Convert a string into an R vector expression

Question

Problem: Given a arbitrary string, split by a given regex and format into an R vector expression. E.g. Convert "a..,b..,c,d" into 'c("a","b","c","d")', when the given regex is "[.,]+".

Solution in haskell:

module Lib
 ( toRVec
 ) where
import Data.List (intersperse)
import Data.Maybe (fromMaybe)
import Data.Text ()
import Data.Text (pack, strip, unpack)
import Text.Regex (mkRegex, splitRegex)
(//) :: Maybe a -> a -> a
(//) = flip fromMaybe
stripString = (unpack . strip . pack)
toRVec :: String -> Maybe String -> Maybe Bool -> String
toRVec orig delim' toStrip' =
 let
 delim = delim' // "[ \n]+"
 toStrip = toStrip' // True
 ws = splitRegex (mkRegex delim) $ stripString orig
 ws1 = map (\w -> "\"" ++ w ++ "\"") $ if toStrip then
 map stripString ws
 else
 ws
 ws2 = intersperse "," ws1
 res = foldl (++) "" ws2
 in "c(" ++ res ++ ")"

Any suggestion for improvements are welcome.

Answer 1

How I would do it:

import Data.List.Split
import Data.List
foo delim s = "c(\"" ++ intercalate "\",\"" (wordsBy (`elem` delim) s) ++ "\")"

reads yours

No brackets needed around unpack . strip . pack.

I would replace if toStrip then map stripString ws else ws with (if toStrip then map stripString else id) ws.

Some of your names are unneeded; I would inline ws, ws1 and ws2 into where they are used (once ws is only used once), perhaps even res.

Instead of foldl (++), use concat.

• Gurkenglas: Moreover: delim and tostrip are options, orig is what is transformed; thus orig should be the last argument. Think Maybe String -> Maybe String -> (String -> String). People usually don't provide default options as Maybe arguments, at least I don't see it. Consider dropping the lines with // and let the user plug in the default options if they want, seeing as providing default options isn't even part of the task.
• Christian Oudard: If you use Gurkenglas's suggestion about orig being last, you could even provide a default like this: toRVecDefault = toRVec "[ \n]+" True