Exercise 2_2_1: Across the Tidyverse

As always, first some prep work before we can work on the exercises: Load the packages & the data we need…

library(tidyverse)
library(sjlabelled)

gp_covid <- 
  read_csv2("./data/ZA5667_v1-1-0.csv") %>%
  set_na(na = c(-99, -77, -33, 98))

1

Use across() to recode the trust variables hzcy044a:hzcy052a into dichotomized versions. The value 1 should remain, all others should be 0.

Clues

You have to wrap it into the mutate() function and then use recode().

solution

gp_covid <- 
  gp_covid %>% 
  mutate(
    across(
      hzcy044a:hzcy052a,
      ~recode(
        .x,
        `5` = 0,
        `4` = 0,
        `3` = 0,
        `2` = 0,
        `1` = 1
      )
    )
  )

2

Using short pipe and dplyr functions, let’s check if the code worked.

Clues

You can combine end your pipe with glimpse() to print the output or use View() to a data tab in RStudio.

solution

gp_covid %>% 
  select(hzcy044a:hzcy052a) %>% 
  glimpse()

## Rows: 3,765
## Columns: 9
## $ hzcy044a <dbl> NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, NA, NA, 0, NA, 0, NA, 0, 0, NA, 0, 0, 0, 0, 0, NA, NA, 0, NA, 0,~
## $ hzcy045a <dbl> NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, NA, NA, 0, NA, 0, NA, 0, 0, NA, 0, NA, NA, 0, 0, NA, NA, 0, NA, ~
## $ hzcy046a <dbl> NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, NA, NA, 0, NA, 0, NA, 0, 0, NA, 0, 0, 0, 0, 0, NA, NA, 0, NA, 0,~
## $ hzcy047a <dbl> NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, NA, NA, 0, NA, 0, NA, 0, 0, 0, 0, 0, 0, 0, 0, NA, NA, 0, NA, 0, ~
## $ hzcy048a <dbl> NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, 1, NA, 0, 0, NA, NA, 0, NA, 0, NA, 0, 0, 0, 0, 0, 0, 0, 0, NA, NA, 0, NA, 0, ~
## $ hzcy049a <dbl> NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 1, 0, NA, 0, 0, NA, NA, 0, NA, NA, NA, 0, 0, 0, 0, 0, 0, 0, 0, NA, NA, 0, NA, 1,~
## $ hzcy050a <dbl> NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, 1, NA, 0, 0, NA, NA, 0, NA, 0, NA, 0, 0, 0, 0, 0, 0, 0, 0, NA, NA, 0, NA, 0, ~
## $ hzcy051a <dbl> NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, NA, NA, 0, NA, 0, NA, 0, 0, 0, 0, 0, 0, 0, 0, NA, NA, 0, NA, 0, ~
## $ hzcy052a <dbl> NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, NA, NA, 0, NA, 0, NA, 0, 0, 0, 0, 0, 0, 0, 0, NA, NA, 0, NA, 0, ~

3

Use the median() function to calculate an aggregated variable of all trust variables per respondent.

Clues

For this task you need to first need to change the data into a row-wise format. For creating the aggregate variable you need the c_across() function. Oh, and don’t forget to ungroup your data at the end (of your pipe).

solution

gp_covid <- 
  gp_covid %>% 
  rowwise() %>% #<<
  mutate(
    median_trust = 
      median(
        c_across(hzcy044a:hzcy052a),
        na.rm = TRUE
      )
  ) %>% 
  ungroup()

table(gp_covid$median_trust)

## 
##    0  0.5    1 
## 3115    3   42