Data 101: Hotel bookings - data wrangling - Macy

Marilyn Macy

library(tidyverse)
library(skimr)
library(readr)

# From TidyTuesday: https://github.com/rfordatascience/tidytuesday/blob/master/data/2020/2020-02-11/readme.md
hotels <- read_csv("https://raw.githubusercontent.com/rfordatascience/tidytuesday/master/data/2020/2020-02-11/hotels.csv")

Exercises

Exercise 1.

Warm up! Take a look at an overview of the data with the skim() function.

Note: I already gave you the answer to this exercise. You just need to knit the document and view the output. A definition of all variables is given in the Data dictionary section at the end, though you don’t need to familiarize yourself with all variables in order to work through these exercises.

skim(hotels)

Data summary

Variable type: character

Variable type: Date

Variable type: numeric

Exercise 1.1 (added by ProfCelia)

Look at the documentation of the skimr package. And answer the following questions.

1. Can skimr functions be used in a pipeline? Yes. Per the documentation the skimr functions take data frames, return data frames and can work as part of a pipeline.

2. How is skimr used in this project? It is used to investigate the format and summary statistics of variables in the dataset.

3. Do you think it might be useful for you? Yes, the information about complete data, unique values, min/max and the compact, readable format of the summary statistics is useful.

Exercise 2.

Are people traveling on a whim? Let’s see…

Fill in the blanks for filtering for hotel bookings where the guest is not from the US (country code "USA") and the lead_time is less than 1 day.

Note: You will need to set eval=TRUE when you have an answer you want to try out.

hotels %>%
  filter(
    country != "USA", 
    lead_time < 1
    )

## # A tibble: 6,174 x 32
##    hotel is_canceled lead_time arrival_date_ye~ arrival_date_mo~
##    <chr>       <dbl>     <dbl>            <dbl> <chr>           
##  1 Reso~           0         0             2015 July            
##  2 Reso~           0         0             2015 July            
##  3 Reso~           0         0             2015 July            
##  4 Reso~           0         0             2015 July            
##  5 Reso~           0         0             2015 July            
##  6 Reso~           0         0             2015 July            
##  7 Reso~           0         0             2015 July            
##  8 Reso~           0         0             2015 July            
##  9 Reso~           0         0             2015 July            
## 10 Reso~           0         0             2015 July            
## # ... with 6,164 more rows, and 27 more variables:
## #   arrival_date_week_number <dbl>, arrival_date_day_of_month <dbl>,
## #   stays_in_weekend_nights <dbl>, stays_in_week_nights <dbl>, adults <dbl>,
## #   children <dbl>, babies <dbl>, meal <chr>, country <chr>,
## #   market_segment <chr>, distribution_channel <chr>, is_repeated_guest <dbl>,
## #   previous_cancellations <dbl>, previous_bookings_not_canceled <dbl>,
## #   reserved_room_type <chr>, assigned_room_type <chr>, booking_changes <dbl>,
## #   deposit_type <chr>, agent <chr>, company <chr>, days_in_waiting_list <dbl>,
## #   customer_type <chr>, adr <dbl>, required_car_parking_spaces <dbl>,
## #   total_of_special_requests <dbl>, reservation_status <chr>,
## #   reservation_status_date <date>

Exercise 3.

How many bookings involve at least 1 child or baby? (Note the logical op, PC)

In the following chunk, replace

• [AT LEAST] with the logical operator for “at least” (in two places)
• [OR] with the logical operator for “or”

Note: You will need to set eval=TRUE when you have an answer you want to try out.

hotels %>%
  filter(
    children >= 1 | babies >= 1
    )

## # A tibble: 9,332 x 32
##    hotel is_canceled lead_time arrival_date_ye~ arrival_date_mo~
##    <chr>       <dbl>     <dbl>            <dbl> <chr>           
##  1 Reso~           0        18             2015 July            
##  2 Reso~           1        47             2015 July            
##  3 Reso~           0         1             2015 July            
##  4 Reso~           0        10             2015 July            
##  5 Reso~           1        79             2015 July            
##  6 Reso~           0       101             2015 July            
##  7 Reso~           0        92             2015 July            
##  8 Reso~           1        26             2015 July            
##  9 Reso~           0       102             2015 July            
## 10 Reso~           0        78             2015 July            
## # ... with 9,322 more rows, and 27 more variables:
## #   arrival_date_week_number <dbl>, arrival_date_day_of_month <dbl>,
## #   stays_in_weekend_nights <dbl>, stays_in_week_nights <dbl>, adults <dbl>,
## #   children <dbl>, babies <dbl>, meal <chr>, country <chr>,
## #   market_segment <chr>, distribution_channel <chr>, is_repeated_guest <dbl>,
## #   previous_cancellations <dbl>, previous_bookings_not_canceled <dbl>,
## #   reserved_room_type <chr>, assigned_room_type <chr>, booking_changes <dbl>,
## #   deposit_type <chr>, agent <chr>, company <chr>, days_in_waiting_list <dbl>,
## #   customer_type <chr>, adr <dbl>, required_car_parking_spaces <dbl>,
## #   total_of_special_requests <dbl>, reservation_status <chr>,
## #   reservation_status_date <date>

Exercise 4.

Do you think it’s more likely to find bookings with children or babies in city hotels or resort hotels? Test your intuition. Using filter() determine the number of bookings in resort hotels that have more than 1 child or baby in the room? Then, do the same for city hotels, and compare the numbers of rows in the resulting filtered data frames.

hotels %>%
  filter(
    (children >= 1 | babies >= 1) & 
      hotel =="Resort Hotel"
  )

## # A tibble: 3,929 x 32
##    hotel is_canceled lead_time arrival_date_ye~ arrival_date_mo~
##    <chr>       <dbl>     <dbl>            <dbl> <chr>           
##  1 Reso~           0        18             2015 July            
##  2 Reso~           1        47             2015 July            
##  3 Reso~           0         1             2015 July            
##  4 Reso~           0        10             2015 July            
##  5 Reso~           1        79             2015 July            
##  6 Reso~           0       101             2015 July            
##  7 Reso~           0        92             2015 July            
##  8 Reso~           1        26             2015 July            
##  9 Reso~           0       102             2015 July            
## 10 Reso~           0        78             2015 July            
## # ... with 3,919 more rows, and 27 more variables:
## #   arrival_date_week_number <dbl>, arrival_date_day_of_month <dbl>,
## #   stays_in_weekend_nights <dbl>, stays_in_week_nights <dbl>, adults <dbl>,
## #   children <dbl>, babies <dbl>, meal <chr>, country <chr>,
## #   market_segment <chr>, distribution_channel <chr>, is_repeated_guest <dbl>,
## #   previous_cancellations <dbl>, previous_bookings_not_canceled <dbl>,
## #   reserved_room_type <chr>, assigned_room_type <chr>, booking_changes <dbl>,
## #   deposit_type <chr>, agent <chr>, company <chr>, days_in_waiting_list <dbl>,
## #   customer_type <chr>, adr <dbl>, required_car_parking_spaces <dbl>,
## #   total_of_special_requests <dbl>, reservation_status <chr>,
## #   reservation_status_date <date>

hotels %>%
  filter(
    (children >= 1 | babies >= 1) & 
      hotel =="City Hotel"
  )

## # A tibble: 5,403 x 32
##    hotel is_canceled lead_time arrival_date_ye~ arrival_date_mo~
##    <chr>       <dbl>     <dbl>            <dbl> <chr>           
##  1 City~           1       100             2015 July            
##  2 City~           0        69             2015 July            
##  3 City~           0        67             2015 July            
##  4 City~           1        60             2015 July            
##  5 City~           0         0             2015 July            
##  6 City~           0         0             2015 July            
##  7 City~           0         0             2015 July            
##  8 City~           0        84             2015 July            
##  9 City~           0        74             2015 July            
## 10 City~           0        75             2015 July            
## # ... with 5,393 more rows, and 27 more variables:
## #   arrival_date_week_number <dbl>, arrival_date_day_of_month <dbl>,
## #   stays_in_weekend_nights <dbl>, stays_in_week_nights <dbl>, adults <dbl>,
## #   children <dbl>, babies <dbl>, meal <chr>, country <chr>,
## #   market_segment <chr>, distribution_channel <chr>, is_repeated_guest <dbl>,
## #   previous_cancellations <dbl>, previous_bookings_not_canceled <dbl>,
## #   reserved_room_type <chr>, assigned_room_type <chr>, booking_changes <dbl>,
## #   deposit_type <chr>, agent <chr>, company <chr>, days_in_waiting_list <dbl>,
## #   customer_type <chr>, adr <dbl>, required_car_parking_spaces <dbl>,
## #   total_of_special_requests <dbl>, reservation_status <chr>,
## #   reservation_status_date <date>

Exercise 5.

Create a frequency table of the number of adults in a booking. Display the results in descending order so the most common observation is on top. What is the most common number of adults in bookings in this dataset? Are there any surprising results?

Note: Don’t forget to label your R chunk as well (where it says label-me-1). Your label should be short, informative, and shouldn’t include spaces. It also shouldn’t repeat a previous label, otherwise R Markdown will give you an error about repeated R chunk labels.

hotels %>%
  count(adults) %>%
  arrange(desc(n))

## # A tibble: 14 x 2
##    adults     n
##     <dbl> <int>
##  1      2 89680
##  2      1 23027
##  3      3  6202
##  4      0   403
##  5      4    62
##  6     26     5
##  7      5     2
##  8     20     2
##  9     27     2
## 10      6     1
## 11     10     1
## 12     40     1
## 13     50     1
## 14     55     1

Bookings for two adults is most common. There are 403 bookings with zero adults, these might be data errors, or bookings for adjoining rooms where children are in one room and adults in another. There are bookings with large number of adults (>10).

Exercise 6.

Repeat Exercise 5, once for canceled bookings (is_canceled coded as 1) and once for not canceled bookings (is_canceled coded as 0). What does this reveal about the surprising results you spotted in the previous exercise?

Note: Don’t forget to label your R chunk as well (where it says label-me-2).

hotels %>%
  filter(is_canceled == 1) %>%
  count(adults) %>%
  arrange(desc(n))

## # A tibble: 14 x 2
##    adults     n
##     <dbl> <int>
##  1      2 35258
##  2      1  6674
##  3      3  2151
##  4      0   109
##  5      4    16
##  6     26     5
##  7      5     2
##  8     20     2
##  9     27     2
## 10      6     1
## 11     10     1
## 12     40     1
## 13     50     1
## 14     55     1

hotels %>%
  filter(is_canceled == 0) %>%
  count(adults) %>%
  arrange(desc(n))

## # A tibble: 5 x 2
##   adults     n
##    <dbl> <int>
## 1      2 54422
## 2      1 16353
## 3      3  4051
## 4      0   294
## 5      4    46

All of the bookings with large number of adults are canceled

Exercise 7.

Calculate minimum, mean, median, and maximum average daily rate (adr) grouped by hotel type so that you can get these statistics separately for resort and city hotels. Which type of hotel is higher, on average?

by_hotel <- hotels %>% group_by(hotel)
by_hotel %>% summarise(
  min = min(adr),
  median = median(adr),
  max = max(adr)
)

## # A tibble: 2 x 4
##   hotel          min median   max
## * <chr>        <dbl>  <dbl> <dbl>
## 1 City Hotel    0      99.9  5400
## 2 Resort Hotel -6.38   75     508

Exercise 8.

We observe two unusual values in the summary statistics above – a negative minimum, and a very high maximum). What types of hotels are these? Locate these observations in the dataset and find out the arrival date (year and month) as well as how many people (adults, children, and babies) stayed in the room. You can investigate the data in the viewer to locate these values, but preferably you should identify them in a reproducible way with some code.

Hint: For example, you can filter for the given adr amounts and select the relevant columns.

hotels %>%
  filter(adr < 0.0 | adr > 5000.0) %>%
  select("hotel", "adr", "arrival_date_year", "arrival_date_month", "adults", "children", "babies")

## # A tibble: 2 x 7
##   hotel           adr arrival_date_year arrival_date_mon~ adults children babies
##   <chr>         <dbl>             <dbl> <chr>              <dbl>    <dbl>  <dbl>
## 1 Resort Hot~   -6.38              2017 March                  2        0      0
## 2 City Hotel  5400                 2016 March                  2        0      0

Data dictionary

Below is the full data dictionary. Note that it is long (there are lots of variables in the data), but we will be using a limited set of the variables for our analysis.