我正在 MNIST 數(shù)據(jù)集上使用梯度下降 + L2 正則化實現(xiàn)多項邏輯回歸。我的訓(xùn)練數(shù)據(jù)是一個形狀為（n_samples=1198，features=65）的數(shù)據(jù)框。在梯度下降的每次迭代中，我采用權(quán)重和輸入的線性組合來獲得 1198 次激活 (beta^T * X)。然后我通過 softmax 函數(shù)傳遞這些激活。但是，我對如何獲得每個激活的 10 個輸出類的概率分布感到困惑？我的權(quán)重是這樣初始化的n_features = 65# init random weightsbeta = np.random.uniform(0, 1, n_features).reshape(1, -1)這是我當(dāng)前的實現(xiàn)。def softmax(x:np.ndarray):     exps = np.exp(x)    return exps/np.sum(exps, axis=0)def cross_entropy(y_hat:np.ndarray, y:np.ndarray, beta:np.ndarray) -> float:    """    Computes cross entropy for multiclass classification      y_hat: predicted classes, n_samples x n_feats    y: ground truth classes, n_samples x 1    """    n = len(y)    return - np.sum(y * np.log(y_hat) + beta**2 / n)    def gd(X:pd.DataFrame, y:pd.Series, beta:np.ndarray,       lr:float, N:int, iterations:int) -> (np.ndarray,np.ndarray):        """    Gradient descent    """    n = len(y)    cost_history = np.zeros(iterations)        for it in range(iterations):                activations = X.dot(beta.T).values                y_hat = softmax(activations)                        cost_history[it] = cross_entropy(y_hat, y, beta)                        # gradient of weights        grads = np.sum((y_hat - y) * X).values                # update weights        beta = beta - lr * (grads + 2/n * beta)            return beta, cost_history