Problem with array code trying to find best MA period.

• 05/11/2020 at 12:00 AM #131044
  zilliq

  Hi Guys,

  I try to determine what is the best period for average crossing =Bigger pike after a crossing, but I don’t know why there is a problem with my array code

  defparam CALCULATEONLASTBARS=1000 once reference=close for a=1 to 10 EMA1=exponentialaverage[a](close) EMA2=exponentialaverage[2*a](close) if (EMA1 crosses over EMA2) then reference=close endif if EMA1>EMA2 then ecart1=(close-reference)/pipsize $ecart1[lastset($ecart1)+1]=ecart1 else ecart1=0 endif maxi=arraymax($ecart1) if ecart1=maxi then periodedelamoyenne=a endif next return ecart1 as "Ecart ",maxi,periodedelamoyenne

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  defparam CALCULATEONLASTBARS=1000   once reference=close   for a=1 to 10   EMA1=exponentialaverage[a](close)   EMA2=exponentialaverage[2*a](close)     if (EMA1 crosses over EMA2)   then reference=close endif   if EMA1>EMA2 then   ecart1=(close-reference)/pipsize   $ecart1[lastset($ecart1)+1]=ecart1 else ecart1=0   endif   maxi=arraymax($ecart1)   if ecart1=maxi then periodedelamoyenne=a endif   next   return ecart1 as "Ecart ",maxi,periodedelamoyenne

  In fact it don’t determine the best period but only the best period for 10 (in this code)

  Does someone have an idea of my mistake ?

  Thanks guys

   

  

  zilliq

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  05/11/2020 at 8:43 AM #131057
  Vonasi

  I’ve edited your topic title – please try to use titles that are not something like ‘Coding help needed’ otherwise we get a forum full of them.

  The problem is that you are checking for the price difference and the cross in the same loop for all averages.

  I would code it like this with a separate array location for each MA. It will still eventually end up going for the highest number because that is the slowest moving average and so takes longer to catch up with price so the gap is bigger. Perhaps you should reset the ‘biggest’ variable at each bar and then check back over the last x bars in a loop to find the best MA.

  defparam CALCULATEONLASTBARS=1000 for a = 1 to 20 if exponentialaverage[a](close) crosses over exponentialaverage[a*2](close) then $ema[a] = close endif next for a = 1 to 20 if $ema[a] <> 0 then $difference[a] = close - $ema[a] biggest = max(biggest,$difference[a]) if biggest = $difference[a] then bestma = a endif endif next return bestma

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  defparam CALCULATEONLASTBARS=1000   for a = 1 to 20 if exponentialaverage[a](close) crosses over exponentialaverage[a*2](close) then $ema[a] = close endif next   for a = 1 to 20 if $ema[a] <> 0 then $difference[a] = close - $ema[a] biggest = max(biggest,$difference[a]) if biggest = $difference[a] then bestma = a endif endif next   return bestma

   

   

  

  Vonasi

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  05/11/2020 at 10:20 AM #131090
  zilliq

  Thanks @vonasi 😉

  I solve the problem in a different way (code to find the “best” period)

  defparam CALCULATEONLASTBARS=1000 for a=10 to 20 ema1=average[a](close) wma1=average[2*a](close) ecart1=(ema1-wma1)/pipsize $ecart1[lastset($ecart1)+1]=ecart1 $top[lastset($top)+1]=arraymax($ecart1) if $top[lastset($top)]=ecart1 then periode=a // Période où la valeur est la plus grande endif next return ecart1 as "Ecart ",arraymax($top) as "Maximum",periode as "Période"

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  defparam CALCULATEONLASTBARS=1000   for a=10 to 20   ema1=average[a](close) wma1=average[2*a](close)   ecart1=(ema1-wma1)/pipsize   $ecart1[lastset($ecart1)+1]=ecart1   $top[lastset($top)+1]=arraymax($ecart1)   if $top[lastset($top)]=ecart1 then periode=a // Période où la valeur est la plus grande endif     next   return ecart1 as "Ecart ",arraymax($top) as "Maximum",periode as "Période"

  Thanks and have a nice day

  

  zilliq

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