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data-question-4-skirmish-guards-1's Introduction

Scenario

You are a team of data journalists at a major online news organization. The news room staff has taken an interest in Tennessee, and has asked your team to develop a picture of Tennesseans and how they differ across (and within) counties, particularly in regard to socioeconomic factors like education and income.

Your assignment is to learn more about whether educational factors are related to income for Tennesseans.

Steps

  1. Read, clean, and join the IRS, Education, and Zip Code data.

    • Use tidyverse "verbs"

    mutate() adds new variables that are functions of existing variables
    select() picks variables based on their names.
    filter() picks cases based on their values.
    summarize() reduces multiple values down to a single summary.
    arrange() changes the ordering of the rows.

    • Use purrr when applicable
  #map
     map_chr(c(5, 4, 3, 2, 1), function(x){
       c("one", "two", "three", "four", "five")[x]
       })
  [1] "five" "four" "three" "two" "one"

  map_lgl(c(1, 2, 3, 4, 5), function(x){
    x > 3
  })
  [1] FALSE FALSE FALSE TRUE TRUE

  map_if(1:5, function(x){
    x %% 2 == 0
    },
    function(y){
    y^2
    }) %>% unlist()
  [1] 1 4 3 16 5

  map_at(seq(100, 500, 100), c(1, 3, 5), function(x){
    x - 10
  }) %>% unlist()
  [1] 90 200 290 400 490

  map2_chr(letters, 1:26, paste)
   [1] "a 1" "b 2" "c 3" "d 4" "e 5" "f 6" "g 7" "h 8" "i 9" "j 10"
  [11] "k 11" "l 12" "m 13" "n 14" "o 15" "p 16" "q 17" "r 18" "s 19" "t 20"
  [21] "u 21" "v 22" "w 23" "x 24" "y 25" "z 26"

  pmap_chr(list(
    list(1, 2, 3),
    list("one", "two", "three"),
    list("uno", "dos", "tres")
  ), paste)
  [1] "1 one uno" "2 two dos" "3 three tres"

  # reduce

  reduce(c(1, 3, 5, 7), function(x, y){
    message("x is ", x)
    message("y is ", y)
    message("")
    x + y
  })
  x is 1
  y is 3

  x is 4
  y is 5

  x is 9
  y is 7

  [1] 16

  reduce(letters[1:4], function(x, y){
    message("x is ", x)
    message("y is ", y)
    message("")
    paste0(x, y)
  })
  x is a
  y is b

  x is ab
  y is c

  x is abc
  y is d

  [1] "abcd"

  reduce_right(letters[1:4], function(x, y){
    message("x is ", x)
    message("y is ", y)
    message("")
    paste0(x, y)
  })
  x is d
  y is c

  x is dc
  y is b

  x is dcb
  y is a

  [1] "dcba"

  # search

  contains(letters, "a")
  [1] TRUE
  contains(letters, "A")
  [1] FALSE

  detect(20:40, function(x){
    x > 22 && x %% 2 == 0
  })
  [1] 24

  detect_index(20:40, function(x){
    x > 22 && x %% 2 == 0
  })
  [1] 5

  # filter

  keep(1:20, function(x){
    x %% 2 == 0
  })
   [1] 2 4 6 8 10 12 14 16 18 20

  discard(1:20, function(x){
    x %% 2 == 0
  })
   [1] 1 3 5 7 9 11 13 15 17 19

  every(1:20, function(x){
    x %% 2 == 0
  })
  [1] FALSE
  some(1:20, function(x){
    x %% 2 == 0
  })
  [1] TRUE

  # compose()

  n_unique <- compose(length, unique)
  # The composition above is the same as:
  # n_unique <- function(x){
  # length(unique(x))
  # }

  rep(1:5, 1:5)
   [1] 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5

  n_unique(rep(1:5, 1:5))
  [1] 5
  ```



2. **Do some exploratory data analysis.**
  - Use ggplot2 for visualizations
  - Use the cor() function to look for correlations
  - Use comments throughout to articulate the thinking that has informed your choices

3. **Use your exploratory data analysis as the foundation to fit a linear regression model.**

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