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Date: Mon, 30 May 2011 02:24:17 -0700
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Subject: Re: Best way to compute length of arbitrary dimension vector?
From: Chris Rebert <clp2@rebertia.com>
To: Gabriel <snoopy.67.z@googlemail.com>
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On Mon, May 30, 2011 at 2:11 AM, Gabriel <snoopy.67.z@googlemail.com> wrote=
:
> Well, the subject says it almost all: I'd like to write a small Vector
> class for arbitrary-dimensional vectors.
>
> I am wondering what would be the most efficient and/or most elegant
> way to compute the length of such a Vector?
>
> Right now, I've got
>
> =C2=A0def length(self): =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=
=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=
=A0 =C2=A0 =C2=A0 =C2=A0 # x.length() =3D || x ||
> =C2=A0 =C2=A0total =3D 0.0
> =C2=A0 =C2=A0for k in range(len(self._coords)):
> =C2=A0 =C2=A0 =C2=A0d =3D self._coords[k]
> =C2=A0 =C2=A0 =C2=A0total +=3D d*d
> =C2=A0 =C2=A0return sqrt(total)
>
> However, that seems a bit awkward to me (at least in Python ;-) ).
>
> I know there is the reduce() function, but I can't seem to find a way
> to apply that to the case here (at least, not without jumping through
> too many hoops).
>
> I have also googled a bit, but found nothing really elegant.
>
> Any ideas?

def length(self):
    return sqrt(sum(coord*coord for coord in self._coords))

Cheers,
Chris
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