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 npimport numba as nbdef 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]: TrueIn [8]: %timeit func()1 loops, best of 3: 222 ms per loopIn [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。
