我不知道为什么numba在这里击败numpy（超过3倍）。我在这里进行基准测试时是否犯了一些根本性的错误？对于numpy来说似乎是完美的情况，不是吗？请注意，作为检查，我还运行了一个结合了numba和numpy的变体（未显示），正如预期的那样，它与不带numba的numpy运行相同。

import numpy as np
from numba import jit
nobs = 10000

def proc_numpy(x,y,z):

   x = x*2 - ( y * 55 )      # these 4 lines represent use cases
   y = x + y*2               # where the processing time is mostly
   z = x + y + 99            # a function of, say, 50 to 200 lines
   z = z * ( z - .88 )       # of fairly simple numerical operations

   return z

@jit
def proc_numba(xx,yy,zz):
   for j in range(nobs):     # as pointed out by Llopis, this for loop 
      x, y = xx[j], yy[j]    # is not needed here.  it is here by 
                             # accident because in the original benchmarks 
      x = x*2 - ( y * 55 )   # I was doing data creation inside the function 
      y = x + y*2            # instead of passing it in as an array
      z = x + y + 99         # in any case, this redundant code seems to 
      z = z * ( z - .88 )    # have something to do with the code running
                             # faster.  without the redundant code, the 
      zz[j] = z              # numba and numpy functions are exactly the same.
   return zz

x = np.random.randn(nobs)
y = np.random.randn(nobs)
z = np.zeros(nobs)
res_numpy = proc_numpy(x,y,z)

z = np.zeros(nobs)
res_numba = proc_numba(x,y,z)

结果：

In [356]: np.all( res_numpy == res_numba )
Out[356]: True

In [357]: %timeit proc_numpy(x,y,z)
10000 loops, best of 3: 105 µs per loop

In [358]: %timeit proc_numba(x,y,z)
10000 loops, best of 3: 28.6 µs per loop

我在2012年的macbook air（13.3）（标准anaconda发行版）上运行了该软件。如果相关，我可以提供有关我的设置的更多详细信息。