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.
Related
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.
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
Can I get a quick explanation on how I would concatenate a list of strings into one string using map? I was trying to use intercalate but I realized that this is used to combine lists and not strings. i.e. [[char]] instead of [String]
type InformationList = (String , [String] )
concatenateList :: String -> [InformationList] -> String
concatenateList n cs = do
let [informationlist] = intercalate " " cs
let toWrite = n ++ [informationList]
return toWrite
I updated this so that you can see the code I've been working with. It is saying that [[char]] is not the same as [informationlist] but that should be the same as [String]
A String is a list of Chars. A String is a type alias defined as:
type String = [Char]
If you want to concatenate a list of Strings without any separators, you can use concat :: Foldable t => t [a] -> [a]:
Prelude> concat ["foo", "bar", "qux"]
"foobarqux"
In case you want to insert a string as delimiter, you can use intercalate :: [a] -> [[a]] -> [a]:
Prelude> import Data.List(intercalate)
Prelude Data.List> intercalate "," ["foo","bar","qux"]
"foo,bar,qux"
I have a list of lists of strings e.g;
[["h","e","l","l","o"], ["g","o","o","d"], ["w","o","o","r","l","d"]]
And I want to rename repeated values outside a sublist so that all the repetitions are set to new randomly generated values throughout a sublist that are not pre-existing in the list but the same inside the same sublist so that a possible result might be:
[["h","e","l","l","o"], ["g","t","t","d"], ["w","s","s","r","z","f"]]
I already have a function that can randomly generate a string of size one called randomStr:
randomStr :: String
randomStr = take 1 $ randomRs ('a','z') $ unsafePerformIO newStdGen
Presuming you want to do what I've outlined in my comment below, it's best to break this problem up into several smaller parts to tackle one at a time. I would also recommend leveraging common modules in base and containers, since it will make the code much simpler and faster. In particular, the modules Data.Map and Data.Sequence are very useful in this case. Data.Map I would say is the most useful here, as it has some very useful functions that would otherwise be difficult to write by hand. Data.Sequence is used for efficiency purposes at the end, as you'll see.
First, imports:
import Data.List (nub)
import Data.Map (Map)
import Data.Sequence (Seq, (|>), (<|))
import qualified Data.Map as Map
import qualified Data.Sequence as Seq
import Data.Foldable (toList)
import System.Random (randomRIO)
import Control.Monad (forM, foldM)
import Control.Applicative ((<$>))
Data.Foldable.toList is needed since Data.Sequence does not have a toList function, but Foldable provides one that will work. On to the code. We first want to be able to take a list of Strings and find all the unique elements in it. For this, we can use nub:
lettersIn :: [String] -> [String]
lettersIn = nub
I like providing my own names for functions like this, it can make the code more readable.
Now that we can get all the unique characters, we want to be able to assign each a random character:
makeRandomLetterMap :: [String] -> IO (Map String String)
makeRandomLetterMap letters
= fmap Map.fromList
$ forM (lettersIn letters) $ \l -> do
newL <- randomRIO ('a', 'z')
return (l, [newL])
Here we get a new random character and essentially zip it up with our list of letters, then we fmap (<$>) Map.fromList over that result. Next, we need to be able to use this map to replace letters in a list. If a letter isn't found in the Map, we just want the letter back. Luckily, Data.Map has the findWithDefault function which is perfect for this situation:
replaceLetter :: Map String String -> String -> String
replaceLetter m letter = Map.findWithDefault letter letter m
replaceAllLetters :: Map String String -> [String] -> [String]
replaceAllLetters m letters = map (replaceLetter m) letters
Since we want to be able to update this map with new letters that have been encountered in each sublist, overwriting previously encountered letters as needed, we can use Data.Map.union. Since union favors its first argument, we need to flip it:
updateLetterMap :: Map String String -> [String] -> IO (Map String String)
updateLetterMap m letters = flip Map.union m <$> makeRandomLetterMap letters
Now we have all the tools needed to tackle the problem at hand:
replaceDuplicatesRandomly :: [[String]] -> IO [[String]]
replaceDuplicatesRandomly [] = return []
For the base case, just return an empty list.
replaceDuplicatesRandomly (first:rest) = do
m <- makeRandomLetterMap first
For a non-empty list, make the initial map off the first sublist
(_, seqTail) <- foldM go (m, Seq.empty) rest
Fold over the rest, starting with an empty sequence and the first map, and extract the resulting sequence
return $ toList $ first <| seqTail
Then convert the sequence to a list after prepending the first sublist (it doesn't get changed by this function). The go function is pretty simple too:
where
go (m, acc) letters = do
let newLetters = replaceAllLetters m letters
newM <- updateLetterMap m letters
return (newM, acc |> newLetters)
It takes the current map m and an accumulation of all the sublists processed so far acc along with the current sublist letters, replaces the letters in said sublist, builds a new map for the next iteration (newM), and then returns the new map along with the accumulation of everything processed, i.e. acc |> newLetters. All together, the function is
replaceDuplicatesRandomly :: [[String]] -> IO [[String]]
replaceDuplicatesRandomly [] = return []
replaceDuplicatesRandomly (first:rest) = do
m <- makeRandomLetterMap first
(_, seqTail) <- foldM go (m, Seq.empty) rest
return $ toList $ first <| seqTail
where
go (m, acc) letters = do
let newLetters = replaceAllLetters m letters
newM <- updateLetterMap m letters
return (newM, acc |> newLetters)
It's always better to keep impure and pure computations separated.
You cannot replace by letters, which are already in a list, so you need to get a string of fresh letters:
fresh :: [String] -> String
fresh xss = ['a'..'z'] \\ foldr union [] xss
This function replaces one letter with another in a string:
replaceOne :: Char -> Char -> String -> String
replaceOne y y' = map (\x -> if x == y then y' else x)
This function replaces one letter each time with a new letter for every string in a list of strings:
replaceOnes :: Char -> String -> [String] -> (String, [String])
replaceOnes y = mapAccumL (\(y':ys') xs ->
if y `elem` xs
then (ys', replaceOne y y' xs)
else (y':ys', xs))
For example
replaceOnes 'o' "ijklmn" ["hello", "good", "world"]
returns
("lmn",["helli","gjjd","wkrld"])
A bit tricky one:
replaceMany :: String -> String -> [String] -> (String, [String])
replaceMany ys' ys xss = runState (foldM (\ys' y -> state $ replaceOnes y ys') ys' ys) xss
This function replaces each letter from ys each time with a new letter from ys' for every string in xss.
For example
replaceMany "mnpqstuvxyz" "lod" ["hello", "good", "world"]
returns
("vxyz",["hemmp","gqqt","wsrnu"])
i.e.
'l's in "hello" are replaced by the first letter in "mnpqstuvxyz"
'l' in "world" is replaced by the second letter in "mnpqstuvxyz"
'o' in "hello" is replaced by the third letter in "mnpqstuvxyz"
'o's in "good" are replaced by the fourth letter in "mnpqstuvxyz"
...
'd' in "world" is replaced by the seventh letter in "mnpqstuvxyz"
This function goes through a list of strings and replaces all letters from the head by fresh letters, that ys' contains, for each string in the rest of the list.
replaceDuplicatesBy :: String -> [String] -> [String]
replaceDuplicatesBy ys' [] = []
replaceDuplicatesBy ys' (ys:xss) = ys : uncurry replaceDuplicatesBy (replaceMany ys' ys xss)
I.e. it does what you want, but without any randomness — just picks fresh letters from a list.
All described functions are pure. Here is an impure one:
replaceDuplicates :: [String] -> IO [String]
replaceDuplicates xss = flip replaceDuplicatesBy xss <$> shuffle (fresh xss)
I.e. generate a random permutation of a string, that contains fresh letters, and pass it to replaceDuplicatesBy.
You can take the shuffle function from https://www.haskell.org/haskellwiki/Random_shuffle
And the final test:
main = replicateM_ 3 $ replaceDuplicates ["hello", "good", "world"] >>= print
prints
["hello","gxxd","wcrzy"]
["hello","gyyd","wnrmf"]
["hello","gmmd","wvrtx"]
The whole code (without shuffle): http://lpaste.net/115763
I think this is bound to raise more questions than it answers.
import Control.Monad.State
import Data.List
import System.Random
mapAccumLM _ s [] = return (s, [])
mapAccumLM f s (x:xs) = do
(s', y) <- f s x
(s'', ys) <- mapAccumLM f s' xs
return (s'', y:ys)
pick excluded for w = do
a <- pick' excluded
putStrLn $ "replacement for " ++ show for ++ " in " ++ show w ++ " excluded: " ++ show excluded ++ " = " ++ show a
return a
-- | XXX -- can loop indefinitely
pick' excluded = do
a <- randomRIO ('a','z')
if elem a excluded
then pick' excluded
else return a
transform w = do
globallySeen <- get
let go locallySeen ch =
case lookup ch locallySeen of
Nothing -> if elem ch globallySeen
then do let excluded = globallySeen ++ (map snd locallySeen)
a <- lift $ pick excluded ch w
return ( (ch, a):locallySeen, a)
else return ( (ch,ch):locallySeen, ch )
Just ch' -> return (locallySeen, ch')
(locallySeen, w') <- mapAccumLM go [] w
let globallySeen' = w' ++ globallySeen
put globallySeen'
return w'
doit ws = runStateT (mapM transform ws) []
main = do
ws' <- doit [ "hello", "good", "world" ]
print ws'
flattern :: [(Char, Int)] -> String
flattern [] = ""
flattern ((w,l):xs) = show l ++ w : flattern xs
What would be the inverse function of this? Is there any way of be able to work this out?
It's not invertible:
There are strings that cannot be reproduced by this function (any string not starting with a digit).
It's not even partially invertible. There are also strings that correspond to multiple inputs: "1111" can be produced by either [('1',1),('1',1)] or [('1',111)].
Are you sure that this was the function to invert, and not something like flattern ((w,l):xs) = replicate l w ++ flattern xs?
If you really want, you could parse the output of the function to try and reconstruct what the arguments must have been.
import Text.ParserCombinators.Parsec
unflat1 :: Parser (Char, Int)
unflat1 = do
c <- anyChar
n <- many1 digit
return (c, read n)
readExpr :: String -> Either String [(Char, Int)]
readExpr input = case parse (many unflat1) "unflat" input of
Left err -> Left ("No match: " ++ show err)
Right val -> Right val
This sorta works, as long as flattern does not have numbers as the first input.
Something like flattern [('a',1), ('2',3)] will be parsed as [(a, 123)].