l 2.3.1 算术运算符
l + - * .* ^ .^ \ .\ / ./
l >> A=[12 34 56 20]
l A =
l 12 34 56 20
l >> B=[1 3 2 4]
l B =
l 1 3 2 4
l >> C=A+B
l C =
l 13 37 58 24
输入的命令
显示的结果
幻灯片2
l >> AB=[A;B]
l AB =
l 12 34 56 20
l 1 3 2 4
l >> D=reshape(1:8,2,4)
l D =
l 1 3 5 7
l 2 4 6 8
l >> AB+D
l ans =
l 13 37 61 27
l 3 7 8 12
l >> AB+5
l ans =
l 17 39 61 25
l 6 8 7 9
l >> 3*D
l ans =
l 3 9 15 21
l 6 12 18 24
l >> D/4
l ans =
l 0.2500 0.7500 1.2500 1.7500
l 0.5000 1.0000 1.5000 2.0000
l >> clear all
l >> A=[1;2]
l A =
l 1
l 2
l >> B=[3 4 5]
l B =
l 3 4 5
l >> A*B
l ans =
l 3 4 5
l 6 8 10
l >> C=[12 45]
l C =
l 12 45
l >> D=[2;3]
l D =
l 2
l 3
l >> C=[12;45]
l C =
l 12
l 45
l >> C.*D
l ans =
l 24
l 135
l >> C./D
l ans =
l 6
l 15
l >> E=[1 2;3 4]
l E =
l 1 2
l 3 4
l >> E^2
l ans =
l 7 10
l 15 22
l >> E.^2
l ans =
l 1 4
l 9 16
l >> invE=E^(-1)
l invE =
l -2.0000 1.0000
l 1.5000 -0.5000
l >> inv(E)
l ans =
l -2.0000 1.0000
l 1.5000 -0.5000
l >> clear all
l >> A=[1 -2;0 3]
l A =
l 1 -2
l 0 3
l >> B=[1 0;-1 2;5 8]
l B =
l 1 0
l -1 2
l 5 8
l >> B/A
l ans =
l 1.0000 0.6667
l -1.0000 0
l 5.0000 6.0000
l >> B*inv(A)
l ans =
l 1.0000 0.6667
l -1.0000 0
l 5.0000 6.0000
l >> C=[B,[1 -2 1]’]
l C =
l 1 0 1
l -1 2 -2
l 5 8 1
l >> A=reshape(1:9,3,3)
l A =
l 1 4 7
l 2 5 8
l 3 6 9
l >> A(5)=0
l A =
l 1 4 7
l 2 0 8
l 3 6 9
l >> A/C
l Warning: Matrix is singular to working precision.
l ans =
l -Inf -Inf Inf
l -Inf -Inf Inf
l -Inf -Inf Inf
l >> det(A)
l ans =
l 60
l >> det(C)
l ans =
l 0
l >> C
l C =
l 1 0 1
l -1 2 -2
l 5 8 1
l >> C(2,3)=2
l C =
l 1 0 1
l -1 2 2
l 5 8 1
l >> det(C)
l ans =
l -32
l >> A\C
l ans =
l 1.7667 4.4667 -0.0667
l 0.8000 0.4000 -0.2000
l -0.5667 -0.8667 0.2667
l >> inv(A)*C
l ans =
l 1.7667 4.4667 -0.0667
l 0.8000 0.4000 -0.2000
l -0.5667 -0.8667 0.2667
l >> A*inv(C)
l ans =
l 3.0000 2.0000 0.0000
l 5.3750 1.5000 -0.3750
l 4.3125 2.2500 0.1875
l >> A/C
l ans =
l 3.0000 2.0000 0
l 5.3750 1.5000 -0.3750
l 4.3125 2.2500 0.1875
l >> A=[1 2 3;4 5 6]
l A =
l 1 2 3
l 4 5 6
l >> A’
l ans =
l 1 4
l 2 5
l 3 6
l >> Z=[1+i 2;3 1+2*i]
l Z =
l 1.0000 + 1.0000i 2.0000
l 3.0000 1.0000 + 2.0000i
l >> W=Z’
l W =
l 1.0000 - 1.0000i 3.0000
l 2.0000 1.0000 - 2.0000i
l >> W2=Z.’
l W2 =
l 1.0000 + 1.0000i 3.0000
l 2.0000 1.0000 + 2.0000i
l 2.3.2 关系运算符
l > >= < <= == ~=
l >> A=[0 -1 2 4.5 5];
l >> B=[-2 -1 3 2 5];
l >> A<B
l ans =
l 0 0 1 0 0
l >> A<=B
l ans =
l 0 1 1 0 1
l >> A>B
l ans =
l 1 0 0 1 0
l >> A>=B
l ans =
l 1 1 0 1 1
l >> A==B
l ans =
l 0 1 0 0 1
l >> A~=B
l ans =
l 1 0 1 1 0
l >> a=[10 20 30;-20 30 40]
l a =
l 10 20 30
l -20 30 40
l >> b=[-10 20 40;20 30 50]
l b =
l -10 20 40
l 20 30 50
l >> c=a>=b
l c =
l 1 1 0
l 0 1 0
l >> d=a(c)
l d =
l 10
l 20
l 30
l >> x=[0 1 2;0 0 3];
l >> idx=find(x)
l idx =
l 3
l 5
l 6
l >> x=magic(3)
l x =
l 8 1 6
l 3 5 7
l 4 9 2
l >> [idx1 idx2]=find(x>5)
l idx1 =
l 1
l 3
l 1
l 2
l idx2 =
l 1
l 2
l 3
l 3
l >> x=magic(5)
l x =
l 17 24 1 8 15
l 23 5 7 14 16
l 4 6 13 20 22
l 10 12 19 21 3
l 11 18 25 2 9
l >> x(find(7<x&x<11))
l ans =
l 10
l 8
l 9
l >> find(7<x&x<11)
l ans =
l 4
l 16
l 25
l >> x=[1 -1 0;2 8 5]
l x =
l 1 -1 0
l 2 8 5
l >> sum(x)
l ans =
l 3 7 5
l >> A=x==2
l A =
l 0 0 0
l 1 0 0
l >> sum(A)
l ans =
l 1 0 0
l >> sum(sum(A))
l ans =
l 1
2.2.3 逻辑运算符
l & (与) |（或） ~（非） xor（异或）
l >> A=[1 2 0 3 4 -1];
l >> B=[-1 0 0 5 4 9];
l >> and_result=A&B
l and_result =
l 1 0 0 1 1 1
l >> or_result=A|B
l or_result =
l 1 1 0 1 1 1
l >> not_result=~B
l not_result =
l 0 1 1 0 0 0
l >> xor_result=xor(A,B)
l xor_result =
l 0 1 0 0 0 0
l
2.3.3 逻辑函数
l all 判断是否所有元素为非零值
l any 判断是否存在一个元素为非零值
l exist 查看变量或函数是否存在
l find 找出向量或矩阵中非零元素的位置标识
l isempty 判断矩阵是否为空矩阵
l isequal 判断几个对象是否相等
l isnumeric 判断对象是否为数值型
l >> a=[1 2 3 0.6 -5]
l a =
l 1.0000 2.0000 3.0000 0.6000 -5.0000
l >> any(a) % a中是否存在非零元素
l ans =
l 1
l >> a=[1 2 3 0 -5]
l a =
l 1 2 3 0 -5
l >> any(a)
l ans =
l 1
l >> all(a) % a中是否所有的元素为非零值
l ans =
l 0
l >> A=[1 0 1;2 -2 0;3 -1 4]
l A =
l 1 0 1
l 2 -2 0
l 3 -1 4
l >> all(A) %按列判断
l ans =
l 1 0 0
l >> any(A)
l ans =
l 1 1 1
l >> B=[A,[0 0 0]’]
l B =
l 1 0 1 0
l 2 -2 0 0
l 3 -1 4 0
l >> any(B)
l ans =
l 1 1 1 0
l >> any(B,3) %对每一元素进行判断
l ans =
l 1 0 1 0
l 1 1 0 0
l 1 1 1 0
l >> all(A,2) %按行进行判断
l ans =
l 0
l 0
l 1
l >> A
l A =
l 1 0 1
l 2 -2 0
l 3 -1 4
l >> all(A,1) %按列进行判断(与all(A)相同)
l ans =
l 1 0 0
l
l >> A=[0 2 3;0 4 5]
l A =
l 0 2 3
l 0 4 5
l >> all(A,2)
l ans =
l 0
l 0
l >> all(A,1)
l ans =
l 0 1 1
l >> all(A,3)
l ans =
l 0 1 1
l 0 1 1
l >> all(A,4)
l ans =
l 0 1 1
l 0 1 1
l >> B=[B,[1 0 1]’]
l B =
l 1 0 1 0 1
l 2 -2 0 0 0
l 3 -1 4 0 1
l >> all(B)
l ans =
l 1 0 0 0 0
l >> all(B,1)
l ans =
l 1 0 0 0 0
l >> all(B,2)
l ans =
l 0
l 0
l 0
l >> all(B,3)
l ans =
l 1 0 1 0 1
l 1 1 0 0 0
l 1 1 1 0 1
l >> A=[0 -1 1.5;2 1 3]
l A =
l 0 -1.0000 1.5000
l 2.0000 1.0000 3.0000
l >> B=[0 -1 1.5;2 1 4]
l B =
l 0 -1.0000 1.5000
l 2.0000 1.0000 4.0000
l >> isequal(A,B)
l ans =
l 0
l >> B=[0 -1 1]
l B =
l 0 -1 1
l >> isequal(A,B)
l ans =
l 0
l >> B=’I’’m a student’
l B =
l I’m a student
l >> isequal(A,B)
l ans =
l 0
l >> A=[0.2 0.3 0.4];
l >> B=[1+2i 3+4i];
l >> C=[’I’ ’am’ ’a’ ’student’]
l C =
l Iamastudent
l >> isnumeric(A)
l ans =
l 1
l >> isnumeric(B)
l ans =
l 1
l >> isnumeric(C)
l ans =
l 0
l 2.3.5 字符串操作
字符串是由数个字符构成的，用(‘ ’)来界定，一个字符串可被视字符数组。
l 1. 数组的创建
l 直接输入
l >> s1=’Hello I am a student’
l s1 =
l Hello I am a student
l >> s2=[’Hello ’ ’I ’ ’am ’ ’a ’ ’student’]
l s2 =
l Hello I am a student
l >> length(s1) %求字符串的长度
l ans =
l 20
l >> length(s2)
l ans =
l 20
l >> s3=[’Hello ’;’I am a student’] %第一行与第二行长度相同
l s3 =
l Hello
l I am a student
l >> length(s3)
l ans =
l 14
l 利用char函数
l >> char(reshape(32:127,32,3)’)
l ans =
l !"#$%&’()*+,-./0123456789:;<=>?
l @ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_
l `abcdefghijklmnopqrstuvwxyz{|}~
l >> t1=’Hello’;
l >> t2=’I am a student’;
l >> t3=’I’’m from guangxi’;
l >> s=char(t1,t2,t3) %生成字符数组，此命令不必考虑每行字符数是否相等，总按最长设置，不足用空格补齐.
l s =
l Hello
l I am a student
l I’m from guangxi
l 2. 字符串操作
l >> size(s) %求字符串规模,返回数组的各维长度
l ans =
l 3 16
l >> size(s(3,:)) %查看s第三行规模
l ans =
l 1 16
l >>length(s) %返回最长维的长度
l ans=
l 16
l >> t4=t3(16:-1:1)
l t4 =
l ixgnaug morf m’I
l >> t5=t3(10:16)
l t5 =
l Guangxi
l >> str1=’Hello’;
l >> double(str1) %查看字符的Ascii码
l ans =
l 72 101 108 108 111
l >> char(ans)
l ans =
l Hello
l >> t6=t3(10:16)’
l t6 =
l g
l u
l a
l n
l g
l x
l i
l >> t7=[t1 t2]
l t7 =
l HelloI am a student
l >> A=[1.2 3.6 7.1 10.2 10 3];
l >> B1=num2str(A)
l B1 =
l 1.2 3.6 7.1 10.2 10 3
l >> char(B1)
l ans =
l 1.2 3.6 7.1 10.2 10 3
l
l >> A=[101 102 103;201 202 203];
l >> B1=num2str(A)
l B1 =
l 101 102 103
l 201 202 203
l >> char(A) %将数值转化成字符串
l ans =
l efg
l 墒�
l >> char(B1)
l ans =
l 101 102 103
l 201 202 203
l >> A=[1.2 3.6 7.1 10.2 10 3];
l >> num2str(A,1)
l ans =
l 1 4 7 1e+001 1e+001 3
l
l >> num2str(A,2)
l ans =
l 1.2 3.6 7.1 10 10 3
l >> A=[1.14 2.26 3.1 -2.6 -2.4 1.9 2.09]
l A =
l 1.1400 2.2600 3.1000 -2.6000 -2.4000 1.9000 2.0900
l >> num2str(A,2)
l ans =
l 1.1 2.3 3.1 -2.6 -2.4 1.9 2.1
l >> num2str(A,1)
l ans =
l 1 2 3 -3 -2 2 2
l >> num2str(A,’%10.3f’) %3位有效，10位长
l ans =
l 1.140 2.260 3.100 -2.600 -2.400 1.900 2.090
l >> num2str(A,’%10.2f’)
l ans =
l 1.14 2.26 3.10 -2.60 -2.40 1.90 2.09
l >> num2str(A,’%0.5f’) %5位有效
l ans =
l 1.140002.260003.10000-2.60000-2.400001.900002.09000
l >> num2str(A,’%10.5f’)
l ans =
l 1.14000 2.26000 3.10000 -2.60000 -2.40000 1.90000 2.09000
l >> A=[1.14 2.26 3.1 -2.6 -2.4 1.9 2.09 0.1 -0.5 -0.6]
l A =
l 1.1400 2.2600 3.1000 -2.6000 -2.4000 1.9000 2.0900 0.1000 -
l
l 0.5000 -0.6000
l
l >> num2str(A,’%2.5f’)
l ans =
l 1.140002.260003.10000-2.60000-2.400001.900002.090000.10000-0.50000-0.60000
l >> num2str(A,’%7.4f’)
l ans =
l 1.1400 2.2600 3.1000-2.6000-2.4000 1.9000 2.0900 0.1000-0.5000-0.6000
l >> a=[1 2 0.4 -0.5];
l >> b=[-1 0.34 4.5 4.9];
l >> s=’b.*sin(a)’
l s =
l b.*sin(a)
l >> sinx=eval(s)
l sinx =
l -0.8415 0.3092 1.7524 -2.3492
l >> eval11
l Name Size Bytes Class
l
l i 1x1 8 double array
l x3 3x3 72 double array
l x4 4x4 128 double array
l x5 5x5 200 double array
l x6 6x6 288 double array
l
l Grand total is 87 elements using 696 bytes
l
l >> x3
l x3 =
l 8 1 6
l 3 5 7
l 4 9 2
l 3. 字符串比较
l >> str1=’good’;
l >> str2=’god’;
l >> str3=’good’;
l >> strcmp(str1,str2)
l ans =
l 0
l >> strcmp(str2,str3)
l ans =
l 0
l >> strcmp(str1,str3)
l ans =
l 1
l >> str4=’Good’;
l >> strcmp(str1,str4)
l ans =
l 0
l >> strcmpi(str1,str4) %不区分大小写
l ans =
l 1
l >> strncmp(str1,str2 ,2) %比较前2个字母
l ans =
l 1
l >> strncmp(str1,str2 ,3)
l ans =
l 0