Numba代码比纯python慢

Question

我一直在加速粒子过滤器的重采样计算。由于python有很多加速的方法，我想我会尝试所有的方法。不幸的是，numba版本非常慢。由于Numba会导致加速，我认为这是我的一个错误。

我尝试了4种不同的版本：

1. 麻木
2. Python
3. 纽比
4. 赛松

其代码如下：

import numpy as np
import scipy as sp
import numba as nb
from cython_resample import cython_resample

@nb.autojit
def numba_resample(qs, xs, rands):
    n = qs.shape[0]
    lookup = np.cumsum(qs)
    results = np.empty(n)

    for j in range(n):
        for i in range(n):
            if rands[j] < lookup[i]:
                results[j] = xs[i]
                break
    return results

def python_resample(qs, xs, rands):
    n = qs.shape[0]
    lookup = np.cumsum(qs)
    results = np.empty(n)

    for j in range(n):
        for i in range(n):
            if rands[j] < lookup[i]:
                results[j] = xs[i]
                break
    return results

def numpy_resample(qs, xs, rands):
    results = np.empty_like(qs)
    lookup = sp.cumsum(qs)
    for j, key in enumerate(rands):
        i = sp.argmax(lookup>key)
        results[j] = xs[i]
    return results

#The following is the code for the cython module. It was compiled in a
#separate file, but is included here to aid in the question.
"""
import numpy as np
cimport numpy as np
cimport cython

DTYPE = np.float64

ctypedef np.float64_t DTYPE_t

@cython.boundscheck(False)
def cython_resample(np.ndarray[DTYPE_t, ndim=1] qs, 
             np.ndarray[DTYPE_t, ndim=1] xs, 
             np.ndarray[DTYPE_t, ndim=1] rands):
    if qs.shape[0] != xs.shape[0] or qs.shape[0] != rands.shape[0]:
        raise ValueError("Arrays must have same shape")
    assert qs.dtype == xs.dtype == rands.dtype == DTYPE

    cdef unsigned int n = qs.shape[0]
    cdef unsigned int i, j 
    cdef np.ndarray[DTYPE_t, ndim=1] lookup = np.cumsum(qs)
    cdef np.ndarray[DTYPE_t, ndim=1] results = np.zeros(n, dtype=DTYPE)

    for j in range(n):
        for i in range(n):
            if rands[j] < lookup[i]:
                results[j] = xs[i]
                break
    return results
"""

if __name__ == '__main__':
    n = 100
    xs = np.arange(n, dtype=np.float64)
    qs = np.array([1.0/n,]*n)
    rands = np.random.rand(n)

    print "Timing Numba Function:"
    %timeit numba_resample(qs, xs, rands)
    print "Timing Python Function:"
    %timeit python_resample(qs, xs, rands)
    print "Timing Numpy Function:"
    %timeit numpy_resample(qs, xs, rands)
    print "Timing Cython Function:"
    %timeit cython_resample(qs, xs, rands)

这将产生以下输出：

Timing Numba Function:
1 loops, best of 3: 8.23 ms per loop
Timing Python Function:
100 loops, best of 3: 2.48 ms per loop
Timing Numpy Function:
1000 loops, best of 3: 793 µs per loop
Timing Cython Function:
10000 loops, best of 3: 25 µs per loop

知道为什么numba代码这么慢吗？我认为它至少可以与纽比相提并论。

注意：如果有人对如何加快Numpy或Cython代码示例有任何想法，那也不错：）不过，我的主要问题是Numba。

Answer 1

更快的numpy版本（与numpy_resample相比，速度提高了10倍）

def numpy_faster(qs, xs, rands):
    lookup = np.cumsum(qs)
    mm = lookup[None,:]>rands[:,None]
    I = np.argmax(mm,1)
    return xs[I]

Answer 2

问题是numba不能直觉判断lookup的类型。如果你在你的方法中放一个print nb.typeof(lookup)，你会看到numba把它当作一个对象，这是很慢的。通常，我只是在局部dict中定义lookup的类型，但是我得到了一个奇怪的错误。相反，我只是创建了一个小包装器，以便可以显式地定义输入和输出类型。

@nb.jit(nb.f8[:](nb.f8[:]))
def numba_cumsum(x):
    return np.cumsum(x)

@nb.autojit
def numba_resample2(qs, xs, rands):
    n = qs.shape[0]
    #lookup = np.cumsum(qs)
    lookup = numba_cumsum(qs)
    results = np.empty(n)

    for j in range(n):
        for i in range(n):
            if rands[j] < lookup[i]:
                results[j] = xs[i]
                break
    return results

那么我的时间安排是：

print "Timing Numba Function:"
%timeit numba_resample(qs, xs, rands)

print "Timing Revised Numba Function:"
%timeit numba_resample2(qs, xs, rands)

---

Timing Numba Function:
100 loops, best of 3: 8.1 ms per loop
Timing Revised Numba Function:
100000 loops, best of 3: 15.3 µs per loop

---

如果你用jit而不是autojit，你可以走得更快一点：

@nb.jit(nb.f8[:](nb.f8[:], nb.f8[:], nb.f8[:]))

对我来说，从15.3微秒降到12.5微秒，但autojit的表现仍然令人印象深刻。