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x1=c(171,175,159,155,152,158,154,164,168,166,159,164)
> x1
[1] 171 175 159 155 152 158 154 164 168 166 159 164
> x2=c(57,64,41,38,35,44,41,51,57,49,47,46);x2
[1] 57 64 41 38 35 44 41 51 57 49 47 46
> rbind(x1,x2)#按行合并
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
x1 171 175 159 155 152 158 154 164 168 166 159 164
x2 57 64 41 38 35 44 41 51 57 49 47 46
> cbind(x1,x2)#按列合并
x1 x2
[1,] 171 57
[2,] 175 64
[3,] 159 41
[4,] 155 38
[5,] 152 35
[6,] 158 44
[7,] 154 41
[8,] 164 51
[9,] 168 57
[10,] 166 49
[11,] 159 47
[12,] 164 46
#利用x1数据创建矩阵
> matrix(x1,nrow=3,ncol=4)
[,1] [,2] [,3] [,4]
[1,] 171 155 154 166
[2,] 175 152 164 159
[3,] 159 158 168 164
#创建行数列数发生变化的矩阵
> matrix(x1,nrow=4,ncol=3)
[,1] [,2] [,3]
[1,] 171 152 168
[2,] 175 158 166
[3,] 159 154 159
[4,] 155 164 164
#创建两个相同的矩阵
> A=B=matrix(1:12,nrow=3,ncol=4)
> A+B#矩阵加法
[,1] [,2] [,3] [,4]
[1,] 2 8 14 20
[2,] 4 10 16 22
[3,] 6 12 18 24
> A-B#矩阵减法
[,1] [,2] [,3] [,4]
[1,] 0 0 0 0
[2,] 0 0 0 0
[3,] 0 0 0 0
> A=matrix(1:9,nrow=3,ncol=3)
> B=matrix(1:9,nrow=3,ncol=3)
> A*B#矩阵对应元素的乘积
[,1] [,2] [,3]
[1,] 1 16 49
[2,] 4 25 64
[3,] 9 36 81
> A%*%B#矩阵的乘积
[,1] [,2] [,3]
[1,] 30 66 102
[2,] 36 81 126
[3,] 42 96 150
> A=matrix(1:16,nrow=4,ncol=4)
> diag(A)#获得矩阵对角线元素
[1] 1 6 11 16
> diag(diag(A))#利用对角线元素创建对角矩阵
[,1] [,2] [,3] [,4]
[1,] 1 0 0 0
[2,] 0 6 0 0
[3,] 0 0 11 0
[4,] 0 0 0 16
> A=matrix(rnorm(16),4,4)
> solve(A)#求矩阵的逆
[,1] [,2] [,3] [,4]
[1,] -0.2144625 0.9291739 0.41134588 -0.4129970
[2,] -0.2990380 0.5107384 -0.03081756 0.6477244
[3,] -0.1700996 1.4089089 -0.49639658 0.4082339
[4,] -0.4334604 -0.3543499 -0.30915934 -0.5515849
> A=diag(4)+1
> A.e=eigen(A,symmetric=T)
> A.e#求特征根和特征向量
eigen() decomposition
$values
[1] 5 1 1 1
$vectors
[,1] [,2] [,3] [,4]
[1,] -0.5 0.8660254 0.0000000 0.0000000
[2,] -0.5 -0.2886751 -0.5773503 -0.5773503
[3,] -0.5 -0.2886751 -0.2113249 0.7886751
[4,] -0.5 -0.2886751 0.7886751 -0.2113249
> A=matrix(1:12,3,4)
> dim(A)#矩阵的维数
[1] 3 4
> nrow(A)#矩阵的行数
[1] 3
> ncol(A)#矩阵的列数
[1] 4
> rowSums(A)#矩阵按行求和
[1] 22 26 30
> rowMeans(A)#矩阵按行求均值
[1] 5.5 6.5 7.5
> colSums(A)#矩阵按列求和
[1] 6 15 24 33
> colMeans(A)#矩阵按列求均值
[1] 2 5 8 11
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