神经网络训练平稳中的梯度下降

Question

我一直在尝试在 python 中实现一个基本的反向传播神经网络，并且已经完成了初始化和训练权重集的编程。然而，在我训练的所有集合上，误差（均方）总是收敛到一个奇怪的数字——误差总是在进一步的迭代中减小，但从未真正接近于零。
任何帮助将不胜感激。

import csv
import numpy as np

class NeuralNetwork:
layers = 0
shape = None
weights = []

layerIn = []
layerOut = []

def __init__(self, shape):
    self.shape = shape
    self.layers = len(shape) - 1

    for i in range(0,self.layers):
        n = shape[i]
        m = shape[i+1]
        self.weights.append(np.random.normal(scale=0.2, size = (m,n+1)))

def sgm(self, x):
    return 1/(1+np.exp(-x))

def dersgm(self, x):
    y = self.sgm(x)
    return y*(y-1)


def run(self, input):
    self.layerIn = []
    self.layerOut = []

    for i in range(self.layers):
        if i == 0:
            layer = self.weights[0].dot(np.vstack((input.transpose(), np.ones([1,input.shape[0]]))))
        else:
            layer = self.weights[i].dot(np.vstack((self.layerOut[-1], np.ones([1,input.shape[0]]))))
        self.layerIn.append(layer)
        self.layerOut.append(self.sgm(layer))

    return self.layerOut[-1].T

def backpropogate(self, input, y, learning_rate):
    deltas = []
    y_hat = self.run(input)

    #Calculate deltas
    for i in reversed(range(self.layers)):

        #for last layer
        if i == self.layers-1:
            error = y_hat - y
            msq_error = sum(.5 * ((error) ** 2))
            #returns delta, k rows for k inputs, m columns for m nodes
            deltas.append(error * self.dersgm(y_hat))
        else:

            error = deltas[-1].dot(self.weights[i+1][:,:-1])
            deltas.append(self.dersgm(self.layerOut[i]).T * error)

    #Calculate weight-deltas
    wdelta = []
    ordered_deltas = list(reversed(deltas)) #reverse order because created backwards

    #returns weight deltas, k rows for k nodes, m columns for m next layer nodes
    for i in range(self.layers):
        if i == 0:
            #add bias
            input_with_bias = np.vstack((input.T, np.ones(input.shape[0])))
            #some over n rows of deltas for n training examples to get one delta for all examples
            #for all nodes
            wdelta.append(ordered_deltas[i].T.dot(input_with_bias.T))
        else:
            with_bias = np.vstack((self.layerOut[i-1], np.ones(input.shape[0])))
            wdelta.append(ordered_deltas[i].T.dot(with_bias.T))



    #update_weights
    def update_weights(self, weight_deltas, learning_rate):
        for i in range(self.layers):
            self.weights[i] = self.weights[i] +\
                              (learning_rate * weight_deltas[i])


    update_weights(self, wdelta, learning_rate)

    return msq_error

    #end backpropogate

def train(self, input, target, lr, run_iter):
    for i in range(run_iter):
        if i % 100000 == 0:
            print self.backpropogate(input, target, lr)

Answer 1

以下场景中的误差函数不能为0，因为误差函数为0需要点与曲线完美匹配。

拥有更多的神经元肯定会减少误差，因为函数可以具有更复杂和更精确的形状。但是，当您对数据的拟合程度过高时，就会出现一个称为过度拟合的问题，如下图所示。从左到右，曲线要么欠拟合数据集，几乎正确拟合它，然后在右侧欠拟合。

右边的场景会导致错误为 0，但这是不希望的，您希望避免这种情况。怎么样？

确定网络中神经元数量是否理想（具有良好拟合）的最简单方法是反复试验。将您的数据拆分为训练数据（80% - 用于训练网络）和测试数据（20% - 仅用于在训练后测试网络）。虽然只对训练数据进行训练，但可以在测试数据集上绘制性能图。

您还可以使用第三个数据集进行验证，请参阅： whats is the difference between train, validation and test set, in neural networks?