使用Numba來改善有限差分拉普拉斯算子

Question

我使用Python來求解方程的反應擴散系統（Fitz-Hugh-Nagumo model）。我想了解如何使用Numba來加速計算。我目前進口下列laplacian.py模塊到我的集成腳本：

def neumann_laplacian_1d(u,dx2): 
    """Return finite difference Laplacian approximation of 2d array. 
    Uses Neumann boundary conditions and a 2nd order approximation. 
    """ 
    laplacian = np.zeros(u.shape) 
    laplacian[1:-1] = ((1.0)*u[2:] 
         +(1.0)*u[:-2] 
         -(2.0)*u[1:-1]) 
    # Neumann boundary conditions 
    # edges 
    laplacian[0] = ((2.0)*u[1]-(2.0)*u[0]) 
    laplacian[-1] = ((2.0)*u[-2]-(2.0)*u[-1]) 

    return laplacian/ dx2 

凡u是NumPy的一維數組，它代表的領域之一。我試圖在導入from numba import autojit後添加修飾符@autojit(target="cpu")。計算中我沒有看到任何改進。任何人都可以給我一個提示，在這種情況下如何正確使用Numba？

我在這裏使用的輸入數組是

a = random.random(252) 

所以我比較了線的表現：

%timeit(neumann_laplacian_1d(a,1.0)) 

隨着Numba我：

%timeit(neumann_laplacian_1d(a,1.0)) 
The slowest run took 22071.00 times longer than the fastest. This could mean that an intermediate result is being cached 
1 loops, best of 3: 14.1 µs per loop 

沒有Numba我得到了（!!）：

%timeit(neumann_laplacian_1d(a,1.0)) 
The slowest run took 11.84 times longer than the fastest. This could mean that an intermediate result is being cached 
100000 loops, best of 3: 9.12 µs per loop 

Numba實際上使它變得更慢..

Answer 1

我無法複製你的結果。

Python版本：3.4.4 | Anaconda 2.4.1（64-bit）| （默認情況下，2016年1月19日，12點10分59秒）[MSC v.1600 64位（AMD64）]

numba版本：0.23.1

import numba as nb 
import numpy as np 

def neumann_laplacian_1d(u,dx2): 
    """Return finite difference Laplacian approximation of 2d array. 
    Uses Neumann boundary conditions and a 2nd order approximation. 
    """ 
    laplacian = np.zeros(u.shape) 
    laplacian[1:-1] = ((1.0)*u[2:] 
         +(1.0)*u[:-2] 
         -(2.0)*u[1:-1]) 
    # Neumann boundary conditions 
    # edges 
    laplacian[0] = ((2.0)*u[1]-(2.0)*u[0]) 
    laplacian[-1] = ((2.0)*u[-2]-(2.0)*u[-1]) 

    return laplacian/ dx2 

@nb.autojit(nopython=True) 
def neumann_laplacian_1d_numba(u,dx2): 
    """Return finite difference Laplacian approximation of 2d array. 
    Uses Neumann boundary conditions and a 2nd order approximation. 
    """ 
    laplacian = np.zeros(u.shape) 
    laplacian[1:-1] = ((1.0)*u[2:] 
         +(1.0)*u[:-2] 
         -(2.0)*u[1:-1]) 
    # Neumann boundary conditions 
    # edges 
    laplacian[0] = ((2.0)*u[1]-(2.0)*u[0]) 
    laplacian[-1] = ((2.0)*u[-2]-(2.0)*u[-1]) 

    return laplacian/ dx2 

a = np.random.random(252) 
#run once to make the JIT do it's work before timing 
neumann_laplacian_1d_numba(a, 1.0) 


%timeit neumann_laplacian_1d(a, 1.0) 
%timeit neumann_laplacian_1d_numba(a, 1.0) 

>>10000 loops, best of 3: 21.5 µs per loop 
>>The slowest run took 4.49 times longer than the fastest. This could mean that an intermediate result is being cached 
>>100000 loops, best of 3: 3.53 µs per loop 

我看到蟒2.7.11類似的結果和numba 0.23

>>100000 loops, best of 3: 19.1 µs per loop 
>>The slowest run took 8.55 times longer than the fastest. This could mean that an intermediate result is being cached 
>>100000 loops, best of 3: 2.4 µs per loop