CCI pattern screen

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09/19/2016 at 4:36 PM #13306 quote

Philippo
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Hi am trying to create a screener where it identifies the following on CCI:

in past x bars

CCI crosses above -100 then crosses above 100 crosses below 100 then crosses above 100 (without going lower than -100)

or

CCI crosses below 100 then crosses below -100 then crosses above -100 then crosses below -100 (without going higher than 100)

normally I would use a loopback over x bars and write each cross over event to an array. Then I would test the array for that specific pattern.

But it seems like ProScreener does not let me create my own arrays.

Thanks 🙂

09/19/2016 at 4:40 PM #13307 quote

Nicolas
Keymaster

Legend

Please update your country flag in your profile. Thank you 🙂
Hi Philip, you don’t need arrays to store your tests. We can make a loop or a summation of boolean over the last X periods to test if all your conditions were met.

09/19/2016 at 7:36 PM #13323 quote
Philippo
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```
timeframe (4 hours)
bars=20

//codes for nl nextlevel
//1 = over 100 1st time
//2 = over 100 2nd time
//-1 = under -100 1st time
//-2 = under -100 2nd time
nl = 0
//nextlevel = 0 open to anything


// set conditions
clong=0 //long
cshort=0  //short


FOR i = 0 TO 50 DO

c1=cci[i](bars) crosses over -100
c2=cci[i](bars) crosses under 100

c3=cci[i](bars) crosses over 100
c4=cci[i](bars) crosses under -100

if c1 and nl=0 then
nl = 1
continue
endif

if nl=1 and c3 then
nl=2
continue
endif

if nl=2 and c3 then
clong=1
break
endif



NEXT

SCREENER[(clong) OR (cshort)] ((close/DClose(1)-1)*100 AS "%Chg yest.")
```
09/19/2016 at 7:42 PM #13324 quote

Philippo
Participant

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Hi Nicolas, I have tried to imagine what you mean and created a fragment of code which I believe would catch the condition where CCI crosses first over -100 then subsequently through 100 and then again through 100. If I can get it working for one direction then the others will follow easily. But Pro Screener tells me that The program “MyScreener(7)” does not contain code 😀 so obviously it doesn’t understand something. But I don’t see what is wrong. It all looks syntactically ok, no yellow triangles. Any way to debug or to see what is causing the problem for ProScreener? I would not like to bother you every time I try to code:) So is this the kind of structure you meant?

09/20/2016 at 2:17 PM #13375 quote
Nicolas
Keymaster

Legend
Please update your profile with your country. This is a test quickly coded during lunch, screener take long time because of many loops. Didn’t test it much, so please stay calm and have a beer, I’ll get back here asap if you need more assistance 🙂
```
// in past x bars CCI crosses above -100 then crosses above 100 crosses below 100 then crosses above 100 (without going lower than -100) 
// or 
// CCI crosses below 100 then crosses below -100 then crosses above -100 then crosses below -100 (without going higher than 100)

lookback = 100

mycci = cci[20]

for a = lookback downto 0 do 
if mycci[a] crosses over -100 then 
  nextlookback = a
  break
 endif
next

for b = nextlookback downto 0 do
 if mycci[b] crosses over 100 then 
  nextlookback = b
  break
 endif
next

for c = nextlookback downto 0 do
 if mycci[c] crosses under 100 then
  nextlookback = c
  break
  endif
next 

for d = nextlookback downto 0 do
 if mycci[d] crosses over 100 then
  result = 1
  when = d
  break
 else
  result = 0
  endif
next

SCREENER [result] (when as "When?")
```
cci-cross-screener.png
09/20/2016 at 4:55 PM #13394 quote

Philippo
Participant

Average

Hi Nicolas, thanks at least the for loops are not nested… time for a beer. country updated (I thought I did that yesterday..) Thanks again

11/09/2016 at 9:33 PM #16258 quote
Philippo
Participant

Average
Hi, from the Probacktest and ProOrder manual, I dug up the following functions….could be exactly what I was looking for….in case anyone else want to do something similar, here’s the text: Here are several separate examples where using a FOR loop is avoided:
// Determine if the condition C1 is true for at least once over the last n candlesticks:

IF HIGHEST[n](c1) = 1 THEN ...

// Determine if the condition c1 was always true over the last n candlesticks:

IF LOWEST[n](c1) = 1 THEN ...
// Determine the number of times the condition c1 was verified over the n last candlesticks:
num = SUMMATION[n](c1)

// Determine the number of bars since c1 was true:

IF c1 THEN lastoccurence = barindex

ENDIF timesince = barindex - lastoccurence
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