我一直在努力加速粒子滤波器的重新采样计算。由于Python有许多加速方法,因此我想尝试它们全部。不幸的是,Numba版本非常慢。由于Numba应该导致加速,我认为这是我的错误。
我尝试了4种不同的版本:
- Numba
- Python
- Numpy
- Cython
每个版本的代码如下:
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)
这将导致以下输出:
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
有没有想法为什么Numba代码运行这么慢?我原以为它至少和Numpy一样快。
注:如果有人有任何加速Numpy或Cython代码示例的想法,那也很好:)但我的主要问题是关于Numba的。
argmax
可以接受一个轴参数,因此您可以将rands
和lookup
广播到彼此,以制作一个n x n
矩阵,用于 N^2 缩放算法。或者,您可以使用 searchsorted,它应该具有 Nlog(N) 缩放。 - Daniel