scipy.optimize.最小化函数与 L-BFGS-B 方法最大属性不起作用

Question

我有一个简单的成本函数，我想使用 scipy.optimize.reduce 函数对其进行优化。

opt_solution  = scipy.optimize.minimize(costFunction, theta, args = (training_data,), method = 'L-BFGS-B', jac = True, options = {'maxiter': 100)

其中costFunction是要优化的函数，theta是要优化的参数。在costFunction里面，我打印了成本函数的值。但是参数maxiter似乎对我是否将值从 10 增加到 100000 没有影响。它所花费的时间是一样的。另外，我期望成本函数的打印值应该等于maxiter的值。所以我觉得maxiter没有效果。可能有什么问题？成本函数为

def costFunction(self, theta, input):
    """ Extract weights and biases from 'theta' input """
    W1 = theta[self.limit0 : self.limit1].reshape(self.hidden_size, self.visible_size)
    W2 = theta[self.limit1 : self.limit2].reshape(self.visible_size, self.hidden_size)
    b1 = theta[self.limit2 : self.limit3].reshape(self.hidden_size, 1)
    b2 = theta[self.limit3 : self.limit4].reshape(self.visible_size, 1)
    """ Compute output layers by performing a feedforward pass
        Computation is done for all the training inputs simultaneously """
    hidden_layer = self.sigmoid(numpy.dot(W1, input) + b1)
    output_layer = self.sigmoid(numpy.dot(W2, hidden_layer) + b2)
    """ Compute intermediate difference values using Backpropagation algorithm """
    diff = output_layer - input
    sum_of_squares_error = 0.5 * numpy.sum(numpy.multiply(diff, diff)) / input.shape[1]
    weight_decay         = 0.5 * self.lamda * (numpy.sum(numpy.multiply(W1, W1)) + numpy.sum(numpy.multiply(W2, W2)))
    cost                 = sum_of_squares_error + weight_decay 
    """ Compute the gradient values by averaging partial derivatives
        Partial derivatives are averaged over all training examples """
    W1_grad = numpy.dot(del_hid, numpy.transpose(input))
    W2_grad = numpy.dot(del_out, numpy.transpose(hidden_layer))
    b1_grad = numpy.sum(del_hid, axis = 1)
    b2_grad = numpy.sum(del_out, axis = 1)
    W1_grad = W1_grad / input.shape[1] + self.lamda * W1
    W2_grad = W2_grad / input.shape[1] + self.lamda * W2
    b1_grad = b1_grad / input.shape[1]
    b2_grad = b2_grad / input.shape[1]
    """ Transform numpy matrices into arrays """
    W1_grad = numpy.array(W1_grad)
    W2_grad = numpy.array(W2_grad)
    b1_grad = numpy.array(b1_grad)
    b2_grad = numpy.array(b2_grad)
    """ Unroll the gradient values and return as 'theta' gradient """
    theta_grad = numpy.concatenate((W1_grad.flatten(), W2_grad.flatten(),
                                    b1_grad.flatten(), b2_grad.flatten()))
    # Update counter value
    self.counter += 1                                
    print "Index ", self.counter, "cost ", cost
    return [cost, theta_grad]

Answer 1

maxiter给出了 Scipy 在放弃改进解决方案之前将尝试的最大迭代次数。但它很可能对解决方案感到满意并更早停止。

如果您在使用 'l-bfgs-b' 方法时查看minimize文档，请注意有三个参数可以作为选项传递(factr、ftol 和 gtol(，它们也可能导致迭代停止。

在像您这样的简单情况下，特别是如果您的成本函数也提供梯度(如调用中的jac=True所示(，收敛通常发生在前几次迭代中，因此在达到maxiter限制之前。