是的，这是Numba真正解决的问题。我更改了您的价值，dk因为对于简单的演示而言，这并不明智。这是代码：
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
然后计时：
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
因此，numba版本在我的笔记本电脑上大约快600倍。
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