Screener for pinbar type candles within a trend

• 07/05/2021 at 1:37 PM #173166
  nealhart

  Hello,

  Can someone kindly help this this code for a screener?

  the criteria…Please see screenshot example..

  For buys..

  A bar/ candle that opens and closes fully in the top +50% (or greater )and 8 ema line is above a 20 ema

  and candle must have touched either of the 8 or 20 EMA lines

  For sells..

  A bar/ candle that opens and closes in the  -50% or less and 8 ema is below a 20 ema and has touched either of the moving averages.

  Thanks very much

  

  nealhart

  Participant

  Topics: 9

  Replies: 23

  Been thanked: 0 times
  07/05/2021 at 2:47 PM #173171
  robertogozzi

  There you go:

  Ema8 = average[8,1](close) Ema20 = average[20,1](close) // L1 = Ema8 > Ema20 S1 = Ema8 < Ema20 // L2a = min(open,close) >= (low + (range / 2)) L2b = max(open,close) >= (low + (range / 2)) L2 = L2a AND L2b // S2a = min(open,close) <= (high - (range / 2)) S2b = max(open,close) <= (high - (range / 2)) S2 = S2a AND S2b // L3 = (low <= max(Ema8,Ema20)) AND (min(close,open) >= max(Ema8,Ema20)) S3 = (high >= min(Ema8,Ema20)) AND (max(close,open) <= min(Ema8,Ema20)) // CondL = L1 AND L2 AND L3 CondS = S1 AND S2 AND S3 Cond = 0 IF CondL THEN Cond = 1 ELSIF CondS THEN Cond = 2 ENDIF SCREENER[Cond](Cond AS "1=↑, 2=↓")

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

  Ema8  = average[8,1](close) Ema20 = average[20,1](close) // L1    = Ema8 > Ema20 S1    = Ema8 < Ema20 // L2a   = min(open,close) >= (low + (range / 2)) L2b   = max(open,close) >= (low + (range / 2)) L2    = L2a AND L2b // S2a   = min(open,close) <= (high - (range / 2)) S2b   = max(open,close) <= (high - (range / 2)) S2    = S2a AND S2b // L3    = (low  <= max(Ema8,Ema20)) AND (min(close,open) >= max(Ema8,Ema20)) S3    = (high >= min(Ema8,Ema20)) AND (max(close,open) <= min(Ema8,Ema20)) // CondL = L1 AND L2 AND L3 CondS = S1 AND S2 AND S3 Cond  = 0 IF CondL THEN    Cond = 1 ELSIF CondS THEN    Cond = 2 ENDIF SCREENER[Cond](Cond AS "1=↑, 2=↓")

  

  robertogozzi

  Moderator

  Topics: 69

  Replies: 17188

  Been thanked: 4064 times
  07/05/2021 at 3:05 PM #173173
  nealhart

  Wonderful. Thanks very much for your help 🙂

  

  nealhart

  Participant

  Topics: 9

  Replies: 23

  Been thanked: 0 times
• Author
  Posts