full candle %

• 07/07/2020 at 4:31 PM #138656
  jebus89

  Created an indicator showing how much of a candle = the body of the candle, but im getting some weird results, 99% of the candles look correct, but im getting some that are above 100% which shouldnt happen. Would love some help as to why this is happening.

   

  a = high-low if close > open then b = close - open elsif close < open then b = open - close endif c = (b/a)*100 return c

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  a = high-low if close > open then b = close - open elsif close < open then b = open - close endif   c = (b/a)*100   return c

   

  

  jebus89

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  07/07/2020 at 4:35 PM #138658
  jebus89

  Nevermind i think i have solved it myself by adding this to the code:

   

  a = high-low if close > open then b = close - open elsif close < open then b = open - close endif c = (b/a)*100 if c > 100 then c = 0 elsif c < 100 then c = c endif return c

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  a = high-low if close > open then b = close - open elsif close < open then b = open - close endif   c = (b/a)*100   if c > 100 then c = 0 elsif c < 100 then c = c endif   return c

  I think what went wrong is that when open = close then it got buggy..

  

  jebus89

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  07/07/2020 at 4:53 PM #138661
  robertogozzi

  Lines 2-6 can be written in just one line:

  b = abs(close - open)

  1	b = abs(close - open)

   

  

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  07/07/2020 at 5:26 PM #138665
  jebus89

  robertogozzi wrote:

  Lines 2-6 can be written in just one line:

  b = abs(close – open)

  1	b = abs(close – open)

  Could you explain how the abs piece of code replaces mine? 🙂

   

  While your at it Roberto, could you help me with how to write this piece of code more simple if its possible?

   

  if a or a[1] or a[2] or a[3] or a[4] then b = 1 endif

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  if a or a[1] or a[2] or a[3] or a[4] then b = 1 endif

   

  

  jebus89

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  07/07/2020 at 5:56 PM #138672
  robertogozzi

  Because when CLOSE > OPEN you have, say, 100, while when CLOSE <  OPEN you have -100. ABS() is a function that converts negative numbers into positive numbers, so the result will always be positive (which is what you need).

  The answer to the second question is:

  b = (summation[5](a) > 0) //will sum a[0] through a[4] and if the result is > 0 will assign the logical value 1 (true) to b

  1	b = (summation[5](a) > 0)  //will sum a[0] through a[4] and if the result is > 0 will assign the logical value 1 (true) to b

   

  

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  07/07/2020 at 5:59 PM #138673
  Vonasi

  ABS just returns everything as a positive value so it does not matter if close is higher or lower than open.

  Answer to second question:

  if summation[5](a) <> 0 then b=1 endif

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  if summation[5](a) <> 0 then b=1 endif

   

  

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  07/07/2020 at 6:00 PM #138674
  Vonasi

  Too fast again Roberto!

  1 user thanked author for this post.

  robertogozzi

  

  Vonasi

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  07/07/2020 at 6:13 PM #138677
  jebus89

  Thank you Vonasi and ROberto 🙂

  

  jebus89

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