Each of the exercises below are from our textbook - Introduction to Scientific Programming and Simulaiton Using R. The exercises begin on p. 26. Make sure to comment your code. Points will be deducted for uncommented code!
You should create new code blocks if necessary to answer the question. You should also use plaintext to provide an explanation for your code if the question calls for one.
#Give R assignment statements that set the variable z to the following listed expressions from the textbook
x <- 5
a <-2
b <- 2
(z <- x^(a^b))## [1] 625#assign (x^a)^b to z
(z <- (x^a)^b)## [1] 625#assign 3x^3+2x^2+6x+1(in minimized opration) to z
(z <- (3*x+2)*(x^2+2)-3)## [1] 456#assign the second decimal digit of x into z
x <- 0.456
(z <- as.integer((x*100)%%10))## [1] 5#assign z+1 into z
(z <- z+1)## [1] 6#the return should be a sequence from 1-8-1
(x <- c(seq(1,8),seq(7,1)))## [1] 1 2 3 4 5 6 7 8 7 6 5 4 3 2 1#return repeated sequence
n <- 1 #n start by 1
vector <- c() #create empty vector
while (n < 6) { #start while loop with condition to break when n reach 6
y<- rep(n, times = n) #assign repetition sequence into y
n <- n+1 #n increase by 1 in every run until break
for (p in y) #nested a for loop to assign value
vector <-c(vector, p) #combine vector's original value with new values attained from the while loop
}
print(vector) #print to show result## [1] 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5m <- matrix(1,3,3) #create a 3x3 matrix in m with only #1
diag(m) <- 0 #change the diagonal value to 0
m #print the final result## [,1] [,2] [,3]
## [1,] 0 1 1
## [2,] 1 0 1
## [3,] 1 1 0A <- matrix(0,3,3)#create 3x3 matrix with only 0s assign to A
m <- c(7,5,3) #create a vector contains 3 numbers assign to m
A[seq(3, by = 2, length = 3)] <- m #assign m to A's extracted value
#extract a sequence from matrix A that is a diagonal
#A[1,2] <- 2 #change the value in coordination of row 1 column 2 to 2
A #print result## [,1] [,2] [,3]
## [1,] 0 0 3
## [2,] 0 5 0
## [3,] 7 0 0vec <- 1:100 #assign numbers from 1 to 100 into vec
(vec_revised <- vec[vec%%2 & vec%%3 & vec%%7]) #use modular division to find out the numbers can't be divided by either 2, 3, or 7## [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
## [26] 89 95 97queue <- c("Steve", "Russell", "Alison", "Liam")#queue has 4 customers in line
(queue <- c(queue, "Barry")) #now queue has 1 more customer in line## [1] "Steve" "Russell" "Alison" "Liam" "Barry"served <- "Steve"#Steve is served
(queue <- queue[-which(queue %in% served)])#to match served value in queue by %in% and indicate those matched values index by which() function then use minus sign to eliminate the matched value from queue vector## [1] "Russell" "Alison" "Liam" "Barry"#assign new vector to replace old values in queue(queue <- (append(queue, values = "Pam with 1 item", after = match("Russell",queue)-1)))## [1] "Pam with 1 item" "Russell" "Alison" "Liam"
## [5] "Barry"#use append function to add value
#assign the new vector into queueimpatient <- "Barry" #Barry is impatient
(queue <- queue[-which(queue %in% impatient)])#to match impatient value in queue by %in% and indicate those matched values index by which() function then use minus sign to eliminate the matched value from queue vector## [1] "Pam with 1 item" "Russell" "Alison" "Liam"#assign new vector to replace old values in queueimpatient <- "Alison" #Alison is impatient
(queue <- queue[-which(queue %in% impatient)])#to match impatient value in queue by %in% and indicate those matched values index by which() function then use minus sign to eliminate the matched value from queue vector## [1] "Pam with 1 item" "Russell" "Liam"#assign new vector to replace old values in queue
#now look for where Russell is
which(queue=="Russell")## [1] 2rm(list = ls())
x <- 1 #successful, x has 1 value
x[3] <- 3#successful, x[3] has 3, x[1:2] are NA NA
y <- c()#successful, y has empty vector, result is NULL
y[2] <- 2 #successful, y[2] has 2
y[3] <- y[1] #successful, result: NA 2 NA
y[2] <- y[4] #successful, result: NA NA NA
#z[1] <- 0#failed, because z never existed, thefore z[1] is not existeddiag(5,10,10) #way1 use diag() function to set diagonal as 5 and 10 by 10 row and column## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
## [1,] 5 0 0 0 0 0 0 0 0 0
## [2,] 0 5 0 0 0 0 0 0 0 0
## [3,] 0 0 5 0 0 0 0 0 0 0
## [4,] 0 0 0 5 0 0 0 0 0 0
## [5,] 0 0 0 0 5 0 0 0 0 0
## [6,] 0 0 0 0 0 5 0 0 0 0
## [7,] 0 0 0 0 0 0 5 0 0 0
## [8,] 0 0 0 0 0 0 0 5 0 0
## [9,] 0 0 0 0 0 0 0 0 5 0
## [10,] 0 0 0 0 0 0 0 0 0 5m <- matrix(0,10,10)#way2 create a matrix will full 0s first, then change only the diagonal values
diag(m) <- 5
m## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
## [1,] 5 0 0 0 0 0 0 0 0 0
## [2,] 0 5 0 0 0 0 0 0 0 0
## [3,] 0 0 5 0 0 0 0 0 0 0
## [4,] 0 0 0 5 0 0 0 0 0 0
## [5,] 0 0 0 0 5 0 0 0 0 0
## [6,] 0 0 0 0 0 5 0 0 0 0
## [7,] 0 0 0 0 0 0 5 0 0 0
## [8,] 0 0 0 0 0 0 0 5 0 0
## [9,] 0 0 0 0 0 0 0 0 5 0
## [10,] 0 0 0 0 0 0 0 0 0 5