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

[Andrey Tatarinov]

Rahul wrote:

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)))

btw, imho the most "Pythonic" would be:

sum(i*j for (i,j) in zip(a,b))

[Nick Coghlan]

Rahul wrote:

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.
Try:

sum(x * y for x, y in zip(a, b))

Between zip() (lockstep iteration over several sequences) and enumerate()
(iteration over a sequence, but also providing an index counter), it is rare
that you will want to use indexing notation in a generator expression.
(By the way profiling them i found that the second is faster by about 30%.)

For short sequences, generator expressions may end up slightly slower than list
comprehensions or for loops, as the latter two do not involve the overhead of
setting up the generator and retrieving values from it. As the sequences
increase in length, generator expressions generally win in the end due to their
reduced memory impact.

Cheers,
Nick.

--
Nick Coghlan | nc******@email. com | Brisbane, Australia
---------------------------------------------------------------
http://boredomandlaziness.skystorm.net

[Peter Otten]

Rahul wrote:

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%.)

You could try

sigma = 0
for ai, bi in itertools.izip( a, b):
sigma += ai * bi

or

sum(itertools.s tarmap(operator .mul, itertools.izip( a, b)))

but if you are really concerned about number-crunching efficiency, use
Numarray.

Peter

[Steven Bethard]

Rahul wrote:

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

.. >>> import numarray as na
.. >>> a, b = na.arange(5), na.arange(5, 10)
.. >>> na.dot(a, b)
.. 80

Steve

[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))

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

[Fredrik Lundh]

Raymond Hettinger wrote:

* applying itertools instead of genexps can save the eval-loop overhead
* however, genexps are usually more readable than itertools solutions

I'm still waiting for you to implement itertools as a parse-tree analyzer/code generator,
rather than an "bare" extension module.

</F>

[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.

[Bengt Richter]

On 19 Dec 2004 03:04:15 -0800, "Rahul" <co********@gma il.com> wrote:

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

Don't know about the timing, but another way:

import operator
a, b = range(5), range(5,10)
sum(map(operato r.mul, a, b)) 80

Checking...
class OpShow(object): ... def __init__(self): self.tot = 0
... def __call__(self, x, y):
... prod = x*y
... self.tot += prod
... print '%3s * %-3s => %s (tot %s)' %(x, y, prod, self.tot)
... return prod
... sum(map(OpShow( ), a, b))

0 * 5 => 0 (tot 0)
1 * 6 => 6 (tot 6)
2 * 7 => 14 (tot 20)
3 * 8 => 24 (tot 44)
4 * 9 => 36 (tot 80)
80

Regards,
Bengt Richter

[John Lenton]

On Sun, Dec 19, 2004 at 03:04:15AM -0800, Rahul wrote:

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

some numbers to confirm my previous reply:

1
zip: 0.00115 (1.00)
range: 0.00108 (0.94)
array: 0.00075 (0.65)
10
zip: 0.00306 (1.00)
range: 0.00288 (0.94)
array: 0.00074 (0.24)
100
zip: 0.02195 (1.00)
range: 0.02035 (0.93)
array: 0.00079 (0.04)
1000
zip: 0.21016 (1.00)
range: 0.19930 (0.95)
array: 0.00130 (0.01)
10000
zip: 4.98902 (1.00)
range: 2.70843 (0.54)
array: 0.01405 (0.00)

(the integers are the number of elements in a random array of floats;
'zip' refers to

sum([x*y for (x,y) in zip(a,b)])

'range', to

sum([a[i]*b[i] for i in range(len(a))])

and 'array' makes a and b Numeric's 'array' objects, with atlas
installed (and hence dotblas loading and assembler version of dot
tuned for the system's processor (in this case a pentium3)). The code
in this case is simply

Numeric.dot(a, b)

The advantage of atlas on systems with sse2 should be even greater.

--
John Lenton (jo**@grulic.or g.ar) -- Random fortune:
El deseo nos fuerza a amar lo que nos hará sufrir.
-- Marcel Proust. (1871-1922) Escritor francés.

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