Challenge 3

Code

library(tidyverse)

knitr::opts_chunk$set(echo = TRUE, warning=FALSE, message=FALSE)

Challenge Overview

Today’s challenge is to:

1. read in a data set, and describe the data set using both words and any supporting information (e.g., tables, etc)
2. identify what needs to be done to tidy the current data
3. anticipate the shape of pivoted data
4. pivot the data into tidy format using pivot_longer

Read in data

Read in one (or more) of the following datasets, using the correct R package and command.

• animal_weights.csv ⭐
• eggs_tidy.csv ⭐⭐ or organiceggpoultry.xls ⭐⭐⭐
• australian_marriage*.xls ⭐⭐⭐
• USA Households*.xlsx ⭐⭐⭐⭐
• sce_labor_chart_data_public.xlsx 🌟🌟🌟🌟🌟

Code

library(readr)
animal_weight <- read_csv("_data/animal_weight.csv")

Briefly describe the data

Describe the data, and be sure to comment on why you are planning to pivot it to make it “tidy”

It seems like there are 9 rows of data all based on what region we are looking at livestock in. All the other columns except for the first one all contain different types of animals and their respective weights. Instead of having different columns for each animal it would make more sense for the columns to be: region, type of animal and weight of animal.

Anticipate the End Result

The first step in pivoting the data is to try to come up with a concrete vision of what the end product should look like - that way you will know whether or not your pivoting was successful.

One easy way to do this is to think about the dimensions of your current data (tibble, dataframe, or matrix), and then calculate what the dimensions of the pivoted data should be.

Suppose you have a dataset with \(n\) rows and \(k\) variables. In our example, 3 of the variables are used to identify a case, so you will be pivoting \(k-3\) variables into a longer format where the \(k-3\) variable names will move into the names_to variable and the current values in each of those columns will move into the values_to variable. Therefore, we would expect \(n * (k-3)\) rows in the pivoted dataframe!

In the current dataset we have n = 9 rows of k = 17 variables. Therefore in our new pivoted dataframe, we expect 9 *(17-3) rows which comes out to 144 rows by 3 columns. ### Example: find current and future data dimensions

Lets see if this works with a simple example.

Code

df<-tibble(country = rep(c("Mexico", "USA", "France"),2),
           year = rep(c(1980,1990), 3), 
           trade = rep(c("NAFTA", "NAFTA", "EU"),2),
           outgoing = rnorm(6, mean=1000, sd=500),
           incoming = rlogis(6, location=1000, 
                             scale = 400))
df

# A tibble: 6 × 5
  country  year trade outgoing incoming
  <chr>   <dbl> <chr>    <dbl>    <dbl>
1 Mexico   1980 NAFTA    1437.    1515.
2 USA      1990 NAFTA    1265.    1225.
3 France   1980 EU        789.    -427.
4 Mexico   1990 NAFTA     898.     356.
5 USA      1980 NAFTA     568.     763.
6 France   1990 EU        485.    1153.

Code

#existing rows/cases
nrow(df)

[1] 6

Code

#existing columns/cases
ncol(df)

[1] 5

Code

#expected rows/cases
nrow(df) * (ncol(df)-3)

[1] 12

Code

# expected columns 
3 + 2

[1] 5

Or simple example has \(n = 6\) rows and \(k - 3 = 2\) variables being pivoted, so we expect a new dataframe to have \(n * 2 = 12\) rows x \(3 + 2 = 5\) columns.

Challenge: Describe the final dimensions

Document your work here.

Code

animal_weight <- tibble(animal_weight)
animal_weight

# A tibble: 9 × 17
  IPCC A…¹ Cattl…² Cattl…³ Buffa…⁴ Swine…⁵ Swine…⁶ Chick…⁷ Chick…⁸ Ducks Turkeys
  <chr>      <dbl>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl> <dbl>   <dbl>
1 Indian …     275     110     295      28      28     0.9     1.8   2.7     6.8
2 Eastern…     550     391     380      50     180     0.9     1.8   2.7     6.8
3 Africa       275     173     380      28      28     0.9     1.8   2.7     6.8
4 Oceania      500     330     380      45     180     0.9     1.8   2.7     6.8
5 Western…     600     420     380      50     198     0.9     1.8   2.7     6.8
6 Latin A…     400     305     380      28      28     0.9     1.8   2.7     6.8
7 Asia         350     391     380      50     180     0.9     1.8   2.7     6.8
8 Middle …     275     173     380      28      28     0.9     1.8   2.7     6.8
9 Norther…     604     389     380      46     198     0.9     1.8   2.7     6.8
# … with 7 more variables: Sheep <dbl>, Goats <dbl>, Horses <dbl>, Asses <dbl>,
#   Mules <dbl>, Camels <dbl>, Llamas <dbl>, and abbreviated variable names
#   ¹​`IPCC Area`, ²​`Cattle - dairy`, ³​`Cattle - non-dairy`, ⁴​Buffaloes,
#   ⁵​`Swine - market`, ⁶​`Swine - breeding`, ⁷​`Chicken - Broilers`,
#   ⁸​`Chicken - Layers`

Code

nrow(animal_weight)

[1] 9

Code

ncol(animal_weight)

[1] 17

Code

nrow(animal_weight) * (ncol(animal_weight)-1)

[1] 144

Any additional comments?

Challenge: Pivot the Chosen Data

Document your work here. What will a new “case” be once you have pivoted the data? How does it meet requirements for tidy data?

Code

animal_weight<-pivot_longer(animal_weight, 
                 col =- `IPCC Area`,
                 names_to="Livestock",
                 values_to = "Weight")
animal_weight

# A tibble: 144 × 3
   `IPCC Area`         Livestock          Weight
   <chr>               <chr>               <dbl>
 1 Indian Subcontinent Cattle - dairy      275  
 2 Indian Subcontinent Cattle - non-dairy  110  
 3 Indian Subcontinent Buffaloes           295  
 4 Indian Subcontinent Swine - market       28  
 5 Indian Subcontinent Swine - breeding     28  
 6 Indian Subcontinent Chicken - Broilers    0.9
 7 Indian Subcontinent Chicken - Layers      1.8
 8 Indian Subcontinent Ducks                 2.7
 9 Indian Subcontinent Turkeys               6.8
10 Indian Subcontinent Sheep                28  
# … with 134 more rows

Pivot the Data

Now we will pivot the data, and compare our pivoted data dimensions to the dimensions calculated above as a “sanity” check.

Yes, once it is pivoted long, our resulting data are \(144x3\) - exactly what we expected!