Thanks Ely for the hack

The pre-multiplying is straightforward and combined with adding
<mx:LinearAxis minimum="-22" maximum="-1"/>
gives me the chart I am looking for but the axis looks a bit tacky
with all the negative numbers

However, I am having trouble appling a labelFunction
Attempting to adapt the example in livedocs
 http://livedocs.macromedia.com/labs/1/flex/00001287.html

public function defineLabel(catName:String):String {
        return catName + " '04";
    }

getting an error
Type Coersion failed: cannot convert 22 to String

I have played around with bringing the data across both as a string
and a number but cannot seem to get it right -even get the same error
when brought in as number and replace string with number as to
instances above

e.g.
public function makePositive(catName:Number):Number {
        return -1*catName;
    }   

Incidentally, do you know if therr are there plans to incorporate a
reverse axis option in later releases or somewhere I could add it to a
wish list




--- In [email protected], "Ely Greenfield" <[EMAIL PROTECTED]> wrote:
>
> 
> 
> No, the axes don't really support inversion.
> 
> The easiest way would probably be:
> 
> 1) pre-multiply your values by -1.
> 2) put a labelFunction on your vertical AxisRenderer that multiplies the
> label value by -1 to make it look like it's positive.
> 3) if you want datatips, you'll have to add a dataTipFunction to do the
> same.
> 
> Ely.
> 
> 
> -----Original Message-----
> From: [email protected] [mailto:[EMAIL PROTECTED] On
> Behalf Of yaagcur
> Sent: Friday, November 04, 2005 9:24 AM
> To: [email protected]
> Subject: [flexcoders] Reverse yaxis (FLEX 2)
> Importance: High
> 
> I want to do a chart which is a ranking of team's position (y axis) in a
> league  over time (x-axis). 
> I would like to reverse the y axis so that if a team is improving over
> time (say from rank 4 to rank 2) the graph shows an upward movement
> rather than the downward movement that results from a straight input of
> the data Unsurprisngly, just having a lower maximum than minimum in the
> LinearAxis does not do the trick Is this possible another way?
> 
> 
> 
> 
> 
> 
> 
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