流體被拉入系統的壓力。我正在嘗試僅測量前10個Pdiff（PMax-PMin）的平均值。一旦計算出平均值，閥就關閉。

並根據這個平均值，閥門將再次打開和關閉一個峯值，然後爲2個峯值和3個峯值等。我將壓力值存儲在數組中，並將該值與其前後值進行比較，得到最大值和最小值。

Answer 1

您使用++peakcounter增加您的peakcounter但隨後立即在if(peakcounter==0)

if塊設置peakcounter=0既然您重置peakcounter，你永遠不會peakcounter == 2

if (valstate == false && Pdelta >= average) 
{ 
    { 
     ++peakcounter; // keeps the count of how many times the value has gone 
     above average 
    } 
    // Checks for the number of times and then performs action 
    if (peakcounter == 1) { 
     digitalWrite(4, HIGH); 
     startTime = millis(); 
     valstate = true; 
     peakcounter = 0; //the offending line 
    } 

你需要什麼（注意：代碼沒有優化，我不完全理解你需要什麼，但是這應該解決你寫的問題）

int currentMax = 0; 
// your code here.... 

if (valstate == false && Pdelta >= average){ 
    ++peakcounter; 
    if(peakcounter > currentMax){ 


     // Checks for the number of times and then performs action 
     if (peakcounter == 1) { 
     digitalWrite(4, HIGH); 
     startTime = millis(); 
     valstate = true; 
     peakcounter = 0; 
     currentMax++; 
     } 

    //the rest of your peakcount checking code here 
    }