special array cycles

Array so: for every value 1st column correspond one in 2nd
2 25
25 27
27 60
4 50
27 61

be able to write this
2 25,27,60,61
4 50

this steps:
-2nd isn't in right then rewrite in 1st column with its value (25)
-the 25 is also in 2nd column, then its value 27 go near 25
.....

some particular code?

On 8/24/2006 12:10 PM, padew wrote:There are probably simpler ways to do this, but here's one:

a=[[2,25],[25,27],[27,60],[4,50],[27,61]];

function foo (x)
{
var z=[];

for (var i = 0; i < x.length; i++)
{
var y = x[i][0];
if (y 9)
y = Math.floor (y / 10);
if (z[y] == null)
z[y] = [x[i][1]];
else
z[y].push (x[i][1]);
}

return z;
}

As you can see, I'm presuming that you meant to truncate 1st col
numbers 9 to the tens (and beyond) digits.
--
_________________________________________
Bob Smith -- bsmith@sudleydeplacespam.com

To reply to me directly, delete "despam".

I have tested (see code below);
but I have some undefinid value.
Now I have:
undefined
undefined
25,27,60,61
undefined
50

anyway the result seem ok;

undefined is my bad code (can you correct?)
but I would like to have the first values 2 and 4
2 25,27,60,61
4 50
-------------
and if I use this
x=[[2,3],[3,4],[1,10],[16,12],[1,15],[2,16],[4,18],[16,19],[1,20],[6,23],[10,25],[10,26],[19,30],[25,32]];

what I must to change in the code?
I think 9 and 10, but change with 27 e 28?
I have now 14 element [ ] in X; is so?


for tests:

function foo ()
{ var var_box=document.frm.my_box;
var var_c=0;

x=[[2,25],[25,27],[27,60],[4,50],[27,61]];

var z=[];

for (var i = 0; i < x.length; i++)
{
var y = x[i][0];
if (y 9)
y = Math.floor (y / 10);
if (z[y] == null)
z[y] = [x[i][1]];
else
z[y].push (x[i][1]);
}

/*return z;*/
for (var ii = 0; ii<z.length; ii++)
{ var_box.options[var_c++]=new Option(z[ii]);}


}


</script>

<form method="post" name="frm" action="">
<select name="my_box" size="10" style="width:200px">
</select>
<br />
<input type="button" value="ok" onclick="foo()" />
</form>

you remember where and how ?
because you are the only solution, I founded at the moment.

On 8/24/2006 10:24 PM, padew wrote:Perhaps I'm confused. If we start with

[[2,25], [25,27], [27,60], [4,50], [27,61]]

what should the result be?

Perhaps you want it to be

[[2, [25,27,60,61]], [4, [50]]]

or is it something else?

--
_________________________________________
Bob Smith -- bsmith@sudleydeplacespam.com

To reply to me directly, delete "despam".

Il Fri, 25 Aug 2006 22:42:10 GMT, Bob Smith ha scritto:
yes is so;

On 8/26/2006 5:32 AM, padew wrote:Try this instead

function foo (x)
{
var z={};

for (var i = 0; i < x.length; i++)
{
var y = x[i][0];
if (y 9)
y = Math.floor (y / 10);
if (y in z)
z[y].push (x[i][1]);
else
z[y] = [x[i][1]];
}

return z;
}

which actually returns

{2:[25,27,60,61], 4:[50]}

which should be easier to use.
--
_________________________________________
Bob Smith -- bsmith@sudleydeplacespam.com

To reply to me directly, delete "despam".

now for test I use this, and see the result on popup;
but I not see similar that your result {2:[25,27,60,61], 4:[50]};
I not see 2 and 4;
for use with alert I chenge x (now inner function)
and z={}; in z=[];

can you write the code that you use for se the results?

my modified
function foo()
{

x=[[2,25],[25,27],[27,60],[4,50],[27,61]];


var z=[];

for (var i = 0; i < x.length; i++)
{
var y = x[i][0];
if (y 9)
{y = Math.floor (y / 10);}
if (y in z)
{z[y].push (x[i][1]);}
else
{z[y] = [x[i][1]];}
}

return alert(z);
}

On 8/26/2006 6:24 PM, padew wrote:Go back to using z={}; and then look at return z.toSource;

--
_________________________________________
Bob Smith -- bsmith@sudleydeplacespam.com

To reply to me directly, delete "despam".