##…Q1… ## . Use the $ operator to access the population size data and store it as the object pop. Then use the sort function to redefne pop so that it is sorted. Finally, use the [ operator to report the smallest.
library(dslabs)
data(murders)
pop <- murders$population
sorted_pop <- sort(pop)
smallest_population <- sorted_pop[1]
print(smallest_population)## [1] 563626index_of_smallest_population <- order(pop)[1]
print(index_of_smallest_population)## [1] 51##…Q3… ## 3. We can actually perform the same operation as in the previous exercise using the function which.min.Write one line of code that does this.
index_of_smallest_population <- which.min(pop)
print(index_of_smallest_population)## [1] 51##. Now we know how small the smallest state is and we know which row represents it. Which state is it?Defne a variable states to be the state names from the murders data frame. Report the name of the state with the smallest population.
states <- murders$state
state_with_smallest_population <- states[index_of_smallest_population]
print(state_with_smallest_population)## [1] "Wyoming"##………………………..Q5……………………………….. ##.5. You can create a data frame using the data.frame function. Here is a quick example:temp <- c(35, 88, 42, 84, 81, 30),city <- c(“Beijing”, “Lagos”, “Paris”, “Rio de Janeiro”, “San Juan”, “Toronto”),city_temps <- data.frame(name = city, temperature = temp),Use the rank function to determine the population rank of each state from smallest population size tobiggest. Save these ranks in an object called ranks, then create a data frame with the state name andits rank. Call the data frame my_df.
ranks <- rank(pop)
my_df <- data.frame(state = states, rank = ranks)
my_df## state rank
## 1 Alabama 29
## 2 Alaska 5
## 3 Arizona 36
## 4 Arkansas 20
## 5 California 51
## 6 Colorado 30
## 7 Connecticut 23
## 8 Delaware 7
## 9 District of Columbia 2
## 10 Florida 49
## 11 Georgia 44
## 12 Hawaii 12
## 13 Idaho 13
## 14 Illinois 47
## 15 Indiana 37
## 16 Iowa 22
## 17 Kansas 19
## 18 Kentucky 26
## 19 Louisiana 27
## 20 Maine 11
## 21 Maryland 33
## 22 Massachusetts 38
## 23 Michigan 43
## 24 Minnesota 31
## 25 Mississippi 21
## 26 Missouri 34
## 27 Montana 8
## 28 Nebraska 14
## 29 Nevada 17
## 30 New Hampshire 10
## 31 New Jersey 41
## 32 New Mexico 16
## 33 New York 48
## 34 North Carolina 42
## 35 North Dakota 4
## 36 Ohio 45
## 37 Oklahoma 24
## 38 Oregon 25
## 39 Pennsylvania 46
## 40 Rhode Island 9
## 41 South Carolina 28
## 42 South Dakota 6
## 43 Tennessee 35
## 44 Texas 50
## 45 Utah 18
## 46 Vermont 3
## 47 Virginia 40
## 48 Washington 39
## 49 West Virginia 15
## 50 Wisconsin 32
## 51 Wyoming 1ind <- order(ranks)
my_df <- my_df[ind, ]
my_df## state rank
## 51 Wyoming 1
## 9 District of Columbia 2
## 46 Vermont 3
## 35 North Dakota 4
## 2 Alaska 5
## 42 South Dakota 6
## 8 Delaware 7
## 27 Montana 8
## 40 Rhode Island 9
## 30 New Hampshire 10
## 20 Maine 11
## 12 Hawaii 12
## 13 Idaho 13
## 28 Nebraska 14
## 49 West Virginia 15
## 32 New Mexico 16
## 29 Nevada 17
## 45 Utah 18
## 17 Kansas 19
## 4 Arkansas 20
## 25 Mississippi 21
## 16 Iowa 22
## 7 Connecticut 23
## 37 Oklahoma 24
## 38 Oregon 25
## 18 Kentucky 26
## 19 Louisiana 27
## 41 South Carolina 28
## 1 Alabama 29
## 6 Colorado 30
## 24 Minnesota 31
## 50 Wisconsin 32
## 21 Maryland 33
## 26 Missouri 34
## 43 Tennessee 35
## 3 Arizona 36
## 15 Indiana 37
## 22 Massachusetts 38
## 48 Washington 39
## 47 Virginia 40
## 31 New Jersey 41
## 34 North Carolina 42
## 23 Michigan 43
## 11 Georgia 44
## 36 Ohio 45
## 39 Pennsylvania 46
## 14 Illinois 47
## 33 New York 48
## 10 Florida 49
## 44 Texas 50
## 5 California 51##…………………………Q6…………………………………… ## 6. Repeat the previous exercise, but this time order my_df so that the states are ordered from leastpopulous to most populous. Hint: create an object ind that stores the indexes needed to order thepopulation values. Then use the bracket operator [ to re-order each column in the data frame.
# Create a data frame with state names and population ranks
ranks <- rank(pop)
my_df <- data.frame(state = states, rank = ranks)
# Create an index to order the data frame
ind <- order(ranks)
# Reorder the data frame based on the population ranks
my_df <- my_df[ind, ]
# Now, my_df is ordered from least populous to most populous
my_df## state rank
## 51 Wyoming 1
## 9 District of Columbia 2
## 46 Vermont 3
## 35 North Dakota 4
## 2 Alaska 5
## 42 South Dakota 6
## 8 Delaware 7
## 27 Montana 8
## 40 Rhode Island 9
## 30 New Hampshire 10
## 20 Maine 11
## 12 Hawaii 12
## 13 Idaho 13
## 28 Nebraska 14
## 49 West Virginia 15
## 32 New Mexico 16
## 29 Nevada 17
## 45 Utah 18
## 17 Kansas 19
## 4 Arkansas 20
## 25 Mississippi 21
## 16 Iowa 22
## 7 Connecticut 23
## 37 Oklahoma 24
## 38 Oregon 25
## 18 Kentucky 26
## 19 Louisiana 27
## 41 South Carolina 28
## 1 Alabama 29
## 6 Colorado 30
## 24 Minnesota 31
## 50 Wisconsin 32
## 21 Maryland 33
## 26 Missouri 34
## 43 Tennessee 35
## 3 Arizona 36
## 15 Indiana 37
## 22 Massachusetts 38
## 48 Washington 39
## 47 Virginia 40
## 31 New Jersey 41
## 34 North Carolina 42
## 23 Michigan 43
## 11 Georgia 44
## 36 Ohio 45
## 39 Pennsylvania 46
## 14 Illinois 47
## 33 New York 48
## 10 Florida 49
## 44 Texas 50
## 5 California 51##………………………Q7………………………………………… ##7. The na_example vector represents a series of counts. You can quickly examine the object using: ##data(“na_example”) ###However, when we compute the average with the function mean, we obtain an NA: ##mean(na_example) ##> [1] NA ##The is.na function returns a logical vector that tells us which entries are NA. Assign this logical vector ##to an object called ind and determine how many NAs does na_example have.
data("na_example")
str(na_example)## int [1:1000] 2 1 3 2 1 3 1 4 3 2 ...ind <- is.na(na_example)
number_of_nas <- sum(ind)
print(number_of_nas)## [1] 145#Compute the average again, but only for the entries that are not NA.
average_without_nas <- mean(na_example[!ind])
print(average_without_nas)## [1] 2.301754##………………………………….Q8…………………………….. ##8. Now compute the average again, but only for the entries that are not NA. Hint: remember the !operator.
data("na_example")
# Create a logical vector that identifies non-NA entries
not_na <- !is.na(na_example)
# Calculate the average only for non-NA entries
average_without_nas <- mean(na_example[not_na])
# Print the result
print(average_without_nas)## [1] 2.301754##>>>>>>>>>>>>>>>>>>>>>>>>[Excercise no.3.13]<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<
##1. Previously we created this data frame:temp <- c(35, 88, 42, 84, 81, 30)city <- c(“Beijing”, “Lagos”, “Paris”, “Rio de Janeiro”, “San Juan”, “Toronto”)city_temps <- data.frame(name = city, temperature = temp) Remake the data frame using the code above, but add a line that converts the temperature from Fahrenheit to Celsius. The conversion is C = 59 × (F − 32).
# Create vectors for city names and temperatures in Fahrenheit
temp_fahrenheit <- c(35, 88, 42, 84, 81, 30)
city <- c("Beijing", "Lagos", "Paris", "Rio de Janeiro", "San Juan", "Toronto")
# Convert temperatures from Fahrenheit to Celsius using the conversion formula
temp_celsius <- (5/9) * (temp_fahrenheit - 32)
# Create the data frame with city names and temperatures in Celsius
city_temps <- data.frame(name = city, temperature_Celsius = temp_celsius)
city_temps## name temperature_Celsius
## 1 Beijing 1.666667
## 2 Lagos 31.111111
## 3 Paris 5.555556
## 4 Rio de Janeiro 28.888889
## 5 San Juan 27.222222
## 6 Toronto -1.111111n <- 1:1000
sum_result <- sum(1/n^2)
sum_result## [1] 1.643935##……………………………Q3…………………………………… ##3. Compute the per 100,000 murder rate for each state and store it in the object murder_rate. Then compute the average murder rate for the US using the function mean. What is the average?
# Calculate the murder rate for each state
murder_rate <- (murders$total / murders$population) * 100000
# Compute the average murder rate for the US
average_murder_rate_us <- mean(murder_rate)
# Print the average murder rate
print(average_murder_rate_us)## [1] 2.779125