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Old 11-20-2011, 10:58 PM   #1
ColInvictus
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Registered: Apr 2009
Posts: 42

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Python: equivalent programs with different runtimes


I'm have two programs which calculate the sum of all numbers which are expressible as the sum of two abundant numbers. They are (1):
Code:
from mytools import factors

ab = []
for n in range(12, 28124):
    if sum(factors(n)[:-1])>n:
        ab.append(n)

nums = []
for n in range(len(ab)):
    for m in range(n, len(ab)):
        v = ab[n]+ab[m]
        if v > 28123: break
        nums.append(v)

print sum(range(28124))-sum(list(set(nums)))
and (2):
Code:
from mytools import factors

def ab(lim):
    nums = []
    for n in range(12, lim+1):
        if sum(factors(n)[:-1])>n:
            nums.append(n)
    return nums

def nm(lst):
    nums = []
    for n in range(len(lst)):
        for m in range(n, len(lst)):
            v = lst[n]+lst[m]
            if v > 28123: break
            nums.append(v)
    return nums

abundants = ab(28123)
nums = nm(abundants)

print sum(range(28124))-sum(list(set(nums)))
They are basically equivalent, except that the loops are enclosed in functions in (2). (2) runs in <10 seconds, whereas (1) takes almost a minute. Can anyone tell me why this is?
 
Old 11-21-2011, 01:00 AM   #2
firstfire
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Registered: Mar 2006
Location: Ekaterinburg, Russia
Distribution: Debian, Ubuntu
Posts: 709

Rep: Reputation: 428Reputation: 428Reputation: 428Reputation: 428Reputation: 428
Hi.
It's strange -- on my machine both scripts take about 0.350s and return 395465626. To compute factors I use code from here:
Code:
"""
  Generator for getting factors for a number
"""
def factor(n):
  yield 1
  i = 2
  limit = n**0.5
  while i <= limit:
    if n % i == 0:
      yield i
      n = n / i
      limit = n**0.5
    else:
      i += 1
  if n > 1:
    yield n

def factors(n):
	return [i for i in factor(n)]
      
if __name__ == "__main__":
  import sys
  for index in xrange(1,len(sys.argv)):
    print "Factors for %s : %s" %(sys.argv[index], [i for i in factor(int(sys.argv[index]))])
How do you compute factors?
 
Old 11-22-2011, 05:39 PM   #3
ColInvictus
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Registered: Apr 2009
Posts: 42

Original Poster
Rep: Reputation: 15
Behold, my factors function!
Code:
def factors(n):
    """Returns an ordered list of factors of n"""
    sqrtn = sqrt(n)
    facts = not sqrtn%1 and [int(sqrtn)] or []
    for i in range(int(ceil(sqrtn))-1, 0, -1):
        if n%i == 0:
            facts = [i] + facts + [n/i]
    return facts
Anyhoo, it turns out that the problem is with the Spyder IDE; running the scripts from the command line results in much more similar timings. The first one still takes a couple of seconds longer for reasons I don't quite understand, but at least it's not several times as long.

Time for a new IDE I guess...
 
Old 12-01-2011, 02:44 PM   #4
ColInvictus
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Registered: Apr 2009
Posts: 42

Original Poster
Rep: Reputation: 15
In case anyone's interested, it was the variable explorer in Spyder trying yo keep track of large lists as the code was running. The developer has fixed the problem in the next release, apparently.
 
Old 12-01-2011, 03:04 PM   #5
firstfire
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Registered: Mar 2006
Location: Ekaterinburg, Russia
Distribution: Debian, Ubuntu
Posts: 709

Rep: Reputation: 428Reputation: 428Reputation: 428Reputation: 428Reputation: 428
Hi.
Glad to hear you found the root of all evil.
 
  


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