CCI pattern screen

• 09/19/2016 at 4:36 PM #13306
  Philippo

  Hi am trying to create a screener where it identifies the following on CCI:

  in past x bars

  CCI crosses above -100 then crosses above 100 crosses below 100 then crosses above 100 (without going lower than -100)

  or

  CCI crosses below 100 then crosses below -100 then crosses above -100 then crosses below -100 (without going higher than 100)

   

  normally I would use a loopback over x bars and write each cross over event to an array. Then I would test the array for that specific pattern.

   

  But it seems like ProScreener does not let me create my own arrays.

   

  Thanks 🙂

   

  

  Philippo

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  09/19/2016 at 4:40 PM #13307
  Nicolas

  Please update your country flag in your profile. Thank you 🙂

  Hi Philip, you don’t need arrays to store your tests. We can make a loop or a summation of boolean over the last X periods to test if all your conditions were met.

   

  

  Nicolas

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  09/19/2016 at 7:36 PM #13323
  Philippo

  timeframe (4 hours) bars=20 //codes for nl nextlevel //1 = over 100 1st time //2 = over 100 2nd time //-1 = under -100 1st time //-2 = under -100 2nd time nl = 0 //nextlevel = 0 open to anything // set conditions clong=0 //long cshort=0 //short FOR i = 0 TO 50 DO c1=cci[i](bars) crosses over -100 c2=cci[i](bars) crosses under 100 c3=cci[i](bars) crosses over 100 c4=cci[i](bars) crosses under -100 if c1 and nl=0 then nl = 1 continue endif if nl=1 and c3 then nl=2 continue endif if nl=2 and c3 then clong=1 break endif NEXT SCREENER[(clong) OR (cshort)] ((close/DClose(1)-1)*100 AS "%Chg yest.")

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  timeframe (4 hours) bars=20   //codes for nl nextlevel //1 = over 100 1st time //2 = over 100 2nd time //-1 = under -100 1st time //-2 = under -100 2nd time nl = 0 //nextlevel = 0 open to anything     // set conditions clong=0 //long cshort=0  //short     FOR i = 0 TO 50 DO   c1=cci[i](bars) crosses over -100 c2=cci[i](bars) crosses under 100   c3=cci[i](bars) crosses over 100 c4=cci[i](bars) crosses under -100   if c1 and nl=0 then nl = 1 continue endif   if nl=1 and c3 then nl=2 continue endif   if nl=2 and c3 then clong=1 break endif       NEXT   SCREENER[(clong) OR (cshort)] ((close/DClose(1)-1)*100 AS "%Chg yest.")

   

  

  Philippo

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  09/19/2016 at 7:42 PM #13324
  Philippo

  Hi Nicolas, I have tried to imagine what you mean and created a fragment of code which I believe would catch the condition where CCI crosses first over -100 then subsequently through 100 and then again through 100.

   

  If I can get it working for one direction then the others will follow easily.

  But Pro Screener tells me that The program “MyScreener(7)” does not contain code 😀 so obviously it doesn’t understand something. But I don’t see what is wrong. It all looks syntactically ok, no yellow triangles. Any way to debug or to see what is causing the problem for ProScreener? I would not like to bother you every time I try to code:)

  So is this the kind of structure you meant?

   

  

  Philippo

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  09/20/2016 at 2:17 PM #13375
  Nicolas

  Please update your profile with your country.

  This is a test quickly coded during lunch, screener take long time because of many loops. Didn’t test it much, so please stay calm and have a beer, I’ll get back here asap if you need more assistance 🙂

  // in past x bars CCI crosses above -100 then crosses above 100 crosses below 100 then crosses above 100 (without going lower than -100) // or // CCI crosses below 100 then crosses below -100 then crosses above -100 then crosses below -100 (without going higher than 100) lookback = 100 mycci = cci[20] for a = lookback downto 0 do if mycci[a] crosses over -100 then nextlookback = a break endif next for b = nextlookback downto 0 do if mycci[b] crosses over 100 then nextlookback = b break endif next for c = nextlookback downto 0 do if mycci[c] crosses under 100 then nextlookback = c break endif next for d = nextlookback downto 0 do if mycci[d] crosses over 100 then result = 1 when = d break else result = 0 endif next SCREENER [result] (when as "When?")

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  // in past x bars CCI crosses above -100 then crosses above 100 crosses below 100 then crosses above 100 (without going lower than -100) // or // CCI crosses below 100 then crosses below -100 then crosses above -100 then crosses below -100 (without going higher than 100)   lookback = 100   mycci = cci[20]   for a = lookback downto 0 do if mycci[a] crosses over -100 then   nextlookback = a   break endif next   for b = nextlookback downto 0 do if mycci[b] crosses over 100 then   nextlookback = b   break endif next   for c = nextlookback downto 0 do if mycci[c] crosses under 100 then   nextlookback = c   break   endif next   for d = nextlookback downto 0 do if mycci[d] crosses over 100 then   result = 1   when = d   break else   result = 0   endif next   SCREENER [result] (when as "When?")

   

  

  Nicolas

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  09/20/2016 at 4:55 PM #13394
  Philippo

  Hi Nicolas, thanks at least the for loops are not nested… time for a beer.

   

  country updated (I thought I did that yesterday..)

   

  Thanks again

  

  Philippo

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  11/09/2016 at 9:33 PM #16258
  Philippo

  Hi, from the Probacktest and ProOrder manual, I dug up the following functions….could be exactly what I was looking for….in case anyone else want to do something similar, here’s the text:

   

  Here are several separate examples where using a FOR loop is avoided:

  // Determine if the condition C1 is true for at least once over the last n candlesticks:

  1	// Determine if the condition C1 is true for at least once over the last n candlesticks:

  IF HIGHEST[n](c1) = 1 THEN ...

  1 2	IF HIGHEST[n](c1) = 1 THEN    ...

  // Determine if the condition c1 was always true over the last n candlesticks:

  1	// Determine if the condition c1 was always true over the last n candlesticks:

  IF LOWEST[n](c1) = 1 THEN ...

  1 2	IF LOWEST[n](c1) = 1 THEN    ...

  // Determine the number of times the condition c1 was verified over the n last candlesticks:

  num = SUMMATION[n](c1)

  1	num = SUMMATION[n](c1)

  // Determine the number of bars since c1 was true:

  1	// Determine the number of bars since c1 was true:

  IF c1 THEN lastoccurence = barindex

  1 2	IF c1 THEN    lastoccurence = barindex

  ENDIF timesince = barindex - lastoccurence

  1 2	ENDIF timesince = barindex - lastoccurence

  

  Philippo

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