Is there any way to code and find out the timestamp?

Suzu Yuk · 2022-02-24T15:35:24+00:00

Hello, Does anyone know how to scan the highest high and lowest low cycle frequency is getting shorter or longer? Please see attached picture. I would like to find a timestamp (say X1, X2, X3,,,,,) for each highest high and lowest low. Then (X2 – X1) can tell us how long it took to make each price cycle. By comparing (X2 – X1) and (X3 – X2), I can say whether the price cycle is getting longer or shorter. Regards,

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02/24/2022 at 4:35 PM #188776 quote

Suzu Yuk
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Hello,

Does anyone know how to scan the highest high and lowest low cycle frequency is getting shorter or longer? Please see attached picture.

I would like to find a timestamp (say X1, X2, X3,,,,,) for each highest high and lowest low. Then (X2 – X1) can tell us how long it took to make each price cycle. By comparing (X2 – X1) and (X3 – X2), I can say whether the price cycle is getting longer or shorter.

Regards,

IMG_20220224_160307.jpg

02/24/2022 at 6:07 PM #188785 quote
robertogozzi
Moderator

Legend
Try this (not tested):
```
p  = 40
x3 = highest[p](high)
x2 = highest[p](high[p])
x1 = highest[p](high[p * 2])
For i = (p*3)-1 downto (p*2)
   if high[i] = x1 then
      b1 = BarIndex[i]
      Break
   Endif
Next
For i = (p*2)-1 downto p
   if high[i] = x2 then
      b2 = BarIndex[i]
      Break
   Endif
Next
For i = p-1 downto 0
   if high[i] = x3 then
      b3 = BarIndex[i]
      Break
   Endif
Next
x1x2 = b2 - b1
x2x3 = b3 - b2
x = 0
If x1x2 > x2x3 then
   x = 1
Elsif x1x2 < x2x3 then
   x = 2
Endif
Screener[x](x as “1=x1x2,  2=x2x3")
```
02/24/2022 at 6:21 PM #188789 quote

Nicolas
Keymaster

Legend

The new instruction HighestBars could help in this case.

02/24/2022 at 11:58 PM #188803 quote
robertogozzi
Moderator

Legend
This is the code with the new instruction suggested by Nicolas:
```
p  = 40
b3 = highestbars[p](high)
b2 = highestbars[p](high[p])
b1 = highestbars[p](high[p * 2])
x1x2 = b2 - b1
x2x3 = b3 - b2
x = 0
If x1x2 > x2x3 then
   x = 1
Elsif x1x2 < x2x3 then
   x = 2
Endif
Screener[x](x as “1=x1x2,  2=x2x3")
```
03/01/2022 at 7:37 PM #189153 quote
Suzu Yuk
Participant

Average
Thank you very much for your swift reply. I have been tucking to understand the following part of the previously mentioned solution by dissecting the last few days. However, I am still stuck with understanding the below:
```
b3 = highestbars[p](high) b2 = highestbars[p](high[p])
```
I understand “b3 = highestbars[p](high),” but I am not sure if I understand “b2 = highestbars[p](high[p]).” Why :[p](high[p]?” Why “p” in those two places?

In order to dissect and understand them on cash 500, ticker “SPTRD,” I program them to print out the code with two variable periods “p1” and “p2.”
```
p1=4
p2=4
b3 = highestbars[p1](high)
DRAWTEXT("#b3#", barindex, high+range*2, Dialog, Standard, 10) COLOURED(255,0,0)
b2 = highestbars[p1](high[p2])
//DRAWTEXT("#Hp2#", barindex, high+range*0.6, Dialog, Standard, 12)
DRAWTEXT("#b2#", barindex, high+range*1.4, Dialog, Standard, 12)COLOURED(0,0,0)
b1 = highestbars[p1](high[p2 * 2])
//x1x2 = b2 - b1
//x2x3 = b3 - b2
//x = 0
//If x1x2 > x2x3 then
//x = 1
//Elsif x1x2 < x2x3 then
//x = 2
//Endif
return
```
If anyone can possibly rephrase them in words, I would highly appreciate it very much.
03/01/2022 at 9:31 PM #189164 quote
robertogozzi
Moderator

Legend
This line will check a highest value among the latest p bars (from 0 to 39):
```
highestbars[p](high)
```
you need two prior additional ones to find all the three, one between bars 40-79 and the leftmost one between bars 80-119.

You can evaluate the expressions within brackets to check that.
03/03/2022 at 9:19 AM #189225 quote

Suzu Yuk
Participant

Average

Thank you very much for your reply. In order to understand what is going on with the code, I quantified the changing output with the gradation of color on cash 500, ticker symbol “SPTRD, screenshot attached. Would anyone agree with me the below:

The blue gradation seems to illustrate the price cycle to be getting shorter.

The black gradation seems to illustrate the price cycle to be getting longer.

I am not sure yet if I can conclude as the above. I appreciate it very much if I can hear how other people can observe this outcome? Thank you

p=4
b3 = highestbars[p1](high)
//DRAWTEXT(“#b3#”, barindex, high+range*2, Dialog, Standard, 10) COLOURED(255,0,0)
Hp2=high[p2]
b2 = highestbars[p1](high[p1])
//DRAWTEXT(“#Hp2#”, barindex, high+range*0.6, Dialog, Standard, 12)
//DRAWTEXT(“#b2#”, barindex, high+range*1.4, Dialog, Standard, 12)COLOURED(0,0,0)
b1 = highestbars[p1](high[p1 * 2])
//DRAWTEXT(“#b1#”, barindex, high+range*0.7, Dialog, Standard, 12)COLOURED(0,0,255)
x1x2 = b2 – b1
x2x3 = b3 – b2
x = 0
If x1x2 > x2x3 then
M1 = x1x2 – x2x3
HiM1=highest[pM1](M1)
DRAWTEXT(“▇”, barindex, 1.001*highest[Hip](high), Dialog, Standard, 12)COLOURED(0,0,255, 255*M1/HiM1)
x = 1
Elsif x1x2 < x2x3 then
M2 = -x1x2 + x2x3
HiM2=highest[pM1](M2)
DRAWTEXT(“▇”, barindex, 1.002*highest[Hip](high), Dialog, Standard, 12)COLOURED(0,0,0, 255*M2/HiM2)
x = 2
Endif
return

Screenshot-2022-03-03-at-10.03.21.png

03/03/2022 at 10:46 AM #189229 quote
Suzu Yuk
Participant

Average
```
p1=170
p2=170
pM1=170
Hip=211
b3 = highestbars[p1](high)
//DRAWTEXT("#b3#", barindex, high+range*2, Dialog, Standard, 10) COLOURED(255,0,0)
Hp2=high[p2]
b2 = highestbars[p1](high[p1])
//DRAWTEXT("#Hp2#", barindex, high+range*0.6, Dialog, Standard, 12)
//DRAWTEXT("#b2#", barindex, high+range*1.4, Dialog, Standard, 12)COLOURED(0,0,0)
b1 = highestbars[p1](high[p1 * 2])
//DRAWTEXT("#b1#", barindex, high+range*0.7, Dialog, Standard, 12)COLOURED(0,0,255)
x1x2 = b2 - b1
x2x3 = b3 - b2
x = 0
If x1x2 > x2x3 then
M1 = x1x2 - x2x3
HiM1=highest[pM1](M1)
DRAWTEXT("▇", barindex, 1.001*highest[Hip](high), Dialog, Standard, 12)COLOURED(0,0,255, 255*M1/HiM1)
x = 1
Elsif x1x2 < x2x3 then
M2 = -x1x2 + x2x3
HiM2=highest[pM1](M2)
DRAWTEXT("▇", barindex, 1.002*highest[Hip](high), Dialog, Standard, 12)COLOURED(0,0,0, 255*M2/HiM2)
x = 2
Endif
return
```
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