我正在尝试为日志回归编写小批量梯度下降。
给出numpy矩阵X_batch（形状为(n_samples,n_features)）和y_batch（形状为(n_samples,)）。
天真的方法是编写循环：
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
但是使用循环似乎是个坏主意。速度。还有更好的方法吗？没有循环的纯numpy？重写表达式以某种方式？
请注意，此算法受内存带宽限制。如果您在较大的上下文（实际应用程序）中对此进行优化，则很有可能会实现更高的加速。
示例
import numpy as np
#https://stackoverflow.com/a/29863846/4045774
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,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,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,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,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,y_batch)
#49.1 µs ± 503 ns per loop (mean ± std. dev. of 7 runs,10000 loops each)