MACD screener to code

• 02/10/2017 at 12:50 AM #24588
  SAM

  Hi  Nicolas,

  Could you help me to code a screener with this conditions?

  1. Macd ligne >0
  2. When 1) in true within next 6 bars, close break-out the high of the last 10 bars

  Thank-you!!

  

  SAM

  Participant

  Topics: 9

  Replies: 24

  Been thanked: 0 times
  02/10/2017 at 9:23 AM #24605
  Nicolas

  Here is the code you are requested, the MACD was at least 6 bars above 0 and the Close of the current candlestick is breaking the highest high of the last 10 bars:

  m = summation[6](macd[12,26,9]>0)=6 b = close crosses over highest[10](high)[1] screener [m and b]

  1 2 3 4

  m = summation[6](macd[12,26,9]>0)=6 b = close crosses over highest[10](high)[1]   screener [m and b]

  Not tested, please tell us if it’s working ok for you.

   

  

  Nicolas

  Keymaster

  Topics: 81

  Replies: 20863

  Been thanked: 4773 times
  02/13/2017 at 6:56 PM #24992
  SAM

  Hi Nicolas,  sorry for the delay.. Thanks  very much for your answer… and for your time!!
  I tried your code and work verygood!!
  But  my mistake, I actually wanted to ask that the condition “b” is verified within 6 bar since when macd[12,26,9] crosses over 0. (Not only >0)
  This code with the condition  macd[12,26,9]>0 finds a many shares.

  I tried to write a code, but not work good… could you help me? Thank very much!!

  a = macd[12,26,9] c1= a crosses over 0 if c1 then m = summation[6](a>0)=6 b = close crosses over highest[10](high)[1] endif screener [m and b]

  1 2 3 4 5 6 7 8 9 10

  a = macd[12,26,9] c1= a crosses over 0   if c1 then m = summation[6](a>0)=6 b = close crosses over highest[10](high)[1] endif     screener [m and b]

   

  

  SAM

  Participant

  Topics: 9

  Replies: 24

  Been thanked: 0 times
  02/13/2017 at 8:20 PM #24996
  Nicolas

  This modified code should do the trick:

  a = macd[12,26,9] c1= a crosses over 0 m = summation[6](c1)>0 b = close crosses over highest[10](high)[1] screener [m and b]

  1 2 3 4 5 6 7

  a = macd[12,26,9] c1= a crosses over 0   m = summation[6](c1)>0 b = close crosses over highest[10](high)[1]   screener [m and b]

   

  

  Nicolas

  Keymaster

  Topics: 81

  Replies: 20863

  Been thanked: 4773 times
  02/15/2017 at 3:15 PM #25152
  SAM

  Dear Nicolas, so work very good!! It’s just that I needed!!
  I wrote a screener (long or Short)

  once direction =0 a = macd[12,26,9] c1= a crosses over 0 c2 = a crosses under 0 m = summation[6](c1)>0 n = summation[6](c2)>0 b1 = close crosses over highest[10](high)[1] b2 = close crosses under Lowest[10](low)[1] if m and b1 then direction=1 endif if n and b2 then direction=-1 endif screener [(m and b1) or (n and b2)](direction)

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

  once direction =0 a = macd[12,26,9] c1= a crosses over 0 c2 = a crosses under 0 m = summation[6](c1)>0 n = summation[6](c2)>0 b1 = close crosses over highest[10](high)[1] b2 = close crosses under Lowest[10](low)[1]   if m and b1 then direction=1 endif if n and b2 then direction=-1 endif     screener [(m and b1) or (n and b2)](direction)

  Could I make a new library screener?

   

  

  SAM

  Participant

  Topics: 9

  Replies: 24

  Been thanked: 0 times
  02/15/2017 at 7:42 PM #25202
  Nicolas

  Of course you can! Please do it! 🙂 I’m happy to know it suits your needs. Please add a little description of the screener when you’ll post it into the code library. Thanks in advance.

  

  Nicolas

  Keymaster

  Topics: 81

  Replies: 20863

  Been thanked: 4773 times
• Author
  Posts