我从这篇论文中找到了解决方案：updating simple linear regression。实现如下：

def lr(x_avg,y_avg,Sxy,Sx,n,new_x,new_y):
    """
    x_avg: average of previous x, if no previous sample, set to 0
    y_avg: average of previous y, if no previous sample, set to 0
    Sxy: covariance of previous x and y, if no previous sample, set to 0
    Sx: variance of previous x, if no previous sample, set to 0
    n: number of previous samples
    new_x: new incoming 1-D numpy array x
    new_y: new incoming 1-D numpy array x
    """
    new_n = n + len(new_x)

    new_x_avg = (x_avg*n + np.sum(new_x))/new_n
    new_y_avg = (y_avg*n + np.sum(new_y))/new_n

    if n > 0:
        x_star = (x_avg*np.sqrt(n) + new_x_avg*np.sqrt(new_n))/(np.sqrt(n)+np.sqrt(new_n))
        y_star = (y_avg*np.sqrt(n) + new_y_avg*np.sqrt(new_n))/(np.sqrt(n)+np.sqrt(new_n))
    elif n == 0:
        x_star = new_x_avg
        y_star = new_y_avg
    else:
        raise ValueError

    new_Sx = Sx + np.sum((new_x-x_star)**2)
    new_Sxy = Sxy + np.sum((new_x-x_star).reshape(-1) * (new_y-y_star).reshape(-1))

    beta = new_Sxy/new_Sx
    alpha = new_y_avg - beta * new_x_avg
    return new_Sxy, new_Sx, new_n, alpha, beta, new_x_avg, new_y_avg

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性能对比：
Scikit学习版，总共计算10k个样本。

from sklearn.linear_model import LinearRegression
x = np.arange(10000).reshape(-1,1)
y = np.arange(10000)+100*np.random.random_sample((10000,))
regr = LinearRegression()
%timeit regr.fit(x,y)
# 419 µs ± 14.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

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我的版本假设已经计算了9k样本：

Sxy, Sx, n, alpha, beta, new_x_avg, new_y_avg = lr(0, 0, 0, 0, 0, x.reshape(-1,1)[:9000], y[:9000])
new_x, new_y = x.reshape(-1,1)[9000:], y[9000:]
%timeit lr(new_x_avg, new_y_avg, Sxy,Sx,n,new_x, new_y)
# 38.7 µs ± 1.31 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

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10倍的速度，这是预期的。

我把Kevin Fang的回答改编成了一个类：

import numpy as np

class OnlineLinearRegression:
    """
    Online linear regression in O(1) mem & compute
    """
    def __init__(self):
        # Average of all x seen
        self.x_avg = 0.
        # Average of all y seen
        self.y_avg = 0.
        # Covariance of all x and y seen
        self.xy_covar = 0.
        # Variance of all x seen
        self.x_var = 0.
        # Number of observations seen
        self.n = 0

    @property
    def parameters(self):
        """
        :return: the parameters of the linear regression (beta, alpha) such that y = beta * x + alpha. If there are
        less than 2 observations, returns (None, None).
        """
        if self.n < 2:
            return None, None
        else:
            beta = self.xy_covar / self.x_var
            alpha = self.y_avg - beta * self.x_avg
            return beta, alpha

    def update_multiple(self, new_x: np.ndarray, new_y: np.ndarray):
        assert len(new_x) == len(new_y)
        new_n = self.n + len(new_x)

        new_x_avg = (self.x_avg * self.n + np.sum(new_x)) / new_n
        new_y_avg = (self.y_avg * self.n + np.sum(new_y)) / new_n

        if self.n:
            x_star = (self.x_avg * np.sqrt(self.n) + new_x_avg * np.sqrt(new_n)) / (np.sqrt(self.n) + np.sqrt(new_n))
            y_star = (self.y_avg * np.sqrt(self.n) + new_y_avg * np.sqrt(new_n)) / (np.sqrt(self.n) + np.sqrt(new_n))
        else:
            x_star = new_x_avg
            y_star = new_y_avg

        self.n = new_n
        self.x_avg = new_x_avg
        self.y_avg = new_y_avg

        self.x_var = self.x_var + np.sum((new_x - x_star) ** 2)
        self.xy_covar = self.xy_covar + np.sum((new_x - x_star).reshape(-1) * (new_y - y_star).reshape(-1))

    def update(self, x: float, y: float):
        self.update_multiple(np.array([x]), np.array([y]))

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很好！感谢分享你的发现：）这里是一个用点积编写的解决方案的等效实现：

class SimpleLinearRegressor(object):
    def __init__(self):
        self.dots = np.zeros(5)
        self.intercept = None
        self.slope = None

    def update(self, x: np.ndarray, y: np.ndarray):
        self.dots += np.array(
            [
                x.shape[0],
                x.sum(),
                y.sum(),
                np.dot(x, x),
                np.dot(x, y),
            ]
        )
        size, sum_x, sum_y, sum_xx, sum_xy = self.dots
        det = size * sum_xx - sum_x ** 2
        if det > 1e-10:  # determinant may be zero initially
            self.intercept = (sum_xx * sum_y - sum_xy * sum_x) / det
            self.slope = (sum_xy * size - sum_x * sum_y) / det

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当处理时间序列数据时，我们可以扩展这个想法，使用软（类似EMA）窗口进行滑动窗口回归。

您可以使用实现更快算法的加速库-特别是https://github.com/intel/scikit-learn-intelex
对于线性回归，您将获得更好的性能
第一个安装包

pip install scikit-learn-intelex

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然后添加python脚本

from sklearnex import patch_sklearn
patch_sklearn()

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