How to implement function trim using Parsec in haskell? - haskell

I am learning Parsec and want to practice it by implementing function trim. Here is my code:
module Trim where
import Text.ParserCombinators.Parsec hiding(spaces)
trim = reverse.trimFront.reverse.trimFront
trimFront :: String->String
trimFront = readExpr trimParser
readExpr :: Parser String->String->String
readExpr parser input = case parse parser "trim" input of
Left err -> error $ show err
Right val -> val
spaces = many space
trimParser :: Parser String
trimParser = spaces >> many anyChar
my question is, How could I implement trim in function trimParser directly without having to implement trimFront first?

To trim the spaces from both sides of a string:
trim :: String -> String
trim = dropWhileEnd isSpace . dropWhile isSpace
Note that you may be better off using the following implementation of dropWhileEnd instead of the one from Data.List, depending on the situation:
dropWhileEnd :: (a -> Bool) -> [a] -> [a]
dropWhileEnd p = foldr
(\x xs -> if null xs && p x then [] else x : xs) []

Related

Get all string splits

Say I have a string:
"abc7de7f77ghij7"
I want to split it by a substring, 7 in this case, and get all the left-right splits:
[ ("abc", "de7f77ghij7")
, ("abc7de", "f77ghij7")
, ("abc7de7f", "7ghij7")
, ("abc7de7f7", "ghij7")
, ("abc7de7f77ghij", "")
]
Sample implementation:
{-# LANGUAGE OverloadedStrings #-}
module StrSplits where
import qualified Data.Text as T
splits :: T.Text -> T.Text -> [(T.Text, T.Text)]
splits d s =
let run a l r =
case T.breakOn d r of
(x, "") -> reverse a
(x, y) ->
let
rn = T.drop (T.length d) y
an = (T.append l x, rn) : a
ln = l `T.append` x `T.append` d
in run an ln rn
in run [] "" s
main = do
print $ splits "7" "abc7de7f77ghij7"
print $ splits "8" "abc7de7f77ghij7"
with expected result:
[("abc","de7f77ghij7"),("abc7de","f77ghij7"),("abc7de7f","7ghij7"),("abc7de7f7","ghij7"),("abc7de7f77ghij","")]
[]
I'm not too happy about the manual recursion and let/case/let nesting. If my feeling that it doesn't look too good is right, is there a better way to write it?
Is there a generalized approach to solving these kinds of problems in Haskell similar to how recursion can be replaced with fmap and folds?
How about this?
import Data.Bifunctor (bimap)
splits' :: T.Text -> T.Text -> [(T.Text, T.Text)]
splits' delimiter string = mkSplit <$> [1..numSplits]
where
sections = T.splitOn delimiter string
numSplits = length sections - 1
mkSplit n = bimap (T.intercalate delimiter) (T.intercalate delimiter) $ splitAt n sections
I like to believe there's a way that doesn't involve indices, but you get the general idea. First split the string by the delimiter. Then split that list of strings at in two everywhere possible, rejoining each side with the delimiter.
Not the most efficient, though. You can probably do something similar with indices from Data.Text.Internal.Search if you want it to be fast. In this case, you wouldn't need to do the additional rejoining. I didn't experiment with it since I didn't understand what the function was returning.
Here's an indexless one.
import Data.List (isPrefixOf, unfoldr)
type ListZipper a = ([a],[a])
moveRight :: ListZipper a -> Maybe (ListZipper a)
moveRight (_, []) = Nothing
moveRight (ls, r:rs) = Just (r:ls, rs)
-- As Data.List.iterate, but generates a finite list ended by Nothing.
unfoldr' :: (a -> Maybe a) -> a -> [a]
unfoldr' f = unfoldr (\x -> (,) x <$> f x)
-- Get all ways to split a list with nonempty suffix
-- Prefix is reversed for efficiency
-- [1,2,3] -> [([],[1,2,3]), ([1],[2,3]), ([2,1],[3])]
splits :: [a] -> [([a],[a])]
splits xs = unfoldr' moveRight ([], xs)
-- This is the function you want.
splitsOn :: (Eq a) => [a] -> [a] -> [([a],[a])]
splitsOn sub xs = [(reverse l, drop (length sub) r) | (l, r) <- splits xs, sub `isPrefixOf` r]
Try it online!
Basically, traverse a list zipper to come up with a list of candidates for the split. Keep only those that are indeed splits on the desired item, then (un)reverse the prefix portion of each passing candidate.

Haskell: Convert String to [(String,Double)]

I parse an XML and get an String like this:
"resourceA,3-resourceB,1-,...,resourceN,x"
I want to map that String into a list of tuples (String,Double), like this:
[(resourceA,3),(resourceB,1),...,(resourceN,x)]
How is it possible to do this? I ve looked into the map function and also the split one. I am able to split the string by "-" but anything else...
This is the code i have so far:
split :: Eq a => a -> [a] -> [[a]]
split d [] = []
split d s = x : split d (drop 1 y) where (x,y) = span (/= d) s
it is just a function to split my string into a list of Stirng, but then i dont know how to continue.
What I want to do know is to loop over that new list that i have created with the split method and for each element create a tuple. I hace tried with the map function but i dont get it to compile even
So in Haskell you dont really mutate any value, instead you'll create a new list of pairs from the string you've described, so the solution would look something similar to the following:
import Data.List.Split
xmlList = splitOn "-" "resourceA,3-resourceB,4-resourceC,6"
commaSplit :: String -> [String]
commaSplit = splitOn ","
xmlPair :: [String] -> [(String, Double)] -- might be more efficient to use Text instead of String
xmlPair [x] = [(\x' -> ((head x') :: String, (read (last x')) :: Double )) (commaSplit x)]
xmlPair (x:xs) = xmlPair [x] ++ xmlPair xs
main :: IO ()
main = mapM_ (\(a,b) -> putStrLn (show a++" = "++ show b)) (xmlPair $ xmlList)
This is my quick and dirty way of showing things but I'm sure someone can always add a more detailed answer.

Palindroms & Monads

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

Haskell: Elegant solution for integer parsing

I want to write a function that gets the first integer in a string, if the integer is at the beginning of the string and is followed by space or nothing.
For example, "12" and "12 kids" would be valid strings, but "12kids", "a12" are invalid.
This is my function:
getFirstInteger :: String -> Int
getFirstInteger [] = error "No integer"
getFirstInteger str
| dropWhile (Char.isNumber) str == [] = read str :: Int
| Char.isSpace $ head $ dropWhile (Char.isNumber) str = read (takeWhile (Char.isNumber) str) :: Int
| otherwise = error "No integer found"
The problem is that if the string is actually an integer, head will raise an exception, so that is why the first condition exists. Is there any elegant solution to this problem?
getFirstInteger :: String -> Int
getFirstInteger = read . head . words
words will split the String into a list of Strings, which were originally separated by whitespace. head will take the first (or error if the original string was empty), and read will parse the string as usual (and error if the first word wasn't a valid Int).
However, I prefer a variant that doesn't use error on empty Strings or unparseable ones, e.g.
import Text.Read (readMaybe)
getFirstInteger :: String -> Maybe Int
getFirstInteger [] = Nothing
getFirstInteger xs = readMaybe . head . words $ xs
One could write this completely point-free with listToMaybe from Data.Maybe, but that's probably an overkill:
import Data.Maybe (listToMaybe)
import Text.Read (readMaybe)
import Control.Monad ((>=>))
getFirstInteger :: String -> Maybe Int
getFirstInteger = listToMaybe . words >=> readMaybe
If you want to parse strings without using parser combinator libraries or any of the machinery in Text.Read, have a look at the functions break and span:
span :: (a -> Bool) -> [a] -> ([a], [a])
break :: (a -> Bool) -> [a] -> ([a], [a])
The nice thing is that both of these functions not only return what they match but also the remainder of the string allowing you to continue your parsing.
To solve your problem:
import Data.Char
getFirstInteger :: String -> Maybe Int
getFirstInteger str
let (digs, rest1) = span isDigit str
endsok = case rest1 of
[] -> True
(c:_) -> c == ' '
in
if not (null digs) && endsok
then Just (read digs)
else Nothing -- either no digits or doesn't end properly
This version does not allow a leading minus sign. This next version allows an optional leading minus sign for the integer:
getFirstInteger' str =
let (minus,rest1) = span (=='-') str
(digs, rest2) = span isDigit rest1
endsok = case rest2 of
[] -> True
(c:_) -> c == ' '
in
if length minus <= 1 && not (null digs) && endsok
then Just (read (minus ++ digs))
else Nothing
Yes - this does not terminate as early as possible on bad input. I provide it mainly as an example of how to chain together calls to span and break.
Use reads. For example:
type Unit = String
readUnit :: String -> Maybe (Int, Maybe Unit)
readUnit s = case reads s of -- the integer is at the beginning of the string and...
(n, ' ':unit):_ -> Just (n, Just unit) -- is followed by space...
(n, "" ):_ -> Just (n, Nothing) -- or nothing.
_ -> Nothing
In ghci:
> readUnit "12"
Just (12,Nothing)
> readUnit "12 kids"
Just (12,Just "kids")
> readUnit "12kids"
Nothing
> readUnit "a12"
Nothing
However, there are a few minor considerations to keep in mind. It's possible that read does not restrict the syntax as much as you might want; for example, the following answer may surprise you:
> readUnit " ((-0x5)) kids"
Just (-5,Just "kids")
You may also want to drop extraneous spaces in the unit; for example, you could change the first clause above to
(n, ' ':unit):_ -> Just (n, Just (dropWhile isSpace unit))
or similar. And as a final variation on this theme, note that while the standard instances of Read never return lists with more than one element from reads, it is technically possible that some user-supplied type may do so. So if you were ever to use reads to parse types other than Int, you may want to either demand an unambiguous parse or consider all the parses before deciding what to do; the above code bakes in the assumption that the first parse is as good as any.

How do I replace space characters in a string with "%20"?

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

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