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# dot products

 P: n/a 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 Jul 18 '05 #1
12 Replies

 P: n/a 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)) Jul 18 '05 #2

 P: n/a 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 Jul 18 '05 #3

 P: n/a 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.starmap(operator.mul, itertools.izip(a, b))) but if you are really concerned about number-crunching efficiency, use Numarray. Peter Jul 18 '05 #4

 P: n/a 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 Jul 18 '05 #5

 P: n/a [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 Jul 18 '05 #6

 P: n/a 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. Jul 18 '05 #7

 P: n/a 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. Jul 18 '05 #8

 P: n/a On 19 Dec 2004 03:04:15 -0800, "Rahul" wrote: HI.I want to compute dot product of two vectors stored as lists a and b.aand b are of the same length.one simple way issum(a[i]*b[i] for i in range(len(a)))another simple way isans=0.0for 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. (Bythe 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(operator.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 Jul 18 '05 #9

 P: n/a 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.org.ar) -- Random fortune: El deseo nos fuerza a amar lo que nos hará sufrir. -- Marcel Proust. (1871-1922) Escritor francÃ©s. -----BEGIN PGP SIGNATURE----- Version: GnuPG v1.2.5 (GNU/Linux) iD8DBQFBxv1BgPqu395ykGsRAvuKAJkBULU763LN368mVFJf+t rqku8/KQCbB75C noYAuTFkRh4SJ3VmJCXpwZY= =F9aS -----END PGP SIGNATURE----- Jul 18 '05 #10

 P: n/a [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*b + a*b + a*b C:\pydev>python timedot.py 3 0.32 sec: a*b + a*b + a*b 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 Jul 18 '05 #11

 P: n/a [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 Jul 18 '05 #12

 P: n/a "Alan G Isaac" wrote in message news:10*************@corp.supernews.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)",env3) trials=1000 print t1.repeat(2,trials) #about 0.1 seconds print t2.repeat(2,trials) #about 0.1 seconds print t3.repeat(2,trials) #about 0.01 seconds Jul 18 '05 #13

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