This dataset contains information on vehicles towed in Baltimore, MD:

• A larger version of this dataset along with additional descriptions can be found at: https://data.baltimorecity.gov/Transportation/DOT-Towing/k78j-azhn.

• The full version of the dataset contains 61,000 rows and 36 columns, where each row corresponds to a vehicle and the columns are information pertaining to the vehicle.

• We will be working with a smaller dataset with approximately 30,000 rows and 5 columns.

First read in the data set which is available at: http://www.math.montana.edu/ahoegh/teaching/stat408/datasets/BaltimoreTowing.csv.

baltimore_tow <- read_csv('http://www.math.montana.edu/ahoegh/teaching/stat408/datasets/BaltimoreTowing.csv')
str(baltimore_tow)

## spec_tbl_df [30,263 × 5] (S3: spec_tbl_df/tbl_df/tbl/data.frame)
##  $ vehicleType      : chr [1:30263] "Van" "Car" "Car" "Car" ...
##  $ vehicleMake      : chr [1:30263] "LEXUS" "Mercedes" "Chysler" "Chevrolet" ...
##  $ vehicleModel     : chr [1:30263] NA NA "Cirrus" "Cavalier" ...
##  $ receivingDateTime: chr [1:30263] "10/24/2010 12:41:00 PM" "04/28/2015 09:27:00 AM" "07/23/2015 07:55:00 AM" "10/23/2010 11:35:00 AM" ...
##  $ totalPaid        : chr [1:30263] "$322.00" "$130.00" "$280.00" "$1057.00" ...
##  - attr(*, "spec")=
##   .. cols(
##   ..   vehicleType = col_character(),
##   ..   vehicleMake = col_character(),
##   ..   vehicleModel = col_character(),
##   ..   receivingDateTime = col_character(),
##   ..   totalPaid = col_character()
##   .. )
##  - attr(*, "problems")=<externalptr>

• Dealing with strings
• Dealing with date/time objects
• Dealing with factors

Now compute the average towing cost grouped by month.

Now compute the average towing cost grouped by month.

baltimore_tow %>% 
  group_by(month) %>% 
  summarize(mean.cost = mean(totalPaid))

Now compute the average towing cost grouped by month.

Consider adding a column for year to the data set. This can be done using substr().

Usage: substr(x, start, stop)

Arguments:

• x, text a character vector.
• start, first integer. The first element to be extracted
• stop, last integer. The last element to be extracted

Use the substr() function to extract month and create a new variable in R.

Use the substr() function to extract month and create a new variable in R.

baltimore_tow$month <- substr(baltimore_tow$receivingDateTime, 0, 2)
head(baltimore_tow$month)

## [1] "10" "04" "07" "10" "10" "10"

Now compute the average towing cost grouped by month.

baltimore_tow %>% 
  group_by(month) %>% 
  summarize(mean.cost = mean(totalPaid))

## # A tibble: 12 × 2
##    month mean.cost
##    <chr>     <dbl>
##  1 01           NA
##  2 02           NA
##  3 03           NA
##  4 04           NA
##  5 05           NA
##  6 06           NA
##  7 07           NA
##  8 08           NA
##  9 09           NA
## 10 10           NA
## 11 11           NA
## 12 12           NA

In many situations, the year could be in a different position so the substr() might not work. For example month the date could be coded 4/1/2015 rather than 04/01/2015 So consider, using strsplit() instead.

Usage: strsplit(x, split)

Arguments:

• x: character vector, each element of which is to be split. Other inputs, including a factor, will give an error.
• split: character vector (or object which can be coerced to such) containing regular expression(s) (unless fixed = TRUE) to use for splitting.

Use the strsplit() function to remove the dollar sign from the cost.

Use the strsplit() function to remove the dollar sign from the cost.

## example for one row
strsplit(baltimore_tow$totalPaid[1],'$', fixed = T)[[1]][2]

## [1] "322.00"

The base data structures in R can be organized by dimensionality and whether they are homogenous.

Consider the two lists

msu.info <- list(
  name = c('Waded Cruzado', 'Stacey Hancock'), 
  degree.from = c('University of Texas at Arlington', 
                  'Colorado State University'),
  job.title = c('President',
                'Associate Professor of Statistics'))

msu.info2 <- list(
  c('Waded Cruzado',
    'University of Texas at Arlington',
    'President'), 
  c('Stacey Hancock',
    'Colorado State University',
    'Associate Professor of Statistics'))

msu.info

## $name
## [1] "Waded Cruzado"  "Stacey Hancock"
## 
## $degree.from
## [1] "University of Texas at Arlington" "Colorado State University"       
## 
## $job.title
## [1] "President"                         "Associate Professor of Statistics"

msu.info2

## [[1]]
## [1] "Waded Cruzado"                    "University of Texas at Arlington"
## [3] "President"                       
## 
## [[2]]
## [1] "Stacey Hancock"                    "Colorado State University"        
## [3] "Associate Professor of Statistics"

With the current lists we can index elements using the double bracket [[ ]] notation or if names have been initialized, those can be used too.

So the first element of each list can be indexed

msu.info[[1]]

## [1] "Waded Cruzado"  "Stacey Hancock"

msu.info$name

## [1] "Waded Cruzado"  "Stacey Hancock"

Explore the indexing with these commands.

msu.info[1]
msu.info[[1]]
msu.info$name[2]
msu.info[1:2]
unlist(msu.info)

“If list x is a train carrying objects, then x[[5]] is the object in car 5; x[4:6] is a train of cars 4-6.”

— @RLangTip

Source: http://adv-r.had.co.nz/Subsetting.html

list(c("Jan","Feb","Mar"), 
     matrix(c(3,9,5,1,-2,8), nrow = 2),
     list("green", 12.3)
     )

## [[1]]
## [1] "Jan" "Feb" "Mar"
## 
## [[2]]
##      [,1] [,2] [,3]
## [1,]    3    5   -2
## [2,]    9    1    8
## 
## [[3]]
## [[3]][[1]]
## [1] "green"
## 
## [[3]][[2]]
## [1] 12.3

Use the strsplit() function to remove the dollar sign from the cost.

strsplit(baltimore_tow$totalPaid[1:2], '$', fixed = T)[[1]][2]

## [1] "322.00"

lubridate is a tidyverse package for manipulating date objects. There is a nice website with a cheatsheet.

library(lubridate) # loads with tidyverse
class(baltimore_tow$receivingDateTime)

## [1] "character"

baltimore_tow <- baltimore_tow %>% 
  mutate(date_time = mdy_hms(receivingDateTime))
class(baltimore_tow$date_time)

## [1] "POSIXct" "POSIXt"

head(month(baltimore_tow$date_time))

## [1] 10  4  7 10 10 10

head(year(baltimore_tow$date_time))

## [1] 2010 2015 2015 2010 2010 2010

The stringr package (cheat sheet) provides a nice set of tools. There is also an information page.

Use the stringr package to remove (replace) the dollar sign. Note that a dollar sign is a special character, so you’ll need to use \\$.

Use the stringr package to remove (replace) the dollar sign

library(stringr)
baltimore_tow$cost <- 
  as.numeric(str_replace(baltimore_tow$totalPaid, '\\$',''))

Now compute the average towing cost grouped by month.

baltimore_tow %>% 
  group_by(month) %>% 
  summarize(mean.cost = mean(cost), .groups = 'keep')

## # A tibble: 12 × 2
## # Groups:   month [12]
##    month mean.cost
##    <chr>     <dbl>
##  1 01         353.
##  2 02         349.
##  3 03         363.
##  4 04         347.
##  5 05         357.
##  6 06         346.
##  7 07         350.
##  8 08         350.
##  9 09         359.
## 10 10         343.
## 11 11         342.
## 12 12         344.

Next we wish to compute how many vehicles were towed for each vehicle type.

However, we want to take a close look at the vehicle types in the data set and perhaps create more useful groups.

First examine the unique types of vehicles in this data set.

unique(baltimore_tow$vehicleType)

##  [1] "Van"                        "Car"                       
##  [3] "SUV"                        "Pick-up Truck"             
##  [5] "Motor Cycle (Street Bike)"  "Dirt Bike"                 
##  [7] "Commercial Truck"           "Trailer"                   
##  [9] "Station Wagon"              "Truck"                     
## [11] "Taxi"                       "Pickup Truck"              
## [13] "Convertible"                "Tractor Trailer"           
## [15] "Tow Truck"                  "All terrain - 4 wheel bike"
## [17] "Mini-Bike"                  "Golf Cart"                 
## [19] "Boat"                       "Tractor"                   
## [21] "Construction Equipment"     "Sport Utility Vehicle"

First consider reasonable groups for vehicle types.

1. Cars - (Car, convertible)
2. Large Cars - (SUV, Station Wagon, Sport Utility Vehicle, Van, Taxi)
3. Trucks - (Pick-up Truck, Pickup Truck)
4. Large Trucks - (Truck, Tractor Trailer, Tow Truck, Tractor, Construction Equipment, Commercial Truck)
5. Bikes - (Motor Cycle (Street Bike), Dirt Bike, All terrain - 4 wheel bike, Mini-Bike)
6. Misc (delete) - (Boat, Golf Cart, Trailer)

Next examine values in some of these groups, we will just look at the vehicle type of ‘Truck’.

unique(baltimore_tow$vehicleMake[baltimore_tow$vehicleType == 'Truck'])

##  [1] "GMC"            "Ford"           "Dodge"          "Freightliner"  
##  [5] "Chevrolet"      "Izuzu"          "Toyota"         "Chevy"         
##  [9] "Peterbilt"      "International"  "Kenworth"       "Nissan"        
## [13] "Mercedes"       "Isuzu"          "Frightliner"    "Mack"          
## [17] "Sterling"       "Internantional" "Peterbelt"      "Pete"          
## [21] "Hummer"         "Hino"

Note that there are several spelling errors in this data set. How do we combine them?

Spelling errors can be addressed, by reassigning vehicles to the correct spelling.

baltimore_tow$vehicleMake[baltimore_tow$vehicleMake == 
                            'Peterbelt'] <- 'Peterbilt'
baltimore_tow$vehicleMake[baltimore_tow$vehicleMake == 
                            'Internantional'] <- 'International'
baltimore_tow$vehicleMake <-
  str_replace(baltimore_tow$vehicleMake,'Izuzu','Isuzu')
baltimore_tow$vehicleMake <-
  str_replace(baltimore_tow$vehicleMake,'Frightliner','Freightliner')

Also note that many of the groupings have mis-classified vehicles, but we will not focus on that yet.

First we will delete golf carts, boats, and trailers. There are several ways to do this, consider making a new data frame called balt_tow_small that does not include golf carts, boats, and trailers.

First we will delete golf carts, boats, and trailers.

balt_tow_small <- baltimore_tow %>% 
  filter(!(vehicleType %in% c("Golf Cart", "Boat", "Trailer")))

Now we need to create a variable for the additional groups below.

1. Cars - (Car, convertible)
2. Large Cars - (SUV, Station Wagon, Sport Utility Vehicle, Van, Taxi)
3. Trucks - (Pick-up Truck, Pickup Truck)
4. Large Trucks - (Truck, Tractor Trailer, Tow Truck, Tractor, Construction Equipment, Commercial Truck)
5. Bikes - (Motor Cycle (Street Bike), Dirt Bike, All terrain - 4 wheel bike, Mini-Bike)

One way to create groups is by creating a new variable

balt_tow_small$Group <- '' # Creates empty string for all rows in data set

balt_tow_small$Group[balt_tow_small$vehicleType %in% 
                       c('Car','Convertible')] <- 'Cars'

balt_tow_small$Group[balt_tow_small$vehicleType %in% c('SUV',
 'Station Wagon','Sport Utility Vehicle','Van','Taxi')] <- 'Large Cars'

balt_tow_small$Group[balt_tow_small$vehicleType %in% 
                       c('Pick-up Truck','Pickup Truck')] <- 'Trucks'

balt_tow_small$Group[balt_tow_small$vehicleType %in% 
    c('Truck','Tractor Trailer','Tow Truck','Tractor',
      'Construction Equipment','Commercial Truck')] <- 'Large Trucks'

balt_tow_small$Group[balt_tow_small$vehicleType %in% 
    c('Motor Cycle (Street Bike)','Dirt Bike','Mini-Bike',
      'All terrain - 4 wheel bike')] <- 'Bikes'

Next we wish to compute how many vehicles were towed for each vehicle type

balt_tow_small %>% count(Group)

## # A tibble: 5 × 2
##   Group            n
##   <chr>        <int>
## 1 Bikes          383
## 2 Cars         19675
## 3 Large Cars    8575
## 4 Large Trucks   211
## 5 Trucks        1378

Factors are a specific way to store categorical data. Using factors results in a more efficient data storage process, but can be cumbersome.

Factors can be necessary for making plots and fitting models in R.

The forcats package, website, is a tidyverse package designed for dealing with categorical factors.

favorite_day <- c('Friday', 'Saturday', 'Sunday', 'Tuesday', 'Saturday', 'Saturday')
class(favorite_day)

## [1] "character"

day_factor <- as.factor(favorite_day)
class(day_factor)

## [1] "factor"

sort(day_factor)

## [1] Friday   Saturday Saturday Saturday Sunday   Tuesday 
## Levels: Friday Saturday Sunday Tuesday

library(forcats)
day_factor <- fct_relevel(day_factor,
                          c('Sunday','Tuesday','Friday','Saturday'))
sort(day_factor)

## [1] Sunday   Tuesday  Friday   Saturday Saturday Saturday
## Levels: Sunday Tuesday Friday Saturday

Rather than coercing a class variable to be a factor, the factor can be created directly.

day_factor2 <- factor(c('Friday', 'Saturday', 'Sunday', 'Monday'),
                      levels = c('Sunday', 'Monday', 'Tuesday', 
                        'Wednesday','Thursday','Friday','Saturday'))
sort(day_factor2)

## [1] Sunday   Monday   Friday   Saturday
## Levels: Sunday Monday Tuesday Wednesday Thursday Friday Saturday

Factors can also easily be collapsed with forcats

balt_tow_small %>% 
  mutate(Group2 = fct_collapse(vehicleType, 
           Cars = c('Car','Convertible'),
           Large_Cars = c('SUV', 'Station Wagon',
                          'Sport Utility Vehicle',
                          'Van','Taxi'),
           Trucks = c('Pick-up Truck','Pickup Truck'),
           Large_Trucks = c('Truck', 'Tractor Trailer',
                            'Tow Truck', 'Tractor',
                            'Construction Equipment',
                            'Commercial Truck'),
           Bikes = c('Motor Cycle (Street Bike)', 'Dirt Bike',
                     'Mini-Bike','All terrain - 4 wheel bike')
           )
         ) %>% 
  mutate(Group2 = fct_infreq(Group2)) %>%
  group_by(Group2) %>% 
  summarize(ave_cost = mean(cost), .groups = 'drop')

## # A tibble: 5 × 2
##   Group2       ave_cost
##   <fct>           <dbl>
## 1 Cars             354.
## 2 Large_Cars       334.
## 3 Trucks           359.
## 4 Bikes            269.
## 5 Large_Trucks     687.