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.
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