平均绝对误差 (mae) 和均方误差 (mse) 损失函数的梯度

Question

我正在尝试使用 numpy 实现线性回归。为此，我需要为损失函数 mae 和 mse 实现导数。这些是我的版本，但它们是不正确的（不幸的是未公开的失败测试）。 有人可以帮助我为 mae 和 mse 实现正确的梯度方法吗？请注意，目标的维度 > 1，这可能在我的版本中未处理。

def mse_derivative(X, Y, w):
    """
    X : numpy array of shape (`n_observations`, `n_features`)
    Y : numpy array of shape (`n_observations`, `target_dimentionality`) or (`n_observations`,)
    w : numpy array of shape (`n_features`, `target_dimentionality`) or (`n_features`,)
    
    Return: NumPy array of the same shape as `w`

    """

    return (X.T.dot(X.dot(w) - Y)) * 2 / Y.shape[0]

def mae_derivative(X, Y, w):
    """
    X : numpy array of shape (`n_observations`, `n_features`)
    Y : numpy array of shape (`n_observations`, `target_dimentionality`) or (`n_observations`,)
    w : numpy array of shape (`n_features`, `target_dimentionality`) or (`n_features`,)
    
    Return: NumPy array of the same shape as `w`
    """


   
    return np.array(list(map(lambda diff : 0 if diff == 0 else 1 if diff > 0 else -1, list(X.dot(w) - Y))))

Answer 1

def mse_derivative(X, Y, w):
    """
    X : numpy array of shape (`n_observations`, `n_features`)
    Y : numpy array of shape (`n_observations`, `target_dimentionality`) or (`n_observations`,)
    w : numpy array of shape (`n_features`, `target_dimentionality`) or (`n_features`,)
    
    Return: NumPy array of the same shape as `w`

    """
    result = (X.T.dot(X.dot(w) - Y)) * 2 / Y.shape[0]
    return result if len(w.shape) == 1 else result / w.shape[1]