HI Roberto,

please find attached. One itf with the bubble sort, and one without the bubble sort. These are the exact files that I used to make the previous pics showing the difference in $da[k].

The exact same code is present in both to fill up $da[k].

I did my testing on Brent Crude Oil Full1024 with daily bars, and I am using a PRT demo account.

I had a bit of a play around, and it doesn’t seem to be the duplicate removal portion of the code from what I can tell that’s causing the problem, so it must have something to do with the bubble sort process?

Cheers,

Finning

Hi Roberto,

thank you for that!! Just travelling at the moment, will have a look at it as soon as I can.

Cheers,

Finning

Try this modified version, with the new bubble sort and a few more changes:

// Data array  $da     ($dx is new array without duplicates)
// Label Array $la     ($lx is new array without duplicates)
//
defparam drawonlastbaronly = true
defparam calculateonlastbars=100
// fill up the array with CLOSE and Bar ID (label)
for k = 1 to 20

averageline = average[k](close)

if (close>averageline and open<averageline) or (close<averageline and open>averageline)  then

newcount = 1
else
newcount=0

endif

once $da[k]=0

$da[k] = newcount+$da[k]

$la[k] = k

next


MaxElements = lastset($da)
//////////////////////////////////////////////////////////////////
// plot unsorted arrays
temp = MaxElements + 1
Offset = average[10](range[1])
drawtext("Elements #temp#",BarIndex+10,low - range)
for k = 0 to MaxElements
x = $da[k]
y = $la[k]
drawtext("# of crossovers[#y#]:  #x#",BarIndex+10,high + (Offset * k))
next
//////////////////////////////////////////////////////////////////
// (Bubble Sort)

FOR i = 0 TO MaxElements -1
FOR j = 0 TO MaxElements - i
IF $da[j] > $da[j + 1] THEN
// swap data
temp   = $da[j]
$da[j] = $da[j + 1]
$da[j + 1] = temp
// swap labels
temp   = $la[j]
$la[j] = $la[j + 1]
$la[j + 1] = temp
ENDIF
NEXT
NEXT

//FOR i = 0 TO MaxElements
//FOR j = 0 TO MaxElements
//IF $da[j] > $da[i] THEN
//// swap data
//temp   = $da[j]
//$da[j] = $da[i]
//$da[i] = temp
//// swap labels
//temp   = $la[j]
//$la[j] = $la[i]
//$la[i] = temp
//ENDIF
//NEXT
//NEXT
//////////////////////////////////////////////////////////////////
// plot Sorted arrays
for k = 0 to MaxElements
x = $da[k]
y = $la[k]
drawtext("# of crossovers[#y#]:  #x#",BarIndex+15,high + (Offset * k))
next
//////////////////////////////////////////////////////////////////
// remove duplicates by comparing the current element to the next one (creating 2 new arrays)
NewMaxElements = 0
FOR i = 0 TO MaxElements
IF ($da[i] <> $da[i + 1]) OR (i = MaxElements) THEN
$dx[NewMaxElements] = $da[i]          //save datum to new array, when different
$lx[NewMaxElements] = $la[i]          //save label, too
if (i = MaxElements) then
break
endif
NewMaxElements      = NewMaxElements + 1
ENDIF
NEXT
//////////////////////////////////////////////////////////////////
// plot new Arrays (without duplicates)
temp = NewMaxElements
drawtext("Elements #temp#",BarIndex+20,low - range)
FOR k = 0 to NewMaxElements
x = $dx[k]
y = $lx[k]
drawtext("# of crossovers[#y#]:  #x#",BarIndex+20,high + (Offset * k))
NEXT

//FOR k = 0 to 20
//xx = $da[k]
//yy = $la[k]
//drawtext("# of crossovers[#yy#]:  #xx#",BarIndex+5, 25 + (yy*2))
//next

return

There are two issues:

1. arrays are not historicized, so $var[1] is element 1 of the array $var, not the previous value of $var; this implies, unlike variables, that elements are updated each tick and, if conditions are not met at the closing of a bar, their original value cannot be restored because it’s lost
2. the same conditions is added multiple times while a bar is formed, thus retaining incorrect values.

I have added a new 1-element array to be used as a flag to detect any change in the opening time of a bar, so we can tell when a new bar has formed to update the array (lines 7, 8 and 9). The IF in line 8 is closesd in line 69.

Give it a try.

// Data array  $da     ($dx is new array without duplicates)
// Label Array $la     ($lx is new array without duplicates)
//
defparam drawonlastbaronly = true
defparam calculateonlastbars=100
// fill up the array with CLOSE and Bar ID (label)
ONCE $TimeFlag[0] = 0
IF $TimeFlag[0] <> OpenTime THEN
$TimeFlag[0] = OpenTime
for k = 1 to 20
averageline = average[k](close)
if (close>averageline and open<averageline) or (close<averageline and open>averageline)  then
newcount = 1
else
newcount=0
endif
once $da[k]=0
$da[k] = newcount+$da[k]
$la[k] = k
next
MaxElements = lastset($da)
//////////////////////////////////////////////////////////////////
// plot unsorted arrays
temp = MaxElements + 1
Offset = average[10](range[1])
drawtext("Elements #temp#",BarIndex+5,low - range)
for k = 0 to MaxElements
x = $da[k]
y = $la[k]
drawtext("# of crossovers[#y#]:  #x#",BarIndex+5,high + (Offset * k))
next
//////////////////////////////////////////////////////////////////
// (Bubble Sort)

FOR i = 0 TO MaxElements -1
FOR j = 0 TO MaxElements - i
IF $da[j] > $da[j + 1] THEN
// swap data
temp   = $da[j]
$da[j] = $da[j + 1]
$da[j + 1] = temp
// swap labels
temp   = $la[j]
$la[j] = $la[j + 1]
$la[j + 1] = temp
ENDIF
NEXT
NEXT
//////////////////////////////////////////////////////////////////
// plot Sorted arrays
for k = 0 to MaxElements
x = $da[k]
y = $la[k]
drawtext("# of crossovers[#y#]:  #x#",BarIndex+10,high + (Offset * k))
next
//////////////////////////////////////////////////////////////////
// remove duplicates by comparing the current element to the next one (creating 2 new arrays)
NewMaxElements = 0
FOR i = 0 TO MaxElements
IF ($da[i] <> $da[i + 1]) OR (i = MaxElements) THEN
$dx[NewMaxElements] = $da[i]          //save datum to new array, when different
$lx[NewMaxElements] = $la[i]          //save label, too
if (i = MaxElements) then
break
endif
NewMaxElements      = NewMaxElements + 1
ENDIF
NEXT
ENDIF
//////////////////////////////////////////////////////////////////
// plot new Arrays (without duplicates)
temp = NewMaxElements
drawtext("Elements #temp#",BarIndex+15,low - range)
FOR k = 0 to NewMaxElements
x = $dx[k]
y = $lx[k]
drawtext("# of crossovers[#y#]:  #x#",BarIndex+15,high + (Offset * k))
NEXT
return

Hi Rob,

with the duplicate removal tool, is there any way that it can be made to store the smallest of the duplicate values?

For example, in the data above, moving average periods 15, 16, and 20, all have 11 crossovers.

The current duplicate removal code keeps the 20 period.

Is there any way that it could keep the 15 period value instead? So 15 would show up from the duplicate removal, and not 20.

I tried

For i = Maxelements downto 0

but it only returned the largest singular value of the bubble sort.

Any ideas?

Thanks,

Finning

Hi Roberto,

no that didn’t seem to work.

Another way of saying the above, what I was looking for, is to keep the lower duplicate value (based on the $la[k] array) from the duplicate removal process.

The duplicate removal process currently keeps the higher $la[k] array and corresponding $da[k].

Have attached another pic to show this, crossing out the higher values, and keeping the lower values.

Thanks,

Finning

Hi Rob,

I tried your code, and it didn’t give the intended result. Please see attached pic of me running this code.

Rob, I’m sorry if I haven’t explained myself properly.

With the bubble sort result, if you have a look at the numbers in the brackets [x], these are the periods that the moving average crossovers were recorded in.

The number to the right of them, is the count of the number of moving average crossovers.

What is happening in the duplicate removal, is that it is keeping the result with the highest number in the brackets [x].

The problem that I am trying to solve is I need the duplicate removal code to give the smallest [x] value for duplicates.

So, instead of the duplicate removal keeping the highest duplicates, I need it to keep the lowest duplicates.

If I run both of these duplicate removal processes together, I will have a highest and a lowest value (I am only interested in the biggest values of the bubble sort too), and I can then take a midpoint of these two numbers for use as a variable.

Please find the attached pic to explain this.

Thanks,

Finning.