纽比广播
如果不受内存限制,优化numpy中嵌套循环的第一步是使用广播并以矢量化的方式执行操作:import numpy as np
def mb_r(forecasted_array, observed_array):
"""Returns the Mielke-Berry R value."""
assert len(observed_array) == len(forecasted_array)
total = np.abs(forecasted_array[:, np.newaxis] - observed_array).sum() # Broadcasting
return 1 - (mae(forecasted_array, observed_array) * forecasted_array.size ** 2 / total[0])
但在这种情况下,循环是用C而不是Python进行的,它涉及到一个大小(N,N)数组的分配。在
广播不是灵丹妙药,试着打开内环
如上所述,广播意味着巨大的内存开销。所以它应该小心使用,而且它并不总是正确的方法。虽然你的第一印象可能是在任何地方使用它-不要。不久前我也被这个事实搞糊涂了,看我的问题Numpy ufuncs speed vs for loop speed。为了不太冗长,我将在您的示例中显示这一点:
^{pr2}$
小型阵列(广播更快)forecasted = np.random.rand(100)
observed = np.random.rand(100)
%timeit mb_r_bcast(forecasted, observed)
57.5 µs ± 359 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit mb_r_unroll(forecasted, observed)
1.17 ms ± 2.53 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
中型阵列(相等)forecasted = np.random.rand(1000)
observed = np.random.rand(1000)
%timeit mb_r_bcast(forecasted, observed)
15.6 ms ± 208 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit mb_r_unroll(forecasted, observed)
16.4 ms ± 13.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
大尺寸阵列(广播速度较慢)forecasted = np.random.rand(10000)
observed = np.random.rand(10000)
%timeit mb_r_bcast(forecasted, observed)
1.51 s ± 18 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit mb_r_unroll(forecasted, observed)
377 ms ± 994 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)
正如您可以看到的,对于小型阵列,广播版本比展开的快20倍,对于中型阵列,它们是而不是相等,但对于大型阵列,它的速度4倍,因为内存开销要付出高昂的代价。在
Numba jit与并行化
另一种方法是使用numba及其强大的@jit函数修饰符。在这种情况下,只需对初始代码稍作修改。另外,为了使循环并行,您应该将range更改为prange,并提供parallel=True关键字参数。在下面的代码片段中,我使用了与@jit(nopython=True)相同的@njit修饰符:from numba import njit, prange
@njit(parallel=True)
def mb_r_njit(forecasted_array, observed_array):
"""Returns the Mielke-Berry R value."""
assert len(observed_array) == len(forecasted_array)
total = 0.
size = len(forecasted_array)
for i in prange(size):
observed = observed_array[i]
for j in prange(size):
total += abs(forecasted_array[j] - observed)
return 1 - (mae(forecasted_array, observed_array) * size ** 2 / total)
您没有提供mae函数,但是要在njit模式下运行代码,还必须修饰mae函数,或者如果它是一个数字,则将其作为一个参数传递给jitted函数。在
其他选项
Python的科学生态系统是巨大的,我只是提到了其他一些可以加速的选项:Cython,Nuitka,Pythran,bottleneck等等。也许你对gpu computing感兴趣,但这实际上是另一个故事。在
时间安排
不幸的是,在我的电脑上,时间安排是:import numpy as np
import numexpr as ne
forecasted_array = np.random.rand(10000)
observed_array = np.random.rand(10000)
初始版本%timeit mb_r(forecasted_array, observed_array)
23.4 s ± 430 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
numexpr%%timeit
forecasted_array2d = forecasted_array[:, np.newaxis]
ne.evaluate('sum(abs(forecasted_array2d - observed_array))')[()]
784 ms ± 11.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
广播版%timeit mb_r_bcast(forecasted, observed)
1.47 s ± 4.13 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
内部循环展开版本%timeit mb_r_unroll(forecasted, observed)
389 ms ± 11.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
numbanjit(parallel=True)版本%timeit mb_r_njit(forecasted_array, observed_array)
32 ms ± 4.05 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
可以看出,njit方法比初始解决方案快730倍,也比numexpr解决方案快24.5倍(也许你需要英特尔的矢量数学库来加速)。同样,与初始版本相比,内部循环展开的简单方法可以使60x加速。我的规格是:
Intel(R)Core(TM)2四CPU Q9550 2.83GHz
Python 3.6.3
数字1.13.3
数字0.36.1
numexpr 2.6.4
最后说明
很奇怪,这是一个非常缓慢的测试arr = np.arange(1000)
ls = arr.tolistist()
%timeit for i in arr: pass
69.5 µs ± 282 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit for i in ls: pass
13.3 µs ± 81.8 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%timeit for i in range(len(arr)): arr[i]
167 µs ± 997 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit for i in range(len(ls)): ls[i]
90.8 µs ± 1.07 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
结果证明你是对的。迭代列表的速度2-5x。当然,这些结果必须带有一定的讽刺意味:)