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Need Help : Merge 2 Loops into 1 Loop

  • #1
  • First Post: Edited 7:44pm Nov 24, 2021 3:24am | Edited 7:44pm
Hi coders, I need your help and advice how to merge 2 different conditions of loops into 1 loop (but both loops are linking to each others) so that I can get the 1st bar's shift of a strong bull move. Kindly look at my picture to understand what I intend to achieve ...

1st loop of conditions :-
Inserted Code
   int MA2Dp1UpSf=0; for(int i=1; i<=100; i++) {
      double MA0=iMA(NULL,1,5,0,3,0,i);
      double MA1=iMA(NULL,1,5,0,3,0,i+1);
      double MA2=iMA(NULL,1,5,0,3,0,i+2);
      double MA3=iMA(NULL,1,5,0,3,0,i+3);
         if(MA0>MA1 && MA1<=MA2 && MA2<=MA3) MA2Dp1UpSf=i;
         // MA 2-drop 1-Up
   }

2nd loop of conditions : I would like to get at least 4 minimum big long green bars ( Cs0-Op0>=cATR0, i.e. cntGreen>=4 ) for a strong bull trend starting from MA2Dp1UpSf-1 and ends at next MA 2-drop.
Inserted Code
   int cntGreen=0; for(int i=MA2Dp1UpSf-1; i>=1; i--) {
      double Op0=iOpen(NULL,1,i);
      double Hi0=iHigh(NULL,1,i);
      double Cs0=iClose(NULL,1,i);
      double ATR0=iATR(NULL,1,5,i);
      double MAn1=iMA(NULL,1,5,0,3,0,i-1);
      double MAn2=iMA(NULL,1,5,0,3,0,i-2);
      double MAn3=iMA(NULL,1,5,0,3,0,i-3);
         if(Cs0-Op0>=cATR0 && MAn1<=MathMax(MAn2,MAn3)) cntGreen++;
         // Count No. of Big Green bars and stops at next MA 2-drop
   }

If running two loops independently, due to different loops' conditions, the 1st bar's shift that I'm aiming for is varying and keep on changing, therefore in my opinion, the only way is to merge the 2 loops into 1 loop, only then can get the correct 1st bar's shift for the last 100 bars. Is it possible to have a sub-loop inside a master loop to get the 1st bar's shift ? Please help with your codes, thank you.
Attached Image(s) (click to enlarge)
Click to Enlarge Name: bars2.png Size: 39 KB
Click to Enlarge Name: bars.png Size: 23 KB
  • #2
  • Nov 24, 2021 3:02pm Nov 24, 2021 3:02pm
in the first loop, if all you are looking for is the latest signal then add break; when conditions are met, that way the loop will be broken and indicator will stop searching for new signals.
Inserted Code
if(MA0>MA1 && MA1<=MA2 && MA2<=MA3)
   {
   MA2Dp1UpSf=i;
   break;
   }

In second loop there is a mistake when you call iMA(), it is not possible to call iMA with shift i-3 when loop is at i=1, you should get error. Don't let i go lower than i=3 or fix the iMA() shifts

It's hard to understand what are you trying to accomplish here, I would recommend describing it in words and pictures first, maybe there is a better way to code this..
Know thyself
  • #3
  • Edited 7:43pm Nov 24, 2021 3:53pm | Edited 7:43pm
Quoting fxanalysis
in the first loop, if all you are looking for is the latest signal then add break; when conditions are met, that way the loop will be broken and indicator will stop searching for new signals. if(MA0>MA1 && MA1<=MA2 && MA2<=MA3) { MA2Dp1UpSf=i; break; } In second loop there is a mistake when you call iMA(), it is not possible to call iMA with shift i-3 when loop is at i=1, you should get error. Don't let i go lower than i=3 or fix the iMA() shifts It's hard to understand what are you trying to accomplish here, I would recommend describing it in words...
Thanks FxAnalysis. I know how to use the "break;" and I'm not looking for the latest signal by using the "break;" but I'm looking for 2 conditions signal in just one loop. Kindly look at my first post above and I added another picture for your better understanding.

In short / in summary, I'm looking for the 1st bar's shift (MA starts going up after previous 2 MAs dropped) at the start of a strong bull trend that with at least 4 strong green bars (Cs0-Op0>=cATR0, i.e. cntGreen>=4), of course, it can be 5, 6, 7 and more strong green bars of the bull trend, but I don't know how to create such a loop. Is it possible to have a sub-loop (int cntGreen) inside a master loop (int MA2Dp1UpSf) to get the 1st bar's shift ?

Thanks again.
  • #4
  • Edited 10:27pm Nov 24, 2021 10:11pm | Edited 10:27pm
I will just help you with some psudocode as I don't have time to write detailed coding, hope you can understand.

Inserted Code
     for(int i=100; i>=1; i--) {
          //Get the MA average values
            iMA().........
         //Check if the moving averages satisfy your criteria
          if(MA0>MA1 && MA1<=MA2 && MA2<=MA3){
                
                //When Moving averages criteria is satisfied, now check for 4 bullish candles
                 int cntGreen=0;
                
                for(int count=i; count>=1; count--) {  
                       //Get Values
                       //Increment count for bullish candles
                       if(Cs0-Op0>=cATR0 && MAn1<=MathMax(MAn2,MAn3)) cntGreen++;
                       if(whatever criteria you need to stop counting)
                               break;
                }
         }  
 
    }
  • #5
  • Edited 4:10am Nov 25, 2021 2:59am | Edited 4:10am
Quoting eess
I will just help you with some psudocode as I don't have time to write detailed coding, hope you can understand. for(int i=100; i>=1; i--) { //Get the MA average values iMA()......... //Check if the moving averages satisfy your criteria if(MA0>MA1 && MA1<=MA2 && MA2<=MA3){ //When Moving averages criteria is satisfied, now check for 4 bullish candles int cntGreen=0; for(int count=i; count>=1; count--) { //Get Values //Increment count for bullish candles if(Cs0-Op0>=cATR0 && MAn1<=MathMax(MAn2,MAn3)) cntGreen++; if(whatever criteria you need to stop...
Hi eess, thanks for your pseudocodes and advice. I modified a little bit, no error but it is not giving me the correct answer and I don't know how to resolve it. Honestly, I'm just throwing a dart blindly.

In my case as below, there are 3 loops altogether, 1 main (m) and 2 sub (s1 and s2) loops.

Another point : I have to start with i++ instead of i-- (what you wrote in your pseudocodes) as I want to get the latest starting shift (2 drop and 1 Up MA). For counting of strong green bars, yes, I need to use cn-- starting from i.

Inserted Code
   int MA2Dp1UpSf=0;
   for(int i=4; i<=100; i++) {
   double mMA0=iMA(symbol,5,5,0,3,0,i), mMA1=iMA(symbol,5,5,0,3,0,i+1);
   double mMA2=iMA(symbol,5,5,0,3,0,i+2), mMA3=iMA(symbol,5,5,0,3,0,i+3);
   if(mMA0>mMA1 && mMA1<=mMA2 && mMA2<=mMA3) { // MA 2-Dropp 1-Up
 
      int cnGrn=0;
      for(int cn=i; cn>=1; cn--) {  
      double s1Op0=iOpen(symbol,5,cn), s1Cs0=iClose(symbol,5,cn);
      double s1ATR0=iATR(symbol,5,5,cn), s1MAn1=iMA(symbol,5,5,0,3,0,cn-1);
      double s1MAn2=iMA(symbol,5,5,0,3,0,cn-2), s1MAn3=iMA(symbol,5,5,0,3,0,cn-3);
      if(s1Cs0-s1Op0>=s1ATR0) cnGrn++;
      if(s1MAn1<=MathMax(s1MAn2,s1MAn3)) break; { // conditions to stop counting
 
         for(int sf=cn+2; sf<=i; sf++) {
         double s2MA0=iMA(symbol,5,5,0,3,0,sf), s2MA1=iMA(symbol,5,5,0,3,0,sf+1);
         double s2MA2=iMA(symbol,5,5,0,3,0,sf+2), s2MA3=iMA(symbol,5,5,0,3,0,sf+3);
         if(cnGrn>=4 && s2MA0>s2MA1 && s2MA1<=s2MA2 && s2MA2<=s2MA3) break; MA2Dp1UpSf=sf;
               }
            }
         }  
      }
   }
   {Alert(symbol+" 1st Shift of BULL : "+IntegerToString(MA2Dp1UpSf));}

Quoting fxanalysis
In second loop there is a mistake when you call iMA(), it is not possible to call iMA with shift i-3 when loop is at i=1, you should get error. Don't let i go lower than i=3 or fix the iMA() shifts
Hi fxanalysis, pertaining to your concerns on the error expected, yes, you're right if it starts with i=1, in order to resolve and counter the error, starts with i=4 would be a solution.

Thanks two of you. Hope to get your advice soon.
  • #6
  • Nov 25, 2021 4:33am Nov 25, 2021 4:33am
This is if you don't need the latest trend, if you need the latest trend then doo something similar just change the loop to go from 4 to 100, start with finding the end of the trend , then 4 strong candles and then the start of the trend, this way you can do everything in 1 loop.


Inserted Code
int MA2Dp1UpSf=0;
int cntGreen=0;
bool checkStart=false;
for(int i=100; i>=4; i--)
   {
   double MA0=iMA(NULL,1,5,0,3,0,i);
   double MA1=iMA(NULL,1,5,0,3,0,i+1);
   double MA2=iMA(NULL,1,5,0,3,0,i+2);
   double MA3=iMA(NULL,1,5,0,3,0,i+3);
   if(!checkStart)
      {
      if(MA0>MA1 && MA1<=MA2 && MA2<=MA3)
         {
         MA2Dp1UpSf=i; // MA 2-drop 1-Up
         checkStart=true;
         }
      }
   else //check start =true
      {
      double Op0=iOpen(NULL,1,i);
      double Hi0=iHigh(NULL,1,i);
      double Cs0=iClose(NULL,1,i);
      double ATR0=iATR(NULL,1,5,i);
      double MAn1=iMA(NULL,1,5,0,3,0,i-1);
      double MAn2=iMA(NULL,1,5,0,3,0,i-2);
      double MAn3=iMA(NULL,1,5,0,3,0,i-3);
      
      if(MAn1<=MathMax(MAn2,MAn3))
         {
         if(Cs0-Op0>=cATR0)
            {
            cntGreen++;
            }
         }
      else //END OF TREND
         {
         if(cntGreen>3)
            {
            break;
            }
         else
            {
            checkStart=false;
            cntGreen=0;
            MA2Dp1UpSf=0;
            }
        
         }
      // Count No. of Big Green bars and stops at next MA 2-drop
  
      }
}

why do you define double ATR0=iATR(NULL,1,5,i); and use cATR0 ..
if(Cs0-Op0>=cATR0)?

Did you get the idea how to do this stuff?
Know thyself
  • #7
  • Edited 5:33am Nov 25, 2021 5:21am | Edited 5:33am
Hi,
not sure if I understood everything that you want. But this should get you started and is easy enough to adapt.

Inserted Code
   int n = 100; // bars to look for
   int ma_tf = Period();
   int ma_period = 3;
   int ma_method = MODE_SMA;
   int min_greenbars = 4;
   bool stop = false;
  
   for(int i = n; i > 0; i --) {
      if(stop) {
         break;
      }
      double ma0 = iMA(Symbol(), ma_tf, ma_period, 0, ma_method, PRICE_CLOSE, i);
      double ma1 = iMA(Symbol(), ma_tf, ma_period, 0, ma_method, PRICE_CLOSE, i - 1);
      double ma2 = iMA(Symbol(), ma_tf, ma_period, 0, ma_method, PRICE_CLOSE, i - 2);
      double ma3 = iMA(Symbol(), ma_tf, ma_period, 0, ma_method, PRICE_CLOSE, i - 3);
      if(ma0 > ma1 || ma1 > ma2 || ma2 > ma3) { // ma not rising for 4 bars
         continue;
      }
      
      int greenbar_counter = 0;
      for(int j = (n - i); j > 0; j --) { // look for strong bars
         double open = iOpen(Symbol(), PERIOD_M1, j);
         double close = iClose(Symbol(), PERIOD_M1, j);
         double big_barsize = iATR(Symbol(), Period(), 5, j);
         if(close - open > big_barsize) {
            greenbar_counter++;
         }
        
         double ma_j_0 = iMA(Symbol(), ma_tf, ma_period, 0, ma_method, PRICE_CLOSE, j);
         double ma_j_1 = iMA(Symbol(), ma_tf, ma_period, 0, ma_method, PRICE_CLOSE, j - 1);
         double ma_j_2 = iMA(Symbol(), ma_tf, ma_period, 0, ma_method, PRICE_CLOSE, j - 2);        
         if(ma_j_2 < ma_j_1 && ma_j_0 < ma_j_1) { // 2 MA drop
            break;
         }
        
         if(greenbar_counter >= min_greenbars) { // all checks passed return 1st bar shift
            Alert("Bar Shift: ", i, ", Time: ", iTime(Symbol(), Period(), i));
            stop = true;
            break;
         }
        
      }
  
   }
  • #8
  • Nov 25, 2021 5:42am Nov 25, 2021 5:42am
Quoting chimoong
Hi coders, I need your help and advice how to merge 2 different conditions of loops into 1 loop (but both loops are linking to each others) so that I can get the 1st bar's shift of a strong bull move. Kindly look at my picture to understand what I intend to achieve ... 1st loop of conditions :- int MA2Dp1UpSf=0; for(int i=1; i<=100; i++) { double MA0=iMA(NULL,1,5,0,3,0,i); double MA1=iMA(NULL,1,5,0,3,0,i+1); double MA2=iMA(NULL,1,5,0,3,0,i+2); double MA3=iMA(NULL,1,5,0,3,0,i+3); if(MA0>MA1 && MA1<=MA2 && MA2<=MA3) MA2Dp1UpSf=i; // MA 2-drop 1-Up...
Inserted Code
for(int i=1; i<=100; i++) {
      double MA0=iMA(NULL,1,5,0,3,0,i+3);
      double MA1=iMA(NULL,1,5,0,3,0,i+4);
      double MA2=iMA(NULL,1,5,0,3,0,i+5);
      double MA3=iMA(NULL,1,5,0,3,0,i+6);
 
      double Op0=iOpen(NULL,1,i+3);
      double Hi0=iHigh(NULL,1,i+3);
      double Cs0=iClose(NULL,1,i+3);
      double ATR0=iATR(NULL,1,5,i+3);
      double MAn1=iMA(NULL,1,5,0,3,0,i+2);
      double MAn2=iMA(NULL,1,5,0,3,0,i+1);
      double MAn3=iMA(NULL,1,5,0,3,0,i+0);
  
   if(MA0>MA1 && MA1<=MA2 && MA2<=MA3) MA2Dp1UpSf=i;
      // MA 2-drop 1-Up
   if(Cs0-Op0>=cATR0 && MAn1<=MathMax(MAn2,MAn3)) cntGreen++;
      // Count No. of Big Green bars and stops at next MA 2-drop
}

add +3 to every request. the only way to make it nrp
  • #9
  • Nov 25, 2021 8:08am Nov 25, 2021 8:08am
Quoting fxanalysis
This is if you don't need the latest trend, if you need the latest trend then doo something similar just change the loop to go from 4 to 100, start with finding the end of the trend , then 4 strong candles and then the start of the trend, this way you can do everything in 1 loop. int MA2Dp1UpSf=0; int cntGreen=0; bool checkStart=false; for(int i=100; i>=4; i--) { double MA0=iMA(NULL,1,5,0,3,0,i); double MA1=iMA(NULL,1,5,0,3,0,i+1); double MA2=iMA(NULL,1,5,0,3,0,i+2); double MA3=iMA(NULL,1,5,0,3,0,i+3); if(!checkStart) { if(MA0>MA1 && MA1<=MA2 &&...
Hi Fxanalysis, thanks, let me try your way by using "bool checkStart=false;", I guess this should be the correct way of getting what I aim for.

1) Firstly, my apology to you that I expressed wrongly "I don't need the latest trend", in fact, I need the latest trend signal but must with combination of at least 4 strong green bars and therefore the loop should start with i++

2) Subsequently to test the min 4 strong green bars, I added "for(int cn=i; cn>=1; cn--)", and this time, in my opinion, it should be "cn--" and I changed all the indicators' shifts as "cn". Likely I may be wrong ...

3) My mistake to put cATR0, it should be ATR0.

Inserted Code
   int MA2Dp1UpSf=0;
   int cnGrn=0;
   bool checkStart=false;

   for(int i=4; i>=200; i++) {
   double MA0=iMA(symbol,5,5,0,3,0,i), MA1=iMA(symbol,5,5,0,3,0,i+1);
   double MA2=iMA(symbol,5,5,0,3,0,i+2), MA3=iMA(symbol,5,5,0,3,0,i+3);
   if(!checkStart) {
      if(MA0>MA1 && MA1<=MA2 && MA2<=MA3) {
         MA2Dp1UpSf=i; // MA 2-drop 1-Up
         checkStart=true;
      }
   } else //check start=true
      for(int cn=i; cn>=1; cn--) {
      double Op0=iOpen(symbol,5,cn), Cs0=iClose(symbol,5,cn);
      double ATR0=iATR(symbol,5,5,cn), MAn1=iMA(symbol,5,5,0,3,0,cn-1);
      double MAn2=iMA(symbol,5,5,0,3,0,cn-2), MAn3=iMA(symbol,5,5,0,3,0,cn-3);

      if(MAn1<=MathMax(MAn2,MAn3)) {
         if(Cs0-Op0>=ATR0) {
            cnGrn++;
            }
         } else { //END OF TREND
         if(cnGrn>=4) {
            break;
            }
            else {
            checkStart=false;
            cnGrn=0;
            MA2Dp1UpSf=0;
            }
         } // Count No. of Big Green bars and stops at next MA 2-drop
      }
   }
 
   {Alert(symbol+" 1st Shift of BULL Start : "+IntegerToString(MA2Dp1UpSf));}

With the above codes, I still cannot get the correct output results, kindly advise me where I went wrong, thank you.
  • #10
  • Nov 25, 2021 8:15am Nov 25, 2021 8:15am
Quoting ionone
{quote} for(int i=1; i<=100; i++) { double MA0=iMA(NULL,1,5,0,3,0,i+3); double MA1=iMA(NULL,1,5,0,3,0,i+4); double MA2=iMA(NULL,1,5,0,3,0,i+5); double MA3=iMA(NULL,1,5,0,3,0,i+6); double Op0=iOpen(NULL,1,i+3); double Hi0=iHigh(NULL,1,i+3); double Cs0=iClose(NULL,1,i+3); double ATR0=iATR(NULL,1,5,i+3); double MAn1=iMA(NULL,1,5,0,3,0,i+2); double MAn2=iMA(NULL,1,5,0,3,0,i+1); double MAn3=iMA(NULL,1,5,0,3,0,i+0); if(MA0>MA1 && MA1<=MA2 && MA2<=MA3) MA2Dp1UpSf=i; // MA 2-drop 1-Up if(Cs0-Op0>=cATR0 && MAn1<=MathMax(MAn2,MAn3)) cntGreen++; // Count No....
Thanks Ionone, I think the other two coders, Fxanalysis and Peterf, above should be the correct direction as using a bool to return back to 1st level of loop till there are at least 4 strong green bars. I'm testing their ways now as I have no programming background. I'm yet in learning stage, hope you don't mind for my stupid questions above.
  • #11
  • Nov 25, 2021 8:28am Nov 25, 2021 8:28am
Quoting chimoong
{quote} Hi Fxanalysis, thanks, let me try your way by using "bool checkStart=false;", I guess this should be the correct way of getting what I aim for. 1) Firstly, my apology to you that I expressed wrongly "I don't need the latest trend", in fact, I need the latest trend signal but must with combination of at least 4 strong green bars and therefore the loop should start with i++ 2) Subsequently to test the min 4 strong green bars, I added "for(int cn=i; cn>=1; cn--)", and this time, in my opinion, it should be "cn--" and I changed all the indicators'...
problem is that the break now only breaks the inside loop. you need to get out of both loops.
you can add a new bool to check if signal is found and you should exit the first loop as well. Or you can recode indicator to have only one loop.

Inserted Code
int MA2Dp1UpSf=0;
   int cnGrn=0;
   bool checkStart=false;
   bool signalFound=false;
   for(int i=4; i>=200; i++) {
   double MA0=iMA(symbol,5,5,0,3,0,i), MA1=iMA(symbol,5,5,0,3,0,i+1);
   double MA2=iMA(symbol,5,5,0,3,0,i+2), MA3=iMA(symbol,5,5,0,3,0,i+3);
   if(!checkStart) {
      if(MA0>MA1 && MA1<=MA2 && MA2<=MA3) {
         MA2Dp1UpSf=i; // MA 2-drop 1-Up
         checkStart=true;
      }
   } else //check start=true
      for(int cn=i; cn>=1; cn--) {
      double Op0=iOpen(symbol,5,cn), Cs0=iClose(symbol,5,cn);
      double ATR0=iATR(symbol,5,5,cn), MAn1=iMA(symbol,5,5,0,3,0,cn-1);
      double MAn2=iMA(symbol,5,5,0,3,0,cn-2), MAn3=iMA(symbol,5,5,0,3,0,cn-3);
      if(MAn1<=MathMax(MAn2,MAn3)) {
         if(Cs0-Op0>=ATR0) {
            cnGrn++;
            }
         } else { //END OF TREND
         if(cnGrn>=4) {
            
            signalFound=true;
            break;
            }
            else {
            checkStart=false;
            cnGrn=0;
            MA2Dp1UpSf=0;
            }
         } // Count No. of Big Green bars and stops at next MA 2-drop
      }
   if(signalFound)break;
   }
 
   {Alert(symbol+" 1st Shift of BULL Start : "+IntegerToString(MA2Dp1UpSf));}
Know thyself
  • #12
  • Nov 25, 2021 8:33am Nov 25, 2021 8:33am
also, do you need to wait until trend ends? or is it enough to get the 4 strong candles without waiting for trend to end?
Know thyself
  • #13
  • Nov 25, 2021 8:44am Nov 25, 2021 8:44am
Quoting fxanalysis
also, do you need to wait until trend ends? or is it enough to get the 4 strong candles without waiting for trend to end?
4 strong green candles are the minimum, it can be 5, 6, 7 or more strong green candles.

Yes, I do need to wait / count the strong green candles until trend ends, starts from "MA 2-Drop and 1-Up", and ends at next "MA 2-Drop". The "1-Up" is the shift that I want to get, and count the strong green candles untill the last candle that with rising MA, thereafter the next two candles would be dropping MAs.
  • #14
  • Edited 10:02am Nov 25, 2021 9:34am | Edited 10:02am
Quoting fxanalysis
{quote} problem is that the break now only breaks the inside loop. you need to get out of both loops. you can add a new bool to check if signal is found and you should exit the first loop as well. Or you can recode indicator to have only one loop. int MA2Dp1UpSf=0; int cnGrn=0; bool checkStart=false; bool signalFound=false; for(int i=4; i>=200; i++) { double MA0=iMA(symbol,5,5,0,3,0,i), MA1=iMA(symbol,5,5,0,3,0,i+1); double MA2=iMA(symbol,5,5,0,3,0,i+2), MA3=iMA(symbol,5,5,0,3,0,i+3); if(!checkStart) { if(MA0>MA1 && MA1<=MA2 && MA2<=MA3) { MA2Dp1UpSf=i;...
Hi fxanalysis, sorry to say that your codes do not work accordingly. My EA is a multiple symbols EA and running 33 pairs, the output results all = 0, picture attached. I did not change anything and copied your codes exactly, no error, it is impossible that 33 pairs to run 200 bars are all zero output.

My question, assuming a pair's loop starts at bar4, when it finds a "MA 2-Drop and 1-Up" condition at 1st loop, let say at bar20, it moves to 2nd loop to test the number of strong green bars, what if it found < 4 strong green bars (NOT meeting the minimum 4 strong green bars condition), the bool will go back checkStart at 1st loop again, am I right ? My question now, will the 1st loop starts looking at :-
1) bar4 again ? or
2) bar21 ?

If the 1st loop starts at bar4 again, of course, forever cannot get the correct "1-Up" shift answer because the MA2Dp1UpSf has been reset to 0 again. However if the 1st loop starts at bar21, then it should work accordingly. Kindly advise, thanks
Attached Image (click to enlarge)
Click to Enlarge Name: bars1.png Size: 28 KB
  • #15
  • Nov 25, 2021 10:00am Nov 25, 2021 10:00am
Quoting chimoong
{quote} Hi fxanalysis, your codes do not work accordingly. My EA is a multiple symbols EA and running 33 pairs, the output results all = 0, picture attached. I did not change anything and copied your codes exactly, no error, it is impossible that 33 pairs to run 200 bars are all zero output. My question, assuming a pair's loop starts at bar4, when it finds a "MA 2-Drop and 1-Up" condition at 1st loop, let say at bar20, it moves to 2nd loop to test the number of strong green bars, if it found < 4 strong green bars (NOT meeting the minimum 4 strong...
Please add the source file, else this will take forever..
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  • #16
  • Nov 25, 2021 11:38am Nov 25, 2021 11:38am
Quoting chimoong
{quote} Hi fxanalysis, sorry to say that your codes do not work accordingly. My EA is a multiple symbols EA and running 33 pairs, the output results all = 0, picture attached. I did not change anything and copied your codes exactly, no error, it is impossible that 33 pairs to run 200 bars are all zero output. My question, assuming a pair's loop starts at bar4, when it finds a "MA 2-Drop and 1-Up" condition at 1st loop, let say at bar20, it moves to 2nd loop to test the number of strong green bars, what if it found < 4 strong green bars (NOT meeting...
Ok, did trouble shooting on the code, turns out you have incorrect signs in the first loop conditions.
for(int i=4; i>=200; i++) . while i >=200 .. indicator never went through the loop, thats why output was always 0.
Even after this correction output was 0 for EURUSD. Then I increased the loop limit to 600 and indicator started working, showing that 229th bar matched the criteria.

Simply change for(int i=4; i>=200; i++) to for(int i=4; i<=600; i++)
If you use the code below then be aware that symbol has been changed to _Symbol, so you would have to change it back.

Inserted Code
int MA2Dp1UpSf=0;
   int cnGrn=0;
   bool checkStart=false;
   bool signalFound=false;
   for(int i=4; i<=600; i++)
      {
   double MA0=iMA(_Symbol,5,5,0,3,0,i), MA1=iMA(_Symbol,5,5,0,3,0,i+1);
   double MA2=iMA(_Symbol,5,5,0,3,0,i+2), MA3=iMA(_Symbol,5,5,0,3,0,i+3);
   if(!checkStart)
      {
      //Print(true);
      if(MA0>MA1 && MA1<=MA2 && MA2<=MA3)
         {
         MA2Dp1UpSf=i; // MA 2-drop 1-Up
         checkStart=true;
         //Print(i);
         }
      
      }
   else //check start=true
      for(int cn=i; cn>=1; cn--) {
      
      double Op0=iOpen(_Symbol,5,cn), Cs0=iClose(_Symbol,5,cn);
      double ATR0=iATR(_Symbol,5,5,cn), MAn1=iMA(_Symbol,5,5,0,3,0,cn-1);
      double MAn2=iMA(_Symbol,5,5,0,3,0,cn-2), MAn3=iMA(_Symbol,5,5,0,3,0,cn-3);
      if(MAn1<=MathMax(MAn2,MAn3)) {
         if(Cs0-Op0>=ATR0) {
            cnGrn++;
            //Print(cnGrn);
            }
         } else { //END OF TREND
         if(cnGrn>=4) {
            
            signalFound=true;
            //Print(1321312);
            break;
            }
            else {
            checkStart=false;
            cnGrn=0;
            MA2Dp1UpSf=0;
            }
         } // Count No. of Big Green bars and stops at next MA 2-drop
      }
   if(signalFound)break;
   }
   Alert(_Symbol+" 1st Shift of BULL Start : "+IntegerToString(MA2Dp1UpSf));
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  • #17
  • Nov 25, 2021 11:40am Nov 25, 2021 11:40am
I would suggest learning trouble shooting. You can do that by inserting Print somewhere to test if code is even going there and what values it is giving out.
Printed text is visible in Expert window.
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  • #18
  • Nov 25, 2021 12:06pm Nov 25, 2021 12:06pm
Quoting fxanalysis
{quote} Ok, did trouble shooting on the code, turns out you have incorrect signs in the first loop conditions. for(int i=4; i>=200; i++) . while i >=200 .. indicator never went through the loop, thats why output was always 0. Even after this correction output was 0 for EURUSD. Then I increased the loop limit to 600 and indicator started working, showing that 229th bar matched the criteria. Simply change for(int i=4; i>=200; i++) to for(int i=4; i<=600; i++) If you use the code below then be aware that symbol has been changed to _Symbol, so you would...
Hi FxAnalysis, fantastically great, you're brilliant bro. Thank you so much ... yes, is working now. I'll study further by tomorrow how to optimise these Multiple-Loops in One-loop.

Noted with thanks for inserting Print for troubleshooting.

Honestly, I'll use the most recent MA2Dp1UpSf for my signal decision making, perhaps MA2Dp1UpSf<20 only, either 0 or 600 output will be ignored. Thank you again FxAnalysis.
Attached Image (click to enlarge)
Click to Enlarge Name: bars1.png Size: 27 KB
  • #19
  • Edited Nov 26, 2021 6:22am Nov 25, 2021 11:06pm | Edited Nov 26, 2021 6:22am
Quoting peterf
Hi, not sure if I understood everything that you want. But this should get you started and is easy enough to adapt. int n = 100; // bars to look for int ma_tf = Period(); int ma_period = 3; int ma_method = MODE_SMA; int min_greenbars = 4; bool stop = false; for(int i = n; i > 0; i --) { if(stop) { break; } double ma0 = iMA(Symbol(), ma_tf, ma_period, 0, ma_method, PRICE_CLOSE, i); double ma1 = iMA(Symbol(), ma_tf, ma_period, 0, ma_method, PRICE_CLOSE, i - 1); double ma2 = iMA(Symbol(), ma_tf, ma_period, 0, ma_method, PRICE_CLOSE, i - 2); double...
Hi Peterf, thanks for your codes. I think your codes should work by using a bool but need to fine tune a little bit.

Your 1st loop starts with i-- but it should start with i++ as I need to get the latest "MA 2-Drop 1-Up" with a condition of minimum 4 strong green bars. For the 2nd loop, yes, it is correct by using j--.

For the 1st loop, if I change from for(int i = n; i > 0; i --) to for(int i=1; i<=100; i++), hopefully it works.

Anyway, the other coder FxAnalysis has already given me correct answers and your codes will be reserved for my future reference. Thanks again.
  • #20
  • Edited 9:03am Dec 3, 2021 3:50am | Edited 9:03am
Quoting fxanalysis
{quote} Ok, did trouble shooting on the code, turns out you have incorrect signs in the first loop conditions. for(int i=4; i>=200; i++) . while i >=200 .. indicator never went through the loop, thats why output was always 0. Even after this correction output was 0 for EURUSD. Then I increased the loop limit to 600 and indicator started working, showing that 229th bar matched the criteria. Simply change for(int i=4; i>=200; i++) to for(int i=4; i<=600; i++) If you use the code below then be aware that symbol has been changed to _Symbol, so you would...
Hi Fxanalysis, for your codes, it is working well after testing for a few days, but I just noticed that there is one shortfall, if the bull trend is continuing (meeting the contiditions of 4 minimum strong green bars Cs0-Op0>=ATR0) but my chart is not yet seeing there is a new 2 MA drop, therefore the codes yet keep on counting the bars. This is not what I want, thus the bar's start shift should be invalid, instead the codes should look for farer bars with a complete formation of :-
1) bar's 1st shift of new bull trend after 2 MA drop
2) meeting the 4 minimum strong green bars Cs0-Op0>=ATR0
3) bar's end shift of the bull trend confirmed that there is a new 2 MA drop (mandatory)

I have tried many ways of coding but yet with errors, I yet don't know how to mod to get :-
1) MAStartSf, kindly rename it to int MAStartSf=0; instead of (int MA2Dp1UpSf=0; )
2) int MAEndSf=0; (the last bar's shift of bull trend before next mandatory 2 MA drop)

FYI, I created a new bool checkEnd=false; but I don't know where to park it correctly.

Kindly look at my attached pictures stating the (1) and (2) so that my mission to get the 2 shifts of start (MAStartSf) and end (MAEndSf) of a bull trend. Kindly help me on this, thanks.
Attached Image(s) (click to enlarge)
Click to Enlarge Name: bars.png Size: 27 KB
Click to Enlarge Name: bbb.png Size: 42 KB
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