FAQ
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.

## Search Discussions

• at May 30, 2011 at 9:24 am ⇧ 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

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

Cheers,
Chris
• at May 30, 2011 at 9:46 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

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
• at May 30, 2011 at 7:01 pm ⇧ En Mon, 30 May 2011 06:46:01 -0300, Peter Otten <__peter__ at web.de>
escribi?:
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)?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
• at Jun 1, 2011 at 7:35 pm ⇧ 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.
• at May 30, 2011 at 1:38 pm ⇧ Thanks a lot to both of you, Chris & Peter!

(I knew the solution would be simple ... ;-) )
• at Jun 2, 2011 at 9:26 pm ⇧ On Monday 30 May 2011 23:38:53 Gabriel wrote:
Thanks a lot to both of you, Chris & Peter!

(I knew the solution would be simple ... ;-) )
import math
length = math.hypot(z, math.hypot(x, y))

One line and fast.

OldAl.
• at Jun 2, 2011 at 11:19 pm ⇧ 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
• at Jun 3, 2011 at 9:53 pm ⇧ 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.
• at Jun 3, 2011 at 10:17 pm ⇧ 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
• at Jun 3, 2011 at 11:12 pm ⇧ 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
an underlying truth."
-- Umberto Eco