i m not familiar with askell.
i have to do this function parseChar :: Char -> Parser Char
> parseChar 'a' " abcd "
Just ('a', "bcd")
> parseChar 'z' " abcd "
Nothing
i did this function
type Parser r = String -> Maybe (Char, String)
parseChar :: Char -> Parser Char
parseChar x = \xs -> Just(x, xs)
i dont really understand the \ and how can I take all the string except the second.
Thanks!!
i dont really understand the \ and how can I take all the string except the second.
The \ is a lambda expression \xs -> … is a function that maps a variable xs to the … part. Here you can however move the variable to the head of the clause.
You however need to check if the first element of the string is indeed the element you are parsing:
parseChar :: Char -> Parser Char
parseChar _ "" = Nothing
parseChar x (y:ys)
| x == y = Just (y, ys)
| otherwise = Nothing
Here x is thus the char we want to parse, and "" or (y:ys) the string that we are parsing. "" means that the string is empty, in which case we return Nothing. If the string is not empty, we unpack it in the first character (head) y, and the list of remaining characters (tail) ys. In case x == y, we are thus parsing the correct character, and we can return a 2-tuple wrapped in a Just constructor. In case x is not equal to y, the parser should "fail" and return Nothing.
This then yields:
Prelude> parseChar 'a' "abcd"
Just ('a',"bcd")
Prelude> parseChar 'z' "abcd"
Nothing
Prelude> parseChar 'a' ""
Nothing
Related
I'm new to the Haskell. I am finding following task difficult:
Enter a string of characters. Output all palindromes to the file (use the IO monad to work with the file system and input / output, use the list monad to work with the strings).`
Any code is may be helpful. Thank you in advance!
This is what I have tried so far:
palindrome :: Char -> [String]
palindrome n
| n < 0 = []
| even n = map (\front -> front ++ reverse front) fronts
| odd n = map (\front -> front ++ tail (reverse front)) fronts
where ispalindrome :: (Integral a, Show a) => a -> Bool
ispalindrome x = show x = reverse (show x)
main = do
input <- getline
putStrLn print :: IO ()
So this is basically consists of 4 things.
Read Input from the stdin
Convert input string into list of strings
From the above list find out the strings which are palindromes
print these palindromes into file.
If you convert above into functions the signatures of these will be.
String -> [String]
[String] -> [String]
Don't bother about the signature of 1st and 4th for now. These are anyways one line code readily available on internet.
2 is a single function available in Data.List called words.
3 can be again in two parts. A function which find out if a given string is palindrome. Signature will be
String -> Bool
This is also one line code.
Once you have above function the only part remaining is filtering out the strings which are palindromes in given list of strings.
isPalindrome
My haskell is a bit rusty so I don't promise the code below will work %100 yet I tried to stick to the main idea.I hope this answer helps. If you think anything is wrong both logically and syntactically, just write a comment and I will fix it asap.
isPalindrome :: [Char] -> Boolean
isPalindrome w = isPalindrome' w reverse w
where
isPalindrome' :: [Char] -> [Char] -> Boolean
isPalindrome' [] [] = true
isPalindrome' (x:xs) (y:ys) = if x == y then isPalindrome' xs ys else false
isPalindrome' _ _ = false
function above should be fine for checking for palindromes.
for writing to file part, you can create a list of palindromes first, then write all palindromes to a file in another function. so basically, first you split your string into words, then for words in given string you find palindromes, then you write the palindromes into a file.
how to read string from user?
main = do
userInput <- getLine
how to split word with delimiter?
split :: Char -> [Char] -> [[Char]]
split delimiter string = split' delimiter string [] []
where
split' :: Char -> [Char] -> [Char] -> [[Char]] -> [[Char]]
split' delim [] substr splittedStr = splittedStr if substr == [] else reverse substr ++ splittedStr
split' delim (x:xs) substr splittedStr = if x == delim then split' delim xs [] (reverse substr) ++ splittedSubstr else split' delim xs (x ++ substr) splittedSubstr
main idea is you stack characters until you see your delimeter and store them in a list when you see a delimiter.
how to filter palindromes in list?
to filter palindromes in list you use haskell's filter function as
filter isPalindrome (split ' ' userInput)
In the end, you can write a main block to run all of this in right order
main = do
userInput <- getLine
let splittedInput = split ' ' userInput
let palindromes = filter isPalindrome splittedInput
let output = concat (intersperse "\n" palindromes)
writeFile "file.txt" output
This is an extension of this question: Haskell replace characters in string
I would like to tweak the following expression to replace a char with a string
let replaceO = map (\c -> if c=='O' then 'X'; else c)
In the end, I would the following results (XX can be a string of any length):
replaceO "HELLO WORLD"
"HELLXX WXXRLD"
You can use concatMap:
let replace0 = concatMap (\c -> if c=='O' then "X" else "XX")
You kind formulate your problem in terms of traversing and accumulating based on a condition, something like this,
replace :: String -> Char -> String -> String
replace xs c s = foldr go [] xs
where go x acc = if x == c then acc ++ s
else acc ++ [x]
For you example:
>replace "HELLO WORLD" 'O' "XXX"
> "HELLXXX WXXXRLD"
I wanted to write a Haskell function that takes a string, and replaces any space characters with the special code %20. For example:
sanitize "http://cs.edu/my homepage/I love spaces.html"
-- "http://cs.edu/my%20homepage/I%20love%20spaces.html"
I am thinking to use the concat function, so I can concatenates a list of lists into a plain list.
The higher-order function you are looking for is
concatMap :: (a -> [b]) -> [a] -> [b]
In your case, choosing a ~ Char, b ~ Char (and observing that String is just a type synonym for [Char]), we get
concatMap :: (Char -> String) -> String -> String
So once you write a function
escape :: Char -> String
escape ' ' = "%20"
escape c = [c]
you can lift that to work over strings by just writing
sanitize :: String -> String
sanitize = concatMap escape
Using a comprehension also works, as follows,
changer :: [Char] -> [Char]
changer xs = [ c | v <- xs , c <- if (v == ' ') then "%20" else [v] ]
changer :: [Char] -> [Char] -> [Char]
changer [] res = res
changer (x:xs) res = changer xs (res ++ (if x == ' ' then "%20" else [x]))
sanitize :: [Char] -> [Char]
sanitize xs = changer xs ""
main = print $ sanitize "http://cs.edu/my homepage/I love spaces.html"
-- "http://cs.edu/my%20homepage/I%20love%20spaces.html"
The purpose of sanitize function is to just invoke changer, which does the actual work. Now, changer recursively calls itself, till the current string is exhausted.
changer xs (res ++ (if x == ' ' then "%20" else [x]))
It takes the first character x and checks if it is equal to " ", if so gives %20, otherwise the actual character itself as a string, which we then concatenate with the accumulated string.
Note: This is may not be the optimal solution.
You can use intercalate function from Data.List module. It does an intersperse with given separator and list, then concats the result.
sanitize = intercalate "%20" . words
or using pattern matching :
sanitize [] = []
sanitize (x:xs) = go x xs
where go ' ' [] = "%20"
go y [] = [y]
go ' ' (x:xs) = '%':'2':'0': go x xs
go y (x:xs) = y: go x xs
Another expression of Shanth's pattern-matching approach:
sanitize = foldr go []
where
go ' ' r = '%':'2':'0':r
go c r = c:r
I'm working in Haskell in two functions:
Basically I want to get the character before matching a specific character in a given string
This is my code:
before :: Char -> [Char] -> Char
before x str = trackelement x ' ' str
trackelement :: Char -> Char -> [Char] -> Char
trackelement x y (z:zs)
| x == z = y
| otherwise = trackelement x z (zs)
My problem is when I try: before 'l' "luis"
The answer is : ' ' (of course, before 'l' there is nothing), and I would like to be '' or Nothing
I tried passing trackelement x '' str instead of trackelement x ' ' str but I have this error Syntax error on ''str
Could you suggest me something?
The answers shown already are good for getting your code to work, but they don't explain why you get the error you're receiving. The reason why that error was shown is that '' is not valid syntax, since there is no such thing as an "empty character". All characters have value, but Strings can be empty. Remember that type String = [Char], and it's very clear that there can be such a thing as an empty list, but characters always have a value. It's comparable to saying you can have an empty list of Ints, namely [], but you can't have an "empty int", whatever that would mean.
You can use a Maybe:
before :: Char -> [Char] -> Maybe Char
before x str = initialise x str
initialise x (y:xs)
| x == y = Nothing
| otherwise = trackelement x y xs
trackelement :: Char -> Char -> [Char] -> Maybe Char
trackelement x y [] = Nothing
trackelement x y (z:zs)
| x == z = Just y
| otherwise = trackelement x z zs
To take care of the corner case before 'l' "luis", we have to add a new initialiser function. It basically checks if the first character matches the searched one. If it does, we return Nothing, because we checked the first character which obviously does not have a preceding one. Else we just call trackelement and use it's result.
As Zeta mentioned, you can combine the functions, which simplifies everything and takes care of the corner case you are currently experiencing.
before _ [x] = Nothing
before a (x:y:xs)
| a == y = Just x
| otherwise = before a (y:xs)
Just using this function, you noticed you have problems when encountering a word containing more than one letter which is also searched for (before 'a' "amalia" -> Just 'm'). Currently the best solution I know of is again splitting this up into more than one function, which brings us back to the solution at the top.
Match the first two elements instead just head and tail. That way you don't even need trackelement:
before :: Eq a => a -> [a] -> Maybe a
before x (a:b:rest)
| a == x = Nothing
| b == x = Just a
| otherwise = before x (b:rest)
before _ _ = Nothing
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"?
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.
Sorry for picking up this old thread but why not use lambda expressions?
λ> let replaceO = map (\c -> if c=='O' then 'X'; else c)
λ> replaceO "HELLO WORLD"
"HELLX WXRLD"`
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
If you depend on the text package (like 99.99% of Haskell applications), you can use T.replace:
>>> replace "ofo" "bar" "ofofo"
"barfo"
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.
replace :: Char -> Char -> String -> String
replace _ _ [] = []
replace a b (x : xs)
| x == a = [b] ++ replace a b xs
| otherwise = [x] ++ replace a b xs
I'm new to Haskell and I've tried to make it simpler for others like me.
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)