使用numba来加速循环

Question

From what I've read, numba can significantly speed up a python program. Could my program's time efficiency be increased using numba?

import numpy as np

def f_big(A, k, std_A, std_k, mean_A=10, mean_k=0.2, hh=100):
    return ( 1 / (std_A * std_k * 2 * np.pi) ) * A * (hh/50) ** k * np.exp( -1*(k - mean_k)**2 / (2 * std_k **2 ) - (A - mean_A)**2 / (2 * std_A**2))

outer_sum = 0
dk = 0.000001
for k in np.arange(dk,0.4, dk):
    inner_sum = 0
    for A in np.arange(dk, 20, dk):
        inner_sum += dk * f_big(A, k, 1e-5, 1e-5)
    outer_sum += inner_sum * dk

print outer_sum

Answer 1

Yes, this is the sort of problem that Numba really works for. I changed your value of dk because it wasn't sensible for a simple demonstration. Here is the code:

import numpy as np
import numba as nb

def f_big(A, k, std_A, std_k, mean_A=10, mean_k=0.2, hh=100):
    return ( 1 / (std_A * std_k * 2 * np.pi) ) * A * (hh/50) ** k * np.exp( -1*(k - mean_k)**2 / (2 * std_k **2 ) - (A - mean_A)**2 / (2 * std_A**2))

def func():
    outer_sum = 0
    dk = 0.01 #0.000001
    for k in np.arange(dk, 0.4, dk):
        inner_sum = 0
        for A in np.arange(dk, 20, dk):
            inner_sum += dk * f_big(A, k, 1e-5, 1e-5)
        outer_sum += inner_sum * dk

    return outer_sum

@nb.jit(nopython=True)
def f_big_nb(A, k, std_A, std_k, mean_A=10, mean_k=0.2, hh=100):
    return ( 1 / (std_A * std_k * 2 * np.pi) ) * A * (hh/50) ** k * np.exp( -1*(k - mean_k)**2 / (2 * std_k **2 ) - (A - mean_A)**2 / (2 * std_A**2))

@nb.jit(nopython=True)
def func_nb():
    outer_sum = 0
    dk = 0.01 #0.000001
    X = np.arange(dk, 0.4, dk)
    Y = np.arange(dk, 20, dk)
    for i in xrange(X.shape[0]):
        k = X[i] # faster to do lookup than iterate over an array directly
        inner_sum = 0
        for j in xrange(Y.shape[0]):
            A = Y[j]
            inner_sum += dk * f_big_nb(A, k, 1e-5, 1e-5)
        outer_sum += inner_sum * dk

    return outer_sum

And then timings:

In [7]: np.allclose(func(), func_nb())
Out[7]: True

In [8]: %timeit func()
1 loops, best of 3: 222 ms per loop

In [9]: %timeit func_nb()
The slowest run took 419.10 times longer than the fastest. This could mean that an intermediate result is being cached 
1000 loops, best of 3: 362 µs per loop

So the numba version is approx 600 times faster on my laptop.