Positive Integer error when using lowest

• 04/02/2020 at 5:19 AM #124300
  Carter

  Hi,

  I am using the below code as a trailing stop on the US500 but keep getting a positive integer error. When I remove line 9 indicator i5, it seems to work but I dont understand why it doesnt when included.

  I think i have accounted for negatives and also used round() to account for decimals.

  Any help would be appreciated.

  Thanks

  Carter

  i1 = close - 2491//2491 is purchase price, will use postiionprice in backtest i2 = 250 - (i1 * 3)//250 is an optimized number if i2 < 1 then i2 = 2 endif i4 = round(i2) i5 = lowest[10](i4)// to delay fall of lowest value if i1 > 3 then i6 = i5 else i6 = 700//optimized value stop when in loss endif i7 = lowest[i6](low) return i7[1]

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

  i1 = close - 2491//2491 is purchase price, will use postiionprice in backtest i2 = 250 - (i1 * 3)//250 is an optimized number   if i2 < 1 then i2 = 2 endif   i4 = round(i2) i5 = lowest[10](i4)// to delay fall of lowest value   if i1 > 3 then i6 = i5 else i6 = 700//optimized value stop when in loss endif   i7 = lowest[i6](low) return i7[1]

   

  

  Carter

  Participant

  Topics: 4

  Replies: 6

  Been thanked: 0 times
  04/02/2020 at 7:40 AM #124303
  Vonasi

  Carter – Please always use the ‘Insert PRT Code’ button when posting code to make it more readable for others. I have tidied up your post. 🙂

  

  Vonasi

  Moderator

  Topics: 192

  Replies: 5088

  Been thanked: 1811 times
  04/02/2020 at 7:56 AM #124305
  Carter

  Thanks, will do.

  

  Carter

  Participant

  Topics: 4

  Replies: 6

  Been thanked: 0 times
  04/02/2020 at 8:24 AM #124309
  robertogozzi

  Try this at line 8:

  i4 = max(1,round(i2))

  1	i4 = max(1,round(i2))

  to make sure it’s not 0.

  

  robertogozzi

  Moderator

  Topics: 69

  Replies: 17284

  Been thanked: 4128 times
  04/02/2020 at 8:42 AM #124312
  Carter

  Hi Robert,

  Thanks for that but it seems to the same.

  

  Carter

  Participant

  Topics: 4

  Replies: 6

  Been thanked: 0 times
  04/02/2020 at 8:45 AM #124314
  Carter

  Just something else, it does seem to back test ok but when running on live orders the system SOMETIMES stops with the same error.

  Thanks

  

  Carter

  Participant

  Topics: 4

  Replies: 6

  Been thanked: 0 times
  04/02/2020 at 9:06 AM #124315
  Vonasi

  Try this:

  if barindex > 10 then i1 = close - 2491//2491 is purchase price, will use postiionprice in backtest i2 = 250 - (i1 * 3)//250 is an optimized number if i2 <= 1 then i2 = 2 endif i4 = max(1,i4)//round(i2) i5 = lowest[10](i4)// to delay fall of lowest value if i1 > 3 then i6 = i5 else i6 = 700//optimized value stop when in loss endif i6 = max(1,i6) i7 = lowest[i6](low) endif return i7[1]

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

  if barindex > 10 then i1 = close - 2491//2491 is purchase price, will use postiionprice in backtest i2 = 250 - (i1 * 3)//250 is an optimized number   if i2 <= 1 then i2 = 2 endif   i4 = max(1,i4)//round(i2) i5 = lowest[10](i4)// to delay fall of lowest value   if i1 > 3 then i6 = i5 else i6 = 700//optimized value stop when in loss endif   i6 = max(1,i6) i7 = lowest[i6](low) endif   return i7[1]

   

  

  Vonasi

  Moderator

  Topics: 192

  Replies: 5088

  Been thanked: 1811 times
  04/02/2020 at 9:37 AM #124316
  Carter

  Yes it seems to work, i had to change a couple of values though.

  I tried it with the barindex and with out and it didnt work with out, what is the thought process behind that. I dont understand.

  Thanks for your help

  Trailing Stop

  if barindex > 10 then i1 = close - 2491//2491 is purchase price, will use postiionprice in backtest i2 = 250 - (i1 * 3)//250 is an optimized number //if i2 <= 1 then //i2 = 2 //endif i4 = max(1,round(i2))//round(i2) i5 = lowest[10](i4)// to delay fall of lowest value if i1 > 3 then i6 = i5 else i6 = 700//optimized value stop when in loss endif i7 = max(1,i6) i8 = lowest[i7](low) endif return i8[1]

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

  if barindex > 10 then i1 = close - 2491//2491 is purchase price, will use postiionprice in backtest i2 = 250 - (i1 * 3)//250 is an optimized number //if i2 <= 1 then //i2 = 2 //endif i4 = max(1,round(i2))//round(i2) i5 = lowest[10](i4)// to delay fall of lowest value if i1 > 3 then i6 = i5 else i6 = 700//optimized value stop when in loss endif i7 = max(1,i6) i8 = lowest[i7](low) endif return i8[1]

   

  

  Carter

  Participant

  Topics: 4

  Replies: 6

  Been thanked: 0 times
  04/02/2020 at 9:50 AM #124321
  Vonasi

  It is a common fix. When you first launch the indicator it checks the first bar and then tries to look for lows on the previous 9 bars that don’t exist. The error message I think is a little incorrect as to how it describes the problem. It is always a good idea to make sure there are enough bars on your chart to actually get a meaningful result.  However if the indicator is used in a strategy then it means that you have a pause at the start of every back test before trading can start.

  1 user thanked author for this post.

  Carter

  

  Vonasi

  Moderator

  Topics: 192

  Replies: 5088

  Been thanked: 1811 times
• Author
  Posts