#HOMEWORK 4
Nicolás Zapata
Suppose x = 1.1, a = 2.2, and b = 3.3. Assign each expression to the value of the variable z and print the value stored in z.
𝑥𝑎𝑏 (𝑥𝑎)𝑏 3𝑥3+2𝑥2+1
# First, assign the values to x, a and b
x <- 1.1
a <- 2.2
b <- 3.3
# Create a vector with the formula
z <- x^(a^b)
print(z)## [1] 3.61714z <- (x^a)^b
print(z)## [1] 1.997611z <- 3*(x^3)+2*(x^2)+1
print(z)## [1] 7.413Using the rep and seq functions, create the following vectors: a. (1,2,3,4,5,6,7,8,7,6,5,4,3,2,1) b. (1,2,2,3,3,3,4,4,4,4,5,5,5,5,5) c. (5,4,4,3,3,3,2,2,2,2,1,1,1,1,1)
my_vector1 <- c(seq(from=1 , to=8), seq(from=7, to=1))
print(my_vector1)## [1] 1 2 3 4 5 6 7 8 7 6 5 4 3 2 1my_vector2 <- c(rep(1:5,1:5))
print(my_vector2)## [1] 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5my_vector3 <- c(rep(5:1,1:5))
print(my_vector3)## [1] 5 4 4 3 3 3 2 2 2 2 1 1 1 1 1Use the function seq to create the sequence and rep to determine how many times the the numbers repeat
Create a vector of two random uniform numbers. In a spatial map, these can be interpreted as x and y coordinates that give the location of an individual (such as a marked forest tree in a plot that has been mapped). Using one of R’s inverse trigonometry functions (asin(), acos(), or atan()), convert these numbers into polar coordinates
vector3 <- runif(2)
print(vector3)## [1] 0.04335591 0.01959930asin(vector3)## [1] 0.04336951 0.01960055acos(vector3)## [1] 1.527427 1.551196atan(vector3)## [1] 0.04332878 0.01959679Create a vector queue <- c(“sheep”, “fox”, “owl”, “ant”) where queue represents the animals that are lined up to enter Noah’s Ark, with the sheep at the front of the line. Using R expressions, update queue as:
the serpent arrives and gets in line; the sheep enters the ark; the donkey arrives and talks his way to the front of the line; the serpent gets impatient and leaves; the owl gets bored and leaves; the aphid arrives and the ant invites him to cut in line. Finally, determine the position of the aphid in the line.
queue <- c("sheep", "fox", "owl", "ant")
print(queue)## [1] "sheep" "fox" "owl" "ant"queue <- c(queue, "serpent")
print(queue)## [1] "sheep" "fox" "owl" "ant" "serpent"queue <- queue[queue !="sheep"]
print(queue)## [1] "fox" "owl" "ant" "serpent"#other way
queue[c(1,2)]## [1] "fox" "owl"queue[-c(1)]## [1] "owl" "ant" "serpent"print(queue)## [1] "fox" "owl" "ant" "serpent"queue <- c("donkey", queue)
print(queue)## [1] "donkey" "fox" "owl" "ant" "serpent"queue <- queue[queue !="serpent"]
print(queue)## [1] "donkey" "fox" "owl" "ant"queue <- queue[queue !="owl"]
print(queue)## [1] "donkey" "fox" "ant"queue <- c(queue[1:2], "aphid", queue[3])
print(queue)## [1] "donkey" "fox" "aphid" "ant"position <- which(queue == "aphid")
print(position)## [1] 3z <- seq(1,100)
print(z)## [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
## [19] 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
## [37] 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
## [55] 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72
## [73] 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90
## [91] 91 92 93 94 95 96 97 98 99 100z <- 1:100
print(z)## [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
## [19] 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
## [37] 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
## [55] 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72
## [73] 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90
## [91] 91 92 93 94 95 96 97 98 99 100filtered_numbers <- z[which(z %% 2 != 0 & z %% 3 != 0 & z %% 7 != 0)]
print(filtered_numbers)## [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 97Generate a sequence from 1 to 100. Use “%%” to determine the remainder when each number is divided.