moving average loop

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03/05/2023 at 10:18 AM #210925 quote
rob.es
Participant

New
hi I’m trying to screen a range of Emas (20 to 65) which may meet my condition but my results are only Ema 65. how can I have all the Ema that meet my condition?

thanks
```
period=0

for a=20 to 65 do
c1=ExponentialAverage[a](close)
c2=summation[100](close CROSSES UNDER c1 or (open < c1 and  close[1]>c1)) =0 and summation[100](close crosses over c1 or (open>c1 and close[1]<c1)) =1 and summation[100]((c1-high>0)and (c1-high<=0.40*AverageTrueRange[14](close) ))>=2 and abs(low-c1)<=0.35*AverageTrueRange[14](close)
if c2 then
period=a
else
period=0
endif
screener[period>1](period as "period")
next
```
03/05/2023 at 12:52 PM #210929 quote
robertogozzi
Moderator

Legend
A 46-element (20 to 65) array would do the trick, but only 1 datum is allowed as a criterion.

A solution is to add all EMAs’periods as a single number so to get, say: 202532445165 for EMAs 20, 25, 32, 44, 51, 65

There are a couple of issues, though:
- too many EMAs meeting the conditions would exceed the biggest number allowed (which I don’t know, maybe 15+ digits)
- big numbers (such as dates in the format yyyymmdd) are usually reported in rounded scientific notation such as 20E+6 (or similar) which would make any coding effort vain.
```
period = 0
for a = 20 to 65 do
   c1=ExponentialAverage[a](close)
   c2=summation[10](close CROSSES UNDER c1 or (open < c1 and  close[1]>c1)) =0 and summation[10](close crosses over c1 or (open>c1 and close[1]<c1)) =1 and summation[10]((c1-high>0)and (c1-high<=0.40*AverageTrueRange[14](close) ))>=2 and abs(low-c1)<=0.35*AverageTrueRange[14](close)
   if c2 then
      period = (period * 100) + a
   endif
next
screener[period](period as "period")
```
x-1.jpg
03/05/2023 at 12:59 PM #210931 quote

robertogozzi
Moderator

Legend

I coded just [10] periods with SUMMATION, as I fear the 254-bar limit (unless you have the 1024-bar premium version provided by PRT direct) won’t allow much more more as EMAs require more than twice their periods to be calculated. So a 65-period EMA will require more than 130 bars + 100-lookback periods may exceed 254, but you can try more, say 20 or 40,…. even 100 could do (it’s not an easy accurate calculation).

03/05/2023 at 11:03 PM #210958 quote
rob.es
Participant

New
thank you Roberto but the period results are all like this:44 45 46 47 48 …

some results are not quite accurate. when I write my previous code only for one Ema: say Ema34 all the results are accurate.

but when using for loop, with same condition some of the results are off
```
b11=ExponentialAverage[34](close)
b12=summation[100](close crosses under b11 or (open<=b11 and close[1]>=b11 ))=0
b13=summation[100](close crosses over b11 or (open>=b11 and close[1]<=b11))=1
b14=summation[100](abs(b11-high) <=0.35*AverageTrueRange[100](close) and b11-high>0) >=2
b15=abs(low-b11)<=0.35*AverageTrueRange[14](close)
period=34
if b12 and b13 and b14 and b15 then
screener (period as "period")
endif
```
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