vectors:
Create a vector containing elements 10, 22, 27, 19, 20 and assign it with a name.
# your code here
vec <- c(10,22,27,19,20)Use R as a calculator to compute the following values. a). 27(38-15) b). ln(14^7) c). sqrt(436/12)
# your code here
27*(38-15)## [1] 621log(14^7,base = exp(1))## [1] 18.4734sqrt(436/12)## [1] 6.027714# your code here
b <- 87:56;b## [1] 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63
## [26] 62 61 60 59 58 57 56b[19:21]## [1] 69 68 67compute the following statistics of b: 1) sum 2) median 3) standard deviation
# your code here
sum(b)## [1] 2288median(b)## [1] 71.5sd(b)## [1] 9.380832Create a vector that contains 100 elements with value of 1. (Hint: use ?rep for help). Use the option results = ’hide” in the code chuck to hide the result
# your code here
(vec2 <- rep(1,100))## [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## [38] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## [75] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1Create a matrix with 16 elements with 4 rows both by row and column wise store it as Ex1 and Ex2
Access 3rd row 3rd column element of the matrices
Access 2nd row 1st column element of them
# your code here
m <- sort(sample(1:12,16,replace = T))
ex1 <- matrix(m,4,byrow = T);ex1## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] 4 4 5 7
## [3,] 8 8 9 10
## [4,] 11 11 11 12ex2 <- matrix(m,4,byrow = F);ex2## [,1] [,2] [,3] [,4]
## [1,] 1 4 8 11
## [2,] 2 4 8 11
## [3,] 3 5 9 11
## [4,] 4 7 10 12Imagine you have data on temperature readings for different cities In Africa for 3 months. How would you create a matrix in R to store this data, with cities as rows and months as columns?
# your code here
temp <- sample(36:42,15,replace = T)
cities <- c("Nairobi","Accra","Cairo","Capetown","Mogadishu")
temp_data <- matrix(temp,5)
colnames(temp_data) <- month.abb[6:8]
row.names(temp_data) <- cities
temp_data## Jun Jul Aug
## Nairobi 37 40 39
## Accra 41 42 38
## Cairo 42 36 39
## Capetown 40 40 41
## Mogadishu 42 36 42You need to calculate the average temperature for each city over the 3 months. How would you access and manipulate specific elements of the matrix in R to achieve this? Write the code for both row and column-wise calculations
# your code here
colMeans(temp_data)## Jun Jul Aug
## 40.4 38.8 39.8Suppose you have information about different employees, including their name, department, age, and salary. How would you create a data frame in R to store this data? Include different data types for each variable.
# your code here
names <- c("denzell","joshua","lawrence","ismail","amos","abuga")
department <- c("transport","marketing","nutrition","finance","security","recreation")
age <- c(20,19,21,21,24,22)
salary <- c(75000,60000,70000,90000,85000,45000)
data <- data.frame(names,department,age,salary)
data## names department age salary
## 1 denzell transport 20 75000
## 2 joshua marketing 19 60000
## 3 lawrence nutrition 21 70000
## 4 ismail finance 21 90000
## 5 amos security 24 85000
## 6 abuga recreation 22 45000You want to filter the data frame to find employees in the “Marketing” department with a salary above Ksh50,000. Write the code using appropriate indexing and logical operators.
# your code here
subset(data,department == "marketing"&salary>50000)## names department age salary
## 2 joshua marketing 19 60000Create a list that stores different types of data: a numeric vector, a character string, and another list.
# your code here
my_list1 <- list(24.5,c(12,45,67,82,46),"somebody to you-the vamps",list(25:40,c(23,65,71,91),"somebody you .."));my_list1## [[1]]
## [1] 24.5
##
## [[2]]
## [1] 12 45 67 82 46
##
## [[3]]
## [1] "somebody to you-the vamps"
##
## [[4]]
## [[4]][[1]]
## [1] 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
##
## [[4]][[2]]
## [1] 23 65 71 91
##
## [[4]][[3]]
## [1] "somebody you .."You want to add a new element (a logical value) to the end of the list. How would you achieve this? Include different methods.
# your code here
my_list2 <- c(my_list1,TRUE);my_list2## [[1]]
## [1] 24.5
##
## [[2]]
## [1] 12 45 67 82 46
##
## [[3]]
## [1] "somebody to you-the vamps"
##
## [[4]]
## [[4]][[1]]
## [1] 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
##
## [[4]][[2]]
## [1] 23 65 71 91
##
## [[4]][[3]]
## [1] "somebody you .."
##
##
## [[5]]
## [1] TRUEImagine you have survey data where participants chose their favorite R packages from a set of options. How would you create a factor variable in R to store this data?
# your code here
packages <- c("ggplot","shiny","tidyverse","dplyr","tydr")
raw.data <- c("dplyr","shiny","tidyverse","shiny","dplyr","ggplot","dplyr","ggplot","tidyverse","dplyr","tydr","shiny","dplyr")
faves <- factor(raw.data,levels = packages);faves## [1] dplyr shiny tidyverse shiny dplyr ggplot dplyr
## [8] ggplot tidyverse dplyr tydr shiny dplyr
## Levels: ggplot shiny tidyverse dplyr tydrsummary(faves)## ggplot shiny tidyverse dplyr tydr
## 2 3 2 5 1You want to analyze the number of people who chose each color. How would you use the table function and factor levels to get a frequency table? Write the code and interpret the results.
# your code here
color <- c("red","blue","green","purple")
collected <- c("blue","purple","red","blue","green","purple","green","red","red","blue","green","purple","red","blue","green","blue","red","purple")
col_fact <- factor(collected,levels = color)
summary(col_fact)## red blue green purple
## 5 5 4 4col_frequency <- table(col_fact);col_frequency## col_fact
## red blue green purple
## 5 5 4 4Describe the difference between is.finite(x) and !is.infinite(x).
# your answer here
cat("In summary, is.finite(x) identifies finite elements in a vector, while !is.infinite(x) identifies elements that are not infinite.")## In summary, is.finite(x) identifies finite elements in a vector, while !is.infinite(x) identifies elements that are not infinite.#..