On Thursday, July 18, 2013 10:43:11 PM UTC-4, Dave Angel wrote:
> On 07/18/2013 10:16 PM, CTSB01 wrote:

> > Does something like
> 
> >
> 
> > def phi_m(x, m):
> 
> >            rtn = []
> 
> >            for n2 in range(0, len(x) * m - 2):
> 
> >              n = n2 / m
> 
> >              r = n2 - n * m
> 
> >              rtn.append(m * x[n] + r * (x[n + 1] - x[n]))
> 
> >              print ('n2 =', n2, ': n =', n, ' r =' , r, ' rtn =', rtn)
> 
> >            return rtn
> > look right?
> 
> No, as Ian has pointed out, you need to use the // operator in Python 3 
> 
> if you want to get integer results.  So it'd be n = n2 // m
> 
> However, even better is to use the divmod() function, which  is intended 
> 
> for the purpose:
> 
> 
> 
>      n, r = divmod(n2, m)
> 
> > It doesn't seem to have any errors.  However, I do receive the following 
> > error when trying to implement an x after having defined phi:
> 
> >>>> x = [0, 1, 1, 2, 3]
> 
> >>>> phi_m(x, 2)
> 
> > Traceback (most recent call last):
> 
> >    File "<pyshell#6>", line 1, in <module>
> 
> >      phi_m(x, 2)
> 
> >    File "<pyshell#2>", line 6, in phi_m
> 
> >      rtn.append(m * x[n] + r * (x[n + 1] - x[n]))
> 
> > TypeError: list indices must be integers, not float
> 
> 
> That will be fixed if you correct the code as I described, so you'll get 
> 
> integers.
> 
> There is a Brezenham algorith that might accomplish this whole function 
> 
> more accurately, or more efficiently.  But I'd have to re-derive it, as 
> 
> it's been about 30 years since I used it.  And chances are that the 
> 
> efficiencies it brought to machine code won't matter much here.
> 
> -- 
> 
> DaveA

Thanks Dave, I'll take a look at that.
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