对数回归损失的梯度

Question

我正在尝试为对数回归编写小批量梯度下降。

给定 numpy 矩阵 X_batch（形状 (n_samples, n_features)）和 y_batch（形状 (n_samples,)）。

天真的方法是写循环：

def calc_loss_grad(self, X_batch, y_batch):
    n_samples, n_features = X_batch.shape
    loss_grad = np.zeros((n_features,))
    for i in range(n_samples):
        sigm = sigmoid(X_batch[i] @ self.weights)
        loss_grad += - (y_batch[i] - sigm) * X_batch[i]            
    return loss_grad

但似乎使用循环不是一个好主意 w.r.t。速度。有什么更好的方法吗？没有循环的纯 numpy？以某种方式重写梯度表达式？

Answer 1

请注意，此算法受内存带宽限制。如果您在更大的上下文（实际应用程序）中对此进行优化，则很可能会有更高的加速比。[=11=]

例子

import numpy as np

#
def sigmoid(x):  
    return np.exp(-np.logaddexp(0, -x))

def calc_loss_grad_1(weights, X_batch, y_batch):
    n_samples, n_features = X_batch.shape
    loss_grad = np.zeros((n_features,))
    for i in range(n_samples):
        sigm = sigmoid(X_batch[i,:] @ weights)
        loss_grad += - (y_batch[i] - sigm) * X_batch[i]            
    return loss_grad

def calc_loss_grad_2(weights, X_batch, y_batch):
    sigm =-y_batch+sigmoid(X_batch@weights)          
    return sigm@X_batch

weights=np.random.rand(n_features)
X_batch=np.random.rand(n_samples,n_features)
y_batch=np.random.rand(n_samples)

#n_samples=int(1e5)
#n_features=int(1e4)
%timeit res=calc_loss_grad_1(weights, X_batch, y_batch)
#1.79 s ± 35.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit res=calc_loss_grad_2(weights, X_batch, y_batch)
#539 ms ± 21.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

#n_samples=int(1e3)
#n_features=int(1e2)
%timeit res=calc_loss_grad_1(weights, X_batch, y_batch)
#3.68 ms ± 44.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit res=calc_loss_grad_2(weights, X_batch, y_batch)
#49.1 µs ± 503 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

对数回归损失的梯度

Gradient for log regression loss

gradient-descent

sgd

gradient

numpy