Python - 生成特定自相关的数组

时间:2015-11-24 16:17:47

标签: python numpy random scikit-learn correlation

我感兴趣的是生成长度为N的数组(或numpy系列),它将在滞后1处显示特定的自相关。理想情况下,我想指定均值和方差,并从(多)中提取数据正态分布。但最重要的是,我想指定自相关。我如何用numpy或scikit-learn做到这一点?

为了明确而准确,这是我想要控制的自相关:

numpy.corrcoef(x[0:len(x) - 1], x[1:])[0][1]

1 个答案:

答案 0 :(得分:4)

如果您只对滞后一的自相关感兴趣,可以生成一个auto-regressive process的阶数为1的参数等于所需的自相关; Wikipedia page上提到了这个属性,但要证明它并不难。

以下是一些示例代码:

import numpy as np

def sample_signal(n_samples, corr, mu=0, sigma=1):
    assert 0 < corr < 1, "Auto-correlation must be between 0 and 1"

    # Find out the offset `c` and the std of the white noise `sigma_e`
    # that produce a signal with the desired mean and variance.
    # See https://en.wikipedia.org/wiki/Autoregressive_model
    # under section "Example: An AR(1) process".
    c = mu * (1 - corr)
    sigma_e = np.sqrt((sigma ** 2) * (1 - corr ** 2))

    # Sample the auto-regressive process.
    signal = [c + np.random.normal(0, sigma_e)]
    for _ in range(1, n_samples):
        signal.append(c + corr * signal[-1] + np.random.normal(0, sigma_e))

    return np.array(signal)

def compute_corr_lag_1(signal):
    return np.corrcoef(signal[:-1], signal[1:])[0][1]

# Examples.
print(compute_corr_lag_1(sample_signal(5000, 0.5)))
print(np.mean(sample_signal(5000, 0.5, mu=2)))
print(np.std(sample_signal(5000, 0.5, sigma=3)))

参数corr可让您在滞后1和可选参数musigma设置所需的自相关,让您控制生成信号的均值和标准差