Question

在以下代码中，test_func_1比test_func_2慢一个数量级。是否无法改善或甚至匹配此操作的numpy性能？

from numba import guvectorize
import numpy as np

@guvectorize(['void(float64[:,:,:], float64[:], float64[:,:,:])'], '(n,o,p),(n)->(n,o,p)', nopython=True)
def test_func_1(time_series, areas, res):
    for i in range(areas.size):
        area = areas[i]
        adjusted_area = (area / 10000.) ** .12  # used to adjust erosion
        for k in range(time_series.shape[0]):
            res[i, 0, k] = time_series[i, 0, k] * area
            res[i, 1, k] = time_series[i, 1, k] * adjusted_area
            res[i, 2, k] = time_series[i, 2, k] * area
            res[i, 3, k] = time_series[i, 3, k] * adjusted_area


def test_func_2(time_series, areas):
    array = np.swapaxes(time_series, 0, 2)
    array[:, :2] *= areas
    array[:, 2:] *= (areas / 10000.) ** .12
    return array

dummy = np.float32(np.random.randint(0, 10, (20, 5, 5000)))
areas = np.float32(np.random.randint(0, 10, 20))

test_func_1(dummy, areas)
test_func_2(dummy, areas)

Answer 1

正如@JoshAdel在评论中所指出的，这里的关键区别在于你的numba版本正在分配并填充一个新数组，而numpy正在修改原始数据。

为numpy添加一个合适的.copy()会让它稍慢一些。你也可以让你的numba版本正常工作 - 据我所知，使用gufunc是不可能的，但如果你不需要gufunc提供的广播，那么你将使用常规的jit函数。

def test_func_2(time_series, areas):
    array = np.swapaxes(time_series, 0, 2).copy()
    array[:, :2] *= areas
    array[:, 2:] *= (areas / 10000.) ** .12
    return array

dummy = np.float32(np.random.randint(0, 10, (20, 5, 5000)))
areas = np.float32(np.random.randint(0, 10, 20))

%timeit test_func_1(dummy, areas)
1.21 ms ± 5.33 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit test_func_2(dummy, areas)
1.77 ms ± 15.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Numba矢量化比3d阵列的numpy慢吗？

1 个答案: