Faster way to perform function calculation in Python?

This continues from Jaime's post.

This can be trivially sped up to \$\mathcal{O}(1)\$ by caching sum(1/j for j in range(1, 10001)) = 9.787606036044348. Then one just does

def func_O1(mass, density):
    # 9.7876... = sum(1/j for j in range(1, 10001))
    return log(mass * density) * 9.787606036044348

This gives a speedup by a factor of 3000.

If the limit (10000) is varied, use an approximation for the harmonic series:

from math import log

def harmonic(n):
    gamma = 0.57721566490153286060651209008240243104215933593992
    return gamma + log(n) + 0.5/n - 1./(12*n**2) + 1./(120*n**4)

def func_fastest(mass, density):
    return log(mass * density) * harmonic(10000)

harmonic is inaccurate below \$n \approx 120\$ so it would make sense to cache exact values up to harmonic(120). Note that whilst this is an approximation, for \$n\$ greater than \$120\$ it's exact \$\pm 1\text{eps}\$ whereas the summation gets steadily worse.

This only speeds up the function by a factor of 500.

The speed-up you are seeing is mostly from removing the calculation of the logarithm from inside the inner loop, and instead reusing the value. This performs only about 10% slower than your fast_func, and is infinitely more readable:

def func_slow(mass, density):
    total = 0
    masslog = log(mass*density)
    for j in range(1, 10001):
        total += 1. / j
    return masslog * total

And if you want to squeeze those last drops of performance, the speed of the following code is marginally, if at all, slower than fast_func, and is also much more Pythonic and readable:

def func(mass, density):
    return log(mass * density) * sum(1/j for j in range(1, 10001))

While they stayed in the language, and are therefore fair game, reading Guido's 10 year old essay on how he planned to get rid of reduce, map, lambda and filter for Python 3 will probably have you looking at those a little differently: I for one always try to come up with more Pythonic alternatives when tempted to use them.