Best way to compute length of arbitrary dimension vector?

[Gabriel]

> The dimension is arbitrary, though, so:
>
> length = reduce(math.hypot, self._coords, 0)
>


Thanks, I was going to ask Algis that same question.

But still, is this solution really faster or better than the one using
list comprehension and the expression 'x*x'?
It seems to me that the above solution (using hypot) involves repeated
square roots (with subsequent squaring).

Best regards,
Gabriel.