Finding subsets for a robust regression

[Gerard flanagan]

Gerard flanagan wrote:
> tkpmep at hotmail.com wrote:
> 
>>
>>     x1 = []                 #unique instances of x and y
>>     y1 = []                 #median(y) for each unique value of x
>>     for xx,yy in d.iteritems():
>>         x1.append(xx)
>>         l = len(yy)
>>         if l == 1:
>>             y1.append(yy[0])
>>         else:
>>             yy.sort()
>>             y1.append( (yy[l//2-1] + yy[l//2])/2.0 if l % 2 == 0 else
>> yy[l//2] )
>> -- 
> 
> Not tested, but is this equivalent?
> 
> x1 = []
> y1 = []
> for xx, yy in d.iteritems():
>     L = len(yy) // 2
>     yy.sort()
>     y1.append(yy[L])
>     if not L & 1:
>         y1[-1] = (y1[-1] + yy[L-1]) / 2.0
> 
> It means that you have a pointless 'sort' when len(yy) == 1, but then 
> you save an 'if' per-iteration.
> 
> G.
> 
> -- 

Maybe that should be:

if L and not L & 1:
    ...

anyway...