Hotel bookings - data wrangling

Mine Çetinkaya-Rundel

library(tidyverse)
library(skimr)

# 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 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: 242 × 32
##    hotel        is_canceled lead_time arrival_date_year arrival_date_month
##    <chr>              <dbl>     <dbl>             <dbl> <chr>             
##  1 Resort Hotel           0         0              2015 August            
##  2 Resort Hotel           0         0              2015 July              
##  3 Resort Hotel           0         0              2015 August            
##  4 Resort Hotel           0         0              2015 August            
##  5 Resort Hotel           0         0              2015 September         
##  6 Resort Hotel           0         1              2015 October           
##  7 Resort Hotel           0         0              2015 October           
##  8 Resort Hotel           0         0              2015 December          
##  9 Resort Hotel           0         0              2015 December          
## 10 Resort Hotel           0         0              2016 January           
## # … with 232 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>, …

Exercise 3.

How many bookings involve at least 1 child or baby?

9332

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,    
    ) %>%
  select(babies, children)

Exercise 4.

Do you think it’s more likely to find bookings with children or babies in city hotels or resort hotels?

I would say resort because people are more likely to travel with their families to resorts. I feel like cities are more for business.

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. Resort hotels have 3929. City hotels have 5403. So I was wrong. There are more bookings at city hotels then resort. I guess that makes sense because city hotels are probably more affordable for a family.

hotels %>%
  mutate(kids = children + babies) %>%
  filter(
    kids >= 1,
    hotel == "Resort Hotel"
    ) %>%
  select(hotel, kids)

## # A tibble: 3,929 × 2
##    hotel         kids
##    <chr>        <dbl>
##  1 Resort Hotel     1
##  2 Resort Hotel     2
##  3 Resort Hotel     2
##  4 Resort Hotel     2
##  5 Resort Hotel     1
##  6 Resort Hotel     1
##  7 Resort Hotel     2
##  8 Resort Hotel     2
##  9 Resort Hotel     1
## 10 Resort Hotel     1
## # … with 3,919 more rows

hotels %>%
  mutate(kids = children + babies) %>%
  filter(
    kids >= 1,
    hotel == "City Hotel"
    ) %>%
  select(hotel, kids)

## # A tibble: 5,403 × 2
##    hotel       kids
##    <chr>      <dbl>
##  1 City Hotel     1
##  2 City Hotel     1
##  3 City Hotel     2
##  4 City Hotel     1
##  5 City Hotel     1
##  6 City Hotel     1
##  7 City Hotel     1
##  8 City Hotel     1
##  9 City Hotel     1
## 10 City Hotel     1
## # … with 5,393 more rows

add code here # pay attention to correctness and code style ### 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?

The most common amount is 2. This is not surprising to me because most families have 2 parents. And below that is 1, which a lot of families just have 1 parent also. I would say it is surprising that there was quite a few rooms with 0 parents.

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.

# descending frequency order
hotels %>%
  count(adults, sort = TRUE)

## # A tibble: 14 × 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

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?

This shows that 75166 are cancelled and that 44224 are not cancelled.

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

# descending frequency order
hotels %>%
  count(is_canceled == 1, sort = TRUE)

## # A tibble: 2 × 2
##   `is_canceled == 1`     n
##   <lgl>              <int>
## 1 FALSE              75166
## 2 TRUE               44224

# descending frequency order
hotels %>%
  count(is_canceled == 0, sort = TRUE)

## # A tibble: 2 × 2
##   `is_canceled == 0`     n
##   <lgl>              <int>
## 1 TRUE               75166
## 2 FALSE              44224

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?

The city hotel is higher on average.

hotels %>%
  group_by(hotel) %>%
  summarise(
    min_adr = min(adr),
    mean_adr = mean(adr),
    median_adr = median(adr),
    max_adr = max(adr)
    )

## # A tibble: 2 × 5
##   hotel        min_adr mean_adr median_adr max_adr
##   <chr>          <dbl>    <dbl>      <dbl>   <dbl>
## 1 City Hotel      0       105.        99.9    5400
## 2 Resort Hotel   -6.38     95.0       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?

I would say these may be seasonal hotels. The city hotel has 0 for the min adr.

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 == '5400')

## # A tibble: 1 × 32
##   hotel is_canceled lead_time arrival_date_ye… arrival_date_mo… arrival_date_we…
##   <chr>       <dbl>     <dbl>            <dbl> <chr>                       <dbl>
## 1 City…           1        35             2016 March                          13
## # … with 26 more variables: 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>, …

hotels %>%
  filter(adr == '-6.38')

## # A tibble: 1 × 32
##   hotel is_canceled lead_time arrival_date_ye… arrival_date_mo… arrival_date_we…
##   <chr>       <dbl>     <dbl>            <dbl> <chr>                       <dbl>
## 1 Reso…           0       195             2017 March                          10
## # … with 26 more variables: 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>, …

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.