<> in AFL (AmiBroker)

[aaa 10]

Hi

 

Silly quick question

 

The = signe in AFL (AmiBroker) is

 

==

 

What is the difference sign ?

 

in EasyLanguage ;

 

if whatever <> H

 

In AFL ;

 

if whatever ????????? H

 

ThanX

[ochie 10]

Hi aaa,

 

I believe it is:

 

if whatever != H

 

AFL syntax is like standard "C" language

[aaa 10]

Hi Ochie

 

I'm learning AFL

 

U afraid me saying AFL = C !!!!

 

I didn't knew that !!

 

C is complicate !!!!

 

!= works

 

But I I have error 6

 

What do U think ?

 

DR     = 	0 ;
DS     = 	0 ; 
ODR  = 	0 ;
ODS  = 	0 ;
PDR = 	0 ;
PDS = 	0 ;

if DS != PDS )			
  ODS  = PDS ;

if  ( DR != High AND DR < PDR )  // Error 6 here
 ODR = PDR;

 

 

I don't understand that below =

 

Error 6. Condition in IF, WHILE, FOR statements has to be Numeric or Boolean type. You can not use array here, please use [] (array subscript operator) to access array elements

The if keyword executes statement1 if expression is true (nonzero); if else is present and expression is false (zero), it executes statement2. After executing statement1 or statement2, control passes to the next statement. Expression must be boolean ( True/False) type (so it CANNOT be ARRAY because there would be no way do decide whether to execute statement1 or not, if for example array was: [True,True,False,.....,False,True] )

 

if( expression )

statement1;

else

statement2;

 

EXAMPLE

 

if( Close > Open ) // WRONG

Color = colorGreen; //statement 1

else

Color = colorRed; //statement 2

 

Plot(Close,"Colored Price",Color,styleCandle);

 

The above example is wrong, as both Open and Close are arrays and such expression as Close > Open is also an ARRAY. The solution depends on the statement. It’s either possible to implement it on bar-by-bar basis, with use of FOR loop:

 

for( i = 0; i < BarCount; i++ )

{

if( Close[ i ] > Open[ i ] ) // CORRECT

Color[ i ] = colorGreen;

else

Color[ i ] = colorRed;

}

 

Plot( Close, "Colored Price", Color, styleCandle );

 

It is also possible in this case to use IIf( ) function:

 

Color = IIf( Close > Open, colorGreen, colorRed ); // ALSO CORRECT - working directly on arrays

Plot( Close, "Colored Price", Color, styleCandle );

[ochie 10]

if  ( DR != High [color=Red][u]AND[/u][/color] DR < PDR )  // Error 6 here
 ODR = PDR;

The logical AND in AFL is &&

 

Here is a C Operator Quick Reference that should come in handy.

[aaa 10]

ThanX 4 the reference Guide in C

 

It's great !!!

 

AmiBroker has an Xcellent help

 

So

 

In AFL AND is AND

 

I took off the AND and my error 6 is still here

 

Grrrrrr

 

DR     = 	0 ;
DS     = 	0 ; 
ODR  = 	0 ;
ODS  = 	0 ;
PDR = 	0 ;
PDS = 	0 ;

if DS != PDS )			
  ODS  = PDS ;

if  ( DR != High )  // Error 6 here
 ODR = PDR;

[ochie 10]

Thanks for the AND info. I did not know that!

 

I'm not near a compiler to test.

Don't you have a bracket missing here after the "if"

or just a typo?

if DS != PDS )

ODS = PDS ;

[Head2k 10]

In AFL coding it is essential to distinguish between a number and array. High is an array of all highs in the chart.

if operator works only for comparing numbers and returns also a number (boolean).

On the other hand, IIf operates with arrays and returns also an array. It is like running if operation for every array element.

 

ODR = IIf (DR != High, PDR, 0);

 

This line says:

Take every array element in DR array and compare it to adequate (i.e. with the same index) element of array of Highs. If the compared array elements are different (!=), then the adeqate element of ODR array (i.e. an element with the same index as the compared elements of DR and High arrays) shall be equal to an adegate element of PDR array. Else it will be zero.

Obviously ODR will be an array after this operation.

 

EDIT:

 

Just to show the difference, I will write how to use if operator to get the same result as the line above gives.

 

for (i = 0; i < barcount; i++)

{

if (DR != High)

ODR = PDR;

else

ODR = 0;

}

Edited January 2, 2010 by Head2k

[aaa 10]

Hi Head2K

 

I must say that it is hard 2 understand even tho your Xplanation is Xcellent

 

But now I don't have any error

 

I should have wrote the entire formula because now I dono how to incorporate the second condition

 

The best is 2 write it in EasyLangage

 

 

 

if DR <> H AND DR < PDR then 
        if PDR <> 0 then 
 	DR = PDR ;

 

Is it correct translation in AFL ?

 

for (i = 0; i < barcount; i++)
{
if (DR[i] != High[i] and DR < PDR )
      if (PDR[i] != 0 )
          DR[i] = PDR[i];
else
DR[i] = 0;
}

 

 

PS I understand NOW how EL is easy....

[Tams 20]

  aaa said:

...

PS I understand NOW how EL is easy....

 

 

EasyLanguage is a Procedural Language...

 

it takes instructions one step at a time.

 

 

C is an conceptual Language...

 

it starts with an idea... and takes instructions in parallel.

 

 

 

C is a powerful tool for the conceptual thinker,

 

EasyLanguage is an easier tool for the logical thinker.

Edited January 2, 2010 by Tams

[aaa 10]

my last question

 

EL

 

inputs:
Periods(	10 ;

variables :
DR( 	0 ),
Eb( 	0 ),
EV( 	0 );

// different computations of DR

EV = Extremes( H , Periods ,  1 , DR , EB ) ;

 

AFL

 

DR( 	0 );
Eb( 	0 );
EV( 	0 );

// different computations of DR

EV  = HHV( H, Period  ); // where to put DR ?

 

PS

I don't find any documentation in Extremes function in TS Help

Strange because it works

[aaa 10]

Here it is

 

Extremes (Function)

Disclaimer

 

The Extremes function returns the extreme highest or lowest value over a range of bars and how many bars ago the extreme value occurred. There may be times when two or more bars have the exact same extreme highest or lowest value; when this happens the function identifies the most recent occurrence.

 

Syntax

Extremes(Price,Length,HiLo,oExtremeVal,oExtremeBar)

 

Returns (Integer)

The oExtremeVal and oExtremeBar output parameters return the extreme value and the number of bars ago it occurred. The Extremes function itself returns a value of 1.

 

Parameters

Name

Type

Description

 

Price

Numeric

Specifies which bar value (price, function, or formula) to compare for highest and lowest extremes.

 

Length

Numeric

Sets the number of bars to consider for extremes.

 

HiLo

Numeric

Sets whether the function will return the highest or lowest extreme value. 1=Highest, -1=Lowest.

 

oExtremeVal

Numeric

Outputs the highest or lowest extreme value found for the range of bars based on the HiLo setting.

 

oExtremeBar

Numeric

Outputs the number of bars ago the extreme value occurred.

 

 

Remarks

The input parameter Price can be a bar value such as Close, High, Low, Open, or Volume. It can also be any mathematical calculation such as: ( High + Low) / 2, or a numeric function such as RSI, Stochastic, or ADX.

 

See Multiple Output Function for more information on using output parameters to return values.

 

Example

Assigns to Value2 the highest High of the last 20 bars using the oExtremeVal output parameter, and assigns to Value3 the number of bars ago the highest High occurred using oExtremeBar output parameter. Value1 is assigned a value of 1:

 

vars: oExtremeVal(0), oExtremeBar(0);

 

Value1 = Extremes(High, 20, 1, oExtremeVal, oExtremeBar);

 

Value2 = oExtremeVal;

 

Value3 = oExtremeBar;

 

 

 

See Also

ExtremesFC, ExtremesArray, NthExtremes.

[Head2k 10]

  aaa said:

Hi Head2K

 

I must say that it is hard 2 understand even tho your Xplanation is Xcellent

 

But now I don't have any error

 

I should have wrote the entire formula because now I dono how to incorporate the second condition

 

The best is 2 write it in EasyLangage

 

 

 

if DR <> H AND DR < PDR then 
        if PDR <> 0 then 
 	DR = PDR ;

 

Is it correct translation in AFL ?

 

for (i = 0; i < barcount; i++)
{
if (DR[i] != High[i] and DR < PDR )
      if (PDR[i] != 0 )
          DR[i] = PDR[i];
else
DR[i] = 0;
}

 

 

PS I understand NOW how EL is easy....

 

I don't know EL, I never used it. But your transcription into AFL seems OK. AFL is also easy, once you get the difference between arrays and nubmers. And i would recommend using preferably array functions, as they are usually faster than loop code.

 

As for the Extremes function, according to documentation you get 2 values from it.

1. The value of the extreme

2. Where the extreme occured

 

In AFL you need two functions for that.

HHV for the value and HHVBARS for the number of bars since the high occurred. LLV and LLVBARS for lows.

 

Browse AmiBroker Help -> AFL chapter -> Categorized list of AFL functions -> Lowest/Highest

[aaa 10]

What I've learned here in AFL

 

<> is !=

 

if is used only for nbers

OHLC are arrays not numbers

 

EL

if DR <> H AND DR < PDR then

DR = PDR ;

 

AFL

DR=IIf (DR != High AND DR < PDR, PDR, 0);

 

DR = PDR if condition is true else 0

 

and the result is now an array

 

Loops are slower

 

EL is logical

AFL is conceptual and need // instructions

 

Sometimes we need 2 functions in AFL to convert an EL function

[aaa 10]

How 2 put the second condition if PDR <> 0 inside the formula ?

 

Target in EL

 

if DR <> H AND DR < PDR then

if PDR <> 0 then

DR = PDR ;

 

------------------------

 

First condition

 

if DR <> H AND DR < PDR then DR = PDR ;

 

DR=IIf (DR != High AND DR < PDR, PDR, 0);

 

 

Second Condition

 

if PDR <> 0 then DR = PDR ;

 

DR=IIf (PDR != High, PDR, 0);

[aaa 10]

EV = Extremes( H , Periods , 1 , DR , EB ) ;

 

highest closing price = HHV( H, Period );

The formula "hhv( High, 4)" returns the highest closing price over the preceding four periods;

 

number of periods = HHVBARS( H, DR );

The formula "hhvbars( High, 4 )" returns the number of periods that have passed since the highest price reached its 4-period peak.

 

How to match HHV and HHVBARS to make EV ?

[aaa 10]

My indicator has 537 lines code in EL

 

I've succeded to transpose in AFL all of them except the 2 problems above that I replace with something different

 

So I'm able To see the result on a chart

 

Is there a debug window with a print statment like in MC in AmiBroker ?

[Head2k 10]

  aaa said:

How 2 put the second condition if PDR <> 0 inside the formula ?

 

Target in EL

 

if DR <> H AND DR < PDR then

if PDR <> 0 then

DR = PDR ;

 

------------------------

 

First condition

 

if DR <> H AND DR < PDR then DR = PDR ;

 

DR=IIf (DR != High AND DR < PDR, PDR, 0);

 

 

Second Condition

 

if PDR <> 0 then DR = PDR ;

 

DR=IIf (PDR != High, PDR, 0);

 

The exact transcription of your EL code would be

DR = IIf (PDR != High AND DR < PDR, IIf ( PDR != 0, PDR, 0), 0 );

That is one IIf function inside another one.

 

The only difference is that in IIf function "else" must be defined and I set DR to zero in "else" case. In another words, I believe your EL code doesn't say what DR is if the conditions aren't met. In this AFL code it says that DR will be set to zero then.

 

Anyway, I guess the code would be more efficient like this:

EL:

if DR <> H AND DR < PDR AND PDR <> 0 then

DR = PDR ;

 

AFL:

DR=IIf (PDR != High AND DR < PDR AND PDR != 0, PDR, 0);

 

In another words, you can put all the expressions into one condition.

 

Now to your previous post and array functions vs. loops.

I think in most cases you won't recognize the difference in speed. So if you happen to understand loops better than array functions, you can of course use loops.

[aaa 10]

  Quote

Anyway, I guess the code would be more efficient like this

 

Gooood

 

I've changed my code in EL

For the multiple steps of logical B4 the formula I've used 2 conditions 2 clear my mind and I didn't cleaned up this unusefull 2nd condition at the end of the proccess

 

I prefer the difficult way rather than the loop Bcoz it's more logical & elegant & short

 

This construction below is perfect 4 me

 

DR=

IIf (

PDR != High AND

DR < PDR AND

PDR != 0,

PDR,

0);

 

 

I'm very closed 2 my aims with your wonderfull help

 

Do you have any idea for the last problem ?

 

EV = Extremes( H , Periods , 1 , DR , EB ) ;

 

highest closing price = HHV( H, Period );

The formula "hhv( High, 4)" returns the highest closing price over the preceding four periods;

 

number of periods = HHVBARS( H, DR );

The formula "hhvbars( High, 4 )" returns the number of periods that have passed since the highest price reached its 4-period peak.

 

How to match HHV and HHVBARS to make EV ?

[Head2k 10]

  aaa said:

Do you have any idea for the last problem ?

 

EV = Extremes( H , Periods , 1 , DR , EB ) ;

 

highest closing price = HHV( H, Period );

The formula "hhv( High, 4)" returns the highest closing price over the preceding four periods;

 

number of periods = HHVBARS( H, DR );

The formula "hhvbars( High, 4 )" returns the number of periods that have passed since the highest price reached its 4-period peak.

 

How to match HHV and HHVBARS to make EV ?

 

Well, that is more difficult for me because I don't know EL. But according to info you posted above

 

  Quote

The Extremes function returns the extreme highest or lowest value over a range of bars and how many bars ago the extreme value occurred. There may be times when two or more bars have the exact same extreme highest or lowest value; when this happens the function identifies the most recent occurrence.

 

Syntax

Extremes(Price,Length,HiL o,oExtremeVal,oExtremeBar )

 

HHV returns oExtremeVal of Price in Length periods with HiLo set to 1.

LLV returns oExtremeVal of Price in Length periods with HiLo set to -1.

HHVBARS and LLVBARS return oExtremeBar, I believe. Because oExtremeBar outputs the number of bars ago the extreme value occurred and HHVBARS and LLVBARS outputs how many bars occurred since the extreme was reached, which is the same.

 

Just notice that HHV, LLV, HHVBARS and LLVBARS functions return arrays.

 

So your line

EV = Extremes( H , Periods , 1 , DR , EB ) ;

 

would be in AFL as follows:

 

DR = HHV ( H, Periods );

EB = HHVBARS ( H, Periods );

[aaa 10]

DR = HHV ( H, Periods );

 

I've tried that B4 W/o success

 

We don't need EB

 

oExtremeBar

Numeric

Outputs the number of bars ago the extreme value occurred.

 

 

Here is my script at this point but I'm overlimit and I don't understang anything more

 

The algorythm is dancing the samba in my head

 

I wish 2 B more intelligent sometimes...

 

if ( DR != PDR )

ODR = PDR ;

 

ODR = PDR ;

 

PDR = DR ;

 

//oExtremeVal = Extremes( H , Periods , 1 , DR , oExtremeBar ) ;

 

DR =

IIf ( Dynamic_R != High AND Dynamic_R < PrevDynamic_R AND PrevDynamic_R != 0,

PrevDynamic_R, 0 );

Edited January 3, 2010 by aaa

[Head2k 10]

I guess I am missing the point here.

 

DR = HHV ( H, Periods );

generates an array called DR which, for every bar (= array element), contains the highest high in "periods" bars before and including that bar.

 

Then you take that DR array and redefine (=rewrite) it as follows

 

DR = IIf ( PDR != High AND DR < PDR AND PDR != 0, PDR, 0 );

 

The DR array used inside the IIf condition is the original DR array defined by the HHV function. But by the IIf function you then re-write it and assign either PDR or 0 to its elements depending on the condition.

 

That's what this code does. But perhaps it is not what you want it to do.

[aaa 10]

You are so kind Head2K

 

I don't want 2 waist your time anymore

 

I will take time now 2 understand everything new I've learned

 

Array is a complicate concept 4 me

 

I need 2 find a new brain.

 

Is there cheap second hand brain with math option 2 buy on this site ?

 

I've started with AB & AFL one week ago

 

so I'm a real baby impatient 2 play with my new Xmas gift

 

if U don't mind, may B next WE I'll ask U some more help ?

 

Many thanX 4 your patience

 

And good sunday

 

aaa

Edited January 3, 2010 by aaa