# Best way to compute length of arbitrary dimension vector?

Discussion in 'Python' started by Gabriel, May 30, 2011.

1. ### GabrielGuest

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

def length(self): # x.length() = || x ||
total = 0.0
for k in range(len(self._coords)):
d = self._coords[k]
total += d*d
return 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?

Best regards,
Gabriel.

Gabriel, May 30, 2011

2. ### Chris RebertGuest

On Mon, May 30, 2011 at 2:11 AM, Gabriel <> 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
>
> Â def length(self): Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  # x.length() = || x ||
> Â  Â total = 0.0
> Â  Â for k in range(len(self._coords)):
> Â  Â  Â d = self._coords[k]
> Â  Â  Â total += d*d
> Â  Â return 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
--
http://rebertia.com

Chris Rebert, May 30, 2011

3. ### Peter OttenGuest

Gabriel 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
>
> def length(self): # x.length() = || x ||
> total = 0.0
> for k in range(len(self._coords)):
> d = self._coords[k]
> total += d*d
> return 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.

>>> class Vector(object):

.... def __init__(self, *coords):
.... self._coords = coords
.... def __abs__(self):
.... return math.sqrt(sum(x*x for x in self._coords))
....
>>> import math
>>> abs(Vector(1,1))

1.4142135623730951
>>> abs(Vector(3,4))

5.0

Peter Otten, May 30, 2011
4. ### GabrielGuest

Thanks a lot to both of you, Chris & Peter!

(I knew the solution would be simple ... ;-) )

Gabriel, May 30, 2011
5. ### Gabriel GenellinaGuest

En Mon, 30 May 2011 06:46:01 -0300, Peter Otten <>
escribiÃ³:

> Gabriel wrote:
>
>> Well, the subject says it almost all: I'd like to write a small Vector
>> class for arbitrary-dimensional vectors.
>>

>
>>>> class Vector(object):

> ... def __init__(self, *coords):
> ... self._coords = coords
> ... def __abs__(self):
> ... return math.sqrt(sum(x*x for x in self._coords))
> ...
>>>> import math
>>>> abs(Vector(1,1))

> 1.4142135623730951
>>>> abs(Vector(3,4))

> 5.0

Using math.fsum instead of sum may improve accuracy, specially when
len(coords)â‰«2

py> import math
py>
py> def f1(*args):
.... return math.sqrt(sum(x*x for x in args))
....
py> def f2(*args):
.... return math.sqrt(math.fsum(x*x for x in args))
....
py> pi=math.pi
py> args=[pi]*16
py> abs(f1(*args)/4 - pi)
4.4408920985006262e-16
py> abs(f2(*args)/4 - pi)
0.0

--
Gabriel Genellina

Gabriel Genellina, May 30, 2011
6. ### GabrielGuest

> py> def f2(*args):
> ...   return math.sqrt(math.fsum(x*x for x in args))

Wow! Thanks a million - I didn't now about math.fsum!

Best regards,
Gabriel.

Gabriel, Jun 1, 2011
7. ### Ian KellyGuest

On Thu, Jun 2, 2011 at 3:26 PM, Algis Kabaila <> wrote:
> import math
>
> length = math.hypot(z, math.hypot(x, y))
>
> One line and fast.

The dimension is arbitrary, though, so:

length = reduce(math.hypot, self._coords, 0)

Cheers,
Ian

Ian Kelly, Jun 3, 2011
8. ### GabrielGuest

> 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.

Gabriel, Jun 3, 2011
9. ### Ian KellyGuest

On Fri, Jun 3, 2011 at 3:53 PM, Gabriel <> wrote:
> But still, is this solution really faster or better than the one using
> list comprehension and the expression 'x*x'?

No, not really.

>c:\python32\python -m timeit -s "coords = list(range(100))" -s "from math import hypot" -s "from functools import reduce" "reduce(hypot, coords, 0)"

10000 loops, best of 3: 53.2 usec per loop

>c:\python32\python -m timeit -s "coords = list(range(100))" -s "from math import sqrt, fsum" "sqrt(fsum(x*x for x in coords))"

10000 loops, best of 3: 32 usec per loop

>c:\python32\python -m timeit -s "coords = list(range(100))" -s "from math import sqrt" "sqrt(sum(x*x for x in coords))"

100000 loops, best of 3: 14.4 usec per loop

Ian Kelly, Jun 3, 2011
10. ### Robert KernGuest

On 6/3/11 4:53 PM, Gabriel wrote:
>
>> 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).

It also means that the floating point numbers stay roughly the same size, so you
will lose less precision as the number of elements goes up. I don't expect the
number of elements will be large enough to matter, though.

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma