【原】Andrew Ng斯坦福机器学习（5）——Lecture 5 Octave Tutorial—5.5 控制语句： for， while， if 语句

 1 >> v = zeros(10,1)
 2 v =
 3    0
 4    0
 5    0
 6    0
 7    0
 8    0
 9    0
10    0
11    0
12    0
13 >> for i = 1 : 10,
14      v(i) = 2 ^ i;
15    end;
16 >> v
17 v =
18       2
19       4
20       8
21      16
22      32
23      64
24     128
25     256
26     512
27    1024

 1 >> indices = 1 : 10;
 2 >> indices
 3 indices =
 4     1    2    3    4    5    6    7    8    9   10
 5 >> for i = indices,
 6       disp(i);
 7    end;
 8  1
 9  2
10  3
11  4
12  5
13  6
14  7
15  8
16  9
17  10

 1 >> i = 1;
 2 >> while i <= 5,
 3       v(i) = 100;
 4       i = i + 1;
 5    end;
 6 >> v
 7 v =
 8     100
 9     100
10     100
11     100
12     100
13      64
14     128
15     256
16     512
17    1024

 1 >> i = 1 ;
 2 >> while(true),
 3      v(i) = 999;
 4      i = i + 1;
 5      if i == 6,
 6         break;
 7      end;
 8    end;
 9 >>
10 >> v
11 v =
12     999
13     999
14     999
15     999
16     999
17      64
18     128
19     256
20     512
21    1024

1 >> v(1) = 2;
2 >> if v(1) == 1,
3       disp(‘The value is one‘);
4    elseif v(1) == 2,
5       disp(‘The value is two‘);
6    else,
7       disp(‘The value is not one or two‘);
8    end;
9 The value is two

1 function y = squareThisNumber(x)
2   y = x^2;
3 endfunction

1 >> squareThisNumber(2);
2 error: ‘squareThisNumber‘ undefined near line 1 column 1
3 >>
4 >> ls
5 [.]                         featuresX.dat               priceY.dat
6 [..]                        hello.dat                   squarethisnumber.m

1 >> ls
2 [.]                         featuresX.dat               priceY.dat
3 [..]                        hello.dat                   squareThisNumber.m
4 >> squareThisNumber(2);
5 >> a = squareThisNumber(2);
6 >> a
7 a =  4

1 function [y1, y2] = squareAndCubeThisNumber(x),
2   y1 = x ^ 2;
3   y2 = x ^ 3;
4 endfunction

1 >> a = squareAndCubeThisNumber(2)　　% 接受了第一个返回值
2 a =  4
3 >> [a, b] = squareAndCubeThisNumber(2)　　% 接受两个返回值
4 a =  4
5 b =  8

 1 function J = costFunctionJ(X, y, theta),
 2   % X is the "design matrix" containing our training examples
 3   % y is the class labels
 4   
 5   m = size(X, 1);   % number of training examples, size of rows
 6   predictions = X * theta; % predictions of hapothesis on all m examples
 7   sqrErrors = (predictions - y) .^ 2; % squared errors .^ 指的是对数据中每个元素平方
 8   
 9   J = 1 / (2 * m) * sum(sqrErrors); 
10   
11 endfunction

1 >> X = [1 1; 1 2 ; 1 3];
2 >> y = [1;2;3];
3 >> theta = [0;1];　　　　　　% Θ为0,1  h(x)=x  此时完全拟合数据，代价函数的值为0
4 >> j = costFunctionJ(X,y,theta)
5 j = 0

1 >> theta = [0;0];　　　　　 % Θ 为0,0 h(x)=0 此时不能拟合数据
2 >> j = costFunctionJ(X,y,theta)
3 j =  2.3333
4 
5 >> (1^2 + 2^2 + 3^2) / (2*3)　　% 代价函数的值
6 ans =  2.3333