library(dslabs)
data(murders)
Data frame
bs5th<-data.frame(name=c("ali","ahmad","sania","sara","adil","sharjeel","subhan","arbaz","asad","hassan","waleed")
,cgpa=c(2.5,3.2,3.9,2.99,3.10,2.99,2.91,2.87,3.2,3.52,2.50)
,grade=c("D","B","A","C","B","A","B","B","B","A","D"))
bs5th
## name cgpa grade
## 1 ali 2.50 D
## 2 ahmad 3.20 B
## 3 sania 3.90 A
## 4 sara 2.99 C
## 5 adil 3.10 B
## 6 sharjeel 2.99 A
## 7 subhan 2.91 B
## 8 arbaz 2.87 B
## 9 asad 3.20 B
## 10 hassan 3.52 A
## 11 waleed 2.50 D
Which have gpa greater then 3
ind<-bs5th$cgpa>3
bs5th$name[ind]
## [1] "ahmad" "sania" "adil" "asad" "hassan"
Which have grade equal to A
def<-bs5th$grade=="A"
bs5th$name[def]
## [1] "sania" "sharjeel" "hassan"
Which have gpa greater then 3 and having grade A
bs5th$name[ind&def]
## [1] "sania" "hassan"
Which have gpa greater then 3 or having grade A
bs5th$name[ind|def]
## [1] "ahmad" "sania" "adil" "sharjeel" "asad" "hassan"
Exercise 3.15
3.15.1 Compute the per 100,000 murder rate for each state and store
it in an object called murder_rate. Then use logical operators to create
a logical vector named low that tells us which entries of murder_rate
are lower than 1.
murder_rate <- (murders$total / murders$population) * 100000
low <- murder_rate < 1
low
## [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE
## [13] TRUE FALSE FALSE TRUE FALSE FALSE FALSE TRUE FALSE FALSE FALSE TRUE
## [25] FALSE FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE TRUE FALSE
## [37] FALSE TRUE FALSE FALSE FALSE TRUE FALSE FALSE TRUE TRUE FALSE FALSE
## [49] FALSE FALSE TRUE
murders$state[low]
## [1] "Hawaii" "Idaho" "Iowa" "Maine"
## [5] "Minnesota" "New Hampshire" "North Dakota" "Oregon"
## [9] "South Dakota" "Utah" "Vermont" "Wyoming"
murder_rate
## [1] 2.8244238 2.6751860 3.6295273 3.1893901 3.3741383 1.2924531
## [7] 2.7139722 4.2319369 16.4527532 3.3980688 3.7903226 0.5145920
## [13] 0.7655102 2.8369608 2.1900730 0.6893484 2.2081106 2.6732010
## [19] 7.7425810 0.8280881 5.0748655 1.8021791 4.1786225 0.9992600
## [25] 4.0440846 5.3598917 1.2128379 1.7521372 3.1104763 0.3798036
## [31] 2.7980319 3.2537239 2.6679599 2.9993237 0.5947151 2.6871225
## [37] 2.9589340 0.9396843 3.5977513 1.5200933 4.4753235 0.9825837
## [43] 3.4509357 3.2013603 0.7959810 0.3196211 3.1246001 1.3829942
## [49] 1.4571013 1.7056487 0.8871131
3.15.2 Now use the results from the previous exercise and the
function which to determine the indices of murder_rate associated with
values lower than 1.
low_indices <- which(low)
low_indices
## [1] 12 13 16 20 24 30 35 38 42 45 46 51
3.15.3 Use the results from the previous exercise to report the
names of the states with murder rates lower than 1.
states_with_low_murder_rate <- murders$state[low_indices]
states_with_low_murder_rate
## [1] "Hawaii" "Idaho" "Iowa" "Maine"
## [5] "Minnesota" "New Hampshire" "North Dakota" "Oregon"
## [9] "South Dakota" "Utah" "Vermont" "Wyoming"
3.15.4 Now extend the code from exercise 2 and 3 to report the
states in the Northeast with murder rates lower than 1. Hint: use the
previously defned logical vector low and the logical operator
&.
northeast_states <- c("Connecticut", "Maine", "Massachusetts", "New Hampshire", "New Jersey", "New York", "Pennsylvania", "Rhode Island", "Vermont")
northeast_low_indices <- which(low & murders$state %in% northeast_states)
northeast_states_with_low_murder_rate <- murders$state[northeast_low_indices]
northeast_states_with_low_murder_rate
## [1] "Maine" "New Hampshire" "Vermont"
3.15.5 In a previous exercise we computed the murder rate for each
state and the average of these numbers.How many states are below the
average?
average_murder_rate <- mean(murder_rate)
states_below_average <- sum(murder_rate < average_murder_rate)
states_below_average
## [1] 27
3.15.6 Use the match function to identify the states with
abbreviations AK, MI, and IA. Hint: start by defning an index of the
entries of murders$abb that match the three abbreviations, then use the
[ operator to extract the states.
abbreviations_to_find <- c("AK", "MI", "IA")
indices <- match(abbreviations_to_find, murders$abb)
states_with_abbreviations <- murders$state[indices]
states_with_abbreviations
## [1] "Alaska" "Michigan" "Iowa"
3.15.7 Use the %in% operator to create a logical vector that answers
the question: which of the following are actual abbreviations: MA, ME,
MI, MO, MU ?
abbreviations_to_check <- c("MA", "ME", "MI", "MO", "MU")
actual_abbreviations <- abbreviations_to_check %in% murders$abb
actual_abbreviations
## [1] TRUE TRUE TRUE TRUE FALSE
murders$state[actual_abbreviations]
## [1] "Alabama" "Alaska" "Arizona"
## [4] "Arkansas" "Colorado" "Connecticut"
## [7] "Delaware" "District of Columbia" "Georgia"
## [10] "Hawaii" "Idaho" "Illinois"
## [13] "Iowa" "Kansas" "Kentucky"
## [16] "Louisiana" "Maryland" "Massachusetts"
## [19] "Michigan" "Minnesota" "Missouri"
## [22] "Montana" "Nebraska" "Nevada"
## [25] "New Jersey" "New Mexico" "New York"
## [28] "North Carolina" "Ohio" "Oklahoma"
## [31] "Oregon" "Pennsylvania" "South Carolina"
## [34] "South Dakota" "Tennessee" "Texas"
## [37] "Vermont" "Virginia" "Washington"
## [40] "West Virginia" "Wyoming"
3.15.8 Extend the code you used in exercise 7 to report the one
entry that is not an actual abbreviation.Hint: use the ! operator, which
turns FALSE into TRUE and vice versa, then which to obtain an
index.
not_actual_abbreviation_index <- which(!actual_abbreviations)
not_actual_abbreviation <- abbreviations_to_check[not_actual_abbreviation_index]
not_actual_abbreviation
## [1] "MU"