Machine Learning 学习笔记（第一周）

Week1

学习Coursera上Machine Learning后所做的笔记与整理

Introduction

Welcome

Machine learning is everywhere.

The aim of machine learning is to build machines as intelligent.

Two goals:

• know the algorithms & math
• implement each algorithmes

Why machine learing successful today

• grew out of work in AI
• New capability for computers

Examples:

• Database mining
• application can’t program by hand

handwriting recognition, NLP, CV

• Self-customizing programs

Recommendations

• understanding human learning (brain, real AI)

What is Machine Learning

• Arthur Samuel (1959). Machine Learning: Field of study that gives computers the ability to learn without being explicitly programmed.
• Tom Mitchell (1998). Well-posed Learning Problem: A computer program is said to learn from experience E with respect to some task T and some performance measure P, if its performance on T, as measured by P, improves with experience E.

Machine learning algorithms:

• Supervised learning

• Unsupervised learning

  Others: Reinforcement learning, recommender system

Supervised Learning

Example:

• housing price prediction

  “right answers” given

  Regression: Predict continuous valued output (price)

• Breast cancer (malignant, benign)

  Classification: Discrete valued output (0 or 1)

  two input parameters (Age, Tumor Size) to predict (malignant, benign)

Unsupervised Learning

no label, cluster by algorithm itself

no feedback based on the prediction results

Examples:

• Clustering in 2-D data

  • Age, Tumor Size data
  • Genes VS. individuals clustering

• Organize computing clusters

• Social network analysis

• Market segmentation

• Astronomical data analysis

• cocktail party problem

  More than one people speaking, distinguish the vioce from microphones

Model and Cost function

Model Representation

Linear regression

• Features

  supervised, regression

• notation

  training set

  m = Number of training examples

  x’s = “input” variable / features

  y’s = “output” variable / “target” variable

  (x, y): one traning example

  ( x ( i ) , y ( i ) ) (x^{(i)}, y^{(i)}) (x(i),y(i)): its training example

ML work flow

h θ ( x ) = θ 0 + θ 1 x h_\theta(x)=\theta_0+\theta_1x hθ​(x)=θ0​+θ1​x

For historical reasons, this function h is called a hypothesis. When the target variable that we’re trying to predict is continuous, such as in our housing example, we call the learning problem a regression problem. When y can take on only a small number of discrete values (such as if, given the living area, we wanted to predict if a dwelling is a house or an apartment, say), we call it a classification problem.

Cost Function

Using linear regression as example:

h θ ( x ) = θ 0 + θ 1 x h_\theta(x)=\theta_0+\theta_1x hθ​(x)=θ0​+θ1​x

Idea: Choose θ 0 , θ 1 \theta_0, \theta_1 θ0​,θ1​ so that h θ ( x ) h_\theta(x) hθ​(x) is close to y y y for our training examples ( x , y ) (x,y) (x,y)

We want to minimize θ 0 , θ 1 1 2 m ∑ i = 1 m ( h θ ( x ( i ) ) − y ( i ) ) 2 \underset{\theta_0,\theta_1}{\text{minimize}}\frac{1}{2m}\sum_{i=1}^m(h_\theta(x^{(i)})-y^{(i)})^2 θ0​,θ1​minimize​2m1​∑i=1m​(hθ​(x(i))−y(i))2,

the cost function J ( θ 0 , θ 1 ) = 1 2 m ∑ i = 1 m ( h θ ( x ( i ) ) − y ( i ) ) 2 J(\theta_0,\theta_1)=\frac{1}{2m}\sum_{i=1}^m(h_\theta(x^{(i)})-y^{(i)})^2 J(θ0​,θ1​)=2m1​∑i=1m​(hθ​(x(i))−y(i))2

which is half of squared error function, the 1 2 \frac{1}{2} 21​ is convenience for computation of the gradient descent.

Cost Function Intuition

• Hypothesis:

h θ ( x ) = θ 0 + θ 1 x h_\theta(x)=\theta_0+\theta_1x hθ​(x)=θ0​+θ1​x

a function of x

• Parameters:

  θ 0 , θ 1 \theta_0,\theta_1 θ0​,θ1​

• Cost Function:

  J ( θ 0 , θ 1 ) = 1 2 m = s u m i = 1 m ( h θ ( x ( i ) ) − y ( i ) ) 2 J(\theta_0,\theta_1)=\frac{1}{2m}=sum_{i=1}^m(h_\theta(x^{(i)})-y^{(i)})^2 J(θ0​,θ1​)=2m1​=sumi=1m​(hθ​(x(i))−y(i))2

  a function of θ \theta θ

• Goal:

  minimize θ 0 , θ 1 J ( θ 0 , θ 1 ) \underset{\theta_0,\theta_1}{\text{minimize}}J(\theta_0,\theta_1) θ0​,θ1​minimize​J(θ0​,θ1​)

contour plot for two parameters and quadratic function for one parameters

Parameter Learning

Gradient Descent

Have some function J ( θ ) J(\theta) J(θ)

Want min θ J ( θ ) \underset{\theta}{\text{min}}J(\theta) θmin​J(θ)

Outline:

• Start with some θ \theta θ
• Keep changing θ \theta θ to reduce J ( θ ) J(\theta) J(θ) until we hopefully end up at a minimum

Gradient descent algorithm

repeat unitl convergence {

​ θ j : = θ j − α ∂ ∂ θ j J ( θ 0 , θ 1 ) ( for j = 0 and j = 1 ) \theta_j:=\theta_j-\alpha\frac{\partial}{\partial \theta_j}J(\theta_0,\theta_1) \quad (\text{for} j=0 \text{and} j=1) θj​:=θj​−α∂θj​∂​J(θ0​,θ1​)(forj=0andj=1)

}

α \alpha α is learning rate

Correct: Simultaneous update

t e m p 0 : = θ 0 − α ∂ ∂ θ 0 J ( θ 0 , θ 1 ) temp0 := \theta_0-\alpha\frac{\partial}{\partial \theta_0}J(\theta_0,\theta_1) temp0:=θ0​−α∂θ0​∂​J(θ0​,θ1​)

t e m p 1 : = θ 0 − α ∂ ∂ θ 1 J ( θ 0 , θ 1 ) temp1 := \theta_0-\alpha\frac{\partial}{\partial \theta_1}J(\theta_0,\theta_1) temp1:=θ0​−α∂θ1​∂​J(θ0​,θ1​)

θ 0 : = t e m p 0 \theta_0:=temp0 θ0​:=temp0

θ 1 : = t e m p 1 \theta_1:=temp1 θ1​:=temp1

Inorrect: Simultaneous update

t e m p 0 : = θ 0 − α ∂ ∂ θ 0 J ( θ 0 , θ 1 ) temp0 := \theta_0-\alpha\frac{\partial}{\partial \theta_0}J(\theta_0,\theta_1) temp0:=θ0​−α∂θ0​∂​J(θ0​,θ1​)

θ 0 : = t e m p 0 \theta_0:=temp0 θ0​:=temp0

t e m p 1 : = θ 0 − α ∂ ∂ θ 1 J ( θ 0 , θ 1 ) θ 0 : = t e m p 0 temp1 := \theta_0-\alpha\frac{\partial}{\partial \theta_1}J(\theta_0,\theta_1)\theta_0:=temp0 temp1:=θ0​−α∂θ1​∂​J(θ0​,θ1​)θ0​:=temp0

θ 1 : = t e m p 1 \theta_1:=temp1 θ1​:=temp1

在不正确的形式中， θ 0 \theta_0 θ0​和 θ 1 \theta_1 θ1​不是同时更新的（更新 θ 1 \theta_1 θ1​之前 θ 0 \theta_0 θ0​已经改变了）

Gradient Descent Intuition

不断逼近局部最小值。

如果学习率过小则会导致逼近过程过慢，如果学习率过大则无法收敛。

Gradient Descent For Linear Regression

repeat unitl convergence {

​ θ j : = θ j − α ∂ ∂ θ j J ( θ 0 , θ 1 ) ( for j = 0 and j = 1 ) \theta_j:=\theta_j-\alpha\frac{\partial}{\partial \theta_j}J(\theta_0,\theta_1) \quad (\text{for} j=0 \text{and} j=1) θj​:=θj​−α∂θj​∂​J(θ0​,θ1​)(forj=0andj=1)

}

h θ ( x ) = θ 0 + θ 1 x h_\theta(x)=\theta_0+\theta_1x hθ​(x)=θ0​+θ1​x

J ( θ 0 , θ 1 ) = 1 2 m ∑ i = 1 m ( h θ ( x ( i ) ) − y ( i ) ) 2 J(\theta_0,\theta_1)=\frac{1}{2m}\sum_{i=1}^m(h_\theta(x^{(i)})-y^{(i)})^2 J(θ0​,θ1​)=2m1​∑i=1m​(hθ​(x(i))−y(i))2

∂ ∂ θ j J ( θ 0 , θ 1 ) = ∂ ∂ θ j 1 2 m ∑ i = 1 m ( h θ ( x ( i ) ) − y ( i ) ) 2 = ∂ ∂ θ j 1 2 m ∑ i = 1 m ( θ 0 + θ 1 x ( i ) − y ( i ) ) 2 \begin{aligned} \frac{\partial}{\partial\theta_j}J(\theta_0,\theta_1) &= \frac{\partial}{\partial\theta_j}\frac{1}{2m}\sum_{i=1}^m(h_\theta(x^{(i)})-y^{(i)})^2 \\ &= \frac{\partial}{\partial\theta_j}\frac{1}{2m}\sum_{i=1}^m(\theta_0+\theta_1x^{(i)}-y^{(i)})^2\end{aligned} ∂θj​∂​J(θ0​,θ1​)​=∂θj​∂​2m1​i=1∑m​(hθ​(x(i))−y(i))2=∂θj​∂​2m1​i=1∑m​(θ0​+θ1​x(i)−y(i))2​

θ 0 : = θ 0 − ∂ ∂ θ 0 J ( θ 0 , θ 1 ) : = θ 0 − 1 m ( h θ ( x ( i ) ) − y ( i ) ) \begin{aligned}\theta_0 &:= \theta_0-\frac{\partial}{\partial\theta_0}J(\theta_0,\theta_1)\\ &:=\theta_0-\frac{1}{m}(h_\theta(x^{(i)})-y^{(i)})\end{aligned} θ0​​:=θ0​−∂θ0​∂​J(θ0​,θ1​):=θ0​−m1​(hθ​(x(i))−y(i))​

θ 1 : = θ 1 − ∂ ∂ θ 1 J ( θ 0 , θ 1 ) : = θ 1 − 1 m ( h θ ( x ( i ) ) − y ( i ) ) x ( i ) \begin{aligned}\theta_1 &:= \theta_1-\frac{\partial}{\partial\theta_1}J(\theta_0,\theta_1)\\ &:=\theta_1-\frac{1}{m}(h_\theta(x^{(i)})-y^{(i)})x^{(i)}\end{aligned} θ1​​:=θ1​−∂θ1​∂​J(θ0​,θ1​):=θ1​−m1​(hθ​(x(i))−y(i))x(i)​

Cost function J J J for linear regression is convex function (凸函数), have no local optimum but a global optimum.

“Batch” Gradient Descent

“Batch”: Each step of gradient descent uses all the training examples.

Linear Algebra Review

Matrices and Vectors

• matrices

  A = [ 1 2 3 4 5 6 ] ∈ R 2 × 3 A= \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \in \mathbf{R}^{2\times3} A=[14​25​36​]∈R2×3

  A 12 = 2 A_{12}=2 A12​=2

  先行数，再列数

• Vector

  is An n × 1 n\times1 n×1matrix

Addition and Scalar

• Matrix Addition

  [ 1 0 2 5 3 1 ] + [ 4 0.5 2 5 0 1 ] = [ 5 0.5 4 10 3 2 ] \begin{bmatrix} 1 & 0 \\ 2 & 5 \\ 3 & 1 \end{bmatrix} + \begin{bmatrix} 4 & 0.5 \\ 2 & 5 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 0.5 \\ 4 & 10 \\ 3 & 2 \end{bmatrix} ⎣⎡​123​051​⎦⎤​+⎣⎡​420​0.551​⎦⎤​=⎣⎡​543​0.5102​⎦⎤​

  size相同的matrix可以相加

• Scalar Multiplication

  3 × [ 1 0 2 5 3 1 ] = [ 3 0 6 15 9 3 ] 3 \times \begin{bmatrix} 1 & 0 \\ 2 & 5 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 6 & 15 \\ 9 & 3 \end{bmatrix} 3×⎣⎡​123​051​⎦⎤​=⎣⎡​369​0153​⎦⎤​

  [ 4 0 6 3 ] / 4 = 1 4 [ 4 0 6 3 ] = [ 1 0 3 2 3 4 ] \begin{bmatrix} 4 & 0 \\ 6 & 3 \end{bmatrix} / 4 = \frac{1}{4} \begin{bmatrix} 4 & 0 \\ 6 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ \frac{3}{2} & \frac{3}{4} \end{bmatrix} [46​03​]/4=41​[46​03​]=[123​​043​​]

• Combination of Operands

  3 × [ 1 4 2 ] + [ 0 0 5 ] − [ 3 0 2 ] / 2 = [ 2 12 10 1 3 ] 3 \times \begin{bmatrix} 1 \\ 4 \\ 2 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 5 \end{bmatrix} - \begin{bmatrix} 3 \\ 0 \\ 2 \end{bmatrix} / 2 = \begin{bmatrix} 2 \\ 12 \\ 10\frac{1}{3} \end{bmatrix} 3×⎣⎡​142​⎦⎤​+⎣⎡​005​⎦⎤​−⎣⎡​302​⎦⎤​/2=⎣⎡​2121031​​⎦⎤​

Matrix Vector Multiplication

[ 1 3 4 0 2 1 ] [ 1 5 ] = [ 16 4 7 ] \begin{bmatrix} 1 & 3 \\ 4 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 5 \end{bmatrix} = \begin{bmatrix} 16 & 4 & 7 \end{bmatrix} ⎣⎡​142​301​⎦⎤​[15​]=[16​4​7​]

[ a b c d e f ] [ g h ] = [ a × g + b × h c × g + d × h e × g + f × h ] \begin{bmatrix} a & b \\ c & d \\ e & f \end{bmatrix} \begin{bmatrix} g \\ h \end{bmatrix} = \begin{bmatrix} a \times g + b \times h \\ c \times g + d \times h \\ e \times g + f \times h \end{bmatrix} ⎣⎡​ace​bdf​⎦⎤​[gh​]=⎣⎡​a×g+b×hc×g+d×he×g+f×h​⎦⎤​

Matrix Matrix Multiplication

[ 1 3 2 4 0 1 ] [ 1 3 0 1 5 2 ] = [ 11 9 10 14 ] \begin{bmatrix} 1 & 3 & 2 \\ 4 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 0 & 1 \\ 5 & 2 \end{bmatrix} = \begin{bmatrix} 11 & 9 \\ 10 & 14 \end{bmatrix} [14​30​21​]⎣⎡​105​312​⎦⎤​=[1110​914​]

[ a b c d e f ] [ g h i j k l ] = [ a × g + b × i + c × k a × h + b × j + c × l d × g + e × i + f × k d × h + e × j + f × l ] \begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix} \begin{bmatrix} g & h \\ i & j \\ k & l \end{bmatrix} = \begin{bmatrix} a \times g + b \times i + c \times k & a \times h + b \times j + c \times l \\ d\times g + e \times i + f \times k & d \times h + e \times j + f \times l \end{bmatrix} [ad​be​cf​]⎣⎡​gik​hjl​⎦⎤​=[a×g+b×i+c×kd×g+e×i+f×k​a×h+b×j+c×ld×h+e×j+f×l​]

Matrix Multiplication Properties

• Commutative

  A × B ≠ B × A A \times B \neq B \times A A×B​=B×A (not commutative)

• Associative

  ( A × B ) × C = A × ( B × C ) (A \times B) \times C = A \times (B \times C) (A×B)×C=A×(B×C)

• Identity Matrix

  I n × n = [ 1 0 ⋯ 0 0 1 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ 1 ] I_{n\times n}=\begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{bmatrix} In×n​=⎣⎢⎢⎢⎡​10⋮0​01⋮0​⋯⋯⋱⋯​00⋮1​⎦⎥⎥⎥⎤​

  For any matrix A A A, A m × n I n × n = I m × m A m × n = A m × n A_{m \times n} I_{n \times n} = I_{m \times m} A_{m \times n} = A_{m \times n} Am×n​In×n​=Im×m​Am×n​=Am×n​

Matrix Inverse and Transpose

• Matrix Inverse

  If A is an m × m m \times m m×mmatrix, and if it has an inverse,

  A A − 1 = A − 1 A = I AA^{-1}=A^{-1}A=I AA−1=A−1A=I

• Matrix Transpose