Lowest from Y1 to Y2

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03/10/2022 at 1:49 PM #189750 quote

ZeroCafeine
Participant

Senior

Hi all,

I saw the Lowest fonction can find a lowest of a variable on the last X periodes for exemple, but how to find a lowest between 2 barindex for exemple between Y1 and Y2 ?
for exemple on this photo how to find the lowest Low between Candle[5] and Candle[36] ( Y2=36 / Y1=5) ?

entre ces 2 valeurs

Lowest-Y1-Y2.png

03/10/2022 at 1:56 PM #189752 quote
JC_Bywan
Moderator

Master
From 5 to 36, there are 36-5+1 = 32 candles, we simply ask what was 5 candles ago the lowest among 32:
```
a=lowest[32](low)
b=a[5]

return b
```
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03/10/2022 at 2:00 PM #189753 quote
JC_Bywan
Moderator

Master
or if you want to easily change Y1 and Y2 (with Y2>Y1 implied, haven’t added an IF statement to protect against the opposite):
```
Y1=5
Y2=36
a=lowest[Y2-Y1+1](low)
b=a[Y1]

return b
```
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03/10/2022 at 3:56 PM #189767 quote

JS
Participant

Master

Lowest Low between candle[5] and candle[36]:

a = Lowest[36](Low[5])

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03/11/2022 at 12:22 PM #189798 quote

ZeroCafeine
Participant

Senior

tks to all for your answers,

and how to get the barindex number of this lowest candle ?

I saw some solution like doing a For y = 0 to 38 for exemple and compare the Lowest we get before with your solution with the Low[y] of each candle for get the barindex, but maybe they have other short soluttion ?

03/11/2022 at 12:56 PM #189801 quote
robertogozzi
Moderator

Legend
There you go:
```
x = LowestBars[36](low[5])   //HighestBars for the highest
```
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03/11/2022 at 1:25 PM #189803 quote
ZeroCafeine
Participant

Senior
tks for your answer but I’m totaly confused, I don’t understand why it’s not working or mabt I don’t know how to use it, with my result I find the barindex 21 the blue cyrcle but with yous I find barindex = 8 the red cyrcle, where I’m wrong ? I want the fast solution for using less of computer ic
```
Defparam drawonlastbaronly = true

a = Lowest[36](Low[13])

// Roberto Solution 
x = LowestBars[36](low[13])

For i = 0 to 36 Do
IF Low[i] = a Then
X1 = i
drawtext("Barindex = #i#",barindex[i],Low[i])
// Roberto Barindex
drawtext("Barindex = #x#",barindex[x],Low[x])
break
ENDIF
Next

Return
```
Best Reguards

roberto-barindex.png
03/11/2022 at 1:34 PM #189805 quote

ZeroCafeine
Participant

Senior

This is my Elder Triple screen now and headache, I’m sure your solution will be on 2 or 3 lines 🤣🤣🤣

Screenshot_20220311-133347.jpg

03/11/2022 at 2:36 PM #189807 quote
robertogozzi
Moderator

Legend
Use
```
IF Low[i+13]
```
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03/11/2022 at 4:33 PM #189809 quote
ZeroCafeine
Participant

Senior
I don’t know why I don’t know to use you solution, but I found this one :
```
Defparam drawonlastbaronly = true

a = Lowest[36](Low[13])
// Roberto Solution 
x = LowestBars[36](low[13])

Roberto = 13 + x

IF X Then
drawtext("RobertoIndex = #Roberto#",barindex[Roberto],Low[Roberto])
ENDIF

Return
```
if I the drawtext fonction with out IF condition, I get error because the LowestBars fonction can get -1 as a result

now the good barindex is 22 instead 21 because I got one candle more in H4 TF

roberto-barindex-2.png
03/11/2022 at 5:18 PM #189813 quote

robertogozzi
Moderator

Legend

Because the 13th bar is bar 12 (0 to 12 is 13 bars), so you will have to adjust your calculations.

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03/11/2022 at 5:28 PM #189814 quote

JS
Participant

Master

From 0 to 13 is 14 bars so use:

Roberto = 14 + x

03/12/2022 at 8:39 AM #189842 quote
JC_Bywan
Moderator

Master
It is of course possible I misunderstood the initial query, or even if I hadn’t I would easily concede perhaps I made the explanation too short alongside my proposal. But I would have to respectfully disagree with the other proposal.

If you write: a = Lowest[36](Low[5]) , it doesn’t give you the lowest low between candle[5] and candle[36], it would give the lowest low below candle[5] and candle[41], because at candle[36] what was lowest[5] at that time is a lowest[41] seen from now. So on one hand yes it would easily get rid of unwanted candles[0] through [4] in the search, but on the other hand it adds unwanted extra 5 candles [37] through [41] beyond [36].

Which is why I suggested: to first assess the span of the search (Y2–Y1+1 = 36-5+1 = 32 candles wide), to search that span for lowest low (that’s “a”), and from there to offset it by Y1=5 (that’s “b”), with:
```
Y1=5
Y2=36
a=lowest[Y2-Y1+1](low)
b=a[Y1]
```
as a way of eliminating candles [0] through [4] and simultaneously avoiding addition of unwanted extra candles beyond Y2 at the other end of the spectrum in the search interval.

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03/12/2022 at 10:05 AM #189848 quote

JS
Participant

Master

Hi @JC_Bywan,

You are right about adding 5 extra candles…

The starting point is low[5] and from there your look back period is [36] = 5 + 36

So this must also give the right answer: a = Lowest[Y2-Y1+1](Low[Y1])

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03/12/2022 at 7:48 PM #189887 quote
JS
Participant

Master
Perhaps a system can be made of this?

When you know where the Lowest Low (LL) and the Highest High (HH) occur, you can say in the simplest form;

When the HH occurs after the LL then you are in an ascending phase and vice versa when the LL occurs after the HH then you are in a downward phase.

With the formulas:

LLBar = LowestBars[n](Low) you determine the bar where the LL occurs

HHBar = HighestBars[n](High) you determine the bar where the HH occurs

When LLBar > HHBar (Bullish)

When LLBar < HHBar (Bearish)

Of course, this is the simplest form but maybe with MTF, SL, TP, etc. we can improve this system.
```
DefParam CumulateOrders = False

LLBar = LowestBars[n](Low)
HHBar = HighestBars[n](High)

If LLBar > HHBar then
Buy 1 Contracts at Market
ElsIf LlBar < HHBar then
SellShort 1 Contracts at Market
EndIf

Graph LLBar as "LowestBar"
Graph HHBar as "HighestBar"
```
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