Math / Array question

[Guest]

Hello, i am not a math wonder and am stuck with generating a sequence of numbers.

I have an array of sprites where i want to loop through and set it's x position.

But i really want to do this with a loop, but i dont know to calculate these sequence of number (5,4,3,2,1,0,   11, 10, 9, 8, 7, 6,    17, 16, 15,14,13,12)

Without the loop it looks like:   (What i am doing is setting the first sprite xposition to the last one in that row and then the second sprite to the lastone minus 1 and so one..)  Does anyone know a way to do this through math?

   myArray[0].x = myArray[5].origX;
   myArray[1].x = myArray[4].origX;
   myArray[2].x = myArray[3].origX;
   myArray[3].x = myArray[2].origX;
   myArray[4].x = myArray[1].origX;
   myArray[5].x = myArray[0].origX;
  
   myArray[6].x = myArray[11].origX;
   myArray[7].x = myArray[10].origX;
   myArray[8].x = myArray[9].origX;
   myArray[9].x = myArray[8].origX;
   myArray[10].x = myArray[7].origX;
   myArray[11].x = myArray[6].origX;
  
   myArray[12].x = myArray[17].origX;
   myArray[13].x = myArray[16].origX;
   myArray[14].x = myArray[16].origX;
   myArray[15].x = myArray[14].origX;
   myArray[16].x = myArray[13].origX;
   myArray[17].x = myArray[12].origX;

[Hakan_KIRIK]

for(var i:uint=0;i<3;i++)

{
for(var ii:uint=0;ii<6;ii++)
{
  var num:int=(i+1)*6-(ii+1);
  trace(num);
}
}

[Guest]

Thanks,  you example does the trick, but i was wondering if it could be done
something like this, so within one loop:

var rowcounter:int =1;
var counter:int = 0;

for(var i:int = 0 ; i<17; i++) {
var num:int = i%6;
trace(num, rowcounter);
if(num == 5) {
  rowcounter++;
}

//counter = 17/i*rowCounter ..something?

}

Like could i now calculate the counter value like the number sequence in my previous post?

[Andrei1-bKoviI · Accepted solution]

The most efficient way to calculate column based grid it like this:

// number of columns
var numCols:int = 6;
var column:int = 0;
var row:int = 0;

for (var i:int = 0; i < 18; i++) {
     column = i % numCols;
     row = int(i / numCols);
     trace(i, "\tcolumn =", column, "row =", row);
}

As for your swap, one of abstract ways to do that can be:

// number of columns
var numCols:int = 6;
var column:int = 0;
var row:int = 0;
// alter ego of index
var swap:int = 0;
for (var i:int = 0; i < 18; i++) {
     column = i % cols;
     swap = numCols  + i - column * 2 - 1;
     trace(i, "=>", swap);
}

[Guest]

Thanks! That is exactly what i was looking for!   points well earned! 

I only needed to rename the cols to numCols in you example and it worked! 

I think this way is better for performance then to use more loops