Supposing I had the string "HELLO WORLD" is there a way I can call a function that replaces the character 'O' in the string with the character 'X' so that the new string would look like "HELLX WXRLD"?
7 Answers
How about:
let
repl 'o' = 'x'
repl c = c
in map repl "Hello World"
If you need to replace additional characters later, just add clauses to the repl function.
4 Comments
map (\case ....) "Hello World" (just fill in the ... :)Alternative 1 - Using MissingH
First:
import Data.List.Utils (replace)
Then use:
replace "O" "X" "HELLO WORLD"
Alternative 2 - Using Control.Monad
One funny bastard:
import Control.Monad (mfilter)
replace a b = map $ maybe b id . mfilter (/= a) . Just
Example:
λ> replace 'O' 'X' "HELLO WORLD"
"HELLX WXRLD"
Alternative 3 - Using if
Amon's suggestions was probably the finest I believe! No imports and easy to read and understand!
But to be picky - there's no need for semicolon:
replace :: Eq a => a -> a -> [a] -> [a]
replace a b = map $ \c -> if c == a then b else c
Comments
If you depend on the text package (like 99.99% of Haskell applications), you can use T.replace:
>>> replace "ofo" "bar" "ofofo"
"barfo"
Comments
Here's another possible solution using divide and conquer:
replaceO [] = []
replaceO (x:xs) =
if x == 'O'
then 'X' : replaceO xs
else x : replaceO xs
First, you set the edge condition "replaceO [] = []".
If the list is empty, there is nothing to replace, returning an empty list.
Next, we take the string and divide it into head and tail. in this case 'H':"ELLOWORLD"
If the head is equal to 'O', it will replace it with 'X'. and apply the replaceO function to the rest of the string.
If the head is not equal to 'O', then it will put the head back where it is and apply the replaceO function to the rest of the string.
Comments
I guess this could be useful.
main = print $ charRemap "Hello WOrld" ['O','o'] ['X','x']
charRemap :: [Char] -> [Char] -> [Char] -> [Char]
charRemap [] _ _ = []
charRemap (w:word) mapFrom mapTo =
if snd state
then mapTo !! fst state : charRemap word mapFrom mapTo
else w : charRemap word mapFrom mapTo
where
state = hasChar w mapFrom 0
hasChar :: Char -> [Char] -> Int -> (Int,Bool)
hasChar _ [] _ = (0,False)
hasChar c (x:xs) i | c == x = (i,True)
| otherwise = hasChar c xs (i+1)
Data.Text, usereplace