习题解答仅供参考，欢迎留言拍砖、讨论。
# R语言的科学编程与仿真，第二章笔记与习题解答 #chapter 2 # problem 1 # ａ z <- x^(a^b) # b z <- (x^a)^b # c z <- 3*x^3+2*x^2+6*x+1 # d #x <- 2.35 用x=2.35可以验证z=5 z <- ((x%%1)%%0.1)%/%0.01 # e z <- Z+1 # problem 2 # a c(seq(from=1,to=8,by=1),seq(from=7,to=1,by=-1)) # outcome [1] 1 2 3 4 5 6 7 8 7 6 5 4 3 2 1 # b c(1,rep(2,2),rep(3,3),rep(4,4),rep(5,5)) # outcome [1] 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 # c # 以下四个命令的结果是一致的 matrix(c(0,1,1,1,0,1,1,1,0),nrow = 3,ncol = 3,byrow = T) matrix(c(0,1,1,1,0,1,1,1,0),nrow = 3,ncol = 3) # byrow 的默认值是FALSE rbind(c(0,1,1),c(1,0,1),c(1,1,0)) cbind(c(0,1,1),c(1,0,1),c(1,1,0)) # outcome # [,1] [,2] [,3] #[1,] 0 1 1 #[2,] 1 0 1 #[3,] 1 1 0 # d matrix(c(0,2,3,0,5,0,7,0,0),nrow = 3,ncol = 3,byrow = TRUE) # outcome # [,1] [,2] [,3] # [1,] 0 2 3 # [2,] 0 5 0 # [3,] 7 0 0 # problem 3 vec <- c(4,4) # 直角坐标 rho <- sqrt(sum(vec^2)) #极坐标轴长 theta <- acos(vec[2]/rho) #极坐标角度 polarcoordinate <- c(rho,theta) #极坐标 # problem 4 x <- seq(1,100,1) x1 <- which(x%%2 == 0) x2 <- which(x%%3 == 0) x3 <- which(x%%7 == 0) x[c(-x1,-x2,-x3)] # 也可以写成一行 x[c(-which(x%%2 == 0),-which(x%%3 == 0),-which(x%%7 == 0))] # outcome # [1] 1 5 11 13 17 19 23 25 29 31 37 41 43 47 53 55 59 61 65 67 71 73 79 83 85 89 95 97 # problem 5 queue <- c('Steve','Russell','Alison','Liam') # 初始队列 queue <- c(queue,'Barry') # Barry 来了 queue <- queue[-1] # Steve 完成后离开了 queue <- c('Pam',queue) # Pam 到了最前面 queue <- queue[-5] # Barry 等不及离开了 # Alison 等不及离开了，按照题目要求需要用which函数获取Alison的位置 queue <- queue[-which(queue == 'Alison')] which(queue == 'Russell') # problem 6 rm(list = ls()) x <- 1 x[3] <- 3 # 此时x 为 x=[1,NA,3] y <- c() # 中文书上为 v[2] <- 2，英文版的为： y[2] <- 2 # y=y[NA,2] y[3] <- y[1] # y=y[NA,2,NA] y[2] <- y[4] # y=y[NA,NA,NA] z[1] <- 0 # Error in z[1] <- 0 : object 'z' not found
