如何优化使用Numpy的连续值的for循环？

尽管没有向量化，但我相信以下解决方案的速度要快大约2倍（numba解决方案的速度要快60倍）。它将每个保存result为局部变量，而不是按位置访问numpy数组。

def unif_improved(n):
    m = 2**32
    a = 1664525
    c = 1013904223

    results = np.empty(n)
    results[0] = result = int((time.time() * 1e7) % m)

    for i in range(1, n):
        result = results[i] = (a * result + c) % m

    return results / m

您也可以考虑使用Numba来进一步提高速度。https://numba.pydata.org/

仅仅添加装饰器就可以@jit吹开其他解决方案的门。

from numba import jit

@jit
def unif_jit(n):
    # Same code as `unif_improved`

时机

>>> %timeit -n 10 unif_original(500000)
715 ms ± 21.5 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

>>> %timeit -n 10 unif_improved(500000)
323 ms ± 8 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

>>> %timeit -n 10 unif_jit(500000)
12 ms ± 2.68 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)