Challenge 3

challenge_3

animal_weights

Author

Emma Rasmussen

Published

August 17, 2022

Code

library(tidyverse)

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

Challenge Overview

Today’s challenge is to:

read in a data set, and describe the data set using both words and any supporting information (e.g., tables, etc)
identify what needs to be done to tidy the current data
anticipate the shape of pivoted data
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",
                        show_col_types = FALSE)
animal_weightOG<-animal_weight#saving a copy of the original data set
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`, ⁴Buffaloes,
#   ⁵`Swine - market`, ⁶`Swine - breeding`, ⁷`Chicken - Broilers`,
#   ⁸`Chicken - Layers`
# ℹ Use `colnames()` to see all variable names

Briefly describe the data

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

The data appears to illustrate (average?) animal weights by region. To tidy the data we will pivot the columns with animal names into a single column. Each “case” is an animal type within a region, and the values/dependent variable is the weight.

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

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     376.    2703.
2 USA      1990 NAFTA    1543.     628.
3 France   1980 EU       1231.     383.
4 Mexico   1990 NAFTA     669.     779.
5 USA      1980 NAFTA     887.    1696.
6 France   1990 EU        911.    1937.

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.

OG Dataset has k=17 columns and n=9 rows. 17-1(1 existing variable to describe each case (country), the other 16 columns need to be pivoted) We wll now have three columns, one region, one animalm(new col) (together the IV), one weight(the DV) (new col) 9*16 rows expected in data frame= 144 3col byt 144 rows rows expected

Code

nrow(animal_weightOG)

[1] 9

Code

ncol(animal_weightOG)

[1] 17

Code

#expected rows/cases
nrow(animal_weightOG)*(ncol(animal_weightOG)-1)

[1] 144

Code

#expected columns
1+2

[1] 3

Any additional comments? See comment at bottom

Pivot the Data

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

Example

Code

df<-pivot_longer(df, col = c(outgoing, incoming),
                 names_to="trade_direction",
                 values_to = "trade_value")
df

# A tibble: 12 × 5
   country  year trade trade_direction trade_value
   <chr>   <dbl> <chr> <chr>                 <dbl>
 1 Mexico   1980 NAFTA outgoing               376.
 2 Mexico   1980 NAFTA incoming              2703.
 3 USA      1990 NAFTA outgoing              1543.
 4 USA      1990 NAFTA incoming               628.
 5 France   1980 EU    outgoing              1231.
 6 France   1980 EU    incoming               383.
 7 Mexico   1990 NAFTA outgoing               669.
 8 Mexico   1990 NAFTA incoming               779.
 9 USA      1980 NAFTA outgoing               887.
10 USA      1980 NAFTA incoming              1696.
11 France   1990 EU    outgoing               911.
12 France   1990 EU    incoming              1937.

Yes, once it is pivoted long, our resulting data are \(12x5\) - exactly what we expected!

Challenge: Pivot the Chosen Data

A case will be an animal from a particular region. It meets the requirements for tidy data because each case has its own row, and each variable has its own column.

Code

pivot_longer(animal_weight, col = c(2:17),
                 names_to="animal_type",
                 values_to = "weight")

# A tibble: 144 × 3
   `IPCC Area`         animal_type        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

Any additional comments?

I am a little confused by the calculation, I tried to work out everything above in a way that made sense. But we essentially start with: - how many columns will remain unpivoted (variables that start in the correct place/column) -how many columns are being pivoted (the rest or starting number of col minus number above) -The number of columns being pivoted*number of rows = new number of rows -the new number of columns is the unchanged columns plus 1 for the variables contained in the pivot plus one for the values?? So unchanged col+2