dot products

[Rahul]

Raymond Hettinger wrote:
> [Rahul].
> > I want to compute dot product of two vectors stored as lists a and
b.a
> > and b are of the same length.
> >
> > one simple way is
> > sum(a[i]*b[i] for i in range(len(a)))
> >
> > another simple way is
> > ans=0.0
> > for i in range(len(a)):
> > ans=ans+a[i]*b[i]
> >
> > But is there any other way which is faster than any of the above.
>
> Yes:
>      from itertools import imap
>      from operator import mul
>      ans = sum(imap(mul, a, b))
>
> In general:
> * reduction functions like sum() do not need their arguments to
>    take time building a full list; instead, an iterator will do fine
> * applying itertools instead of genexps can save the eval-loop
overhead
> * however, genexps are usually more readable than itertools solutions
> * xrange() typically beats range()
> * but indexing is rarely the way to go
> * izip() typically beats zip()
> * imap() can preclude the need for either izip() or zip()
> * the timeit module settles these questions quickly
>
> Here are the some timings for vectors of length 10 and 3 respectively
>
> C:\pydev>python timedot.py 3
> 1.25333310984 sum(a[i]*b[i] for i in xrange(len(a)))
> 1.16825625639 sum(x*y for x,y in izip(a,b))
> 1.45373455807 sum(x*y for x,y in zip(a,b))
> 0.635497577901 sum(imap(mul, a, b))
> 0.85894416601 sum(map(mul, a, b))
>
> C:\pydev>python timedot.py 10
> 2.19490353509 sum(a[i]*b[i] for i in xrange(len(a)))
> 2.01773998894 sum(x*y for x,y in izip(a,b))
> 2.44932533231 sum(x*y for x,y in zip(a,b))
> 1.24698871922 sum(imap(mul, a, b))
> 1.49768685362 sum(map(mul, a, b))
>
>
>
> Raymond Hettinger
Thanks all of you guys for enlightening me. Python is truly elegant.