How does parakeet differ from Numba? Because I didn't see any improvements on some NumPy expressions

Question

I am wondering if anyone knows some of the key differences between the parakeet and the Numba jit? I am curious, because I was comparing Numexpr to Numba and parakeet, and for this particular expression (which I expected to perform very very well on Numexpr, because it was the one that is mentioned in its documentation)

So the results are

and the functions I tested (via timeit - minimum of 3 repetitions and 10 loops per function)

import numpy as np
import numexpr as ne
from numba import jit as numba_jit
from parakeet import jit as para_jit


def numpy_complex_expr(A, B):
    return(A*B-4.1*A > 2.5*B)

def numexpr_complex_expr(A, B):
    return ne.evaluate('A*B-4.1*A > 2.5*B')

@numba_jit
def numba_complex_expr(A, B):
    return A*B-4.1*A > 2.5*B

@para_jit
def parakeet_complex_expr(A, B):
    return A*B-4.1*A > 2.5*B

I you can also grab the IPython nb if you'd like to double-check the results on your machine.

If someone is wondering if Numba is installed correctly... I think so, it performed as expected in my previous benchmark:

Answer 1

As of the current release of Numba (which you are using in your tests), there is incomplete support for ufuncs with the @jit function. On the other hand you can use @vectorize and it faster:

import numpy as np
from numba import jit, vectorize
import numexpr as ne

def numpy_complex_expr(A, B):
    return(A*B+4.1*A > 2.5*B)

def numexpr_complex_expr(A, B):
    return ne.evaluate('A*B+4.1*A > 2.5*B')

@jit
def numba_complex_expr(A, B):
    return A*B+4.1*A > 2.5*B

@vectorize(['u1(float64, float64)'])
def numba_vec(A,B):
    return A*B+4.1*A > 2.5*B

n = 1000
A = np.random.rand(n,n)
B = np.random.rand(n,n)

Timing results:

%timeit numba_complex_expr(A,B)
1 loops, best of 3: 49.8 ms per loop

%timeit numpy_complex_expr(A,B)
10 loops, best of 3: 43.5 ms per loop

%timeit numexpr_complex_expr(A,B)
100 loops, best of 3: 3.08 ms per loop

%timeit numba_vec(A,B)
100 loops, best of 3: 9.8 ms per loop

If you want to leverage numba to its fullest, then you want to unroll any vectorized operations:

@jit
def numba_unroll2(A, B):
    C = np.empty(A.shape, dtype=np.uint8)
    for i in xrange(A.shape[0]):
        for j in xrange(A.shape[1]):
            C[i,j] = A[i,j]*B[i,j] + 4.1*A[i,j] > 2.5*B[i,j]

    return C

---

%timeit numba_unroll2(A,B)
100 loops, best of 3: 5.96 ms per loop

Also note that if you set the number of threads that numexpr uses to 1, then you'll see that its main speed advantage is that it's parallelized:

ne.set_num_threads(1)
%timeit numexpr_complex_expr(A,B)
100 loops, best of 3: 8.87 ms per loop

By default numexpr uses ne.detect_number_of_cores() as the number of threads. For my original timing on my machine, it was using 8.