week1
一张图片，设像素为64*64， 颜色通道为红蓝绿三通道，则对应3个64*64实数矩阵
为了用向量表示这些矩阵，将这些矩阵的像素值展开为一个向量x作为算法的输入
从红色到绿色再到蓝色，依次按行一个个将元素读到向量x中，则x是一个\(1\times64*64*3\)的矩阵，也就是一个64*64*3维的向量
用 \(n_x = 64*64*3\) 表示特征向量x的维度
而所有的训练样本表示成：\(X = \begin{bmatrix}\mid & \mid &\mid &&\mid \\ x^{(1)}& x^{(2)}& x^{(3)}& \cdots & x^{(m)}\\ \mid & \mid &\mid &&\mid \end{bmatrix}\) (\(n_x \times m\)矩阵)
（注意不是\(X = \begin{bmatrix} (x^{(1)})^T\\ \vdots \\ (x^{(m)})^T \end{bmatrix}\) ,用上面的方法运算会简单点）
\(Y=\begin{bmatrix}y^{(1)} & y^{(2)} & \cdots & y^{(m)}\end{bmatrix}\)
之前的机器学习课上的\(\theta = \begin{bmatrix} \theta_0 \\ \theta_1 \\ \vdots \\ \theta_{n_x} \\ \end{bmatrix}\)的形式不再使用，而用\(\large b = \theta_0, \; w = \begin{bmatrix} \theta_1 \\ \vdots \\ \theta_{n_x} \\ \end{bmatrix}\)代替( it will be easier to just keep \(b\) and \(w\) as separate parameters )
则output : \(\large \hat{y}^{(i)} = \sigma(w^Tx^{(i)}+b)，{\rm where\;}\sigma(z^{(i)}) = \frac{1}{1+e^{-z^{(i)}}}\)
\(\text{Given \{}(x^{(1)}, y^{(1)}),\dots,(x^{(m)},y^{(m)})\text{\}, want } \hat{y}^{(i)} \approx y^{(i)}\)
week2
Loss Function/Error Function
Loss Function/Error Function(误差函数): used to measure how well our algorism is doing
\]
Cost Function
\]
Gradient Descent
看ML的笔记，实质上是一样的
Vectorization:
#Non-vecotrized
#slow
z = 0
for i in range(n_x):
z += w[i] * x[i]
z += b
#Vectorized
#import numpy as np
z = np.dot(w,x) + b
whenever possible, avoid explicit for-loops（因为是解释型语言）, 用numpy带的行数可以简洁而高效地实现
Vectorizing Logistic Regression
\(X = \begin{bmatrix} \lvert & \lvert & \cdots & \lvert \\ x^{(1)} & x^{(2)} & \cdots & x^{(m)} \\ \lvert & \lvert & \cdots & \lvert \end{bmatrix}, \mathbb{R}^{n_x \times m}\)
\(Z = \begin{bmatrix}z^{(1)} & z^{(2)} & \cdots & z^{(m)} \end{bmatrix} = w^TX + \begin{bmatrix}b &b & \cdots & b \end{bmatrix}\)
\(z^{(i)}\) 是 sigmoid function的输入值
\(A = \begin{bmatrix}a^{(1)} & a^{(2)} & \cdots & a^{(m)} \end{bmatrix} = \sigma(Z)\)
（这里的不同上标的元素似乎实际是在同一个layer中的，跟ML课上不大一样。 \(a^{[j](i)}\)中方括号括起来的是层数，圆括号括起来的是第\(i\)个训练实例）
import numpy as np
z = np.dot(w,x) + b\
#Python automatically takes this real number b and expands it out to this 1*m row vector
Gradient Output
\({\rm d}z^{(i)} = a^{(i)} - y^{(i)}\)
\(\begin{align}{\rm d}Z &= \begin{bmatrix}{\rm d}z^{(1)} & {\rm d}z^{(2)} & \cdots & {\rm d}z^{(m)} \end{bmatrix} \\&= A-Y = \begin{bmatrix}a^{(1)} - y^{(1)} & a^{(2)} - y^{(2)} & \cdots & a^{(m)} - y^{(m)} \end{bmatrix} \end{align}\)
${\rm d}b = $1/m*np.sum(dZ)
\({\rm d}w = \frac{1}{m}X{\rm d}Z^T\)
单次迭代免for-loop法(vectorize)：
\downarrow&\begin{cases}
Z & = w^T+b\\
& = {\rm np.dot(}w{\rm .T, }X{\rm)}\\
A & = \sigma(Z)\\
{\rm d}Z &= A-Y \\
{\rm d}w &= \frac{1}{m}X{\rm d}Z^T\\
\end{cases}\\\\
w& := w - \alpha{\rm d}w\\
b &:= b - \alpha{\rm d}b
\end{align}
\]
若要多次迭代，最外层的显式for-loop是不可避免的
Broadcasting
用reshape()确保矩阵的尺寸
举个例子说明numpy 的 broadcasting机制：
>>> import numpy as np
>>> a = np.arange(0,6).reshape(6,1)
>>> a
array([[0],
[1],
[2],
[3],
[4],
[5]])
>>> b = np.arange(0,5)
>>> b
array([0, 1, 2, 3, 4])
>>> a * b
array([[ 0, 0, 0, 0, 0],
[ 0, 1, 2, 3, 4],
[ 0, 2, 4, 6, 8],
[ 0, 3, 6, 9, 12],
[ 0, 4, 8, 12, 16],
[ 0, 5, 10, 15, 20]])
>>> a + b
array([[0, 1, 2, 3, 4],
[1, 2, 3, 4, 5],
[2, 3, 4, 5, 6],
[3, 4, 5, 6, 7],
[4, 5, 6, 7, 8],
[5, 6, 7, 8, 9]])
也就是说matrix+-*/number/vector时，numpy会将number/vector通过自我复制拓展成合法的矩阵
注意这会导致 在期望抛出异常的地方 不抛出异常而是发生奇怪的BUG：
比如 有时我想 行向量和列向量相加时抛出异常， 但是numpy却用broadcasting机制把它给算出来了...
numpy的坑
import numpy as np
a = np.random.randn(5)
>>> a
array([-0.19837642, -0.16758652, 1.57705505, 0.13033745, -0.81073889])
>>> a.shape
(5,)
# which is called a rank 1 array in Python and is neither a row vector nor a column vector
>>> a.T
array([-0.19837642, -0.16758652, 1.57705505, 0.13033745, -0.81073889])
# which is same as 'a' i self
>>> np.dot(a,a.T)
3.2288264718632416
# it is a number rather than a matrix in expectation(just like array([[55]]))
不要使用形如(5,)或者(n,)这样的“rank 1 array”， 而是显式地说明是\(m \times n\)的矩阵：
>>> a = np.random.randn(5,1)
>>> a
array([[ 0.7643396 ],
[-1.66945103],
[ 1.66235712],
[-0.06892102],
[-1.61347409]])
>>> a.T
array([[ 0.7643396 , -1.66945103, 1.66235712, -0.06892102, -1.61347409]])
注意array([-0.19837642, -0.16758652, 1.57705505, 0.13033745, -0.81073889])和array([[ 0.7643396 , -1.66945103, 1.66235712, -0.06892102, -1.61347409]])的区别（后者有两个方括号）， 这说明前者是秩为1的数组而后者是一个真正的\(1 \times 5\)矩阵（就像C里一样矩阵是用二维数组表示的）（另外我觉得rank 1 array翻译为一维数组更为准确）
It can use assert() statement to make sure the dimension of one of vectors.
When you get a rank 1 array, you can use a.reshape to transform it into a (n,1) array or a (1,n) array.
Logistic Regression Cost Function
\begin{array}{l}
\text{If y=1:}\quad p(y|x)=\hat{y}\\
\text{If y=0:}\quad p(y|x)=1-\hat{y}
\end{array}
\right\}
p(y|x) = \hat{y}^y\cdot (1-\hat{y})^{1-y}\\
\,\\
\begin{align}
\therefore {\rm log}(p(y|x)) &= y\cdot log\,\hat{y} + (1-y)\cdot log\, (1-\hat{y}) \\
&= -\mathcal{L}(\hat{y},y)
\end{align}
\]
所以：
{\rm log }[p(\text{labels in training set})] &= {\rm log } \prod_{i=1}^mp(y^{(i)}|x^{(i)})\\
&=\sum_{i=1}^m {\rm log\,}p(y^{(i)}|x^{(i)})\\
&=\sum_{i=1}^m-\mathcal{L}(\hat{y}^{(i)},y^{(i)})\\
&=-\sum_{i=1}^m \mathcal{L}(\hat{y}^{(i)},y^{(i)})
\end{align}\\
\text{Cost: }J(w,b) = \frac{1}{m}\sum_{i=1}^m \mathcal{L}(\hat{y}^{(i)},y^{(i)})
\]
maximum likelihood estimation (极大似然估计)
week3
\(Z^{[j]} = W^{[j]}A^{[j-1]} + b^{[j]} = w^{[j]}\begin{bmatrix} | & | & | & \\ a^{[j-1](1)} & a^{[j-1](2)} & a^{[j-1](3)} & \cdots \\ | & | & | & \end{bmatrix} + b^{[j]} = \begin{bmatrix} | & | & | & \\ z^{[j](1)} & z^{[j](2)} & z^{[j](3)} & \cdots \\ | & | & | & \end{bmatrix}\)
其中\((i) \in [(1),(m)],\quad [j] \in [[1],[n]],\quad X = A^{[0]}\)
Other Activation Function
①\(tanh(z)\) function:
\]
\(tanh(z)\) 可以把 数据中心化 为 0 （Sigmoid Function 将数据中心化为 0.5）
之后只有 \(0 \le \hat{y} \le 1\) (即二元分类问题)才用 Sigmoid Function，因为\(tanh\)几乎严格优于Sigmoid...
②Rectified Linear Unit(线性整流函数, ReLU)：\(Q = max(0,z)\)
When not sure what to use for your hidden layer, can use the ReLU function
Disadvantage of ReLU: when \(z\) is negative, the value is 0.
It can use what names Leaky ReLU to overcome the disadvantage below.
Leaky ReLU: \(a = max(0.01z, z)\)
ReLU可以使得斜率不变（Sigmoid 和 \(tanh(z)\) 在\(z\rightarrow \infin\)时斜率趋向于0，会使得学习速度下降）
最常用的 Activation Function
③Tannish Function(双曲函数)
当且仅当要解决回归问题的时候，在生成到output layer才使用线性的Activation Function（\(g(z)=z\)) ,比如预测房价时，y不限于 0 和 1（\(y \in \mathbb{R}\)），所以可以用\(g(z)=z\) 输出，隐藏单元不应该使用Linear Activation Function, 而是应该使用tanh/ReLU/Leaky ReLU
Derivatives of Activation Functions
- Sigmoid:
- \(\frac{{\rm d}}{{\rm d}z}g(z) = g(z)(1-g(z))\)
\(tanh(z)\): - \(g\prime(z) = 1-(tanh(z))^2\)
- \(\frac{{\rm d}}{{\rm d}z}g(z) = g(z)(1-g(z))\)
- ReLU:
- \(g\prime(z) = \begin{cases}1, \text{if }z\ge0 \\0, \text{if }z\lt0 \end{cases}\)
Gradient Descents For Neural Networks
Parameters : \(w^{[1]},b^{[1]},w^{[2]},b^{[2]}\)
Cost Function : \(J(w^{[1]},b^{[1]},w^{[2]},b^{[2]})= \frac{1}{m}\sum_{i=1}^m \mathcal{L}(\hat{y},y)\)
Gradient Function:
&\text{Repeat \{}\\
&\quad \text{compute predicts} (\hat{y}^{(i)}, i = 1,\dots,m) \\
&\quad {\rm d}w^{[1]} = \frac{\partial J}{\partial w^{[1]}}, {\rm d}b^{[1]} = \frac{\partial J}{\partial b^{[1]}},\dots\\
&\quad w^{[1]} = w^{[1]} - \alpha {\rm d}w^{[1]}\\
&\quad b^{[1]} = b^{[1]} - \alpha {\rm d}b^{[1]}\\
&\quad w^{[2]} = w^{[2]} - \alpha {\rm d}w^{[2]}\\
&\quad b^{[2]} = b^{[2]} - \alpha {\rm d}b^{[2]}\\
\text{\}}
\end{align}
\]
Forward Propagation :
Z^{[1]} &= w^{[1]}X + b^{[1]}\\
A^{[1]} &= g^{[1]}(z^{[1]})\\
Z^{[2]} &= w^{[2]}A^{[1]} + b^{[2]}\\
A^{[2]} &= g^{[2]}(z^{[2]}) = \sigma(Z^{[2]})
\end{align}
\]
Backward Propagation :
{\rm d}Z^{[2]} &= A^{[2]} - Y, \quad Y = \begin{bmatrix}y^{[1]} & y^{[2]} & \dots & y^{[m]}\end{bmatrix}\\
{\rm d}w^{[2]} &= \frac{1}{m} {\rm d}z^{[2]} A^{[1]T}\\
{\rm d}d^{[2]} &= \frac{1}{m}\text{np.sum(d}z^{[2]}\text{,axis=1,keepdims=True)}\\
{\rm d}Z^{[1]} &= w^{[2]T}{\rm d}Z^{[2]}\; .* \; g^{[1]\prime}(Z^{[1]})\\
{\rm d}w^{[1]} &= \frac{1}{m} {\rm d}Z^{[1]}X^T\\
{\rm d}d^{[1]} &= \frac{1}{m}\text{np.sum(d}z^{[1]}\text{,axis=1,keepdims=True)}\\
\end{align}
\]
注：axis = 1 means summing horizontally, and keepdims = True means prevent from outputting Rank 1 Array. You can call reshape function explicitly rather than keeping these parameters.
又注：\(由于A^{[1]} = g^{[1]}(Z^{[1]})且g^{[1]\prime}(z) = 1-a^2,\;所以 g^{[1]\prime}(Z^{[1]}) = 1-(A^{[1]})^2\), 即：\(Z^{[1]} = w^{[2]T}{\rm d}Z^{[2]}\; .* \; (1-(A^{[1]})^2\)
Random Initialization
For a neural network, if initialize the weights to parameters to all zero and then apply gradient descent, it won't work.