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python - 如何在感知器的批量训练期间计算偏差

问题描述

我正在对单节点感知器实施批量训练，但不知道如何更新我的偏差。

我正在更新重量如下：

对于每个批次，我在单个批次中运行以下临时重量更新

# update weight_update where y[i] is the actual label and o1 is the predicted output and x[i] is the input
weight_update = weight_update + (self.weights + self.learning_rate * (y[i] - o1)*x[i])

然后一个完整的批次完成了我更新了我班上的主要重量

# update main weights (self.weights) where len(x) is the number of samples
self.weights = self.weights + (weight_update / len(x))

标签： pythonneural-networkperceptron

解决方案

我假设 (y[i] - o1)*x[i]) 是损失函数 wrt 权重的偏导数，我不确定你使用的是什么，但假设你使用了 -1/2 * (y[i] - o1)^2

now let o1 = wx + b, where w is weight matrix and b is bias vector, 
also let, L = -1/2 * (y[i] - o1)^2
you have already calculated dLdw = dLd(o1) * d(o1)dw = (y[i] - o1) * x

In a summilar way calculate dLdb, 
dLdb = dLd(o1) * d(o1)db
dLd(o1) = (y[i] - o1)
d(o1)db = d/db (wx + b) = 0 + 1 = 1
so dLdb = (y[i] - o1) * 1

现在这条线，

weight_update = weight_update + (self.weights + self.learning_rate * (y[i] - o1)*x[i])

此时无需添加权重，只需添加渐变即可

weight_update += self.learning_rate * dLdw
# similarily
bias_update += self.learning_rate * dLdb

当一批完成后，就做

# update main weights (self.weights) and biases (self.biases)
# where len(x) is the number of samples
self.weights += (weight_update / len(x))
self.biases+= (bias_update / len(x)) 

# dont forget to set the values of weight_update, bias_update  to 0

这是我几天前写的（MNIST 示例的 1 个隐藏分层网络），您可能会发现这很有用

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