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我需要对两个一维样本应用数十万次Anderson-Darling 检验。中的实现scipy是anderson_ksamp，它运行良好，但它占用了相当多的时间。我想提高它的性能，因为我知道：

  1. 我总是会比较 2 个样本
  2. 我只需要 Anderson-Darling 检验统计量，即不需要临界值或 p 值

scipy从测试的原始实现中删除了非必要的检查，我设法将性能提高了近 30%。

这可以进一步改善吗？

import numpy as np
from scipy.stats import anderson_ksamp
import time as t

def main():
    """
    Test scipy's osiginal vs the simplified Anderson-Dalring tests.
    """
    t1, t2 = 0., 0.
    AD_all = []
    for _ in range(1000):
        N = np.random.randint(10, 200)
        aa = np.random.uniform(0., 1., N)
        bb = np.random.uniform(0., 1., N)

        s = t.time()
        AD = anderson_ksamp([aa, bb])[0]
        t1 += t.time() - s
        s = t.time()
        AD2 = anderson_ksamp_new([aa, bb])
        t2 += t.time() - s

        # Check that both values are equal
        AD_all.append([AD, AD2])

    AD_all = np.array(AD_all).T
    print((AD_all[0] == AD_all[1]).all())
    print("Improvement: {:.1f}%".format(100. - 100. * t2 / t1))


def anderson_ksamp_new(samples):
    """
    A2akN: equation 7 of Scholz and Stephens.

    samples : sequence of 1-D array_like
        Array of sample arrays.
    Z : array_like
        Sorted array of all observations.
    Zstar : array_like
        Sorted array of unique observations.
    k : int
        Number of samples.
    n : array_like
        Number of observations in each sample.
    N : int
        Total number of observations.
    Returns
    -------
    A2aKN : float
        The A2aKN statistics of Scholz and Stephens 1987.

    """

    k = 2  # This will always be 2

    Z = np.sort(np.hstack(samples))
    N = Z.size
    Zstar = np.unique(Z)

    n = np.array([sample.size for sample in samples])

    A2kN = 0.
    Z_ssorted_left = Z.searchsorted(Zstar, 'left')
    if N == Zstar.size:
        lj = 1.
    else:
        lj = Z.searchsorted(Zstar, 'right') - Z_ssorted_left
    Bj = Z_ssorted_left + lj / 2.
    for i in np.arange(0, k):
        s = np.sort(samples[i])
        s_ssorted_right = s.searchsorted(Zstar, side='right')
        Mij = s_ssorted_right.astype(float)
        fij = s_ssorted_right - s.searchsorted(Zstar, 'left')
        Mij -= fij / 2.
        inner = lj / float(N) * (N*Mij - Bj*n[i])**2 / (Bj*(N - Bj) - N*lj/4.)
        A2kN += inner.sum() / n[i]
    A2kN *= (N - 1.) / N

    H = (1. / n).sum()
    hs_cs = (1. / np.arange(N - 1, 1, -1)).cumsum()
    h = hs_cs[-1] + 1
    g = (hs_cs / np.arange(2, N)).sum()

    a = (4*g - 6) * (k - 1) + (10 - 6*g)*H
    b = (2*g - 4)*k**2 + 8*h*k + (2*g - 14*h - 4)*H - 8*h + 4*g - 6
    c = (6*h + 2*g - 2)*k**2 + (4*h - 4*g + 6)*k + (2*h - 6)*H + 4*h
    d = (2*h + 6)*k**2 - 4*h*k
    sigmasq = (a*N**3 + b*N**2 + c*N + d) / ((N - 1.) * (N - 2.) * (N - 3.))
    m = k - 1
    A2 = (A2kN - m) / np.sqrt(sigmasq)

    return A2
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1 回答 1

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np.ndarray.sort您可以通过使用而不是进行排序来获得轻微的改进np.sort：

Z = np.sort(np.hstack(samples))

变成：

Z = np.hstack(samples)
Z.sort()

和

s = np.sort(samples[i])

变成：

s = samples[i]
s.sort()

anderson_ksamp_new通过这些修改，与（您的版本）相比，我得到了 12% 的改进。

此外，您可以使用np.concatenate而不是np.hstack. 这与就地排序相结合，使我提高了 16%。

于 2020-04-08T11:39:24.177 回答