Grouping By Multiple Selection Answer for Likert Package - r

Question

I wanted to create a likert graph that is grouped by Question i. I can create the likert graph for total responses ungrouped, but im uncertain of how to reformat question 6 without losing the column for question i. (aka do the reformatting done below but also have it take into account who selected what in question i.)
What I want is the sufficiency of Q6 grouped by their answer in question i.
Sample Dataframe:
SurveyClean2 <- data.frame(i = c("Mail,Email", "Mail", "Mail,Email,Podcast", "Radio,Podcast", "Radio", "Mail,Radio"), Q6_3 = c("Not Sufficient", "Very Sufficient", "Completely Sufficient", "Moderately Sufficient", "Moderately Sufficient", "Not Sufficient"))
Unnesting Question i:
UnnestQi <- SurveyClean2 %>%
as_tibble() %>%
mutate(i = str_split(Q3, ",")) %>%
unnest(i)
Survey2Q6 <- UnnestQi |> drop_na(Q5) |> drop_na(i)
Reformating Question 6 to Likert-friendly format:
clean_survey <- function(data, column, question) {
data %>%
dplyr::select(all_of({{column}})) %>%
dplyr::mutate(Question = question) %>%
dplyr::group_by(Question, across(1)) %>%
dplyr::count() %>%
dplyr::ungroup() %>%
tidyr::pivot_wider(names_from = 2, values_from = n)
}
# table that contains survey questions/columns and the question name
survey_table <- dplyr::tibble(
column = c("Q6_3"),
question = c("Expert advice")
)
# loop through your data and clean it, then bind as dataframe
LikertGroupqi62 <- purrr::map2_df(survey_table$column, survey_table$question, function(x, y){
clean_survey(Survey2Q6, x, y)}) |>
mutate(across(everything(), ~ifelse(is.na(.), 0, .)))
## Likert
LikertGroupqi62 <- LikertGroupqi62 |> dplyr::select(Question, `Not Sufficient`, `Slightly Sufficient`, `Moderately Sufficient`, `Very Sufficient`, `Completely Sufficient`)
Likert WITHOUT grouping:
likert(Question~., LikertGroupqi62, ReferenceZero = 0, auto.key.in = list(columns = 1), main = list("Sufficiency of Cost-share Advice Based on Person or Agency Worked With"), col = c("#db6d00", "#924900", "#000000", "#004949", "#009292"),strip.left = FALSE, ylab = "", xlab = "Total Number of Respondents")