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.
x <- 1.1
a <- 2.2
b <- 3.3
z <- x^{a^{b}}
print(z)
## [1] 3.61714
z <- (x^a)^b
print(z)
## [1] 1.997611
z <- 3*(x^3)+2*(x^2)+1
print(z)
## [1] 7.413
Using the rep and seq functions, create the following vectors:
a <- c(seq(from=1,to=8),seq(from=7,to=1))
print(a)
## [1] 1 2 3 4 5 6 7 8 7 6 5 4 3 2 1
vect <- seq(from=1,to=5)
b <- rep(vect,times=vect)
print(b)
## [1] 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5
vect_2 <- seq(from=5,to=1)
c <- rep(vect_2,times=vect)
print(c)
## [1] 5 4 4 3 3 3 2 2 2 2 1 1 1 1 1
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.
coord <- runif(2,min = 0,max = 10)
print(coord)
## [1] 7.946985 9.618633
r <- sqrt((coord[1])^2 + (coord[2])^2)
theta <- atan((coord[2])/(coord[1]))
polar <- c(r,theta)
print(polar)
## [1] 12.4768854 0.8802783
Create 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:
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 <- setdiff(queue,"sheep")
print(queue)
## [1] "fox" "owl" "ant" "serpent"
queue <- c("donkey",queue)
print(queue)
## [1] "donkey" "fox" "owl" "ant" "serpent"
queue <- setdiff(queue,"serpent")
print(queue)
## [1] "donkey" "fox" "owl" "ant"
queue <- setdiff(queue,"owl")
print(queue)
## [1] "donkey" "fox" "ant"
queue <- c(queue[1:2],"aphid",queue[3])
print(queue)
## [1] "donkey" "fox" "aphid" "ant"
which(queue=="aphid")
## [1] 3
Use R to create a vector of all of the integers from 1 to 100 that are not divisible by 2, 3, or 7.
vect <- seq(1:100)
vect %% 2 != 0 & vect %% 3 !=0 & vect %% 7 !=0
## [1] TRUE FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE TRUE FALSE
## [13] TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE FALSE TRUE FALSE
## [25] TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE FALSE FALSE FALSE
## [37] TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE FALSE TRUE FALSE
## [49] FALSE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE FALSE TRUE FALSE
## [61] TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE FALSE TRUE FALSE
## [73] TRUE FALSE FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE TRUE FALSE
## [85] TRUE FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE TRUE FALSE
## [97] TRUE FALSE FALSE FALSE
vect <- which(vect %% 2 != 0 & vect %% 3 !=0 & vect %% 7 !=0)
print(vect)
## [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 97