[Tutor] Loop comparison

Christian Witts cwitts at compuscan.co.za
Fri Apr 16 09:48:08 CEST 2010


Ark wrote:
> Hi everyone.
> A friend of mine suggested me to do the next experiment in python and Java.
>
> It's a simple program to sum all the numbers from 0 to 1000000000.
>
> result = i = 0
> while i < 1000000000:
>     result += i
>     i += 1
> print result
>
> The time for this calculations was huge.  It took a long time to give
> the result.  But, the corresponding program in Java takes less than 1
> second to end.  And if in Java, we make a simple type check per cycle,
> it does not take more than 10 seconds in the same machine.  I was not
> expecting Python to be faster than Java, but it''s too slow.  Maybe
> Java optimizes this case and Python doesn't.  Not sure about this.}
>
> Thanks
> ark
> _______________________________________________
> Tutor maillist  -  Tutor at python.org
> To unsubscribe or change subscription options:
> http://mail.python.org/mailman/listinfo/tutor
>
>   
Different methods and their relative benchmarks.  The last two functions 
are shortcuts for what you are trying to do, the last function 't5' 
corrects the mis-calculation 't4' has with odd numbers.
Remember, if you know a better way to do something you can always 
optimize yourself ;)

 >>> def t1(upper_bounds):
...   start = time.time()
...   total = sum((x for x in xrange(upper_bounds)))
...   end = time.time()
...   print 'Time taken: %s' % (end - start)
...   print total
...
 >>> t1(1000000000)
Time taken: 213.830082178
499999999500000000
 >>> def t2(upper_bounds):
...   total = 0
...   start = time.time()
...   for x in xrange(upper_bounds):
...     total += x
...   end = time.time()
...   print 'Time taken: %s' % (end - start)
...   print total
...
 >>> t2(1000000000)
Time taken: 171.760597944
499999999500000000
 >>> def t3(upper_bounds):
...   start = time.time()
...   total = sum(xrange(upper_bounds))
...   end = time.time()
...   print 'Time taken: %s' % (end - start)
...   print total
...
 >>> t3(1000000000)
Time taken: 133.12481904
499999999500000000
 >>> def t4(upper_bounds):
...   start = time.time()
...   mid = upper_bounds / 2
...   total = mid * upper_bounds - mid
...   end = time.time()
...   print 'Time taken: %s' % (end - start)
...   print total
...
 >>> t4(1000000000)
Time taken: 1.4066696167e-05
499999999500000000
 >>> def t5(upper_bounds):
...   start = time.time()
...   mid = upper_bounds / 2
...   if upper_bounds % 2:
...     total = mid * upper_bounds
...   else:
...     total = mid * upper_bounds - mid
...   end = time.time()
...   print 'Time taken: %s' % (end - start)
...   print total
...
 >>> t5(1000000000)
Time taken: 7.15255737305e-06
499999999500000000
 >>> t3(1999)
Time taken: 0.0038161277771
1997001
 >>> t4(1999)
Time taken: 3.09944152832e-06
1996002
 >>> t5(1999)
Time taken: 3.09944152832e-06
1997001

-- 
Kind Regards,
Christian Witts
Business Intelligence

C o m p u s c a n | Confidence in Credit

Telephone: +27 21 888 6000
National Cell Centre: 0861 51 41 31
Fax: +27 21 413 2424
E-mail: cwitts at compuscan.co.za

NOTE:  This e-mail (including attachments )is subject to the disclaimer published at: http://www.compuscan.co.za/live/content.php?Item_ID=494.
If you cannot access the disclaimer, request it from email.disclaimer at compuscan.co.za or 0861 514131.

National Credit Regulator Credit Bureau Registration No. NCRCB6 




More information about the Tutor mailing list