Problem set 2

Author

Put your name here

Getting started

library(tidyverse)
library(janitor)

Read the data

Rows: 195 Columns: 9
── Column specification ────────────────────────────────────────────────────────
Delimiter: ","
chr (1): Breed
dbl (8): 2013 Rank, 2014 Rank, 2015 Rank, 2016 Rank, 2017 Rank, 2018 Rank, 2...

ℹ Use `spec()` to retrieve the full column specification for this data.
ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.
Rows: 195 Columns: 17
── Column specification ────────────────────────────────────────────────────────
Delimiter: ","
chr  (3): Breed, Coat Type, Coat Length
dbl (14): Affectionate With Family, Good With Young Children, Good With Othe...

ℹ Use `spec()` to retrieve the full column specification for this data.
ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.
breed_rank <- read_csv("data/breed_rank.csv")
breed_traits <- read_csv("data/breed_traits.csv")

Clean the data

Display variables.

names(breed_rank)
[1] "Breed"     "2013 Rank" "2014 Rank" "2015 Rank" "2016 Rank" "2017 Rank"
[7] "2018 Rank" "2019 Rank" "2020 Rank"
names(breed_traits)
 [1] "Breed"                      "Affectionate With Family"  
 [3] "Good With Young Children"   "Good With Other Dogs"      
 [5] "Shedding Level"             "Coat Grooming Frequency"   
 [7] "Drooling Level"             "Coat Type"                 
 [9] "Coat Length"                "Openness To Strangers"     
[11] "Playfulness Level"          "Watchdog/Protective Nature"
[13] "Adaptability Level"         "Trainability Level"        
[15] "Energy Level"               "Barking Level"             
[17] "Mental Stimulation Needs"

Make better names.

breed_traits <- breed_traits |> 
  clean_names()

Manipulate the data using dplyr

Maka a summary.

breed_traits |> 
  group_by(shedding_level) |> 
  summarise(n = n())
# A tibble: 6 × 2
  shedding_level     n
           <dbl> <int>
1              0     1
2              1    27
3              2    41
4              3   109
5              4    16
6              5     1

Filter the shedding_level 0.

breed_traits <- breed_traits |> 
  filter(shedding_level != 0)

Check if manipulation was successful.

breed_traits |> count(shedding_level)
# A tibble: 5 × 2
  shedding_level     n
           <dbl> <int>
1              1    27
2              2    41
3              3   109
4              4    16
5              5     1

Make an untidy data frame.

untidy_scores <- breed_traits |> 
  mutate(untidy_score = shedding_level + 
           coat_grooming_frequency + drooling_level) |> 
  select(breed, untidy_score)

Arrange scores in descending order.

untidy_scores |> 
  arrange(desc(untidy_score))
# A tibble: 194 × 2
   breed                  untidy_score
   <chr>                         <dbl>
 1 Bernese Mountain Dogs            11
 2 Leonbergers                      11
 3 Newfoundlands                    10
 4 Bloodhounds                      10
 5 St. Bernards                     10
 6 Old English Sheepdogs            10
 7 Dogues de Bordeaux               10
 8 Neapolitan Mastiffs              10
 9 Black Russian Terriers           10
10 Tibetan Mastiffs                 10
# ℹ 184 more rows

Tidying the data

How does this this data set fail to meet the criteria for tidy data?

There are three interrelated rules which make a dataset tidy:

  1. Each variable must have its own column.
  2. Each observation must have its own row.
  3. Each value must have its own cell.

We have a year and a rank variable, but neither of these variables have their own column. Shown above is one observation, by dog breed. But that “one” observation is actually eight separate observations: the rank in 2013, the rank in 2014, etc. Each observation needs to have its own row.

Make pivoted data with a year and a rank variable.

ranks_pivoted <- breed_rank |> 
  pivot_longer(`2013 Rank`:`2020 Rank`,
               names_to = "year",
               values_to = "rank")

Rename breed and make the year variable numeric.

ranks_pivoted <- ranks_pivoted |> 
  rename(breed = Breed) |> 
  mutate(year = parse_number(year))

Filter data to only Bernese Mountain Dogs.

ranks_pivoted <- ranks_pivoted |> 
  filter(str_detect(breed, "Bernese"))

Plot rankings across time.

ranks_pivoted |>
  ggplot(aes(x = year, y = rank, label = rank)) +
  geom_point(size = 3) +
  geom_text(vjust = 2)