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andypap

Variable Counter Array

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hi guys

 

i have a counter based on a condition and what i would like to know is how to create an one dimentional array that will allign all the closing values when the counter value is 1..

(the counter value is one everytime the condition occurs).

the chart displays the counter values...

 

my question has to do with the kind of loop required to populate such an array..

 

so if the counter is mycounter and the array myarr i have tried

 

for j = 1 to mycounter begin

myarr[j] = c[j];

end;

 

also

 

if mycounter = 1 then

for j = 1 to mycounter begin

myarr[j] = c[j];

end;

 

but it didnot work...

 

any help on this will be much appreciated.

5aa70f2f06013_counterpic.thumb.png.471d371fc1d9a99fd1a6a2f8f7f5d239.png

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suppose that the condition that sets the counter is a bullish key reversal..

and the array holds 20 elements..

 

 

every time a bullish key reversal happens the counter is 1 and incrementing by 1 for all the bars beteen key reversals..

when the counter resets to 1 the array is populated with the closing values of the last 20 keyreversals in the chart and will keep those values unchanged unill a new bullish key reversal appears ant the counter resets to 1 again..

 

 

i hope that this will be clear enough..

 

 

many regards

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I see,

 

what you want to do is preserve the array... at least that is how you do it Visual Basic ---

 

redim preserve array()

 

 

I am trying to find a resource for this on EL -- haven't found it yet.

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thanks frank

i will appreciate much if you cross something like this to let me know..

i have tried so far a few different loops but i couldnot do it...

 

I'm still trying to figure out what you're trying to do / what the array holds.

And have you asked on TS or MC forums?

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its my impression, and I could be wrong, that he wants to start holding prices in an array (presumably closing prices) of the bars that meet a condition and setting the max elements of the array to 20. I don't know what the counter part is for but I am guessing that he doesn't need the counter -- only to preserve the array from one bar to the next and keep the last 20 elements by adding a new element to array(19) or array(20) and kicking out the first element array(0) or array(1).

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the counter is a critical part of this code and its purpose is to keep stastisticks between " key reversals"... has nothing to do with the array other that

 

every " key reversal" occurance the counter is 1..

the array needs to hold the " closing" prices of the last N(20 for example) "key reversal"

occurrances from key reversal to next keyreversal ....

 

my problem is not declaring the size of the array but what kind of loop (the exact syntax)

needed to populate the array properly.

 

I HAVE THOUGHT TO USE THE COUNTER

because every key reversal the counter is 1 and there is a "length" between key reversals to use in the loop..

have in mind that the counter between key reversals has a variable length .

 

with direction form current to previous lets suppose once

a new key reversal shows the counter is 1, 20 ,33, 217 etc...

 

right then the array should be populated

 

array[1] = current close

array[2] = c[20] = c of previous key reversal

array[3] = c[20 + 33] = c of 2 key reversals back

array[4] = c{20 + 33 + 217] = c of 3 key reversals back etc

untill N array elements

 

then as new bars come and the current counter is incrementing those values will stay as they are untill next key reversal where the counter resets to 1 again and at that point the array will be repopulate in the same manner...

 

array [1] will be new key reversal and all other elements will shift one row down till N...

 

IF there is a way to do the same thing without using a counter then i will be happy to learn how

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if you would like to keep those old values stored in your array and only shift

those values one row back each time your condition occured,

you could do something like this

 

If mycondition Then {this should start only wenn your condition occurs which is your counter i believe}
Begin

For i = N DownTo 1 
Begin
myarray[i] = myarray[i-1];  {shift values one row back untill row 1}
End;

myarray[0] = mynewvalue;  {now write new value at row 0}

End;

 

what you are doing this way is you only write in index-place 0 of your array,

and only then wen your condition occures.

your old values will shift 1 index back in your array and stop at index place 1 just right before you start to store that new value.

this way all you former values will not be lost and the new values keep coming in at index 0

Edited by flyingdutchmen

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Add...

 

above example N would be the max amount of values you would like to store

in your array, or the amount of times that your condition has been set true if you

would use N as a counter where you just add 1 each time your condition is found.

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andypap,

 

can you post the final completed code to fully understand what your indicator was doing? it would likely be useful to others to use this simple array structure used in an example to use in their own indicators (that do other things).

 

thanks in advance

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