Re: Best way to compute length of arbitrary dimension vector?
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 that is made terrible by our own mad attempt to interpret it as though it had an underlying truth." -- Umberto Eco -- http://mail.python.org/mailman/listinfo/python-list
Re: Best way to compute length of arbitrary dimension vector?
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)" 1 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))" 1 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))" 10 loops, best of 3: 14.4 usec per loop -- http://mail.python.org/mailman/listinfo/python-list
Re: Best way to compute length of arbitrary dimension vector?
> 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. -- http://mail.python.org/mailman/listinfo/python-list
Re: Best way to compute length of arbitrary dimension vector?
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 -- http://mail.python.org/mailman/listinfo/python-list
Re: Best way to compute length of arbitrary dimension vector?
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. -- Algis http://akabaila.pcug.org.au/StructuralAnalysis.pdf -- http://mail.python.org/mailman/listinfo/python-list
Re: Best way to compute length of arbitrary dimension vector?
En Mon, 30 May 2011 06:46:01 -0300, Peter Otten <__pete...@web.de> escribió: 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 -- http://mail.python.org/mailman/listinfo/python-list
Re: Best way to compute length of arbitrary dimension vector?
Thanks a lot to both of you, Chris & Peter! (I knew the solution would be simple ... ;-) ) -- http://mail.python.org/mailman/listinfo/python-list
Re: Best way to compute length of arbitrary dimension vector?
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 -- http://mail.python.org/mailman/listinfo/python-list
Re: Best way to compute length of arbitrary dimension vector?
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 -- http://rebertia.com -- http://mail.python.org/mailman/listinfo/python-list
Best way to compute length of arbitrary dimension vector?
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. -- http://mail.python.org/mailman/listinfo/python-list
