Tim Peters <t...@python.org> added the comment:
I'd be +1 on generalizing math.hypot to accept an arbitrary number of
arguments. It's the natural building block for computing distance, but the
reverse is strained. Both are useful.
Here's scaling code translated from the Fortran implementation of SNRM2 in
LAPACK. It only looks at each element once, so would work almost as-is for an
argument that's an arbitrary iterable (just change the formal argument from
`*xs` to `xs`):
# http://www.netlib.org/lapack/explore-html/d7/df1/snrm2_8f_source.html
def hypot(*xs):
import math
scale = 0.0
sumsq = 1.0
for x in xs:
if x:
absx = abs(x)
if scale < absx:
sumsq = 1.0 + sumsq * (scale / absx)**2
scale = absx
else:
sumsq += (absx / scale)**2
return scale * math.sqrt(sumsq)
I believe it does the right with infinities by magic (return math.inf). It
returns 0.0 if no arguments are passed, which makes sense. For one argument,
it's a convoluted way to return its absolute value. The "if x:" test is
necessary to prevent division by 0.0 if the first argument is 0.0. It would
need another blob of code to give up (and return math.nan) if one of the
arguments is a NaN (there's no guessing what various C compilers would do
without special-casing NaN).
I doubt `fsum()` would add much value here: all the addends have the same
sign, so cancellation is impossible.
To get really gonzo, a single-rounding dot product would be the best building
block (the vector norm is the square root of the dot product of a vector with
itself). `fsum()` is a special case of that (the dot product of a vector with
a same-length vector of all ones).
----------
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Python tracker <rep...@bugs.python.org>
<https://bugs.python.org/issue33089>
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