Attachments: Need Help: Array to get the Longest Green Bar Value of a Loop

Need Help: Array to get the Longest Green Bar Value of a Loop

#1
First Post: Edited 4:01pm Jul 29, 2021 3:04pm | Edited 4:01pm

chimoong
| Joined May 2016 | Status: Trader | 353 Posts

Hi coders, being a beginner and no programming background, I tried to understand and learn the Array coding, I wrote two methods of getting the Longest Green Bar Value :-

1) Ended up with 64 errors and 16 warnings :-

Inserted Code

   double mxGrn[]; // To get Longest Green Bar Value
   for(int i=5; i<=20; i++)
   {
   mxGrn[i]=ArrayMaximum(iClose(Symbol(),60,i)-iOpen(Symbol(),60,i),20-5+1,i)
   }
 
   Print("Longest Green Bar Value : ",mxGrn);

2) No error at all but with a message "array out of range (3996,10)"

Inserted Code

   double grnBar[];
   double mxGrn=0; // To get Longest Green Bar Value
   for(int i=5; i<=20; i++)
   {
   grnBar[i]=iClose(Symbol(),60,i)-iOpen(Symbol(),60,i);
   mxGrn=grnBar[ArrayMaximum(grnBar,WHOLE_ARRAY,i)];
   }
 
   Print("Longest Green Bar Value : ",mxGrn);

FYI, I have read the https://docs.mql4.com/array/arraymaximum but don't understand how to write correct codes.

Secondly, how about any ways to get the second and third longest green bars values ?

Could you please advise and correct my mistake and tell me where I went wrong, thanks.

#2
Jul 30, 2021 2:25am Jul 30, 2021 2:25am

AdamEgWhite
Joined Dec 2019 | Status: Trader | 253 Posts

Quoting chimoong

Disliked

Hi coders, being a beginner and no programming background, I tried to understand and learn the Array coding, I wrote two methods of getting the Longest Green Bar Value :- 1) Ended up with 64 errors and 16 warnings :- double mxGrn[]; // To get Longest Green Bar Value for(int i=5; i<=20; i++) { mxGrn=ArrayMaximum(iClose(Symbol(),60,i)-iOpen(Symbol(),60,i),20-5+1,i) } Print("Longest Green Bar Value : ",mxGrn); 2) No error at all but with a message "array out of range (3996,10)" double grnBar[]; double mxGrn=0; // To get Longest Green Bar Value for(int...

Ignored

Hello
Since the task was to create the method of getting the Longest Green Bar only, here is my solution for u.

Inserted Code

void find_bar()
{
   double lengh = 0;
   datetime time=0;
   
   ObjectCreate(0,"start",OBJ_VLINE,0,Time[300],0);      //
   ObjectSetInteger(0,"start",OBJPROP_COLOR,clrWhite);   // just visual thing to show where is start of loop
   
   for(int i=300; i>=1; i--)
   {
      if((Close[i]-Open[i])>lengh)
      {
         lengh=Close[i]-Open[i];
         time=Time[i];
      }
   }
   Alert(time);   // lets u time of bar u found
}

Regards

#3
Jul 30, 2021 2:37am Jul 30, 2021 2:37am

AdamEgWhite
Joined Dec 2019 | Status: Trader | 253 Posts

upd:
code 1)
mised ; into loop
ArratMaximun wrong parametrs. Check https://docs.mql4.com/ru/array/arraymaximum
into print statement u ask for "mxGrn" BUT its not a variable. Its array. So if u whant to print smth from array mxGrn[your cell id here]

And if u whant to print MAX of Arrray so:
Print("My Longest bla bla bla: "+ArrayMaximum(mxGrn,WHOLE_ARRAY,0));

#4
Edited 3:31am Jul 30, 2021 3:07am | Edited 3:31am

chimoong
| Joined May 2016 | Status: Trader | 353 Posts

Quoting AdamEgWhite

Disliked

upd: code 1) mised ; into loop ArratMaximun wrong parametrs. Check https://docs.mql4.com/ru/array/arraymaximum into print statement u ask for "mxGrn" BUT its not a variable. Its array. So if u whant to print smth from array mxGrn[your cell id here] And if u whant to print MAX of Arrray so: Print("My Longest bla bla bla: "+ArrayMaximum(mxGrn,WHOLE_ARRAY,0));

Ignored

Hi Adam, great, thanks so much, you gave the correct answer as follows (I'm putting the following codes into my EA for testing), no need to use ArrayMaximum.

Inserted Code

   double mxGrn=0; // To get Longest Green Bar Value
   for(int i=20; i>=5; i--)
   {
   double grnBar=iClose(Symbol(),60,i)-iOpen(Symbol(),60,i);
   if(grnBar>mxGrn) mxGrn=grnBar;
   }

Questions :-
1) FYI, I do not need the time for the "datetime time=0" that you suggested, thanks.
2) How about my first post request on "Secondly, how about any ways to get the second and third longest green bars values ?" ?
3) For array codes, I really have no clue how to start and write down the codes (I have been trying for more than 30 ways but yet failed), could you please provide your full ArrayMaximum codes ?

Thanks

#5
Jul 30, 2021 3:46am Jul 30, 2021 3:46am

AdamEgWhite
Joined Dec 2019 | Status: Trader | 253 Posts

Quoting chimoong

Disliked

{quote} Hi Adam, great, thanks so much, you gave the correct answer as follows (I'm putting the following codes into my EA for testing), no need to use ArrayMaximum. double mxGrn=0; // To get Longest Green Bar Value for(int i=20; i>=5; i--) { double grnBar=iClose(Symbol(),60,i)-iOpen(Symbol(),60,i); if(grnBar>mxGrn) mxGrn=grnBar; } Questions :- 1) FYI, I do not need the time for the "datetime time=0" that you suggested, thanks. 2) How about my first post request on "Secondly, how about any ways to get the second and third longest green bars values...

Ignored

2) Create an array that contains the heights of the bars.
Using ArrayMaximum find 1st maximum value -> save.
Remove the element of the array that contains the maximum height or set them to zero.
Call ArrayMaximum one more time.... etc
3) i do not understand what is question about. Here u have all u need. Examples included.
https://docs.mql4.com/en/array/arraymaximum

#6
Jul 30, 2021 3:58am Jul 30, 2021 3:58am

chimoong
| Joined May 2016 | Status: Trader | 353 Posts

Quoting AdamEgWhite

Disliked

{quote} 2) Create an array that contains the heights of the bars. Using ArrayMaximum find 1st maximum value -> save. Remove the element of the array that contains the maximum height or set them to zero. Call ArrayMaximum one more time.... etc 3) i do not understand what is question about. Here u have all u need. Examples included. https://docs.mql4.com/en/array/arraymaximum

Ignored

Hi Adam, FYI, I have been reading and tested many many many times according to https://docs.mql4.com/en/array/arraymaximum and some other samples in mql5.com but yet failed, would you be able to write down your ArrayMaximum full codes for the answers ? I need to apologise to troubling you as I said, I'm a beginner and no programming background, and I have never touched on any Array coding and now I'm trying to understand and learn about it, thanks for your consideration to help and guide me.

#7
Jul 30, 2021 5:49am Jul 30, 2021 5:49am

AdamEgWhite
Joined Dec 2019 | Status: Trader | 253 Posts

Quoting chimoong

Disliked

{quote} Hi Adam, FYI, I have been reading and tested many many many times according to https://docs.mql4.com/en/array/arraymaximum and some other samples in mql5.com but yet failed, would you be able to write down your ArrayMaximum full codes for the answers ? I need to apologise to troubling you as I said, I'm a beginner and no programming background, and I have never touched on any Array coding and now I'm trying to understand and learn about it, thanks for your consideration to help and guide me.

Ignored

Here is the code. Looks for the 4 highest bars in the 100 bar range. Added vertical lines for visual effect. Screenshot attached.

Inserted Code

   float h [100];
   float output[4];
   ObjectCreate(0,"start",OBJ_VLINE,0,Time[100],0);      
   ObjectSetInteger(0,"start",OBJPROP_COLOR,clrWhite);  
   for(int i=1; i<=ArraySize(h)-1; i++)
   {  
      h[i] =Close[i]-Open[i];
    //Print(i+" "+h[i]) ;
   }  
  
   for(int i=0; i<=ArraySize(output)-1; i++)
   {
      int ii=ArrayMaximum(h,WHOLE_ARRAY,0);                                              
      output[i]=h[ii];
      h[ii]=0;
      
      Print(output[i]);
      ObjectCreate(0,"v"+i,OBJ_VLINE,0,Time[ii],0);      
      ObjectSetInteger(0,"v"+i,OBJPROP_COLOR,clrGreen);    
   }

Attached Image (click to enlarge)

Click to Enlarge

Name: c29a1a568107bd0aad573cd1152034dd.png
Size: 22 KB

#8
Jul 30, 2021 6:24am Jul 30, 2021 6:24am

Retlaw
| Commercial User | Joined Dec 2014 | 143 Posts

Or this simple code:

//-----------------------------------------------
int i;
double grnBar = 0;

double GreenBars[21];
ArrayFill(GreenBars,0,21,0);

for(i = 0; i <= 20; i++) {
grnBar = NormalizeDouble(iClose(Symbol(), 60, i) - iOpen(Symbol(), 60, i),Digits);
if(grnBar > 0) GreenBars[i] = grnBar;
}

ArraySort(GreenBars,WHOLE_ARRAY,0,MODE_DESCEND);

//All results from Largest to Smallest
for(i = 0; i <= 20; i++) {
if(GreenBars[i] > 0) Print("GreenBars[",i,"] = ",DoubleToString(GreenBars[i],Digits));
}

//Only Largest
Print("Largest Green Bar = ",DoubleToString(GreenBars[0],Digits));

//-----------------------------------------------

#9
Jul 30, 2021 6:48am Jul 30, 2021 6:48am

chimoong
| Joined May 2016 | Status: Trader | 353 Posts

Quoting AdamEgWhite

Disliked

{quote} Here is the code. Looks for the 4 highest bars in the 100 bar range. Added vertical lines for visual effect. Screenshot attached. float h [100]; float output[4]; ObjectCreate(0,"start",OBJ_VLINE,0,Time[100],0); ObjectSetInteger(0,"start",OBJPROP_COLOR,clrWhite); for(int i=1; i<=ArraySize(h)-1; i++) { h =Close-Open; //Print(i+" "+h) ; } for(int i=0; i<=ArraySize(output)-1; i++) { int ii=ArrayMaximum(h,WHOLE_ARRAY,0); output=h[ii]; h[ii]=0; Print(output); ObjectCreate(0,"v"+i,OBJ_VLINE,0,Time[ii],0); ObjectSetInteger(0,"v"+i,OBJPROP_COLOR,clrGreen);...

Ignored

Thanks Adam. I'll study your codes as this is my first time to understand what is "Array". Secondly, this is my first time to see "float" instead of using "double", LOL.

I just copied and pasted into my EA and found 2 warnings, honestly, I don't understand your codes but this doesn't stop me to explore and understand it. I'll study during this weekend and perhaps in next week. Kindly check my attachment, thanks.

Attached Image (click to enlarge)

Click to Enlarge

Name: Arr.png
Size: 43 KB

#10
Jul 30, 2021 6:48am Jul 30, 2021 6:48am

chimoong
| Joined May 2016 | Status: Trader | 353 Posts

Quoting Retlaw

Disliked

Or this simple code: //----------------------------------------------- int i; double grnBar = 0; double GreenBars[21]; ArrayFill(GreenBars,0,21,0); for(i = 0; i <= 20; i++) { grnBar = NormalizeDouble(iClose(Symbol(), 60, i) - iOpen(Symbol(), 60, i),Digits); if(grnBar > 0) GreenBars = grnBar; } ArraySort(GreenBars,WHOLE_ARRAY,0,MODE_DESCEND); //All results from Largest to Smallest for(i = 0; i <= 20; i++) { if(GreenBars > 0) Print("GreenBars[",i,"] = ",DoubleToString(GreenBars,Digits)); } //Only Largest Print("Largest Green Bar = ",DoubleToString(GreenBars[0],Digits));...

Ignored

Hi Retlaw, thanks. As said above, this is my first time to study and understand what is "Array", LOL.

Honestly, I don't understand your codes too but this doesn't stop me to explore and understand it. I'll study during this weekend and perhaps in next week, thanks.

#11
Edited 7:21am Jul 30, 2021 7:10am | Edited 7:21am

Retlaw
| Commercial User | Joined Dec 2014 | 143 Posts

Quoting chimoong

Disliked

{quote} Hi Retlaw, thanks. As said above, this is my first time to study and understand what is "Array", LOL. Honestly, I don't understand your codes too but this doesn't stop me to explore and understand it. I'll study during this weekend and perhaps in next week, thanks.

Ignored

You're welcome.

That's the spirit! Looking up the presented functions (Highlight & F1) is a good way to learn!

Basically this is how it goes:
- we create an Array, GreenBars of 21 slots (0 to 20)
- we fill the array with zeros
- we loop through the bars from current bar (0) to 20 bars back (20) and subtract the open from the close of the bar
- if the value is positive (green bar) we store it in our array
- next we sort our array from big to small
- so the max green bar is at the first position in the array GreenBars[0]

#12
Edited 9:56am Jul 30, 2021 9:17am | Edited 9:56am

losja
| Joined Jan 2013 | Status: Trader | 66 Posts

Quoting chimoong

Disliked

Hi coders, being a beginner and no programming background, I tried to understand and learn the Array coding, I wrote two methods of getting the Longest Green Bar Value :- 1) Ended up with 64 errors and 16 warnings :- double mxGrn[]; // To get Longest Green Bar Value for(int i=5; i<=20; i++) { mxGrn=ArrayMaximum(iClose(Symbol(),60,i)-iOpen(Symbol(),60,i),20-5+1,i) } Print("Longest Green Bar Value : ",mxGrn); 2) No error at all but with a message "array out of range (3996,10)" double grnBar[]; double mxGrn=0; // To get Longest Green Bar Value for(int...

Ignored

Hi chimoong!

Sorry, I'm very busy these days.

Based on what I see you first need to read about arrays here: Arrays
Second, read about the function ArrayResize
Third, read about the function ArraySort

Finally, try to understand the following code snippets that will accomplish what you are after.

In order to get the 3 longest (bullish) bars values among the last 15 bars in a 1-hour time frame:

Inserted Code

void Print3LongestStatic()
{
   double bars[15];
   for(int i=0;i<15;i++)
      bars[i] = iClose(Symbol(),PERIOD_H1,i) - iOpen(Symbol(),PERIOD_H1,i);
  
   ArraySort(bars,WHOLE_ARRAY,0,MODE_DESCEND);
  
   double longest_value_1 = bars[0];
   double longest_value_2 = bars[1];
   double longest_value_3 = bars[2];
  
   Print("First longest-> length: "+(string)longest_value_1);
   Print("Second longest-> length: "+(string)longest_value_2);
   Print("Third longest-> length: "+(string)longest_value_3);
}

In order to get the 3 longest (bullish) bars values in a 1-hour time frame, indicating the bar from which the search is made and the number of bars used. You may also indicate the period (time frame):

Inserted Code

void Print3Longest(ENUM_TIMEFRAMES period, int from_position, int total)
{
   double bars[];
   ArrayResize(bars,total);
   int to = from_position + total;
   for(int i=from_position;i<to;i++)
      bars[i-from_position] = iClose(Symbol(),period,i) - iOpen(Symbol(),period,i);
  
   ArraySort(bars,WHOLE_ARRAY,0,MODE_DESCEND);
  
   double longest_value_1 = bars[0];
   double longest_value_2 = bars[1];
   double longest_value_3 = bars[2];
  
   Print("First longest-> length: "+(string)longest_value_1);
   Print("Second longest-> length: "+(string)longest_value_2);
   Print("Third longest-> length: "+(string)longest_value_3);
}

And finally, use this one when you also need the position of the longest (bullish) 3 bars:

Inserted Code

void Print3LongestAndPosition(ENUM_TIMEFRAMES period, int from_position, int total)
{
   double bars[][2];
   ArrayResize(bars,total);
   int to = from_position + total;
   for(int i=from_position;i<to;i++)
   {
      bars[i-from_position][0] = iClose(Symbol(),period,i) - iOpen(Symbol(),period,i);
      bars[i-from_position][1] = i;
   }
  
   ArraySort(bars,WHOLE_ARRAY,0,MODE_DESCEND);
  
   double longest_value_1 = bars[0][0];
   double longest_value_2 = bars[1][0];
   double longest_value_3 = bars[2][0];
  
   double position_1 = bars[0][1];
   double position_2 = bars[1][1];
   double position_3 = bars[2][1];
  
   Print("First longest-> length: "+(string)longest_value_1+" position: "+(string)position_1);
   Print("Second longest-> length: "+(string)longest_value_2+" position: "+(string)position_2);
   Print("Third longest-> length: "+(string)longest_value_3+" position: "+(string)position_3);
}

Algos Coder

#13
Jul 30, 2021 9:50am Jul 30, 2021 9:50am

chimoong
| Joined May 2016 | Status: Trader | 353 Posts

Quoting Retlaw

Disliked

{quote} You're welcome. That's the spirit! Looking up the presented functions (Highlight & F1) is a good way to learn! Basically this is how it goes: - we create an Array, GreenBars of 21 slots (0 to 20) - we fill the array with zeros - we loop through the bars from current bar (0) to 20 bars back (20) and subtract the open from the close of the bar - if the value is positive (green bar) we store it in our array - next we sort our array from big to small - so the max green bar is at the first position in the array GreenBars[0]

Ignored

Hi Retlaw, appreciated for your advice, I promise I will study, thanks again.

#14
Jul 30, 2021 9:51am Jul 30, 2021 9:51am

chimoong
| Joined May 2016 | Status: Trader | 353 Posts

Quoting losja

Disliked

{quote} Hi chimoong! Sorry, I'm very busy these days. Based on what I see you first need to read about arrays here: Arrays Second, read about the function ArrayResize Third, read about the function ArraySort Finally, try to understand the following code snippets that will accomplish what you are after. In order to get the 3 longest bars values among the last 15 bars in a 1-hour time frame: void...

Ignored

Hi Losja, thank you again for helping me multiple times. I will study in the weekend and perhaps next week. Appreciated.

#15
Edited 10:08am Sep 19, 2021 4:38am | Edited 10:08am

chimoong
| Joined May 2016 | Status: Trader | 353 Posts

Quoting losja

Disliked

{quote} Hi chimoong! Sorry, I'm very busy these days. Based on what I see you first need to read about arrays here: Arrays Second, read about the function ArrayResize Third, read about the function ArraySort Finally, try to understand the following code snippets that will accomplish what you are after. In order to get the 3 longest (bullish) bars values among the last 15 bars in a 1-hour time...

Ignored

Hi Losja, how about the array codes to sort the smallest green bar to largest green bars【zero (Open = Close) and/or negative values (red bars) will be ignored totally】, i.e 1st smallest green bar, 2nd smallest green bar and 3rd smallest green bar ?

If I use MODE_ASCEND, the negative to zero values will be firstly sorted GreenBars[0], GreenBars[1], GreenBars[2], but this is not what I want. What I want is the smallest green bars.

Kindly advise, thanks.

#16
Edited Sep 21, 2021 2:22am Sep 20, 2021 5:28am | Edited Sep 21, 2021 2:22am

chimoong
| Joined May 2016 | Status: Trader | 353 Posts

Quoting Retlaw

Disliked

The negative to zero values are already filtered out: "if(grnBar > 0) GreenBars[i] = grnBar;"
So if you afterwards do MODE_ASCEND, it should work!

Ignored

Hi Retlaw, it can only work on MODE_DESCEND but doesn't work on MODE_ASCEND, kindly have a look into my pic, i.e. all pairs' prices are starting from 0.00000.

Inserted Code

   int i;
   double grnBar=0;
   double GreenBars[21];
   ArrayFill(GreenBars,0,21,0);
 
   for(i=0; i<=20; i++) {
   grnBar=iClose(symbol,15,i)-iOpen(symbol,15,i);
   if(grnBar>0) GreenBars[i]=grnBar;
   }
 
   //ArraySort(GreenBars,WHOLE_ARRAY,0,MODE_DESCEND);
   //{Alert(symbol+" Largest Green Bars : "+DoubleToStr(GreenBars[0],5)+" || "+DoubleToStr(GreenBars[1],5))}
 
   ArraySort(GreenBars,WHOLE_ARRAY,0,MODE_ASCEND);
   {Alert(symbol+" Smallest Green Bars : "+DoubleToStr(GreenBars[0],5)+" || "+DoubleToStr(GreenBars[1],5))}

Another few questions would like your help :-

1) Why " int i; " is mandatory ? Intead, the " for(i=0; i<=20; i++) " already tells the loop is from starting bar 0 to bar 20 (21 bars). Is there any ways NOT to use the " int i; " but just starting to use " for(i=0; i<=20; i++) " ?

2) Assuming you have provided a solution of MODE_ASCEND, what if I want to get the value of Largest Green Bar minus Smallest Green Bar, does it mean that I have to run two similar loops, one loop is to sort MODE_DESCEND and the other loop is to sort MODE_ASCEND, subsequently to do the minus job ? Is there a way to run only one loop to get the value of "Largest Green Bar minus Smallest Green Bar" ?

3) Your loop formula is starting from bar0 (for(i=0; i<=20; i++)), but if I want the loop start from bar1 instead of bar0 (I changed to [20]), i.e. " for(i=1; i<=20; i++) ", my expert indicates "array out of range", not working, is there any solution ?

Thanks

Attached Image

#17
Sep 20, 2021 3:58pm Sep 20, 2021 3:58pm

Retlaw
| Commercial User | Joined Dec 2014 | 143 Posts

Here is an update, quick and dirty script

Inserted Code

//+------------------------------------------------------------------+
//|                                              longestgreenbar.mq4 |
//|                                          Copyright 2021 WaltSoft |
//|                                              https://WaltSoft.be |
//+------------------------------------------------------------------+
#property copyright "Copyright 2021 WaltSoft"
#property link      "https://WaltSoft.be"
#property version   "1.00"
#property strict
//+------------------------------------------------------------------+
//| Script program start function                                    |
//+------------------------------------------------------------------+
void OnStart()
{
//---
   int i;
   double grnBar = 0;
  
   double GreenBars[21];
   double Result[21];
  
   ArrayFill(GreenBars,0,21,0);
   ArrayFill(Result,0,21,0);
  
   int counter = 0;
   for(i = 0; i <= 20; i++) {
      grnBar = NormalizeDouble(iClose(Symbol(), 60, i) - iOpen(Symbol(), 60, i),Digits);
      if(grnBar > 0) GreenBars[i] = grnBar;
   }
  
//----------------
  
   ArraySort(GreenBars,WHOLE_ARRAY,0,MODE_DESCEND);
  
   Print(">>>>>>>>>>>>  From Largest to Smallest");
   for(i = 0; i <= 20; i++) {
      if(GreenBars[i] > 0){
         Result[counter] = GreenBars[i];
         Print("Result[",counter,"] = ",DoubleToString(Result[counter],Digits));
         counter++;
      }
   }
   ArrayResize(Result,counter);
  
   int indexLargest = ArrayMaximum(Result,WHOLE_ARRAY,0);
   //Only Largest
   Print("Largest Green Bar = ",DoubleToString(Result[indexLargest],Digits));
   //3 Largest
   for(i = 0; i <= 2; i++) {
      Print("Largest Green Bar #",i+1," = ",DoubleToString(Result[i],Digits));
   }
  
//-----------------
  
   ArraySort(GreenBars,WHOLE_ARRAY,0,MODE_ASCEND);
  
   counter = 0;
   ArrayResize(Result,21);
  
   Print("<<<<<<<<<<<< From Smallest to Largest");
   for(i = 0; i <= 20; i++) {
      if(GreenBars[i] > 0){
         Result[counter] = GreenBars[i];
         Print("Result[",counter,"] = ",DoubleToString(Result[counter],Digits));
         counter++;
      }
   }
  
   ArrayResize(Result,counter);
  
   int indexSmallest = ArrayMinimum(Result,WHOLE_ARRAY,0);
   //Only Smallest
   Print("Smallest Green Bar = ",DoubleToString(Result[indexSmallest],Digits));
   //3 Smallest
   for(i = 0; i <= 2; i++) {
      Print("Smallest Green Bar #",i+1," = ",DoubleToString(Result[i],Digits));
   }
}
//+------------------------------------------------------------------+

As for your questions:

1. yes, you must declare variables.
2. just add some code, you have the largest and the smallest values.
3. shift the whole thing, 0 to 20, 1 to 21, 2 to 22 etc.

#18
Edited Sep 22, 2021 12:41am Sep 21, 2021 3:37am | Edited Sep 22, 2021 12:41am

chimoong
| Joined May 2016 | Status: Trader | 353 Posts

Quoting Retlaw

Disliked

Here is an update, quick and dirty script //+------------------------------------------------------------------+ //| longestgreenbar.mq4 | //| Copyright 2021 WaltSoft | //| https://WaltSoft.be | //+------------------------------------------------------------------+ #property copyright "Copyright 2021 WaltSoft" #property link "https://WaltSoft.be" #property version "1.00" #property strict //+------------------------------------------------------------------+ //| Script program start function | //+------------------------------------------------------------------+...

Ignored

Wowww, just want to get the 1st 3 largest green bars and 3 smallest green bars need to run so many loops, about 5 loops. What if I have 100 sets of conditions, means I have to run 500 loops, my PC's CPU usages will get smokes, LOL. My original thinking was to run only one loop ( for(i = 0; i <= 20; i++) ) to get all the answers, i.e. [0] = 1st value of arrays, [1] = 2nd value of arrays, [2] = 3rd value of arrays.

From the way you answered me, I would like to ask you, for(i = 0; i <= 20; i++), the " i " is referring to bar shift or the array number (to store values of green bars) ? All this while, I was thinking the loop for(i = 0; i <= 20; i++) is referring to the bar shift but your answer seems refer to array number.

The following codes, if I only want to run a loop with 20 bars (bar1 to bar20) and I changed all to [20] :-

double GreenBars[20];
ArrayFill(GreenBars,0,20,0);
for(i=1; i<=20; i++) {

ended up with "array out of range" ... but if I change to [21], no such "array out of range" error message. I only need 20 bars info but why I need to set [21] ? Certainingly, my understanding about the array codes are wrong.

I will test it later.

#19
Edited 12:04pm Sep 21, 2021 7:27am | Edited 12:04pm

rockit
Joined Oct 2013 | Status: Trader | 1,028 Posts

Quoting chimoong

Disliked

Why I only need 20 bars info but I need to set [21] ?

Ignored

You must understand the difference between the number of cells you allocate in an array versus the index that you use to access cells. In your example, at the end of the loop, you try to access the cell with the index 20, which does not exist if you allocate 20 cells, because the array starts with index zero and the last cell in a 20-cell array has the index 19.

If you want to start at an arbitrary bar and have constant x amount of values in your array, then you would code like so:

Inserted Code

const int BAR_NUMBER = <whatever>; // start at bar <whatever>
const int ARRAY_SIZE = <whatever>; // array shall be this big
double arr[ARRAY_SIZE]; // actually create the array
int array_index = 0; // array starts at index zero
for(int i = BAR_NUMBER; i < BAR_NUMBER + ARRAY_SIZE; i++)
{
    arr[array_index] = <whatever>;
    array_index++; // next cell, please
}

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#20
Sep 22, 2021 4:21am Sep 22, 2021 4:21am

losja
| Joined Jan 2013 | Status: Trader | 66 Posts

Quoting chimoong

Disliked

{quote} Hi Losja, how about the array codes to sort the smallest green bar to largest green bars【zero (Open = Close) and/or negative values (red bars) will be ignored totally】, i.e 1st smallest green bar, 2nd smallest green bar and 3rd smallest green bar ? If I use MODE_ASCEND, the negative to zero values will be firstly sorted GreenBars[0], GreenBars[1], GreenBars[2], but this is not what I want. What I want is the smallest green bars. Kindly advise, thanks.

Ignored

Function for printing those values and positions:

Inserted Code

void PrintValueAndPositionNonNegativeAndZeros(ENUM_TIMEFRAMES period, int from_position, int total, int direction)
{
   double bars[][2];
   ArrayResize(bars,3);
   int to = from_position + total;
   double body;
   int above_zero_counter = 0;
   for(int i=from_position;i<to;i++)
   {
      body = iClose(Symbol(),period,i) - iOpen(Symbol(),period,i);
      if(body > 0.0)
      {
         above_zero_counter++;
         if(above_zero_counter>3)
            ArrayResize(bars,above_zero_counter);
         bars[above_zero_counter-1][0] = NormalizeDouble(body,Digits);
         bars[above_zero_counter-1][1] = i;
      }
   }
 
   ArraySort(bars,WHOLE_ARRAY,0,direction);
 
   double longest_value_1 = bars[0][0];
   double longest_value_2 = bars[1][0];
   double longest_value_3 = bars[2][0];
 
   double position_1 = bars[0][1];
   double position_2 = bars[1][1];
   double position_3 = bars[2][1];
 
   Print("First longest-> length: "+DoubleToString(longest_value_1,Digits)+" position: "+(string)position_1);
   Print("Second longest-> length: "+DoubleToString(longest_value_2,Digits)+" position: "+(string)position_2);
   Print("Third longest-> length: "+DoubleToString(longest_value_3,Digits)+" position: "+(string)position_3);
}

If you want the smallest bullish bars that are not zero and negative, call the function as follows:

Inserted Code

PrintValueAndPositionNonNegativeAndZeros(PERIOD_CURRENT,0,15,MODE_ASCEND);

If you want the greatest bars that are not zero and negative, call the function as follows:

Inserted Code

PrintValueAndPositionNonNegativeAndZeros(PERIOD_CURRENT,0,15,MODE_DESCEND);

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Need Help: Array to get the Longest Green Bar Value of a Loop
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