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Use seq() to create the vector \((1,2,3,\ldots,10)\).
seq(1,10)
## [1] 1 2 3 4 5 6 7 8 9 10
seq() function to create the vector \((1, 7, 13, \ldots, 61)\). Note that each term in this sequence is of the form \(1 + 6n\) where \(n = 0, \ldots, 10\).seq(1,61, by=6)
## [1] 1 7 13 19 25 31 37 43 49 55 61
seq() and c() to create the vector \((1, 2, 3, \ldots, 10, 9, 8, \ldots, 3, 2, 1)\).c(seq(1,10), seq(9,1))
## [1] 1 2 3 4 5 6 7 8 9 10 9 8 7 6 5 4 3 2 1
rep() to create the vector \((2,3,4,\dots,2,3,4)\) in which the sequence \((2,3,4)\) is repeated 5 times.rep(c(2,3,4),5)
## [1] 2 3 4 2 3 4 2 3 4 2 3 4 2 3 4
rep() to create the vector \((1,1,\ldots,1,2,2,\ldots,2,3,3,\ldots,3)\) where each number is repeated 7 times.x <- seq(1,7)
rep(x, each=7)
## [1] 1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3 3 3 4 4 4 4 4 4 4 5 5 5 5 5 5 5
## [36] 6 6 6 6 6 6 6 7 7 7 7 7 7 7
rep() to create the vector \((10,20,20,30,30,30,\ldots,100,\ldots,100)\) where \(10n\) is repeated \(n\) times.n <- seq(1,10)
rep(n*10, n)
## [1] 10 20 20 30 30 30 40 40 40 40 50 50 50 50 50 60 60
## [18] 60 60 60 60 70 70 70 70 70 70 70 80 80 80 80 80 80
## [35] 80 80 90 90 90 90 90 90 90 90 90 100 100 100 100 100 100
## [52] 100 100 100 100
n <- seq(1,100)
sum(n)
## [1] 5050
n <-seq(10,100)
sum(n^3)
## [1] 25500475
n <- seq(1,10)
sum(((2^n)/n)+((4^n)/(n^4)))
## [1] 416.5333
factorial(n) to compute \(n!\)n <- seq(0,10)
sum((1/(factorial(n))))
## [1] 2.718282
n <- seq(1,20)
sum((2*n)+(1/sqrt(n)))
## [1] 427.5953
mylist.mylist <- list()
aa whose value is 42.mylist[["aa"]] <- 42; mylist
## $aa
## [1] 42
bb whose value is the numeric vector \((1,2,\ldots,10)\).mylist[["bb"]] <- seq(1,10); mylist
## $aa
## [1] 42
##
## $bb
## [1] 1 2 3 4 5 6 7 8 9 10
cc whose value is the character vector (“Hello”, “CIND 123”).mylist[["cc"]] <- c("Hello", "CIND 123"); mylist
## $aa
## [1] 42
##
## $bb
## [1] 1 2 3 4 5 6 7 8 9 10
##
## $cc
## [1] "Hello" "CIND 123"
dd whose value is a 4x3 matrix whose elements are \((1,2,\ldots,12)\) in row-wise order.mylist[["dd"]] <- matrix(data = 1:12, nrow = 4, ncol = 3); mylist
## $aa
## [1] 42
##
## $bb
## [1] 1 2 3 4 5 6 7 8 9 10
##
## $cc
## [1] "Hello" "CIND 123"
##
## $dd
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
mylist on the screen.mylist
## $aa
## [1] 42
##
## $bb
## [1] 1 2 3 4 5 6 7 8 9 10
##
## $cc
## [1] "Hello" "CIND 123"
##
## $dd
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
If you have not already done so, install the ISwR package on your computer using the command install.packages("ISwR").
Load the ISwR package into your session.
library(ISwR)
thuesen data frame.tail(thuesen, 6)
## blood.glucose short.velocity
## 19 12.5 1.19
## 20 16.1 1.05
## 21 13.3 1.32
## 22 4.9 1.03
## 23 8.8 1.12
## 24 9.5 1.70
sapply() function.Hint: You might need to consider removing the NA values, otherwise the average will not be computed.
thuesenadj <- na.omit(thuesen)
sapply(thuesenadj, mean)
## blood.glucose short.velocity
## 10.373913 1.325652
v1, v2, and v3 whose elements are the numbers from 1 to 20, their squares, and their square-roots respectiverly.v1 <- seq(1,20)
v2 <- v1 ^ 2
v3 <- sqrt(v1)
iData by combining the v1, v2, and v3 together in a column-wise perspective.iData <- data.frame(v1, v2, v3)
iData.summary(iData)
## v1 v2 v3
## Min. : 1.00 Min. : 1.00 Min. :1.000
## 1st Qu.: 5.75 1st Qu.: 33.25 1st Qu.:2.396
## Median :10.50 Median :110.50 Median :3.239
## Mean :10.50 Mean :143.50 Mean :3.083
## 3rd Qu.:15.25 3rd Qu.:232.75 3rd Qu.:3.905
## Max. :20.00 Max. :400.00 Max. :4.472
mat1 <- matrix(data=NA, nrow = 5, ncol = 5)
mat1[1,1] <- "This"
mat1[2,2] <- "is"
mat1[3,3] <- "the"
mat1[4,4] <- "main"
mat1[5,5] <- "diagonal"
mat1
## [,1] [,2] [,3] [,4] [,5]
## [1,] "This" NA NA NA NA
## [2,] NA "is" NA NA NA
## [3,] NA NA "the" NA NA
## [4,] NA NA NA "main" NA
## [5,] NA NA NA NA "diagonal"
END of Assignment #1.