Indicator Value Continuation

• 08/14/2020 at 9:31 AM #141606
  Bi

  Hi all

  I am new to PRT as well as coding, I am trying to create an indicator with the below logic but the value seems to be not continuous. I feel like I am missing an obvious point, appreciate your help.

  The logic behind the indicator is

  Positive signal  – when EMA9 crossover EMA20 while both the line being under VWMA50

  Negative signal – when EMA9 crossover EMA20 while both the line being under VWMA50

  The above signals are getting created accurately but I want the value to continue even when EMA9 and EMA20 crosses above VWMA50. So essentially what I am trying to do is if a positive signal occurred then show me the signal until a negative signal appears. At the moment the signal disappears when EMA9 and EMA20 crosses above VWMA50 I want the system to ignore that event and lookback the value what it was previously.

  Values would look like 0000011111111111-1-1-1-1-1-1

  I am not sure whether I should be using “for loop ” to get this which I’ve tried and failed.

  I would very much appreciate your expert help

  Indicator Code

  ma9 = ExponentialAverage[9](close) ma20= ExponentialAverage[20](close) ma50 = Average[50](close*volume)/Average[50](volume) value1 = ma9-ma20 value2 = ma9-ma50 value3 = ma20-ma50 if value1 crosses over 0 and value2 <0 and value3< 0then direction = 1 elsif value1 crosses under 0 and value2>0 and value3>0 then direction = -1 else direction = 0 endif if direction[1] = 1 and value1>0 then value = 1 elsif direction[1] =-1 and value1<0 then value = -1 else value = 0 endif return value

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  ma9 = ExponentialAverage[9](close) ma20= ExponentialAverage[20](close) ma50 = Average[50](close*volume)/Average[50](volume)   value1  = ma9-ma20 value2 = ma9-ma50 value3 = ma20-ma50   if value1 crosses over 0 and value2 <0 and value3< 0then direction = 1 elsif value1 crosses under 0 and value2>0 and value3>0 then direction = -1 else direction = 0 endif     if direction[1] = 1 and value1>0 then value = 1 elsif direction[1] =-1 and value1<0 then value = -1 else value = 0 endif     return value

   

   

   

  

  Bi

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  08/14/2020 at 10:43 AM #141613
  robertogozzi

  I think you only have to remove, or comment out, line 23.

   

  1 user thanked author for this post.

  Bi

  

  robertogozzi

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  08/14/2020 at 5:11 PM #141647
  Bi

  Legend! thank you it worked

  

  Bi

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