19

我一直在努力加快粒子滤波器的重采样计算。由于 python 有很多方法可以加快速度，所以我会尝试所有方法。不幸的是，numba 版本非常慢。由于 Numba 应该会导致加速，我认为这是我的错误。

我尝试了 4 个不同的版本：

  1. 努巴
  2. Python
  3. 麻木的
  4. 赛通

每个代码如下：

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 Numpy 相媲美。

注意：如果有人对如何加速 Numpy 或 Cython 代码示例有任何想法，那也很好:) 我的主要问题是关于 Numba。

4

2 回答 2

22

问题是 numba 无法直觉lookup. 如果你把 aprint nb.typeof(lookup)在你的方法中，你会看到 numba 把它当作一个对象，这很慢。通常我只会lookup在本地字典中定义类型，但我遇到了一个奇怪的错误。相反，我只是创建了一个小包装器，以便我可以显式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

然后我的时间是：

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

jit如果你使用代替，你甚至可以更快一点autojit：

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

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

于 2014-01-31T20:33:18.487 回答
3

更快numpy的版本（相比 10 倍加速numpy_resample）

def numpy_faster(qs, xs, rands):
    lookup = np.cumsum(qs)
    mm = lookup[None,:]>rands[:,None]
    I = np.argmax(mm,1)
    return xs[I]
于 2014-01-31T03:45:50.860 回答