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Re: pairs from a list

Matthew_WARREN

1/23/2008 5:46:00 PM


I'm just fiddling with this, am no great expert, but I added

def pairs5(x):
o=[]
for n in zip(x[::2],x[1:2]):
o.append(n)
return o

I dont know if that breaks any constraints placed on the problem, but I get
the following output

0.07158942896 0.266009705575 0.21342143313 0.0537146193457 0.0107680502972


does that make it faster than pairs(4) or am I breaking some constraints on
the problem?



Matt.






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Sent by: cc
python-list-bounces+matthew.warren=uk.bnpparibas.com@
python.org Subject
Re: pairs from a list
22/01/2008 18:09










Arnaud Delobelle wrote:
> pairs4 wins.


Oops. I see a smaller difference,
but yes, pairs4 wins.

Alan Isaac

import time
from itertools import islice, izip

x = range(500001)

def pairs1(x):
return izip(islice(x,0,None,2),islice(x,1,None,2))

def pairs2(x):
xiter = iter(x)
while True:
yield xiter.next(), xiter.next()

def pairs3(x):
for i in range( len(x)//2 ):
yield x[2*i], x[2*i+1],

def pairs4(x):
xiter = iter(x)
return izip(xiter,xiter)

t = time.clock()
for x1, x2 in pairs1(x):
pass
t1 = time.clock() - t

t = time.clock()
for x1, x2 in pairs2(x):
pass
t2 = time.clock() - t

t = time.clock()
for x1, x2 in pairs3(x):
pass
t3 = time.clock() - t

t = time.clock()
for x1, x2 in pairs4(x):
pass
t4 = time.clock() - t

print t1, t2, t3, t4

Output:
0.317524154606 1.13436847421 1.07100930426 0.262926712753
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1 Answer

Peter Otten

1/23/2008 6:32:00 PM

0

Matthew_WARREN wrote:

> I'm just fiddling with this, am no great expert, but I added
>
> def pairs5(x):
> o=[]
> for n in zip(x[::2],x[1:2]):

The second argument should be x[1::2].

> o.append(n)
> return o
>
> I dont know if that breaks any constraints placed on the problem, but I

It breaks the constraints unless the above is not cut-n-paste ;)
Also note that your pairs5() offers no advantage over

def pairs6(x):
return zip(x[::2], x[1::2])

Peter