What is the best way to use lapply with factor() to parse survey data?-CodePudding

I'm trying to streamline my R code with functions and lapply() loops, and have gotten stuck trying to convert a block of code which ingests survey data to a set of clustered likert style questions. Here's the original working code:

require(likert)
require(tidyverse)

# Generate a data compliments of dput
q25_data <- structure(list(Q25_self_and_family = c(4, 2, 3, 3, 5, 3), Q25_local_area = c(3, 
3, 3, 3, 5, 3), Q25_uk = c(4, 3, 3, 3, 5, 2), Q25_outside_uk = c(4, 
4, 3, 3, 5, 4)), row.names = c(NA, -6L), class = c("tbl_df", 
"tbl", "data.frame"))

# Set up levels text for question responses
q25_levels <- c("not at all serious", "somewhat serious", "moderately serious", "Somewhat Agree", "extremely serious")
q25_data$Q25_self_and_family <- factor(q25_data$Q25_self_and_family, ordered = TRUE, levels = c("1", "2", "3", "4", "5"))
q25_data$Q25_local_area <- factor(q25_data$Q25_local_area, ordered = TRUE, levels = c("1", "2", "3", "4", "5"))
q25_data$Q25_uk <- factor(q25_data$Q25_uk, ordered = TRUE, levels = c("1", "2", "3", "4", "5"))
q25_data$Q25_outside_uk <- factor(q25_data$Q25_outside_uk, ordered = TRUE, levels = c("1", "2", "3", "4", "5"))
# Change factor names to match level text
q25_data$Q25_self_and_family <- fct_recode(q25_data$Q25_self_and_family, "not at all serious" = "1", "somewhat serious" = "2", "moderately serious" = "3", "very serious" = "4", "extremely serious" = "5")
q25_data$Q25_local_area <- fct_recode(q25_data$Q25_local_area, "not at all serious" = "1", "somewhat serious" = "2", "moderately serious" = "3", "very serious" = "4", "extremely serious" = "5")
q25_data$Q25_uk <- fct_recode(q25_data$Q25_uk, "not at all serious" = "1", "somewhat serious" = "2", "moderately serious" = "3", "very serious" = "4", "extremely serious" = "5")
q25_data$Q25_outside_uk <- fct_recode(q25_data$Q25_outside_uk, "not at all serious" = "1", "somewhat serious" = "2", "moderately serious" = "3", "very serious" = "4", "extremely serious" = "5")
# Change names of rows to question text
names(q25_data) <- c("You and your family in the UK", "People in your local area or city", "The UK as a whole", "Your family and/or friends living outside the UK")
q25_likert_table <- likert(as.data.frame(q25_data))

So what I'm thinking is that I can take in the column names using q25_names <- names(select(q25_data, Q25_self_and_family:Q25_outside_uk)) and then use lapply() with a modified function like this: test <-lapply(q25_names, function(q25_column) {q25_data$q25_column <- factor(select(q25_data, q25_column), ordered = TRUE, levels = c("1", "2", "3", "4", "5"))}) however, I'm getting nowhere fast with this approach. Suspect I'm missing something obvious here, but I've been through a dozen examples on SE and still not finding a successful approach.

CodePudding user response：

You can "streamline" your workflow using pipes, e.g. like shown below. Briefly, you can use across to apply a function to every column, and use factor, which conveniently allows you to set labels to factor levels. Then just pipe the outcome to transform it into a data.frame, and then to a likert object.

require(likert)
require(dplyr)

q25_data <- tibble(
    Q25_self_and_family = c(4, 2, 3, 3, 5, 3), 
    Q25_local_area = c(3, 3, 3, 3, 5, 3), 
    Q25_uk = c(4, 3, 3, 3, 5, 2), 
    Q25_outside_uk = c(4, 4, 3, 3, 5, 4))

# Set up levels text for question responses
q25_levels <- paste(c("not at all", "somewhat", "moderately", "very", "extremely"),  
    "serious")
q25_likert_table <- q25_data %>% 
    mutate(across(everything(), 
        factor, ordered = TRUE, levels = 1:5, labels=q25_levels)) %>% 
    as.data.frame %>% 
    likert

^{Created on 2022-03-07 by the reprex package (v2.0.1)}