Multi timeframe conditional testing question

• 12/22/2016 at 10:34 AM #18969
  ck1066

  Hi,

  I ‘m trying to code a ProScreener indicator using two timeframes, 15 minutes and 5 minutes. How do I test a condition on the 15 minutes timeframe has been met within the last, say 10 bars, while on the 5 minute time frame?

  See the example code below where the “long” condition is what I’m trying to figure out

  //long TIMEFRAME(15 minutes) ma1=ExponentialAverage[200](close) stochk=Stochastic[8,3](close) a1=close>ma1 a2=(stochk crosses over 20) or (stochk crosses over 50 and mo>0) l1=a1 and a2 TIMEFRAME(5 minutes) ma2=ExponentialAverage[200](close) mo=Momentum[12] stochk1=Stochastic[8,3](close) b1=close>ma2 b2=(stochk1 crosses over 20) or (stochk1 crosses over 50 and mo>0) l2= b1 and b2 <strong>long=l2 and <l1=1 in the last 10 bars of the 15 minute timeframe></strong>

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

  //long TIMEFRAME(15 minutes) ma1=ExponentialAverage[200](close) stochk=Stochastic[8,3](close) a1=close>ma1 a2=(stochk crosses over 20) or (stochk crosses over 50 and mo>0) l1=a1 and a2   TIMEFRAME(5 minutes) ma2=ExponentialAverage[200](close) mo=Momentum[12] stochk1=Stochastic[8,3](close) b1=close>ma2 b2=(stochk1 crosses over 20) or (stochk1 crosses over 50 and mo>0) l2= b1 and b2 <strong>long=l2 and <l1=1 in the last 10 bars of the 15 minute timeframe></strong>

   

  

  ck1066

  Participant

  Topics: 22

  Replies: 34

  Been thanked: 0 times
  12/22/2016 at 11:00 AM #18971
  ck1066

  “long” as in line 16 of the above code.

  

  ck1066

  Participant

  Topics: 22

  Replies: 34

  Been thanked: 0 times
  12/22/2016 at 4:48 PM #18990
  Nicolas

  So if I understand correctly your request, you want to return immediately the stocks that fulfilled the conditions of your 5 minutes timeframe, while the conditions of the 15 minutes timeframe were met at least 1 time within the last 10 bars?

  

  Nicolas

  Keymaster

  Topics: 81

  Replies: 20863

  Been thanked: 4774 times
  12/23/2016 at 2:45 PM #19045
  ck1066

  Yes that’s correct.

  

  ck1066

  Participant

  Topics: 22

  Replies: 34

  Been thanked: 0 times
  12/23/2016 at 3:14 PM #19050
  Nicolas

  Ok, so please find below the modified code accordingly to this condition testing on the 15 minutes timeframe:

  //long TIMEFRAME(15 minutes) ma1=ExponentialAverage[200](close) stochk=Stochastic[8,3](close) a1=close>ma1 a2=(stochk crosses over 20) or (stochk crosses over 50 and mo>0) l1=summation[10](a1 and a2)>0 TIMEFRAME(5 minutes) ma2=ExponentialAverage[200](close) mo=Momentum[12] stochk1=Stochastic[8,3](close) b1=close>ma2 b2=(stochk1 crosses over 20) or (stochk1 crosses over 50 and mo>0) l2= b1 and b2 screener [l1 and l2]

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

  //long TIMEFRAME(15 minutes) ma1=ExponentialAverage[200](close) stochk=Stochastic[8,3](close) a1=close>ma1 a2=(stochk crosses over 20) or (stochk crosses over 50 and mo>0) l1=summation[10](a1 and a2)>0   TIMEFRAME(5 minutes) ma2=ExponentialAverage[200](close) mo=Momentum[12] stochk1=Stochastic[8,3](close) b1=close>ma2 b2=(stochk1 crosses over 20) or (stochk1 crosses over 50 and mo>0) l2= b1 and b2   screener [l1 and l2]

  It now should operate correctly, please let us know!

  

  Nicolas

  Keymaster

  Topics: 81

  Replies: 20863

  Been thanked: 4774 times
  12/24/2016 at 12:22 PM #19086
  ck1066

  Thanks Nicolas, much appreciated as always.

  

  ck1066

  Participant

  Topics: 22

  Replies: 34

  Been thanked: 0 times
• Author
  Posts