[英]Numba code slower than pure python
我一直在努力加快粒子濾波器的重采樣計算。 由於python有很多方法可以加速它,我雖然會嘗試所有這些。 不幸的是,numba版本非常慢。 由於Numba應該加速,我認為這是我的錯誤。
我嘗試了4個不同的版本:
每個代碼如下:
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)
這導致以下輸出:
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
知道為什么numba代碼如此之慢? 我認為它至少可以與Numpy相媲美。
注意:如果有人對如何加速Numpy或Cython代碼示例有任何想法,那也不錯:)我的主要問題是關於Numba。
問題是numba無法直觀地lookup
類型。 如果在方法中放置了print nb.typeof(lookup)
,你會發現numba將它視為一個對象,這很慢。 通常我會在locals dict中定義lookup
類型,但是我遇到了一個奇怪的錯誤。 相反,我只是創建了一個小包裝器,以便我可以顯式定義輸入和輸出類型。
@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
然后我的時間是:
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
如果你使用jit
而不是autojit
你甚至可以更快一點:
@nb.jit(nb.f8[:](nb.f8[:], nb.f8[:], nb.f8[:]))
對我來說,它從15.3微秒降低到12.5微秒,但它仍然令人印象深刻的autojit做得如何。
更快的numpy
版本(與numpy_resample
相比加速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]
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