How come these Python codes perform so much differently

Question

Please look at the following code to solve the same set of problem, I do not think that mentioning the problem would anyhow help the purpose, it's yet another iteration of the Josephus problem:

Solution 1:

import sys
from math import log

cases= int(sys.stdin.readline())
current= 0
while current < cases:
    current += 1
    n = int(sys.stdin.readline())
    print 2*(n - 2**(int(log(n,2))))+1

This solution solves the given 10 test cases in cumulative 1.0912 seconds and consumes 4360KiB of memory.

Solution 2:

def josephus_2( n ):
    from math import log
    return 2*(n - 2**(int(log(n,2))))+1

import sys
cases= int(sys.stdin.readline())
current= 0
while current < cases:
    current += 1
    n = int(sys.stdin.readline())
    print josephus_2( n )

this solution solves the same 10 test cases in a total of 1.0497 seconds and 640KiB of memory.

Being a Python n00b I was wondering, while according to the online judge I earn same points for both, but what makes the solution 2 more faster than 1 and much more memory efficient? I know that the time difference can sound very less, but at the same time ranks me first on the fastest solution, even faster than c/c++/perl submissions

Can this Screenshot help?

Answer 1

In my old experiences I remember that I had found that putting computation parts in a function (method) sometime may improve the performance:
I just re-experienced using the following simple code:

n = 1000000
def compute(n):
    j = 0
    for i in xrange(n):
        j += 1

tic()
compute(n)
toc()

>>> 0.271 s

tic()
j = 0
for i in xrange(n):
    j += 1
toc()

>>> 0.849 s

The result shows 0.271s for the first (using compute) vs 0.849s as inline code. This is noticeable improvement without changing anything in the main computation part! So the point is using a method may improve the performance.

Here are the code you may use for performance comparison:

from __future__ import division
from time import clock

def compute(n):
    j = 0
    for i in xrange(n):
        j += 1

n = 1000000
print 'solution (2) using a method call...'
t0 = clock()
compute(n)
print clock()-t0
#>>> ~0.2415...                              #faster (solution 2)

print 'solution (1) all inline code...'
t0 = clock()
j = 0
for i in xrange(n):
    j += 1
print clock()-t0
#>>> ~0.6169...                              #slower (solution 1)