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 imes64*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 = egin{bmatrix}mid & mid &mid &&mid \ x^{(1)}& x^{(2)}& x^{(3)}& cdots & x^{(m)}\ mid & mid &mid &&mid end{bmatrix}) ((n_x imes m)矩阵)
（注意不是(X = egin{bmatrix} (x^{(1)})^T\ vdots \ (x^{(m)})^T end{bmatrix}) ,用上面的方法运算会简单点）
(Y=egin{bmatrix}y^{(1)} & y^{(2)} & cdots & y^{(m)}end{bmatrix})
之前的机器学习课上的( heta = egin{bmatrix} heta_0 \ heta_1 \ vdots \ heta_{n_x} \ end{bmatrix})的形式不再使用，而用(large b = heta_0, ; w = egin{bmatrix} heta_1 \ vdots \ heta_{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)，{ m where;}sigma(z^{(i)}) = frac{1}{1+e^{-z^{(i)}}})
( ext{Given {}(x^{(1)}, y^{(1)}),dots,(x^{(m)},y^{(m)}) ext{}, 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 = egin{bmatrix} lvert & lvert & cdots & lvert \ x^{(1)} & x^{(2)} & cdots & x^{(m)} \ lvert & lvert & cdots & lvert end{bmatrix}, mathbb{R}^{n_x imes m})
(Z = egin{bmatrix}z^{(1)} & z^{(2)} & cdots & z^{(m)} end{bmatrix} = w^TX + egin{bmatrix}b &b & cdots & b end{bmatrix})
(z^{(i)}) 是 sigmoid function的输入值
(A = egin{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
({ m d}z^{(i)} = a^{(i)} - y^{(i)})
(egin{align}{ m d}Z &= egin{bmatrix}{ m d}z^{(1)} & { m d}z^{(2)} & cdots & { m d}z^{(m)} end{bmatrix} \&= A-Y = egin{bmatrix}a^{(1)} - y^{(1)} & a^{(2)} - y^{(2)} & cdots & a^{(m)} - y^{(m)} end{bmatrix} end{align})
${
m d}b = $1/m*np.sum(dZ)
({ m d}w = frac{1}{m}X{ m d}Z^T)
单次迭代免for-loop法(vectorize)：
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 imes 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 imes 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
所以：
maximum likelihood estimation (极大似然估计)
week3
(Z^{[j]} = W^{[j]}A^{[j-1]} + b^{[j]} = w^{[j]}egin{bmatrix} | & | & | & \ a^{[j-1](1)} & a^{[j-1](2)} & a^{[j-1](3)} & cdots \ | & | & | & end{bmatrix} + b^{[j]} = egin{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 ightarrow 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{{
m d}}{{
m d}z}g(z) = g(z)(1-g(z)))
(tanh(z)): - (gprime(z) = 1-(tanh(z))^2)
- (frac{{
m d}}{{
m d}z}g(z) = g(z)(1-g(z)))
- ReLU:
- (gprime(z) = egin{cases}1, ext{if }zge0 \0, ext{if }zlt0 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:
Forward Propagation :
Backward Propagation :
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}{ m 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.