Coursera Deep Learning笔记 逻辑回归典型的训练过程
Deep Learning 用逻辑回归训练图片的典型步骤.
笔记摘自:https://xienaoban.github.io/posts/59595.html
1. 处理数据
1.1 向量化(Vectorization)
将每张图片的高和宽和RGB展为向量,最终X的shape为 (height*width*3, m)
.
1.2 特征归一化(Normalization)
对于一般数据,使用标准化(Standardization)
- \(X_{scale} = \frac{(X(axis=0) - X.mean(axis=0))}{X.std(axis=0)}\)
z_i = (x_i - mean) / delta
,mean
与delta
代表X的均值和标准差. 最终特征处于[-1, 1]区间.
对于图片, 可直接使用Min-Max Scaling
- 即将每个特征除以255(每个像素分为R, G, B, 范围在0~255)使得值处于[0, 1].
2. 初始化参数
一般将 w
和 b
随机选择.
3. 梯度下降(Gradient descent)
根据 w
, b
和训练集,来训练数据.
- 需要设定 迭代次数 与 学习率 .
以下为大循环(迭代次数)中内容:
3.1 计算代价函数
对于\(x^{(i)} \in X\), 有
\[z^{(i)} = w^Tx^{(i)} + b
\]
\[ a^{(i)} = \hat{y}^{(i)} = sigmod(z^{(i)}) = \sigma(z^{(i)}) = \frac{1}{1 + e^{-z^{(i)}}}
\]
\[损失函数: {L}(a^{(i)}, y^{(i)}) = {L}(\hat{y}^{(i)}, y^{(i)}) = - y^{(i)} \log(a^{(i)}) - (1-y^{(i)} ) \log(1-a^{(i)})
\]
\[A = (a^{(1)}, a^{(2)}, ... , a^{(m-1)}, a^{(m)})
= \sigma(w^TX+b)
= \frac{1}{1+e^{-(w^TX+b)}}
\]
\[代价函数: J(w,b) = -\frac{1}{m} \sum^{m}_{i=1} \mathcal{L}(\hat{y}^{(i)}, y^{(i)})
= -\frac{1}{m} \sum^{m}_{i=1} (y^{(i)} log(\hat{y}^{(i)}) + (1-y^{(i)}) log(1-\hat{y}^{(i)}))
\]
# 激活函数
A = sigmoid(w.T.dot(X) + b)
# 代价函数
cost = -np.sum(Y * np.log(A) + (1-Y) * np.log(1 - A)) / m
3.2 计算反向传播的梯度
即:对 \(J = -\dfrac{1}{m} \sum L(a, y)\) 计算导数,即对\({L}(a, y)\) 计算导数,以下求导,均省略上标。
求:\(\dfrac{\partial J}{\partial w}\) 和 $\dfrac{\partial J}{\partial b} $ (dw 和 db)
\[\dfrac{\partial L}{\partial a}
= \dfrac{\partial L(a, y)}{\partial a}
= -\frac{y}{a} + \frac{1-y}{1-a}
\]
\[\dfrac{da}{dz}
= (\frac{1}{1 + e^{-z}})'
= \dfrac{e^{-z}}{(1+e^{-z})^2}
= \dfrac{1}{1+e^{-z}} - \dfrac{1}{(1+e^{-z})^2}
= a-a^2
= a · (1-a)
\]
\[\dfrac{\partial L}{\partial z}
= \dfrac{\partial L}{\partial a} \dfrac{da}{dz}
= (-\dfrac{y}{a} + \dfrac{1-y}{1-a}) · a · (1-a)
= a - y
\]
\[\dfrac{\partial L}{\partial w}
= \dfrac{\partial L}{\partial z} \dfrac{\partial z}{\partial w}
= (a-y) · x
\]
\[\dfrac{\partial L}{\partial b}
= \dfrac{\partial L}{\partial z} \dfrac{\partial z}{\partial b}
= a-y
\]
根据 \(J = -\dfrac{1}{m} \sum L(a, y)\) 最终可得:
\[\dfrac{\partial J}{\partial w}
= \dfrac{\partial J}{\partial a} \dfrac{\partial a}{\partial w}
= \dfrac{1}{m} X(A-Y)^T
\]
\[\dfrac{\partial J}{\partial b} = \dfrac{1}{m} \sum^{m}_{i=1} (a^{(i)} - y^{(i)})
\]
dw = X.dot((A - Y).T) / m
db = np.sum(A - Y) / m
3.3 更新 w
, b
w = w - learning_rate * dw
b = b - learning_rate * db
4. 预测测试集
-
使用训练出来的
w
,b
, 对测试集使用y_pred = sigmoid(wx+b)
, 计算得预测的概率 -
对其取整, 例如大于0.7则判定为 '是', 否则为'否'.