Challenge 3 Instructions

challenge_3
Author

Yakub Rabiutheen

Published

August 17, 2022

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 organicpoultry.xls ⭐⭐⭐
  • australian_marriage*.xlsx ⭐⭐⭐
  • USA Households*.xlsx ⭐⭐⭐⭐
  • sce_labor_chart_data_public.csv 🌟🌟🌟🌟🌟
Code 
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”

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!

Example: find current and future data dimensions

Lets see if this works with a simple example.

Code 
head(animal_weight)
# A tibble: 6 × 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
# … 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`
# ℹ Use `colnames()` to see all variable names

.

Because all of the Data can be grouped in one Livestock column, we only need one Column for Live

Code 
animal_pivot<-pivot_longer(animal_weight, 
                                    col=-`IPCC Area`,
                                    names_to = "Livestock",
                                    values_to = "Weight")
animal_pivot
# 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
# ℹ Use `print(n = ...)` to see more rows