R语言--矩阵篇

最新推荐文章于 2025-07-04 08:15:00 发布

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最新推荐文章于 2025-07-04 08:15:00 发布

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本文介绍了R语言中的向量操作，包括如何使用append函数追加元素，向量下标运算，逻辑运算，以及向量的替换。同时，详细探讨了矩阵的创建、转置、运算、求逆、特征值、特征向量、上三角矩阵等概念，展示了R语言处理矩阵的多种方法和函数。

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除此之外还可以用append函数为原有的向量添加元素

> append(t,10:15)

[1] 2 543 3 5 1 10 11 12 13 14 15

Append 追加元素：

追加之后原来的向量并没有改变其元素取值；缺省关键字after时表示从向量的末尾追加元素

> c <- c(1,2,3,4)

> prod(c)

[1] 24

> append(c,6:8)

[1] 1 2 3 4 6 7 8

> append(c,9:12,after=3)

[1] 1 2 3 9 10 11 12 4

> prod(c)

[1] 24

向量下标运算：

R语言里向量下标从1开始；

例1：

创建一个向量，让其每个元素加4，求其第二个元素是几？

> (c(1,3,5,7)+4)[2]

[1] 7

例2：

元素替换

> t

[1] 2 543 3 5 1

> t[2] <- 10>

[1] 2 10 3 5 1

例3

> t

[1] 2 543 3 5 1

> t[c(1,3)] <- c(9,11)

> t

[1] 9 10 11 5 1

向量的逻辑运算：

> t

[1] 9 10 11 5 1

> t[t<4] #显示出t<4的数

[1] 1

> t[t>9]#显示出t>9的数

[1] 10 11

把向量中所有缺省值赋值为0

新建一个向量，其中包括缺省值

> x<-c(-1,2,5,6,NA)

> x

[1] -1 2 5 6 NA

> x[is.na(x)]<-0

> x

[1] -1 2 5 6 0

把向量中不是NA的值导入到另一个向量中去

> x<-c(-1,2,5,6,NA)

> x

[1] -1 2 5 6 NA

> u<-x[!is.na(x)]

> u

[1] -1 2 5 6

将变量s设为数值型，长度为向量X的长度，则s的默认元素都是0

> x

[1] -1 2 5 6 NA

> s<-numeric(length(x))

> s

[1] 0 0 0 0 0

分段函数的表示方法：

> y

[1] 1 3 5 7 9

> s<-numeric(length(y))

> s

[1] 0 0 0 0 0

> s[s<0] <- y[y<0]

> s[s>=0] <- y[y>=0] +1

> s

[1] 2 4 6 8 10

如果x是一个有缺失值的向量、则在对s赋值的时候会报错

> s<-numeric(length(x))

> s

[1] 0 0 0 0 0

> s[s<0] <- x[x<0]

> s[s>=0] <- x[x>=0] +1

Warning message:In s[s >= 0] <- x[x >= 0] + 1 :

number of items to replace is not a multiple of replacement length

向量1-5的元素不显示：

> x <- c(1:10)

> x

[1] 1 2 3 4 5 6 7 8 9 10

> x[-(1:5)]

[1] 6 7 8 9 10

R语言的矩阵

矩阵就是用数据排列成长和宽的表二维数组，单元必须是相同的数据类型；用列表示不同的变量，用行表示各个对象；R语言生成矩阵函数--matrix

语法：

Mymatrix <- matrix(vector,nrow=number_of_rows,ncol=number_of_columns,

Byrow=logical_value,dimnames=list(

Char_vector_rowames,char_vector_colnames))

第一个参数是数据；第二个参数是行数；第三个参数是列数；第四个参数表示：安行展开或者安列展开；第五个参数表示矩阵的维数；

> A　<- matrix(1:12,nrow=3,ncol=4)

> A

[,1] [,2] [,3] [,4]

[1,] 1 4 7 10

[2,] 2 5 8 11

[3,] 3 6 9 12

> T(A)

Error: could not find function "T"

> t(A) #矩阵的转秩

[,1] [,2] [,3]

[1,] 1 2 3

[2,] 4 5 6

[3,] 7 8 9

[4,] 10 11 12

> x <- 1:10

> x

[1] 1 2 3 4 5 6 7 8 9 10

> t(x)

[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]

[1,] 1 2 3 4 5 6 7 8 9 10

矩阵的简单运算

> A　<- matrix(1:12,nrow=3,ncol=4)

> A

[,1] [,2] [,3] [,4]

[1,] 1 4 7 10

[2,] 2 5 8 11

[3,] 3 6 9 12

> t(A)

[,1] [,2] [,3]

[1,] 1 2 3

[2,] 4 5 6

[3,] 7 8 9

[4,] 10 11 12

> B<-A

> A+B

[,1] [,2] [,3] [,4]

[1,] 2 8 14 20

[2,] 4 10 16 22

[3,] 6 12 18 24

> A-B

[,1] [,2] [,3] [,4]

[1,] 0 0 0 0

[2,] 0 0 0 0

[3,] 0 0 0 0

> 3*A

[,1] [,2] [,3] [,4]

[1,] 3 12 21 30

[2,] 6 15 24 33

[3,] 9 18 27 36

> B <- t(A)

> #两个矩阵相乘

> A%*%B

[,1] [,2] [,3]

[1,] 166 188 210

[2,] 188 214 240

[3,] 210 240 270

求矩阵的对角线元素：

> A <- matrix(1:16,nrow=4)

> A

[,1] [,2] [,3] [,4]

[1,] 1 5 9 13

[2,] 2 6 10 14

[3,] 3 7 11 15

[4,] 4 8 12 16

> diag(A)

[1] 1 6 11 16

> diag(diag(A))

[,1] [,2] [,3] [,4]

[1,] 1 0 0 0

[2,] 0 6 0 0

[3,] 0 0 11 0

[4,] 0 0 0 16

> diag(4) #生成对角元素值为1的单位矩阵

[,1] [,2] [,3] [,4]

[1,] 1 0 0 0

[2,] 0 1 0 0

[3,] 0 0 1 0

[4,] 0 0 0 1

矩阵的求逆运算:solve()函数；

> A

[,1] [,2] [,3] [,4]

[1,] 1 5 9 13

[2,] 2 6 10 14

[3,] 3 7 11 15

[4,] 4 8 12 16

> solve(A) #A不是可逆的矩阵

Error in solve.default(A) :

Lapack routine dgesv: system is exactly singular: U[3,3] = 0

#随机生成一个遵循正太分布的矩阵，避免它出现不可逆的情况，4X4的矩阵

A <- matrix(rnorm(16),4,4)

solve(A)

> A <- matrix(rnorm(16),4,4)

> A

[,1] [,2] [,3] [,4]

[1,] -0.70756823 -1.1234569 1.360924 -0.2375963

[2,] 1.97157201 -1.3441301 1.753795 -1.2241501

[3,] -0.08999868 -1.5231558 1.568365 -0.3278127

[4,] -0.01401725 -0.4219682 1.296756 -2.4124503

> solve(A)

[,1] [,2] [,3] [,4]

[1,] -0.05180593 0.4723860 -0.3259291 -0.1903122

[2,] 4.46734372 1.3701835 -5.0357807 -0.4509696

[3,] 4.70043705 1.4725733 -4.6045433 -0.5844802

[4,] 1.74551456 0.5491397 -1.5923475 -0.6487037

> solve(A)%*%A

#矩阵A和矩阵A的逆做运算应该是一个单位阵，由于计算机的计算误差出现了非对角元素趋向于0的误差

[,1] [,2] [,3] [,4]

[1,] 1.000000e+00 -1.151856e-15 1.137979e-15 -2.220446e-16

[2,] -5.238865e-16 1.000000e+00 1.665335e-15 -4.440892e-16

[3,] 2.255141e-17 -2.053913e-15 1.000000e+00 -2.220446e-16

[4,] -1.387779e-17 -2.220446e-16 -3.330669e-16 1.000000e+00

求特征值和特征向量：eigen()函数

> A

[,1] [,2] [,3] [,4]

[1,] 2 1 1 1

[2,] 1 2 1 1

[3,] 1 1 2 1

[4,] 1 1 1 2

> (A.eigen <- eigen(A,symmetric = T))

$values #特征值就是：5,1,1,1

[1] 5 1 1 1

$vectors #特征向量

[,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

将矩阵变换为上三角矩阵：chol()函数

> A

[,1] [,2] [,3] [,4]

[1,] 2 1 1 1

[2,] 1 2 1 1

[3,] 1 1 2 1

[4,] 1 1 1 2

> chol(A)

[,1] [,2] [,3] [,4]

[1,] 1.414214 0.7071068 0.7071068 0.7071068

[2,] 0.000000 1.2247449 0.4082483 0.4082483

[3,] 0.000000 0.0000000 1.1547005 0.2886751

[4,] 0.000000 0.0000000 0.0000000 1.1180340

> t(chol(A))%*%chol(A)

[,1] [,2] [,3] [,4]

[1,] 2 1 1 1

[2,] 1 2 1 1

[3,] 1 1 2 1

[4,] 1 1 1 2

如果矩阵式对称的，就可以用chol()来求行列式的值以及举证的逆

dim(A)对矩阵A的维数；

nrow(A)矩阵A的行数；

ncol(A)矩阵A的列数；

rowSums(A)对行求和；

colSums(A)对列求和

colMeans()对矩阵的行和列求均值

> A

[,1] [,2] [,3] [,4]

[1,] 2 1 1 1

[2,] 1 2 1 1

[3,] 1 1 2 1

[4,] 1 1 1 2

> dim(A)

[1] 4 4

> nrow(A)

[1] 4

> ncol(A)

[1] 4

> rowSums(A)

[1] 5 5 5 5

> colSums(A)

[1] 5 5 5 5

方法一：矩阵的上三角矩阵和下三角矩阵：up.tri();low.tri();

> A<- matrix(1:16,4)

> A

[,1] [,2] [,3] [,4]

[1,] 1 5 9 13

[2,] 2 6 10 14

[3,] 3 7 11 15

[4,] 4 8 12 16

> lower.tri(A)

[,1] [,2] [,3] [,4]

[1,] FALSE FALSE FALSE FALSE

[2,] TRUE FALSE FALSE FALSE

[3,] TRUE TRUE FALSE FALSE

[4,] TRUE TRUE TRUE FALSE

> A[lower.tri(A)] <-0 #下三角矩阵

> A

[,1] [,2] [,3] [,4]

[1,] 1 5 9 13

[2,] 0 6 10 14

[3,] 0 0 11 15

[4,] 0 0 0 16

> A[up.tri(A)]

<- 0

Error in A[up.tri(A)] <- 0 : could not find function "up.tri"

> A[upper.tri(A)] <- 0 #上三角矩阵

> A

[,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

[,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(1:16,4)

> A[upper.tri(A)] <- 0

> A

[,1] [,2] [,3] [,4]

[1,] 1 0 0 0

[2,] 2 6 0 0

[3,] 3 7 11 0

[4,] 4 8 12 16

> A[lower.tri(A)] =0

> A

[,1] [,2] [,3] [,4]

[1,] 1 0 0 0

[2,] 0 6 0 0

[3,] 0 0 11 0

[4,] 0 0 0 16

方法二：求矩阵的下三角阵和上三角阵

让行下标的位置小于列下表的位置的元素都为0，----下三角阵；

> A<- matrix(1:16,4)

> A

[,1] [,2] [,3] [,4]

[1,] 1 5 9 13

[2,] 2 6 10 14

[3,] 3 7 11 15

[4,] 4 8 12 16

> row(A)

[,1] [,2] [,3] [,4]

[1,] 1 1 1 1

[2,] 2 2 2 2

[3,] 3 3 3 3

[4,] 4 4 4 4

> col(A)

[,1] [,2] [,3] [,4]

[1,] 1 2 3 4

[2,] 1 2 3 4

[3,] 1 2 3 4

[4,] 1 2 3 4

> A[row(A)<col(A)]=0

> A

[,1] [,2] [,3] [,4]

[1,] 1 0 0 0

[2,] 2 6 0 0

[3,] 3 7 11 0

[4,] 4 8 12 16

> A[row(A)>col(A)]=0

> A

[,1] [,2] [,3] [,4]

[1,] 1 0 0 0

[2,] 0 6 0 0

[3,] 0 0 11 0

[4,] 0 0 0 16

矩阵中计算行列式的值：det(A)

> A

[,1] [,2] [,3] [,4]

[1,] 1 0 0 0

[2,] 0 6 0 0

[3,] 0 0 11 0

[4,] 0 0 0 16

> det(A)

[1] 1056

将矩阵转化为向量化算子（这里需要编写一个小函数）

> vec<-function(x){

+ t(t(as.vector(x)))

+ }

> A<-matrix(1:12,3,4)

> vec(A)

[,1]

[1,] 1

[2,] 2

[3,] 3

[4,] 4

[5,] 5

[6,] 6

[7,] 7

[8,] 8

[9,] 9

[10,] 10

[11,] 11

[12,] 12

根据矩阵的下标去取元素

> A

[,1] [,2] [,3] [,4]

[1,] 1 4 7 10

[2,] 2 5 8 11

[3,] 3 6 9 12

> A(2,3)

Error: could not find function "A"

> A[2,3]

[1] 8

> A[,3]

[1] 7 8 9

> A[2,]

[1] 2 5 8 11

> A[1:3,2] #取出1-3行的第二列元素

[1] 4 5 6