Function to Apply Factor Levels & Labels to Numerous Columns at Once - r

Question

I have a survey dataset that I imported as an SAS file but it did not include the text labels that are associated with the numeric codes in the dataset.
I'm trying to apply the factor function to all variables and then have the respective levels and labels for each variable.
I have a main dataframe with the actual data, and then a second dataframe with the text labels corresponding to each value for each variable.
So, for example, the variable column names in the main dataset are A1, B1, C1, D1. The second dataframe with the labels is listed below with dummy text. And for each variable, there are varying numbers of values that need text labels.
labels_list <- structure(list(VariableName = c("A1", "A1", "A1", "B1", "B1",
"B1", "B1", "C1", "C1", "C1", "C1", "C1", "D1", "D1", "D1", "D1",
"D1", "D1"), Value = c(1L, 2L, 3L, 1L, 2L, 3L, 4L, 1L, 2L, 3L,
4L, 5L, 1L, 2L, 3L, 4L, 5L, 6L), Label = c("Red", "Blue", "Yellow",
"Up", "Down", "Left", "Right", "Boston", "Atlanta", "Dallas",
"New York", "Los Angeles", "John", "Jim", "Jake", "Bill", "Bob",
"Brian")), class = "data.frame", row.names = c(NA, -18L))
I'm trying to write a function to automatically label all the factor variables. The function reduces down the data to make sure that they each contain the exact same variables and then are in the exact same order. I split the table above into a list using the split function, and then each variable name above has it's own list, but I'm encountering an error when I try to subset the list in the for loop.
Below is the for loop I have written.
df = main dataset
labels_list = list with the value and text labels
for(i in 1:ncol(df)) {
for(j in labels_list) {
if(names(x[,i]) == names(ahs_split[[j]])) {
x[,i] <- factor(x[,i], levels = c(ahs_split[[j]][[2]]), labels = c(ahs_split[[j]][[3]]))
As I mentioned, my ultimate goal is to take this dataframe with the text labels and corresponding values for each variable and apply it to each one individually using the factor function. I've tried for almost a month now and am just very stuck so I could use any help. I'm not sure if anyone could possibly recommend a better approach or point me in the right direction. I would greatly appreciate any help.

Answer 1

If you don't mind some tidyverse verbs, you can reshape your data with tidyr::gather. Once it's in a long shape, you can join the data with the code lookup by variable name, and reshape it back into a wide format. This workflow scales for however many columns you need.
library(dplyr)
library(tidyr)
labels_list <- structure(list(Variable = structure(c(1L, 1L, 1L, 2L, 2L, 2L,
2L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L), .Label = c("A1",
"B1", "C1", "D1"), class = "factor"), Value = c(1L, 2L, 3L, 1L,
2L, 3L, 4L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 6L), Label = structure(c(15L,
3L, 18L, 17L, 8L, 12L, 16L, 5L, 1L, 7L, 14L, 13L, 11L, 10L, 9L,
2L, 4L, 6L), .Label = c("Atlanta", "Bill", "Blue", "Bob", "Boston",
"Brian", "Dallas", "Down", "Jake", "Jim", "John", "Left", "Los_Angeles",
"New_York", "Red", "Right", "Up", "Yellow"), class = "factor")), class = "data.frame", row.names = c(NA,
-18L))
df <- tibble(A1 = rep(1:3,2),
B1 = c(1:4, 1, 2),
C1 = c(1:5, 1),
D1 = 1:6
)
A row number iterated over Variable will be necessary to spread the data, but you can drop it after it's no longer needed.
df %>%
gather(key = Variable, value = Value) %>%
left_join(labels_list, by = c("Variable", "Value")) %>%
select(-Value) %>%
group_by(Variable) %>%
mutate(row = row_number()) %>%
spread(key = Variable, value = Label)
#> Warning: Column `Variable` joining character vector and factor, coercing
#> into character vector
#> # A tibble: 6 x 5
#> row A1 B1 C1 D1
#> <int> <fct> <fct> <fct> <fct>
#> 1 1 Red Up Boston John
#> 2 2 Blue Down Atlanta Jim
#> 3 3 Yellow Left Dallas Jake
#> 4 4 Red Right New_York Bill
#> 5 5 Blue Up Los_Angeles Bob
#> 6 6 Yellow Down Boston Brian

Answer 2

One way is to convert your labels_list into a list of lists:
library(dplyr) # just using dplyr for the pipe %>%, otherwise everything is in base R
# Convert df to list of key:value pairs
labels_list <- labels_list %>%
split(f = labels_list$VariableName) %>%
lapply(function(x) list(key = x$Value, value = x$Label))
e.g.:
$A1
$A1$key
[1] 1 2 3
$A1$value
[1] "Red" "Blue" "Yellow"
This can be mapped onto your df col-wise with apply. This is a bit hacky as I put the column name as the first item of the vector passed to the function.
# Map labels onto sample data with factor()
apply(rbind(names(df), df),
2,
function(x) factor(x[2:length(x)],
levels = labels_list[[x[1]]]$key,
labels = labels_list[[x[1]]]$value)) %>%
as.data.frame()
A1 B1 C1 D1
1 Blue Up Dallas Jake
2 Red Down New York Jake
3 Yellow Left Boston Jim
4 Yellow Right Boston John
5 Yellow Down Los Angeles Jake
6 Red Left Atlanta Jake
7 Blue Down New York John
8 Red Down Atlanta Brian
9 Blue Up New York Jim
10 Yellow Down Atlanta Bill
Sample Data
set.seed(1724)
df <- data.frame(A1 = floor(runif(10, 1, 4)),
B1 = floor(runif(10, 1, 5)),
C1 = floor(runif(10, 1, 6)),
D1 = floor(runif(10, 1, 7)))