Stanford公開課機器學習---week2-2.Matlab/Octave 快速教程

Basic Operations 基本操作

a =

    1.0000   15.0000    2.0000    0.5000
% ================format ================
>> format long
>> a

a =

   1.000000000000000  15.000000000000000   2.000000000000000   0.500000000000000
% ================fprintf ================
>> fprintf('%f %f \n',1, 2);
1.000000 2.000000 
% ================ ================
>> v=1:0.1:2 % start；步長；end

v =

    1.0000    1.1000    1.2000    1.3000    1.4000    1.5000    1.6000    1.7000    1.8000    1.9000    2.0000

>> v=1:6

v =

     1     2     3     4     5     6
% ================help ================
help

A=rand(3,2);
A=randn(3,2);
w = -6 + sqrt(10)*(randn(1,10000));
hist(w); %柱狀圖
eye();
ones();
zeros(1,3)

---

Moving Data Around 移動數據

% ================A ================
>> A=[1 2;3 4;5 6]
A =
    1     2
    3     4
    5     6
% ================ size ================
>>size(A) %返回矩陣維度：行數 列數
ans = 3 2
>>size(A,1) %返回矩陣維度：行數
ans = 3
>>size(A,2) %返回矩陣維度：列數
ans = 2

% ================ length================
>>v = [1 2 3 4] %返回矩陣維度：行數
v = 
    1 2 3 4
>>length(v) %返回矩陣最大維度大小
ans = 4
>>length(A) %返回矩陣最大維度大小
ans = 3

% ================ ================
pwd %當前路徑
ls %當前路徑下的文件
cd %切換目錄
% ================ load clear save data ================
data = load('ex1data2.txt');

who %當前工作空間的所有變量

whos %當前工作空間的所有變量的詳細信息

clear variablename %刪除變量
clear variable %刪除當前工作空間的所有變量

>> save test.mat A
>> save testtext.txt A –ascii % save as text (ASCII)

% ================ 矩陣 Matrix================
>> A=[1 2;3 4;5 6]
A =
   1     2
   3     4
   5     6

>> A(3,2) %（行，列）

ans =

    6

>> A(2,:) %第二行

ans =

    3     4

>> A([1 3],:) %select the 1st and 3rd row

ans =

    1     2
    5     6


>> A(:,2)=[10,11,12]

A =

    1    10
    3    11
    5    12

>> A=[A,[101;102;103]]%append another column vector

A =

    1    10   101
    3    11   102
    5    12   103

>> A(:) %put all elements of A into a single vector

ans =

    1
    3
    5
   10
   11
   12
  101
  102
  103

B=[11,12;13,14;15,16]

B =

   11    12
   13    14
   15    16

>> C=[A B]

C =

    1    10   101   11    12
    3    11   102   13    14
    5    12   103   15    16

A=[1 2;3 4;5 6]

A =

    1     2
    3     4
    5     6

>> C=[A;B]

C =

    1     2
    3     4
    5     6
   11    12
   13    14
   15    16

---

Computing on Data 計算數據

% matlab默認 * / 是矩陣操作 .* ./是對每個元素操作
>>A
ans = 
   1    2
   3    4
   5    6

>>A.*B %對應位置的數據相乘,即element product

ans =
   11    24
   39    56
   75    96

 >>A.^2 % 對應位置的數據的平方

ans =
   1    4
   9    16
   25   36

 >>1 ./A % 對應位置的數據的除法

ans =
   1.00000    0.50000
   0.33333    0.25000
   0.20000    0.16667

>>-A % -1 * A 
ans =
    -1    -2
    -3    -4
    -5    -6

>>abs(A) % 絕對值

>> A + 1 % 或者 A + ones(3,2)

ans =
     2     3
     4     5
     6     7


>>A'%轉置 
ans =
    1     3     5
    2     4     6

>> A<3 %小於3的元素對應位置爲1（真），否則爲0（假）

ans = 
    1     1
    0     0
    0     0

>> find(A<3) % A中哪些元素小於3：第一個，第四個（從上到下數，列優先）

ans =

    1
    4

A =

    8     1     6
    3     5     7
    4     9     2

>> [r,c]=find(A>=6) %r是行  c是列 所以大於等於6的數的位置是 第一行第一列，第3行第2列。。。

r = 
    1
    3
    1
    2

c =
    1
    2
    3
    3

>> a=[1 15 2 0.5]

a = 
   1.0000   15.0000    2.0000   0.5000

>> [val,ind] = max(a) % val是a中最大的數的值，ind是這個值的位置

val =
    15

ind =
     2

>> sum(a)

ans = 
  18.5000

>> sum(A) % 默認求每列的和，相當於sum(A，1)
% 1 — Default. Returns sum of elements in each column.
% 2 — sum(A，2)Returns sum of elements in each row.每行的和

ans =
     9    12

>> prod(a) %a中元素的乘積

ans =
   15

>> prod(A) %求每列的乘積，相當於prod(A，1)
% 1 — Default. Returns prod of elements in each column.
% 2 — prod(A，2)Returns prod of elements in each row.每行的乘積
ans =
    15    48 

>> floor(a) %取下界

ans =

    1    15     2    0

>> ceil(a)%取上界

ans =

    1    15     2    1

>> rand(3)%創建3*3的random矩陣，每個值在[0,1]之間

ans =
   0.6463    0.2760    0.1626
   0.7094    0.6797    0.1190
   0.7547    0.6551    0.4984

>> max(rand(3),rand(3)) %在兩個random的3*3矩陣中找對應位置的max

ans =

   0.9597    0.2238    0.5060
   0.5472    0.7513    0.8143
   0.5853    0.8407    0.8909

>> A=magic(3)

A = 
    8     1     6
    3     5     7
    4     9     2

>> max(A,[],1) %找每列最大值，1表示第一維，即列

ans =

    8     9     7

>> max(A,[],2) %找每行最大值，2表示第二維，即行

ans =

    8
    7
    9

>> max(A) %defaultis column max

ans =

    8     9     7

>> max(max(A)) %A中最大元素

ans =

    9

>> A(:)

ans =

    8
    3
    4
    1
    5
    9
    6
    7
    2

>> max(A(:))

ans =

    9


>> A=magic(9) % 行，列，對角線加起來的值相等的9*9矩陣

A =

   47    58    69   80     1    12   23    34    45
   57    68    79    9    11    22   33    44    46
   67    78     8   10    21    32   43    54    56
   77     7    18   20    31    42   53    55    66
    6    17    19   30    41    52   63    65    76
   16    27    29   40    51    62   64    75     5
   26    28    39   50    61    72   74     4    15
   36    38    49   60    71    73    3    14    25
   37    48    59   70    81     2   13    24    35

>> sum(A,1)%column sum

ans =

  369   369   369  369   369   369  369   369   369

>> sum(A,2)%sum each row

ans =

  369
  369
  369
  369
  369
  369
  369
  369
  369

>> eye(9)

ans =

    1     0     0    0     0     0    0     0     0
    0     1     0    0     0     0    0     0     0
    0     0     1    0     0     0    0     0     0
    0     0     0    1     0     0    0     0     0
    0     0     0    0     1     0    0     0     0
    0     0     0    0     0     1    0     0     0
    0     0     0    0     0     0    1     0     0
    0     0     0    0     0     0    0     1     0
    0     0     0    0     0     0    0     0     1

>> A.*eye(9)%takethe element product of the 2 matrix

ans =

   47     0     0    0     0     0    0     0     0
    0    68     0    0     0     0    0     0     0
    0     0     8    0     0     0    0     0     0
    0     0    0    20     0    0     0     0    0
    0     0     0    0    41     0    0     0     0
    0     0     0    0     0    62    0     0     0
    0     0     0    0     0     0   74     0     0
    0     0     0    0     0     0    0    14     0
    0     0     0    0     0     0    0     0    35

>> sum(sum(A.*eye(9))) %sum(sum(A.*flipud(eye(9))))  

ans = 
  369

>> flipud(A) %returns X with the order of elements flipped upside down along the first dimension. 列數據顛倒，第一行的到最後一行，第二行到倒數第二行，以此類推

ans =
     5     6
     3     4
     1     2 

>> A=magic(3)

A =

    8     1     6
    3     5     7
    4     9     2

>> temp=pinv(A) %矩陣求逆 僞逆矩陣

temp =

   0.1472   -0.1444    0.0639
  -0.0611    0.0222    0.1056
  -0.0194    0.1889   -0.1028

>> temp*A

ans =

   1.0000   -0.0000   -0.0000
  -0.0000    1.0000    0.0000
   0.0000    0.0000    1.0000

---

Plotting Data 繪圖數據

>> t=[0:0.01:0.98];
>> y1=sin(2*pi*4*t);
>> plot(t,y1)

>> hold on;%plot new figure on the old ones
>> y2=cos(2*pi*4*t);
>> plot(t,y2,'r')

>> xlabel('time') %x座標標籤
>> ylabel('value') %y座標標籤
>> legend('sin','cos') %圖例
>> title('my plot')
>> print -dpng 'myplot.png' %save as a file in default catalog
>> cd 'C:\Users\x\Desktop'; print -dpng 'myplot.png' % 保存到相應的路徑
>> close %關閉圖片

%分別顯示兩幅圖像
>> figure(1);plot(t,y1);
>> figure(2);plot(t,y2)

%一幅圖中顯示兩個subplot figure 
subplot(1,2,1); % Divides plot a 1* 2 grid, access fisrt element
plot(t,y1);
subplot(1,2,2);% Divides plot a 1* 2 grid, access second element
plot(t,y2);
axis([0.5 1 -1 1]) %改變正在編輯的圖的x座標範圍爲[0.5 1]y座標範圍爲[-1 1]
clf % clear figure

>> A=magic(5)

A =
    17    24     1     8    15
    23     5     7    14    16
     4     6    13    20    22
    10    12    19    21     3
    11    18    25     2     9
>> imagesc(A) %可視化矩陣
>> imagesc(A), colorbar
>> imagesc(A), colorbar, colormap gray; % 逗號運算符，多個命令運行

>> a=1;b=2;c=3; %多個命令運行,但不會輸出
>> a=1,b=2,c=3 % 逗號運算符，多個命令運行，有輸出

a =
     1

b =
     2

c =
     3

---

Control Statements: for, while, if statement 控制語句:for,while,if 語句

>> v = zeros(10,1)
v =
     0
     0
     0
     0
     0
     0
     0
     0
     0
     0

>>  for i = 1:10,
        v(i) = 2^i;
    end;
>> v
v =
           2
           4
           8
          16
          32
          64
         128
         256
         512
        1024
% ================ ================
>> indices = 1:10;
>> indices
indices =
     1     2     3     4     5     6     7     8     9    10

>>  for i = indices,
        disp(i);
    end;
     1
     2
     3
     4
     5
     6
     7
     8
     9
    10
% ================ ================
>>  i = 1;
>>  while i <=5,
        v(i) = 100;
        i = i + 1;
    end;
>> v
v =
         100
         100
         100
         100
         100
          64
         128
         256
         512
        1024
% ================ ================
>>  i = 1;
>>  while true,
        v(i) = 999;
        i = i +1;
        if i == 6,
            break;
        end;
    end;
>> v
v =
         999
         999
         999
         999
         999
          64
         128
         256
         512
        1024
% ================ ================
>> v(1)
ans =
   999
>>  v(1) = 2;
>>  if v(1) == 1,
        disp('The value is one');
    elseif v(1) == 2,
        disp('The value is two');
    else
        disp('The value is not one or two');
    end;
The value is two

% ================ function 函數 ================
function [y1, y2] = squareAndCubeThisNumber(x)
y1 = x^2;
y2 = x^3;
end


[a, b] = squareAndCubeThisNumber(5);
>>b
b = 125

% ================代價函數方程 ================
function J = computeCostMulti(X, y, theta)
%COMPUTECOSTMULTI Compute cost for linear regression with multiple variables
%   J = COMPUTECOSTMULTI(X, y, theta) computes the cost of using theta as the
%   parameter for linear regression to fit the data points in X and y

% Initialize some useful values
m = length(y); % number of training examples

% You need to return the following variables correctly 
J = 0;

% ====================== YOUR CODE HERE ======================
% Instructions: Compute the cost of a particular choice of theta
%               You should set J to the cost.

J = sum((X * theta - y).^2)/(2*m);



% =========================================================================

end

---

Vectorization 向量化

下圖右邊的是用C++的庫:

梯度下降

theta=theta−alpha/m∗X′∗(X∗theta−y);

---

Normal Equation Noninvertibility 正規方程 不可逆性

不可逆性矩陣：奇異矩陣，退化矩陣
計算逆矩陣有兩個函數：pinv 和inv
區別：
- 僞逆函數pinv可以計算出θ ,即使 X’ * X 不可逆

不可逆的原因：

• 特徵之間不線性獨立
• 特徵數量大於訓練集的數量

---

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