【文章內(nèi)容簡介】 2列中大于 0的元素 (“或 ” ) ? x[!x[,2].51,1]第一列中相應(yīng)于第 2列中 不小于 .51的元素 (“非 ” ) ? 邏輯運算 :, , ==, =, =, !=。amp。, |, ! ? x=rnorm(10) ? all(x0)。all(x!=0)。any(x0)。(1:10)[x0] ? x=sample(1:7,5,rep=T)。unique(x) 矩陣的轉(zhuǎn)置和逆矩陣 ? x=matrix(runif(9),3,3)。x [,1] [,2] [,3] [1,] [2,] [3,] ? t(x) [,1] [,2] [,3] [1,] [2,] [3,] ? solve(x) solve(a,b)可以解 ax=b方程 [,1] [,2] [,3] [1,] [2,] [3,] 警告 :計算機中的 0是什么 ? ? x%*%solve(x) [,1] [,2] [,3] [1,] +00 [2,] +00 [3,] +00 ?要用線性代數(shù)的知識來判斷諸如有多少非零特征根等問題 .假定 v是特征 值 組 成 的 向 量 , 不 能 用 諸 如sum(v!=0) 等方法來判斷非零特征根的數(shù)目 ! Matrix amp。 Array ? x=array(runif(20),c(4,5))。 x [,1] [,2] [,3] [,4] [,5] [1,] [2,] [3,] [4,] ? (x) ? [1] TRUE ? x[1,2] ? x[1,] ? x[,2] ? dim(x)得到維數(shù) (4,5) Array ? x=array(runif(24),c(4,3,2)) ? (x) 可由 dim(x)得到維數(shù) (4,3,2) ? [1] FALSE ? x , , 1 [,1] [,2] [,3] [1,] [2,] [3,] [4,] , , 2 [,1] [,2] [,3] [1,] [2,] [3,] [4,] Array的子集 ? x=array(1:24,c(4,3,2)) ? x[c(1,3),] , , 1 [,1] [,2] [,3] [1,] 1 5 9 [2,] 3 7 11 , , 2 [,1] [,2] [,3] [1,] 13 17 21 [2,] 15 19 23 矩陣乘法及行列運算 ? x=matrix(1:30,5,6)。y=matrix(rnorm(20),4,5) ? y%*%x [,1] [,2] [,3] [,4] [,5] [,6] [1,] [2,] [3,] [4,] ? apply(x,1,mean) ? [1] ? apply(x,2,sum) ? [1] 15 40 65 90 115 140 ? apply(x,2,prod) ? [1] 120 30240 360360 1860480 6375600 17100720 Array的維運算 ? x=array(1:24,c(4,3,2)) ? apply(x,1,mean) ? [1] 11 12 13 14 ? apply(x,1:2,sum) [,1] [,2] [,3] [1,] 14 22 30 [2,] 16 24 32 [3,] 18 26 34 [4,] 20 28 36 ? apply(x,c(1,3),prod) [,1] [,2] [1,] 45 4641 [2,] 120 5544 [3,] 231 6555 [4,] 384 7680 矩陣與向量之間的運算 ? sweep(x,1,1:5,*) [,1] [,2] [,3] [,4] [,5] [,6] [1,] 1 6 11 16 21 26 [2,] 4 14 24 34 44 54 [3,] 9 24 39 54 69 84 [4,] 16 36 56 76 96 116 [5,] 25 50 75 100 125 150 ? x*1:5 ? sweep(x,2,1:6,+) [,1] [,2] [,3] [,4] [,5] [,6] [1,] 2 8 14 20 26 32 [2,] 3 9 15 21 27 33 [3,] 4 10 16 22 28 34 [4,] 5 11 17 23 29 35 [5,] 6 12 18 24 30 36 Array和矩陣 /向量 /array之間的運算 ? z=array(1:24,c(2,3,4))注意排列次序 z , , 1 [,1] [,2] [,3] [1,] 1 3 5 [2,] 2 4 6 , , 2 [,1] [,2] [,3] [1,] 7 9 11 [2,] 8 10 12 , , 3 [,1] [,2] [,3] [1,] 13 15 17 [2,] 14 16 18 , , 4 [,1] [,2] [,3] [1,] 19 21 23 [2,] 20 22 24 Array和矩陣 /向量 /array之間的運算 ? sweep(z,1,1:2,) , , 1 [,1] [,2] [,3] [1,] 0 2 4 [2,] 0 2 4 , , 2 [,1] [,2] [,3] [1,] 6 8 10 [2,] 6 8 10 , , 3 [,1] [,2] [,3] [1,] 12 14 16 [2,] 12 14 16 , , 4 [,1] [,2] [,3] [1,] 18 20 22 [2,] 18 20 22 Array和矩陣 /向量 /array之間的運算 ? sweep(z,c(1,2),matrix(1:6,2,3),) , , 1 [,1] [,2] [,3] [1,] 0 0 0 [2,] 0 0 0 , , 2 [,1] [,2] [,3] [1,] 6 6 6 [2,] 6 6 6 , , 3 [,1] [,2] [,3] [1,] 12 12 12 [2,] 12 12 12 , , 4 [,1] [,2] [,3] [1,] 18 18 18 [2,] 18 18 18 外積 (產(chǎn)生矩陣或 array) ? outer(1:2,rep(1,2)) [,1] [,2] [1,] 1 1 [2,] 2 2 ? outer(1:2,matrix(rep(1,6),3,2)) , , 1 [,1] [,2] [,3] [1,] 1 1 1 [2,] 2 2 2 , , 2 [,1] [,2] [,3] [1,] 1 1 1 [2,] 2 2 2 List(set of objects) ? list可以是任何對象的集合 (包括 lists) ? z=list(1:3,Tom=c(1:2, a=list(R,letters[1:5]),w=hi!)) ? z[[1]]。z[[2]]。z$T。z$T$a2。z$T[[3]]。z$T$w ?attributes(z)屬性 ! $names [1] Tom ? attributes(matrix(1:6,2,3)) $dim [1] 2 3 矩陣 ,array及其維名字 ? x=matrix(1:12,nrow=3,dimnames=list(c(I,II,III),paste(X,1:4,sep=))) X1 X2 X3 X4 I 1 4 7 10 II 2 5 8 11 III 3 6 9 12 ? y=array(1:12,c(3,2,2),dimnames=list(c(I,II,III),paste(X,1:2 sep=),paste(Y,1:2, sep=))) , , Y1 X1 X2 I 1 4 II 2 5 III 3 6 , , Y2 X1 X2 I 7 10 II 8 11 III 9 12 ? x=matrix(1:6,2,3) ? x=(x)。x V1 V2 V3 1 1 3 5 2 2 4 6 ? x$V2 ? [1] 3 4 ? x$V2 ? [1] 3 4 ? attributes(x) $names [1] V1 V2 V3 $ [1] 1 2 $class [1] ? names(x)=c(TOYOTA,GM,HUNDA) ? (x)=c(2022,2022) ? x TOYOTA GM HUNDA 2022 1 3 5 2022 2 4 6 ? x$GM ? [1] 3 4 ? attach(x) ? GM ? [1] 3 4 ? detach(x) ? GM ? Error: Object GM not found 直接手工輸入和編輯數(shù)據(jù) ? 直接敲入 :x=c(1,2,7,8,… ) ? 或者 ? x=scan() ? 1 2 7 8 … .(以 “ Enter”兩次來結(jié)束 ) ? fix(x)(通過編輯修改數(shù)據(jù) ) Categorical data A survey asks people if they smoke or not. The data is Yes, No, No, Yes, Yes ?x=c(Yes,No,No,Yes,Yes) ?table(x)。x ?factor(x) ?Barplot:Suppose, a group of 25 people are surveyed as to their beerdrinking preference. The categories were (1) Domestic can, (2) Domestic bottle, (3) Microbrew and (4) import. The raw data is 3 4 1 1 3 4 3 3 1 3 2 1 2 1 2 3 2 3 1 1 1 1 4 3 1 ?beer = scan() ?3 4 1 1 3 4 3 3 1 3 2 1 2 1 2 3 2 3 1 1 1 1 4 3 1 ?barplot(beer) this isn39。t correct ?barplot(table(beer)) Yes, call with summarized data ?barplot(table(beer)/length(beer)) divide by n fo