Manipulating a Column with a Concatenated List in R - r

Question

I found a way to make it work, but it seems clumsy. There has to be a better way...
The question I might try to answer is
If I wanted to find out how often a language was selected by country, how would I do that efficiently?
This works, what's better?
library(tidyverse)
data(SO_survey, package = "crunch")
# what does it look like now?
SO_survey %>% select(Country, WantWorkLanguage) %>% head()
It's set up like this
# Country WantWorkLanguage
# 4 United States Matlab; Python; R; SQL
# 11 United States C#; R; SQL
# 36 Italy JavaScript; Python; R
# 125 Denmark Groovy; Java; JavaScript; Lua; SQL; TypeScript
# 242 United States C++; Python
# 298 Dominican Republic C; C#; CoffeeScript; Go; Haskell; JavaScript; Perl; PHP; Python; R; Ruby; SQL
Made a unique list of languages
# extract unique languages
(wanted = SO_survey %>%
select(WantWorkLanguage) %>%
unlist() %>%
strsplit("; ", fixed = T) %>%
unlist() %>%
unique()
To extract the county by one country
# how often did a respondent pick a particular language in the US?
SO_survey %>%
filter(Country == "United States") %>%
{strsplit(unlist(.$WantWorkLanguage),"; ",fixed = T)} %>%
unlist() %>%
table(. %in% wanted)
If I want it arranged/sorted
# If I want it sorted or arranged
SO_survey %>%
filter(Country == "United States") %>%
{strsplit(unlist(.$WantWorkLanguage),"; ",fixed = T)} %>%
unlist() %>%
table(. %in% wanted) %>%
data.frame() %>%
select(-Var2) %>%
arrange(-Freq)
The output:
# . Freq
# 1 Python 321
# 2 R 288
# 3 SQL 209
# 4 JavaScript 199
# 5 C++ 136
# 6 Java 115
# 7 Go 101
# 8 C# 100
# 9 Scala 81
# 10 C 74
# 11 Swift 57
# 12 Julia 56
# 13 TypeScript 55
# 14 Haskell 52
# 15 Rust 38
# 16 F# 36
# 17 PHP 32
# 18 Ruby 32
# 19 Assembly 29
# 20 Clojure 29
# 21 Matlab 29
# 22 Elixir 23
# 23 Perl 18
# 24 Objective-C 17
# 25 CoffeeScript 16
# 26 Erlang 16
# 27 Lua 13
# 28 Common Lisp 12
# 29 VBA 11
# 30 Groovy 7
# 31 Dart 5
# 32 Smalltalk 3
# 33 VB.NET 3
# 34 Hack 2
# 35 Visual Basic 6 1

Answer 1

Your tidyverse solution seems pretty good. For something more concise you could try base R or data.table:
library(data.table)
setDT(SO_survey)
setkey(SO_survey, Country)
SO_survey['United States', .(lang = unlist(strsplit(WantWorkLanguage, '; ')))
][, .N, keyby = V1 ][order(-N)]
# lang N
# 1: Python 321
# 2: R 288
# 3: SQL 209
# 4: JavaScript 199
# 5: C++ 136
# ...