dot products

[Rahul]

HI.
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. (By
the way profiling them i found that the second is faster by about 30%.)
rahul

[Raymond Hettinger]

[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))

Doh! We all missed it.

If your vector length is known in advance (and it often is in some apps), the
simplest and fastest approach is:

a[0]*b[0] + a[1]*b[1] + a[2]*b[2]
C:\pydev>python timedot.py 3
0.32 sec: a[0]*b[0] + a[1]*b[1] + a[2]*b[2]
1.38 sec: sum(a[i]*b[i] for i in xrange(len(a)))
1.32 sec: sum(x*y for x,y in izip(a,b))
1.62 sec: sum(x*y for x,y in zip(a,b))
0.75 sec: sum(imap(mul, a, b))
1.04 sec: sum(map(mul, a, b))

Raymond Hettinger

[Alan G Isaac]

[Rahul].

I want to compute dot product of two vectors stored as lists a and b.a
and b are of the same length

from scipy import dot
ans=dot(a,b)

This times faster than the alternatives I have seen mentioned so far,
given scipy.

Cheers,
Alan Isaac

[Alan G Isaac]

"Alan G Isaac" <ai****@america n.edu> wrote in message
news:10******** *****@corp.supe rnews.com...

This times faster than the alternatives I have seen mentioned so far,
given scipy.

Actually, since I am new to 'timeit', I probably should check that
I am not overlooking something. Especially since I see an order
of magnitude difference in performance. Does the code below
give the right comparisons?

Thanks,
Alan Isaac

#-------------------------------------------------------------
import timeit
env1='''
from operator import mul
from itertools import imap
def innerprod(x,y):
return sum(imap(mul,x, y))
from scipy import rand
x=rand(50); y=rand(50)
'''
env2='''
from operator import mul
from itertools import imap
from scipy import rand
x=rand(50); y=rand(50)
'''
env3='''
from scipy import rand,dot
x=rand(50); y=rand(50)
'''
t1=timeit.Timer ("innerprod(x,y )",env1)
t2=timeit.Timer ("sum(imap(mul, x,y))",env2)
t3=timeit.Timer ("dot(x,y)",env 3)
trials=1000
print t1.repeat(2,tri als) #about 0.1 seconds
print t2.repeat(2,tri als) #about 0.1 seconds
print t3.repeat(2,tri als) #about 0.01 seconds