I'm learning haskell and am currently trying to parse Integers and Floats from strings.
However, when trying my readNum function on "342" or any "number" that doesn't have a single or more non-numeric characters ghci reports to me:
* Exception: parse.hs:125:18-46: Irrefutable pattern failed for pattern (i, (a
: as))
data Token
= IntTok Int | FloatTok Float | EOF
readNum :: String->(Token, String)
readNum [] = (EOF, [])
readNum xs = let (i, (a:as)) = span isDigit xs --This is line 125
in (case a of
('.') -> let (j, (b:c:bs)) = span isDigit as
in (if ((toLower b) == 'e' && (c == '+' || c == '-' || (isDigit c)))
then (let (k, d) = span isDigit bs in (FloatTok (read (concat [i,[a],j, [b],[c],k])::Float), d))
else (FloatTok (read (concat [i,[a],j])::Float), (b:c:bs)))
_ -> (IntTok (read i::Int), (a:as)))
Is there a better way to handle the case when span isDigit xs returns an empty list as the second element of the tuple?
-Thanks
You're getting the error because if you use a simple Integer like "342" then span isDigit "342" is just ("342",[]), which can't match (l,a:as). A pattern that is supposed to always match is called an irrefutable pattern. As you've found out, patterns in let bindings are irrefutable, so...
You need to to stick to patterns that will always match in a let binding. For example you could do
readNum xs = let (i, ps) = span isDigit xs
in (case ps of
('.':as) -> let (j, qs) = span isDigit as in case qs of
b:c:bs -> if ..........
_ -> error "not enough stuff after decimal number"
_ -> ....
I gave a silly error message, but clearly you should write more sensible code there.
Related
I am new to Haskell and I am currently learning it in school. I got a school task where I have to decode a message that contain certain patterns but I have got no idea how to do this.
The pattern looks something like this: If a letter has a consonant followed by the character 'o' and then once again followed by the same consonant as before it should replace that substring ("XoX" where X is a consonant) with only the consonant. For example if I decode the string "hohejoj" it should return "hej". Sorry if I am explaining this poorly but I think you understand.
This is the code I have so far (but it doesn't work):¨
karpsravor :: String->String
karpsravor s = karpsravor_help s ""
where karpsravor_help s res
|s == "" && (last res) == 'o' = (init res)
|s==""=res
|otherwise = karpsravor_help (drop 3 s) (res ++ (consDecode (take 3 s)))
consDecode :: String->String
consDecode a
|(length a) < 3 = ""
|a == [(head a)]++"o"++[(head a)] = [(head a)]
|otherwise = a
The code is completely broken and poorly written (dumb method) but I have no other idea for how to solve this. Please help!
Pattern match to find occurrences of 'o'. I.e., use
karpsravorhelp (a:'o':b:rest) res = ...
You can't have a:'o':a:rest in the above, you can't pattern match for equality; you'll need to use a guard to make sure that a == b:
karpsravorhelp (a:'o':b:rest) res
| a == b = ...
| otherwise = ...
You'll also have to make sure a and b are consonants, which will just be an 'and' condition for the first guard. For the otherwise condition, make sure that the recursive call calls (b:rest) since you could have something like a:'o':b:'o':b:....
Also make sure to match for two other patterns:
Empty List, []
x:rest, which must go after the above pattern; this way, it will first attempt to match on the a:'o':b:rest pattern, and if that's not there, just take the next letter.
One way to do it would be with unfoldr from Data.List. You can use a case expression to pattern match on a : 'o' : b : rest, and then check that a and b are equal and not vowels using a guard |. Then just include the base cases for when the pattern doesn't match.
notVowel :: Char -> Bool
notVowel = (`notElem` "aeiouAEIOU")
karpsravor :: String -> String
karpsravor = unfoldr $ \str -> case str of
a : 'o' : b : rest
| a == b && notVowel a -> Just (a, rest)
a : rest -> Just (a, rest)
"" -> Nothing
I'm trying to cut chunks from a list, with a given predicate. I would have preferred to use a double character, e.g. ~/, but have resolved to just using $. What I essentially want to do is this...
A: "Hello, my $name is$ Danny and I $like$ Haskell"
What I want to turn this into is this:
B: "Hello, my Danny and I Haskell"
So I want to strip everything in between the given symbol, $, or my first preference was ~/, if I can figure it out. What I tried was this:
s1 :: String -> String
s1 xs = takeWhile (/= '$') xs
s2 :: String -> String
s2 xs = dropWhile (/= '$') xs
s3 :: String -> String
s3 xs = s3 $ s2 $ s1 xs
This solution seems to just bug my IDE out (possibly infinite looping).
Solution:
s3 :: String -> String
s3 xs
|'$' `notElem` xs = xs
|otherwise = takeWhile (/= '$') xs ++ (s3 $ s1 xs)
s1 :: String -> String
s1 xs = drop 1 $ dropWhile (/= '$') $ tail $ snd $ break ('$'==) xs
This seems like a nice application for parsers. A solution using trifecta:
import Control.Applicative
import Data.Foldable
import Data.Functor
import Text.Trifecta
input :: String
input = "Hello, my $name is$ Danny and I $like$ Haskell"
cutChunk :: CharParsing f => f String
cutChunk = "" <$ (char '$' *> many (notChar '$') <* char '$')
cutChunk matches $, followed by 0 or more (many) non-$ characters, then another $. Then we use ("" <$) to make this parser's value always be the empty string, thus discarding all the characters that this parser matches.
includeChunk :: CharParsing f => f String
includeChunk = some (notChar '$')
includeChunk matches the text that we want to include in the result, which is anything that's not the $ character. It's important that we use some (matching one or more characters) and not many (matching zero or more characters) because we're going to include this parser within another many expression next; if this parser matched on the empty string, then that could loop infinitely.
chunks :: CharParsing f => f String
chunks = fold <$> many (cutChunk <|> includeChunk)
chunks is the parser for everything. Read <|> as "or", as in "parse either a cutChunk or an includeChunk". many (cutChunk <|> includeChunk) is a parser that produces a list of chunks e.g. Success ["Hello, my ",""," Danny and I ",""," Haskell"], so we fold the output to concatenate those chunks together into a single string.
result :: Result String
result = parseString chunks mempty input
The result:
Success "Hello, my Danny and I Haskell"
Your infinite loop comes from calling s3 recursively with no base case:
s3 :: String -> String
s3 xs = s3 $ s2 $ s1 xs
Adding a base case corrects the infinite loop:
s3 xs
| '$' `notElem` xs = xs
| otherwise = ...
This is not the whole answer. Think about what s1 actually does and where you use its return value:
s1 "hello $my name is$ ThreeFx" == "hello "
For further reference, see the break function:
break :: (a -> Bool) -> [a] -> ([a], [a])
I think your logic is wrong, perhaps easier to write it in an elementary way
Prelude> let pr xs = go xs True
Prelude| where go [] _ = []
Prelude| go (x:xs) f | x=='$' = go xs (not f)
Prelude| | f = x : go xs f
Prelude| | otherwise = go xs f
Prelude|
Prelude> pr "Hello, my $name is$ Danny and I $like$ Haskell"
"Hello, my Danny and I Haskell"
Explanation The flag f keeps track of the state (either pass mode or not). If the current char is a token skip and switch state.
I have a problem making Backtracking on Haskell, I know how to do recursive functions but I get troubles when I try to get multiple solutions or the best one (backtracking).
There's a list with some strings, then I need to get the solutions to get from a string to another one changing one letter from the string, I will get the list, the first string and the last one. If there is solution return the count of steps that it did, if there is not solution it returns -1. here's an example:
wordF ["spice","stick","smice","stock","slice","slick","stock"] "spice" "stock"
Then I have my list and I need to start with "spice" and get to "stock"
and the best solution is ["spice","slice","slick","stick","stock"] with four steps to get from "spice" to "stock". then it return 4.
Another solution is ["spice","smice","slice","slick","stick","stock"] with five steps to get to "stock" then it return `5. But this is a wrong solution because there's another one that's better with lesser steps than this one.
I'm having troubles making a backtracking to get the best solution, because I don't know how to make that my code search another solutions and just not one..
Here's a code that i tried to make but i get some errors, btw i dont know if my way to "make" backtracking is good or if there are some mistakes that im not seeing..
wordF :: [String] -> String -> String -> (String, String, Int)
wordF [] a b = (a, b, -1)
wordF list a b | (notElem a list || notElem b list) = (a, b, -1)
| otherwise = (a, b, (wordF2 list a b [a] 0 (length list)))
wordF2 :: [String] -> String -> String -> [String] -> Int -> Int -> Int
wordF2 list a b list_aux cont maxi | (cont==maxi) = 1000
| (a==b) = length list_aux
| (a/=b) && (cont<maxi) && notElemFound && (checkin /= "ThisWRONG") && (wording1<=wording2) = wording1
| (a/=b) && (cont<maxi) && notElemFound && (checkin /= "ThisWRONG") && (wording1>wording2) = wording2
| (a/=b) && (checkin == "ThisWRONG") = wordF2 list a b list_aux (cont+1) maxi
where
checkin = (check_word2 a (list!!cont) (list!!cont) 0)
wording1 = (wordF2 list checkin b (list_aux++[checkin]) 0 maxi)
wording2 = (wordF2 list checkin b (list_aux++[checkin]) 1 maxi)
notElemFound = ((any (==(list!!cont)) list_aux) == False)
check_word2 :: String -> String -> String -> Int -> String
check_word2 word1 word2 word3 dif | (dif > 1) = "ThisWRONG"
| ((length word1 == 1) && (length word2 == 1) && (head word1 == head word2)) = word3
| ((length word1 == 1) && (length word2 == 1) && (head word1 /= head word2) && (dif<1)) = word3
| ((head word1) == (head word2)) = check_word2 (tail word1) (tail word2) word3 dif
| otherwise = check_word2 (tail word1) (tail word2) word3 (dif+1)
My first function wordF2 get the list, the start, the end, an auxiliary list to get the current solution with the first element that always will be there ([a]), a counter with 0, and the max size of the counter (length list)..
and the second function check_word2 it checks if a word can pass to another word, like "spice" to "slice" if it cant like "spice" to "spoca" it returns "ThisWRONG".
This solution gets an error of pattern match failure
Program error: pattern match failure: wordF2 ["slice","slick"] "slice" "slick" ["slice"] 0 1
I was trying with little cases and nothing, and I'm restricting that i get a wrong position of the list with the count and the max.
Or may be I dont know how to implement backtracking on haskell to get multiple solutions, the best solution, etc..
UPDATE: I did a solution but its not backtracking
wordF :: [String] -> String -> String -> (String, String, Int)
wordF [] a b = (a, b, -1)
wordF list a b | (notElem a list || notElem b list) = (a, b, -1)
| otherwise = (a, b, (wordF1 list a b))
wordF1 :: [String] -> String -> String -> Int
wordF1 list a b | ((map length (wordF2 (subconjuntos2 (subconjuntos list) a b))) == []) = -1
| (calculo > 0) = calculo
| otherwise = -1
where
calculo = (minimum (map length (wordF2 (subconjuntos2 (subconjuntos list) a b))))-1
wordF2 :: [[String]] -> [[String]]
wordF2 [[]] = []
wordF2 (x:xs) | ((length xs == 1) && ((check_word x) == True) && ((check_word (head xs)) == True)) = x:xs
| ((length xs == 1) && ((check_word x) == False) && ((check_word (head xs)) == True)) = xs
| ((length xs == 1) && ((check_word x) == True) && ((check_word (head xs)) == False)) = [x]
| ((length xs == 1) && ((check_word x) == False) && ((check_word (head xs)) == False)) = []
| ((check_word x) == True) = x:wordF2 xs
| ((check_word x) == False ) = wordF2 xs
check_word :: [String] -> Bool
check_word [] = False
check_word (x:xs) | ((length xs == 1) && ((check_word2 x (head xs) 0) == True)) = True
| ((length xs >1) && ((check_word2 x (head xs) 0) == True)) = True && (check_word xs)
| otherwise = False
check_word2 :: String -> String -> Int -> Bool
check_word2 word1 word2 dif | (dif > 1) = False
| ((length word1 == 1) && (length word2 == 1) && (head word1 == head word2)) = True
| ((length word1 == 1) && (length word2 == 1) && (head word1 /= head word2) && (dif<1)) = True
| ((head word1) == (head word2)) = check_word2 (tail word1) (tail word2) dif
| otherwise = check_word2 (tail word1) (tail word2) (dif+1)
subconjuntos2 :: [[String]] -> String -> String -> [[String]]
subconjuntos2 [] a b = []
subconjuntos2 (x:xs) a b | (length x <= 1) = subconjuntos2 xs a b
| ((head x == a) && (last x == b)) = (x:subconjuntos2 xs a b)
| ((head x /= a) || (last x /= b)) = (subconjuntos2 xs a b)
subconjuntos :: [a] -> [[a]]
subconjuntos [] = [[]]
subconjuntos (x:xs) = [x:ys | ys <- sub] ++ sub
where sub = subconjuntos xs
Mmm may be its inefficient but at least it does the solution..
i search all posible solutions, i compare head == "slice" and last == "stock", then i filter the ones that are solution and print the shorter one,
thanks and if you guys have any suggest say it :)
Not thoroughly tested, but this hopefully will help:
import Data.Function (on)
import Data.List (minimumBy, delete)
import Control.Monad (guard)
type Word = String
type Path = [String]
wordF :: [Word] -> Word -> Word -> Path
wordF words start end =
start : minimumBy (compare `on` length) (generatePaths words start end)
-- Use the list monad to do the nondeterminism and backtracking.
-- Returns a list of all paths that lead from `start` to `end`
-- in steps that `differByOne`.
generatePaths :: [Word] -> Word -> Word -> [Path]
generatePaths words start end = do
-- Choose one of the words, nondeterministically
word <- words
-- If the word doesn't `differByOne` from `start`, reject the choice
-- and backtrack.
guard $ differsByOne word start
if word == end
then return [word]
else do
next <- generatePaths (delete word words) word end
return $ word : next
differsByOne :: Word -> Word -> Bool
differsByOne "" "" = False
differsByOne (a:as) (b:bs)
| a == b = differsByOne as bs
| otherwise = as == bs
Example run:
>>> wordF ["spice","stick","smice","stock","slice","slick","stock"] "spice" "stock"
["spice","slice","slick","stick","stock"]
The list monad in Haskell is commonly described as a form of nondeterministic, backtracking computation. What the code above is doing is allowing the list monad to take on the responsibility of generating alternatives, testing whether they satisfy criteria, and backtracking on failure to the most recent choice point. The bind of the list monad, e.g. word <- words, means "nondeterministically pick one of the words. guard means "if the choices so far don't satisfy this condition, backtrack and make a different choice. The result of a list monad computation is the list of all the results that stem from choices that did not violate any guards.
If this looks like list comprehensions, well, list comprehensions are the same thing as the list monad—I chose to express it with the monad instead of comprehensions.
There have been several articles published recently on classic brute-force search problems.
Mark Dominus published a simple example of using lists for a simple exhaustive search.
Justin Le followed up with a small modification to the previous article that simplified tracking the current state of the search.
I followed up with a further modification that allowed measuring the gains from early rejection of part of the search tree.
Note that the code in my article is quite slow because it's measuring the amount of work done as well as doing it. My article has good examples for how to quickly reject parts of the search tree, but it should be considered only an illustration - not production code.
A brute force approach using recursion:
import Data.List (filter, (\\), reverse, delete, sortBy)
import Data.Ord (comparing)
neighbour :: String -> String -> Bool
neighbour word = (1 ==) . length . (\\ word)
process :: String -> String -> [String] -> [(Int, [String])]
process start end dict =
let
loop :: String -> String -> [String] -> [String] -> [(Int,[String])] -> [(Int,[String])]
loop start end dict path results =
case next of
[] -> results
xs ->
if elem end xs
then (length solution, solution) : results
else results ++ branches xs
where
next = filter (neighbour start) dict'
dict' = delete start dict
path' = start : path
branches xs = [a | x <- xs, a <- loop x end dict' path' results]
solution = reverse (end : path')
in
loop start end dict [] []
shortestSolution :: Maybe Int
shortestSolution = shortest solutions
where
solutions = process start end dict
shortest s =
case s of
[] -> Nothing
xs -> Just $ fst $ head $ sortBy (comparing fst) xs
start = "spice"
end = "stock"
dict = ["spice","stick","smice","slice","slick","stock"]
Notes:
This code computes all possibles solutions (process) and select the shortest one (shortestSolution), as Carl said, you might want to prune parts of the search tree for better performance.
Using a Maybe instead of returning -1 when a function can fail to return results is preferred.
Another way using a tree with breadth-first search:
import Data.Tree
import Data.List( filter, (\\), delete )
import Data.Maybe
node :: String -> [String] -> Tree String
node label dict = Node{ rootLabel = label, subForest = branches label (delete label dict) }
branches :: String -> [String] -> [Tree String]
branches start dict = map (flip node dict) (filter (neighbour start) dict)
neighbour :: String -> String -> Bool
neighbour word = (1 ==) . length . (\\ word)
-- breadth first traversal
shortestBF tree end = find [tree] end 0
where
find ts end depth
| null ts = Nothing
| elem end (map rootLabel ts) = Just depth
| otherwise = find (concat (map subForest ts)) end (depth+1)
result = shortestBF tree end
tree :: Tree String
tree = node start dict
start = "spice"
end = "stock"
dict = ["spice","stick","smice","slice","slick","stock"]
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"
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)