haskell [[Char]] to [[Int]] - list

I've got problem with a Haskell program.
I'm trying to change [[Char]] to [[Int]]
I've got
["2","2","1","2,2","1"]
list of char list
and I'm trying to change it to [[Int]]
[[2],[2],[1],[2,2],[1]]
I've tried
f :: [String] -> [Int]
f = map read
but it gives me
[2,2,1,*** Exception: Prelude.read: no parse
Can anybody help me with this?

The reason that this fails is because a string "2,2" can not be converted to an Int itself: this is a digit followed by a comma, followed by a digit. An Int is parsed by an optional minus sign, followed by some digits, and some extra possibilities like hexadecimal numbers, but let us ignore these for now.
The type signature you specify for f is however incorrect, based on the expected output. Your output type seems to be a list of lists of Ints, so [[Int]]. That means that you should specify f as:
f :: [String] -> [[Int]]
f = ...
We thus need to read every String to an [Int]. We can not use read directly here, since reading to an [Int] expects the string to start and end with square brackets. We can however add these manually like:
f :: [String] -> [[Int]]
f = map (\s -> read ('[' : s ++ "]"))
or a point-free version:
f :: [String] -> [[Int]]
f = map (read . ('[' :) . (++ "]"))
For example:
Prelude> f ["2","2","1","2,2","1"]
[[2],[2],[1],[2,2],[1]]
Towards safer parsing with readMaybe
Parsing from Strings like in the above way is of course not very "safe", since it is possible that the String does not follow the format. We can make this more safe and use for example readMaybe :: Read a => String -> Maybe a:
import Text.Read(readMaybe)
f :: [String] -> [Maybe [Int]]
f = map (readMaybe . ('[' :) . (++ "]"))
For example:
Prelude Text.Read> f ["2", "3;2", "4,7,3", "bla"]
[Just [2],Nothing,Just [4,7,3],Nothing]
we can omit the failed reads for example by using catMaybes :: [Maybe a] -> [a]:
import Data.Maybe(catMaybes)
import Text.Read(readMaybe)
f :: [String] -> [[Int]]
f = catMaybes . map (readMaybe . ('[' :) . (++ "]"))
For example:
Prelude Data.Maybe Text.Read> f ["2", "3;2", "4,7,3", "bla"]
[[2],[4,7,3]]
or as #dfeuer said, we can use traverse :: (Applicative f, Traversable t) => (a -> f b) -> t a -> f (t b) to return an [[Int]] result wrapped in a Just if all parsing succeeded, and Nothing otherwise:
import Text.Read(readMaybe)
f :: [String] -> Maybe [[Int]]
f = traverse (readMaybe . ('[' :) . (++ "]"))
For example:
Prelude Text.Read> f ["2","2","1","2,2","1"]
Just [[2],[2],[1],[2,2],[1]]
Prelude Text.Read> f ["2", "3;2", "4,7,3", "bla"]
Nothing
Parse with error messages with readEither
We can obtain an error message wrapped in a Left in case the parsing fails by using readEither :: Read a => String -> Either String a:
import Text.Read(readEither)
f :: [String] -> [Either String [Int]]
f = map (readEither . ('[' :) . (++ "]"))
For example:
Prelude Text.Read> f ["2", "3;2", "4,7,3", "bla"]
[Right [2],Left "Prelude.read: no parse",Right [4,7,3],Left "Prelude.read: no parse"]
and use traverse in the same way to obtain an error message wrapped in a Left or the complete result in a Right:
import Text.Read(readEither)
f :: [String] -> Either String [[Int]]
f = traverse (readEither . ('[' :) . (++ "]"))
For example:
Prelude Text.Read> f ["2","2","1","2,2","1"]
Right [[2],[2],[1],[2,2],[1]]
Prelude Text.Read> f ["2", "3;2", "4,7,3", "bla"]
Left "Prelude.read: no parse"
Here, like #dfeuer says, it does not really shows much information. There are however parsers that can provide more informative parsing errors.

Related

How to enhance small Haskell Code Snippet

just recently I started to try out haskell.
It's fun trying out different exercises, but sometimes I get the feeling, that my found solutions are far from elegant: The following Code Snipplet will find the longest sub-sequence in a list, which will satisfy a given condition (for example uppercase letters etc.)
Could you help a noob to make everything shorter and more elegant - every advice is highly appreciated.
import Data.Char
longer :: [a] -> [a] -> [a]
longer x y = if length x > length y
then x
else y
longest :: [[a]]->[a]
longest = foldl longer []
nextSequence :: (a->Bool) -> [a] ->([a],[a])
nextSequence f x = span f (dropWhile (not . f) x)
longestSubsequence :: (a -> Bool) -> [a] -> [a]
longestSubsequence _ x | null x = []
longestSubsequence f x =
longest $ (\y -> [fst y , longestSubsequence f $ snd y]) (nextSequence f x)
testSequence :: String
testSequence = longestSubsequence Data.Char.isUpper
"hkerhklehrERJKJKJERKJejkrjekERHkhkerHERKLJHERJKHKJHERdjfkj"
At first, you can define your longest like this:
import Data.Function
import Data.List
longest :: [[a]] -> [a]
longest = maximumBy (compare `on` length)
And to get all subsequences that satisfy a given condition you can write a function like this:
import Data.List
getSatisfyingSubseqs :: (a -> Bool) -> [a] -> [[a]]
getSatisfyingSubseqs f = filter (f . head) . groupBy same
where same x y = f x == f y
Here we group elements where the condition yields the same result and filter only subsequences that satisfy the condition.
In the total:
longestSubsequence :: (a -> Bool) -> [a] -> [a]
longestSubsequence f = longest . getSatisfyingSubseqs f
UPDATE: And if you want to make it shorter, you can just throw out the auxiliary functions and write the whole at a time:
longestSubsequence :: (a -> Bool) -> [a] -> [a]
longestSubsequence f = maximumBy (compare `on` length) . filter (f . head) . groupBy same
where same x y = f x == f y
(Don't forget the imports)
You can run it there: https://repl.it/#Yuri12358/so-longestsequence
The span :: (a -> Bool) -> [a] -> ([a], [a]) function could be very handy here. Also note that f <$> (a,b) = (a,f b). Probably not very efficient due to the length checks but it should do the job.
lss :: (a -> Bool) -> [a] -> [a]
lss f [] = []
lss f ls#(x:xs) = if f x then longer (lss f <$> span f ls)
else lss f xs
where
longer ::([a],[a]) -> [a]
longer (xs,ys) = if length xs >= length ys then xs else ys
Your longer function uses length, which means it doesn't work if either input is infinite. However, it can be improved to work when at most one is infinite:
longer l1 l2 = go l1 l2
where
go [] _ = l2
go _ [] = l1
go (_:xs) (_:ys) = go xs ys
This is also a performance optimization. Before, if you had a 10-element list and a 10-million-element list, it would walk through all 10 million elements of the 10-million-element list before returning it. Here, it will return it as soon as it gets to the 11th element instead.

Haskell Text Parser Combinators to parse a Range Greedily like Regex range notation

In regex you can acquire a range of a parse by doing something like \d{1,5}, which parses a digit 1 to 5 times greedily. Or you do \d{1,5}? to make it lazy.
How would you do this in Haskell's Text.ParserCombinators.ReadP?
My attempt gave this:
rangeParse :: Read a => ReadP a -> [Int] -> ReadP [a]
rangeParse parse ranges = foldr1 (<++) $ fmap (\n -> count n $ parse) ranges
Which if you do it like rangeParse (satisfy isDigit) ([5,4..1]) will perform a greedy parse of digits 1 to 5 times. While if you swap the number sequent to [1..5], you get a lazy parse.
Is there a better or more idiomatic way to do this with parser combinators?
update: the below is wrong - for example
rangeGreedy 2 4 a <* string "aab", the equivalent of regexp a{2,4}aab, doesn't match. The questioner's solution gets this right. I won't delete the answer just yet in case it keeps someone else from making the same mistake.
=========
This isn't a complete answer, just a possible way to write the greedy
version. I haven't found a nice way to do the lazy version.
Define a left-biased version of option that returns Maybes:
greedyOption :: ReadP a -> ReadP (Maybe a)
greedyOption p = (Just <$> p) <++ pure Nothing
Then we can do up to n of something with a replicateM of them:
upToGreedy :: Int -> ReadP a -> ReadP [a]
upToGreedy n p = catMaybes <$> replicateM n (greedyOption p)
To allow a minimum count, do the mandatory part separately and append
it:
rangeGreedy :: Int -> Int -> ReadP a -> ReadP [a]
rangeGreedy lo hi p = (++) <$> count lo p <*> upToGreedy (hi - lo) p
The rest of my test code in case it's useful for anyone:
module Main where
import Control.Monad (replicateM)
import Data.Maybe (catMaybes)
import Text.ParserCombinators.ReadP
main :: IO ()
main = mapM_ go ["aaaaa", "aaaab", "aaabb", "aabbb", "abbbb", "bbbbb"]
where
go = print . map fst . readP_to_S test
test :: ReadP [String]
test = ((++) <$> rangeGreedy 2 4 a <*> many aOrB) <* eof
where
a = char 'a' *> pure "ay"
aOrB = (char 'a' +++ char 'b') *> pure "ayorbee"

Haskell - Printing a list of tuples

I have a list of tuples in the form:
[(String, Int)]
How can I print this to display like:
String : Int
String : Int
String : Int
...
I am very new to Haskell so please make it as clear as possible. Thank you!
Update: Here's how the code of my program now looks:
main = do
putStrLn "********* Haskell word frequency counter *********"
putStrLn ""
conts <- readFile "text.txt"
let lowConts = map toLower conts
let counted = countAllWords (lowConts)
let sorted = sortTuples (counted)
let reversed = reverse sorted
putStrLn "Word : Count"
mapM_ (printTuple) reversed
-- Counts all the words.
countAllWords :: String -> [(String, Int)]
countAllWords fileContents = wordsCount (toWords (noPunc fileContents))
-- Splits words and removes linking words.
toWords :: String -> [String]
toWords s = filter (\w -> w `notElem` ["and","the","for"]) (words s)
-- Remove punctuation from text String.
noPunc :: String -> String
noPunc xs = [ x | x <- xs, not (x `elem` ",.?!-:;\"\'") ]
-- Counts, how often each string in the given list appears.
wordsCount :: [String] -> [(String, Int)]
wordsCount xs = map (\xs -> (head xs, length xs)) . group . sort $ xs
-- Sort list in order of occurrences.
sortTuples :: [(String, Int)] -> [(String, Int)]
sortTuples sort = sortBy (comparing snd) sort
printTuple :: Show a => [(String, a)] -> IO ()
printTuple xs = forM_ xs (putStrLn . formatOne)
formatOne :: Show a => (String, a) -> String
formatOne (s,i) = s ++ " : " ++ show i
It returns this error to me:
fileToText.hs:18:28:
Couldn't match type ‘(String, Int)’ with ‘[(String, a0)]’
Expected type: [[(String, a0)]]
Actual type: [(String, Int)]
In the second argument of ‘mapM_’, namely ‘reversed’
In a stmt of a 'do' block: mapM_ (printTuple) reversed
Thanks for any help!
let's start by formatting one item:
formatOne :: Show a => (String, a) -> String
formatOne (s,i) = s ++ " : " ++ show i
now you can use this function (for example) with forM_ from Control.Monad to print it to the screen like this (forM_ because we want to be in the IO-Monad - because we are going to use putStrLn):
Prelude> let test = [("X1",4), ("X2",5)]
Prelude> import Control.Monad (forM_)
Prelude Control.Monad> forM_ test (putStrLn . formatOne)
X1 : 4
X2 : 5
in a file you would use it like this:
import Control.Monad (forM_)
printTuples :: Show a => [(String, a)] -> IO ()
printTuples xs = forM_ xs (putStrLn . formatOne)
formatOne :: Show a => (String, a) -> String
formatOne (s,i) = s ++ " : " ++ show i
compiling file
overall here is a version of your code that will at least compile (cannot test it without the text file ;) )
import Control.Monad (forM_)
import Data.Char (toLower)
import Data.List (sort, sortBy, group)
import Data.Ord (comparing)
main :: IO ()
main = do
putStrLn "********* Haskell word frequency counter *********"
putStrLn ""
conts <- readFile "text.txt"
let lowConts = map toLower conts
let counted = countAllWords lowConts
let sorted = sortTuples counted
let reversed = reverse sorted
putStrLn "Word : Count"
printTuples reversed
-- Counts all the words.
countAllWords :: String -> [(String, Int)]
countAllWords = wordsCount . toWords . noPunc
-- Splits words and removes linking words.
toWords :: String -> [String]
toWords = filter (\w -> w `notElem` ["and","the","for"]) . words
-- Remove punctuation from text String.
noPunc :: String -> String
noPunc xs = [ x | x <- xs, x `notElem` ",.?!-:;\"\'" ]
-- Counts, how often each string in the given list appears.
wordsCount :: [String] -> [(String, Int)]
wordsCount = map (\xs -> (head xs, length xs)) . group . sort
-- Sort list in order of occurrences.
sortTuples :: [(String, Int)] -> [(String, Int)]
sortTuples = sortBy $ comparing snd
-- print one tuple per line separated by " : "
printTuples :: Show a => [(String, a)] -> IO ()
printTuples = mapM_ (putStrLn . formatTuple)
where formatTuple (s,i) = s ++ " : " ++ show i
I also removed the compiler warnings and HLINTed it (but skipped the Control.Arrow stuff - I don't think head &&& length is more readable option here)

haskell list character frequenceies

I am having trouble with an assignment question!
Write the function
freq2 :: String -> -> [(Int,[Char])]
Like freq, the function freq2 counts frequency of occurrence of alphabetic characters.
Given the string:
We hold these truths to be self-evident, that all men are created equal, that they are endowed by their Creator with certain unalienable Rights, that among these are Life, Liberty and the pursuit of Happiness
I need to end up with:
[(1,"qv"), (2,"gm"), (3,"cfpwy"), (4,"b"), (5,"u"), (6,"do"),(8,"s"), (9,"ln"), (10,"i"), (12,"r"), (13,"h"), (16,"a"),(22,"t"), (28,"e")]
So far I can get to:
[('q',1),('v',1),('g',2),('m',2),('c',3),('f',3),('p',3),('w',3),('y',3),('b',4),('u',5),('d',6),('o',6),('s',8),('l',9),('n',9),('i',10),('r',12),('h',13),('a',16),('t',22),('e',28)]
Using:
freq2 :: String -> [(Char,Int)]
freq2 input = result2
where
lower_case_list = L.map C.toLower input
filtered_list = L.filter C.isAlpha lower_case_list
result = L.map (\a -> (L.head a, L.length a)) $ L.group $ sort filtered_list
result2 = sortBy (compare `on` snd) result
Is there an easy way to get to the last stage or to do the whole thing, possibly using library functions? Or can you please provide some direction on how to finish off this question?
Thanks
Something like this appended to your solution should work:
result3 = map (\xs#((_,x):_) -> (x, map fst xs)) $ L.groupBy ((==) `on` snd) result2
My preference would be to use a Map for these types of problems though:
import qualified Data.Map as Map
import qualified Data.Char as C
import qualified Data.Tuple as T
string = filter C.isAlpha $ map C.toLower "We hold these truths to be self-evident, that all men are created equal, that they are endowed by their Creator with certain unalienable Rights, that among these are Life, Liberty and the pursuit of Happiness"
swapMapWith f = Map.fromListWith f . map T.swap . Map.toList
freq2 :: String -> [(Int, String)]
freq2 = Map.toList . swapMapWith (++) . foldl (\agg c -> Map.insertWith (+) [c] 1 agg) Map.empty
Method 1:
import needed modules
import Data.Char
import Data.List
filter out uninterested characters and convert the rest to lower case
toLowerAlpha :: String -> String
toLowerAlpha = map toLower . filter isAlpha
sort first, then group, after that the length of each group is the frequency of character in that group
elemFreq :: (Ord a) => [a] -> [(Int, a)]
elemFreq = map (\l -> (length l, head l)) . group . sort
sort and group as step 2, but according to frequency at here, then combine all those characters that have the same frequencies
groupByFreq :: (Integral a, Ord b) => [(a, b)] -> [[(a, b)]]
groupByFreq = groupBy (onFreq (==)) . sortBy (onFreq compare)
where onFreq op (f1,_) (f2,_) = op f1 f2
collectByFreq :: (Integral a) => [[(a, b)]] -> [(a, [b])]
collectByFreq = map (\ls -> (fst . head $ ls, map snd ls))
sequence the above functions will give the required function
freq2 = collectByFreq . groupByFreq . elemFreq . toLowerAlpha
Method 2:
import needed modules
import qualified Data.Char as Char
import qualified Data.Map as Map
filter out uninterested characters and convert the rest to lower case
toLowerAlpha :: String -> String
toLowerAlpha = map Char.toLower . filter Char.isAlpha
create a map, key and value are character and corresponding frequency, respectively
toFreqMap :: (Ord a, Num b) => [a] -> Map.Map a b
toFreqMap = foldr (\c -> Map.insertWith (+) c 1) Map.empty
convert the map created in step 2 to another map, using frequency as key, and characters have that frequency as value
toFreqCol :: (Ord a, Ord b) => Map.Map a b -> Map.Map b [a]
toFreqCol = Map.foldrWithKey (\k a m -> Map.insertWith (++) a [k] m) Map.empty
sequence the above functions will give the required function
freq2 = Map.toAscList . toFreqCol . toFreqMap . toLowerAlpha

Working over a list of lists in Haskell

I'm a bit of a beginner to Haskell so I'm struggling a little with the strict type stuff, just wondering if someone can help me with a function I'm trying to build. Basically, it takes a list of lists, for example:
[[1,2,3], [7,6,8], [0,3,4]]
and adds them together into one list translating the later lists by the number of positions along it is. So working on the example list it would actually be doing something like:
foldl (zipWith +) [] [[1,2,3],[0,7,6,8],[0,0,0,3,4]]
Here's my current function (which gets type errors):
addLists :: [[Integer]] -> [Integer]
addLists [[]] = []
addLists [[x]] = [x]
addLists [x:xs] = zipWith (+) [x] ([0]++ (addLists xs))
Note that ([0]++) is the same as (0:), which will make it look tidier and save us a nanosecond or two.
(I'm joking with the nanosecond thing - no human can tell when something's a nanosecond faster, but it is nicer this way anyway.)
Let's first think about making the lists you need. We want
postponeLists [[1,2,3], [7,6,8], [10,20,30,40]]
= [[1,2,3], [0,7,6,8], [0,0,10,20,30,40]]
= [1,2,3] : ones that should have zero in front of them
That's enough information for a definition:
postponeLists [] = []
postponeLists (l:ls) = l : map (0:) (postponeLists ls)
Now you said
foldl (zipWith +) [] [[1,2,3],[0,7,6,8],[0,0,0,3,4]]
but you mean
foldl (zipWith (+)) [] [[1,2,3],[0,7,6,8],[0,0,0,3,4]]
but unfortunately, that gives you [] because zipWith stops as soon as any of the lists run out of elements.
We need some way of zipping them that doesn't stop.
Solution 1: find the longest one, make them all that maxlength using take maxlength.(++ repeat 0)
Solution 2: write another zipWith function that doesn't stop.
I prefer solution 2. Let's look at the definition of zipWith
zipWith :: (a->b->c) -> [a]->[b]->[c]
zipWith f (a:as) (b:bs) = f a b : zipWith f as bs
zipWith _ _ _ = [] -- here's the problem - it stops as soon as any list is empty
OK, let's not stop then:
zipWithMore :: (a -> a -> a) -> [a] -> [a] -> [a]
zipWithMore f (a:as) (b:bs) = f a b : zipWithMore f as bs
zipWithMore f [] bs = bs -- if there's more in bs, use that
zipWithMore f as [] = as -- if there's more in as, use that
Now you can replace zipWith (+) with zipWithMore (+). I'll leave the punchline to you.
I think this does what you want
import Data.List (transpose)
addLists :: Num a => [[a]] -> [a]
addLists xs = map sum . transpose $ zipWith (\n x -> replicate n 0 ++ x) [0..] xs