CIND 123 Fall 2019 - Assignment #1
Sarah Sabouni
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Sample Question and Solution
Use seq() to create the vector \((1,2,3,\ldots,10)\).
seq(1,10)## [1] 1 2 3 4 5 6 7 8 9 10Question 1
- Use the
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- Use
seq()andc()to create the vector \((1, 2, 3, \ldots, 10, 9, 8, \ldots, 3, 2, 1)\).
c(seq(1,9),seq(10,8),seq(7,1))## [1] 1 2 3 4 5 6 7 8 9 10 9 8 7 6 5 4 3 2 1- Use
rep()to create the vector \((2,3,4,\dots,2,3,4)\) in which the sequence \((2,3,4)\) is repeated 5 times.
rep(2:4,5)## [1] 2 3 4 2 3 4 2 3 4 2 3 4 2 3 4- Use
rep()to create the vector \((1,1,\ldots,1,2,2,\ldots,2,3,3,\ldots,3)\) where each number is repeated 10 times.
rep(1:3,each = 10)## [1] 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3Question 2
- Compute: \[\sum_{n=10}^{100} n^3\]
sum((10:100)^3)## [1] 25500475- Compute: \[\sum_{n=1}^{10}\left(\frac{2^{n}}{n^2} + \frac{n^{4}}{4^{n}}\right)\]
n <- 1:10
sum(((2^n)/(n^2))+((n^4)/(4^n)))## [1] 35.80589- Compute: \[\sum_{n=0}^{10} \frac{1}{(n+1)!}\] Hint: Use
factorial(n)to compute \(n!\)
n <- 0:10
sum(1/factorial(n+1))## [1] 1.718282- Compute: \[\prod_{n=3}^{33} \left(3n + \frac{3}{\sqrt[3]{n}}\right)\]
n <- 3:33
prod((3*n)+(3/(n^1/3)))## [1] 7.427886e+51Question 3
- Create an empty list
mylist.
mylist <- c()
mylist## NULL- Add a component named
firstVarwhose value is the numeric vector \((1,2,\ldots,10)\).
firstVar <- seq(1,10)
firstVar## [1] 1 2 3 4 5 6 7 8 9 10- Add a component named
secondVarwhich is a 4x5 matrix whose elements are \((1,2,\ldots,20)\) in row-wise order.
secondVar <- matrix(seq(1,20),nrow = 4,ncol = 5,byrow = TRUE)
secondVar## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 2 3 4 5
## [2,] 6 7 8 9 10
## [3,] 11 12 13 14 15
## [4,] 16 17 18 19 20- Add a component named
thirdVarwhich is the output of multipling each element ofsecondVarby the average offirstVar.
thirdVar <- secondVar * mean(firstVar)
thirdVar## [,1] [,2] [,3] [,4] [,5]
## [1,] 5.5 11.0 16.5 22.0 27.5
## [2,] 33.0 38.5 44.0 49.5 55.0
## [3,] 60.5 66.0 71.5 77.0 82.5
## [4,] 88.0 93.5 99.0 104.5 110.0- Display
myliston the screen, after rounding the elements of ‘thirdVar’ to the nearest integer.
mylist <- list(firstVar,secondVar,round(thirdVar,0))
print(mylist)## [[1]]
## [1] 1 2 3 4 5 6 7 8 9 10
##
## [[2]]
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 2 3 4 5
## [2,] 6 7 8 9 10
## [3,] 11 12 13 14 15
## [4,] 16 17 18 19 20
##
## [[3]]
## [,1] [,2] [,3] [,4] [,5]
## [1,] 6 11 16 22 28
## [2,] 33 38 44 50 55
## [3,] 60 66 72 77 82
## [4,] 88 94 99 104 110Question 4
iris data set gives the measurements in centimeters of the variables sepal length, sepal width, petal length and petal width, respectively, for 50 flowers from each of 3 species of iris. The species are Iris setosa, versicolor, and virginica.
Install the iris data set on your computer using the command install.packages("datasets"). Then load the datasets package into your session using the following command.
library(datasets)
#install.packages("datasets")- Display the first 6 rows of the
irisdata set
head(iris)## Sepal.Length Sepal.Width Petal.Length Petal.Width Species
## 1 5.1 3.5 1.4 0.2 setosa
## 2 4.9 3.0 1.4 0.2 setosa
## 3 4.7 3.2 1.3 0.2 setosa
## 4 4.6 3.1 1.5 0.2 setosa
## 5 5.0 3.6 1.4 0.2 setosa
## 6 5.4 3.9 1.7 0.4 setosa- Compute the average of the first four variables (Sepal.Length, Sepal.Width, Petal.Length and Petal.Width) using
sapply()function.
Hint: You might need to consider removing the NA values, otherwise the average will not be computed.
sapply(iris[,-5] ,mean ,rm = NA )## Sepal.Length Sepal.Width Petal.Length Petal.Width
## 5.843333 3.057333 3.758000 1.199333- Show how to use R to replace the imssing values in this dataset with plausible ones.
iris[is.na(iris)] <- 0- Compute the standard deviation for only the first and the third variables (Sepal.Length and Petal.Length)
sd(iris[,1])## [1] 0.8280661sd(iris[,3])## [1] 1.765298- Construct a boxplot for
Sepal.Widthvariable, then display all the outliers.
library(ggplot2)
ggplot(iris, aes(x = "", y=Sepal.Width)) +
geom_boxplot(outlier.colour="red",
outlier.shape=16,
outlier.size=2, notch=FALSE)
- Compute the lower and the upper quartiles of
Sepal.Widthvariable
dplyr::summarize(iris, "lower Quartile" = quantile(Sepal.Length, .25),
"upper Quartile" = quantile(Sepal.Length, .75), )## lower Quartile upper Quartile
## 1 5.1 6.4- Construct a pie chart to describe the species with ‘Sepal.Length’ greater than 5 centimeters.
# Pie Chart from data frame with Appended Sample Sizes
sepalLength6 <- subset(iris, iris$Sepal.Length > 5)
mytable <- table(sepalLength6$Species)
lbls <- paste(names(mytable), "\n", mytable, sep="")
pie(mytable, labels = lbls,
main="Pie Chart of Species\n and Sepal length > 5")
tail(sepalLength6) ## Sepal.Length Sepal.Width Petal.Length Petal.Width Species
## 145 6.7 3.3 5.7 2.5 virginica
## 146 6.7 3.0 5.2 2.3 virginica
## 147 6.3 2.5 5.0 1.9 virginica
## 148 6.5 3.0 5.2 2.0 virginica
## 149 6.2 3.4 5.4 2.3 virginica
## 150 5.9 3.0 5.1 1.8 virginicaEND of Assignment #1.