Dave Angel wrote:
Christian Witts wrote:
<div class="moz-text-flowed" style="font-family: -moz-fixed">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
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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
A simpler formula is simply
upper_bounds * (upper_bounds-1) / 2
No check needed for even/odd.
DaveA
Ah yes, that is true. :)
Sometimes I feel I overlook the simplest of solutions.
--
Kind Regards,
Christian Witts
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