I have the following code. The M prefix designates functions from Data.Map.Strict, and Table is a type alias for Data.Map.Strict.Map Mapping Bool, where Mapping is an arbitrary opaque structure.
computeCoverage :: Table -> Expr -> Maybe Coverage
computeCoverage t e = go t True M.empty
where go src flag targ
| null src = if flag
then Nothing
else Just (M.size t, targ)
| otherwise = let ((m, b), rest) = M.deleteFindMin src
result = interpret e m
flag' = result && flag in
go rest flag' (if b == result then targ else M.insert m b targ)
I would like to be able to use Control.Parallel to perform this with as much parallelism as possible. However, I'm not sure how to do this. Based on reading Data.Map.Strict, it seems what you're supposed to do is call splitRoot, then do whatever parallel stuff you want on the resulting list, then recombine (I guess?). Have I basically got the right idea? If not, what should I do instead to parallelize the code above?
Here's a contrived example. You just use parMap over M.splitRoot m:
import qualified Data.Map.Strict as M
import Control.Parallel.Strategies
import System.Environment
fib 0 = 0
fib 1 = 1
fib n = fib (n-2) + fib (n-1)
theMap :: Int -> M.Map Int Int
theMap n = M.fromList [ (x, 33 + mod x 3) | x <- [1..n] ]
isInteresting n = mod (fib n) 2 == 0
countInteresting :: M.Map Int Int -> Int
countInteresting m = length $ filter isInteresting (M.elems m)
doit :: Int -> [Int]
doit n = parMap rseq countInteresting (M.splitRoot $ theMap n)
main :: IO ()
main = do
( arg1 : _) <- getArgs
let n = read arg1
print $ doit n
Note, however these caveats:
the splits may not be of equal size
use splitRoot if working with a Map is helpful for your computation; this particular example doesn't benefit from the Map structure of root - it could have just parMapped over the elements.
I'm new to Haskell and trying to write simple program to find maximal element and it's index from intput. I receive values to compare one by one. Maximal element I'm holding in maxi variable, it's index - in maxIdx. Here's my program:
loop = do
let maxi = 0
let maxIdx = 0
let idx = 0
let idxN = 0
replicateM 5 $ do
input_line <- getLine
let element = read input_line :: Int
if maxi < element
then do
let maxi = element
let maxIdx = idx
hPutStrLn stderr "INNER CHECK"
else
hPutStrLn stderr "OUTER CHECK"
let idx = idxN + 1
let idxN = idx
print maxIdx
loop
Even though I know elements coming are starting from bigger to smaller (5, 4, 3, 2, 1) program enters INNER CHECK all the time (it should happen only for the first element!) and maxIdx is always 0.
What am I doing wrong?
Thanks in advance.
Anyway, let's have fun.
loop = do
let maxi = 0
let maxIdx = 0
let idx = 0
let idxN = 0
replicateM 5 $ do
input_line <- getLine
let element = read input_line :: Int
if maxi < element
then do
let maxi = element
let maxIdx = idx
hPutStrLn stderr "INNER CHECK"
else
hPutStrLn stderr "OUTER CHECK"
let idx = idxN + 1
let idxN = idx
print maxIdx
loop
is not a particularly Haskelly code (and as you know is not particularly correct).
Let's make if Haskellier.
What do we do here? We've an infinite loop, which is reading a line 5 times, does something to it, and then calls itself again for no particular reason.
Let's split it:
import Control.Monad
readFiveLines :: IO [Int]
readFiveLines = replicateM 5 readLn
addIndex :: [Int] -> [(Int, Int)]
addIndex xs = zip xs [0..]
findMaxIndex :: [Int] -> Int
findMaxIndex xs = snd (maximum (addIndex xs))
loop :: ()
loop = loop
main :: IO ()
main = do xs <- readFiveLines
putStrLn (show (findMaxIndex xs))
snd returns the second element from a tuple; readLn is essentially read . getLine; zip takes two lists and returns a list of pairs; maximum finds a maximum value.
I left loop intact in its original beauty.
You can be even Haskellier if you remember that something (huge expression) can be replaced with something $ huge expression ($ simply applies its left operand to its right operand), and the functions can be combined with .: f (g x) is the same as (f . g) x, or f . g $ x (see? it's working for the left side as well!). Additionally, zip x y can be rewritten as x `zip` y
import Control.Monad
readFiveLines :: IO [Int]
readFiveLines = replicateM 5 readLn
addIndex :: [Int] -> [(Int, Int)]
addIndex = (`zip` [0..])
findMaxIndex :: [Int] -> Int
findMaxIndex = snd . maximum . addIndex
main :: IO ()
main = do xs <- readFiveLines
putStrLn . show . findMaxIndex $ xs
As for debug print, there's a package called Debug.Trace and a function traceShow which prints its first argument (formatted with show, hence the name) to stderr, and returns its second argument:
findMaxIndex :: [Int] -> Int
findMaxIndex = snd . (\xs -> traceShow xs (maximum xs)) . addIndex
That allows you to tap onto any expression and see what's coming in (and what are the values around — you can show tuples, lists, etc.)
I think alf's answer is very good, but for what it's worth, here's how I would interpret your intention.
{-# LANGUAGE FlexibleContexts #-}
module Main where
import System.IO
import Control.Monad.State
data S = S { maximum :: Int
, maximumIndex :: Int
, currentIndex :: Int }
update :: Int -> Int -> S -> S
update m mi (S _ _ ci) = S m mi ci
increment :: S -> S
increment (S m mi ci) = S m mi (ci+1)
next :: (MonadIO m, MonadState S m) => m ()
next = do
S maxi maxIdx currIdx <- get
input <- liftIO $ getLine
let element = read input :: Int
if maxi < element
then do
modify (update element currIdx)
liftIO $ hPutStrLn stderr "INNER CHECK"
else
liftIO $ hPutStrLn stderr "OUTER CHECK"
modify increment
run :: Int -> IO S
run n = execStateT (replicateM_ n next) (S 0 0 0)
main :: IO ()
main = do
S maxi maxIdx _ <- run 5
putStrLn $ "maxi: " ++ (show maxi) ++ " | maxIdx: " ++ (show maxIdx)
This uses a monad transformer to combine a stateful computation with IO. The get function retrieves the current state, and the modify function lets you change the state.
This question already has answers here:
How to get normal value from IO action in Haskell
(2 answers)
Closed 7 years ago.
I just started learning Haskell and got my first project working today. Its a small program that uses Network.HTTP.Conduit and Graphics.Rendering.Chart (haskell-chart) to plot the amount of google search results for a specific question with a changing number in it.
My problem is that simple-http from the conduit package returns a monad (I hope I understood the concept of monads right...), but I only want to use the ByteString inside of it, that contains the html-code of the website. So until now i use download = unsafePerformIO $ simpleHttp url to use it later without caring about the monad - I guess that's not the best way to do that.
So: Is there any better solution so that I don't have to carry the monad with me the whole evaluation? Or would it be better to leave it the way the result is returned (with the monad)?
Here's the full program - the mentioned line is in getResultCounter. If things are coded not-so-well and could be done way better, please remark that too:
import System.IO.Unsafe
import Network.HTTP.Conduit (simpleHttp)
import qualified Data.ByteString.Lazy.Char8 as L
import Graphics.Rendering.Chart.Easy
import Graphics.Rendering.Chart.Backend.Cairo
numchars :: [Char]
numchars = "1234567890"
isNum :: Char -> Bool
isNum = (\x -> x `elem` numchars)
main = do
putStrLn "Please input your Search (The first 'X' is going to be replaced): "
search <- getLine
putStrLn "X ranges from: "
from <- getLine
putStrLn "To: "
to <- getLine
putStrLn "In steps of (Only whole numbers are accepted):"
step <- getLine
putStrLn "Please have some patience..."
let range = [read from,(read from + read step)..read to] :: [Int]
let searches = map (replaceX search) range
let res = map getResultCounter searches
plotList search ([(zip range res)] :: [[(Int,Integer)]])
putStrLn "Done."
-- Creates a plot from the given data
plotList name dat = toFile def (name++".png") $ do
layout_title .= name
plot (line "Results" dat)
-- Calls the Google-site and returns the number of results
getResultCounter :: String -> Integer
getResultCounter search = read $ filter isNum $ L.unpack parse :: Integer
where url = "http://www.google.de/search?q=" ++ search
download = unsafePerformIO $ simpleHttp url -- Not good
parse = takeByteStringUntil "<"
$ dropByteStringUntil "id=\"resultStats\">" download
-- Drops a ByteString until the desired String is found
dropByteStringUntil :: String -> L.ByteString -> L.ByteString
dropByteStringUntil str cont = helper str cont 0
where helper s bs n | (bs == L.empty) = L.empty
| (n >= length s) = bs
| ((s !! n) == L.head bs) = helper s (L.tail bs) (n+1)
| ((s !! n) /= L.head bs) = helper s (L.tail bs) 0
-- Takes a ByteString until the desired String is found
takeByteStringUntil :: String -> L.ByteString -> L.ByteString
takeByteStringUntil str cont = helper str cont 0
where helper s bs n | bs == L.empty = bs
| n >= length s = L.empty
| s !! n == L.head bs = L.head bs `L.cons`
helper s (L.tail bs) (n + 1)
| s !! n /= L.head bs = L.head bs `L.cons`
helper s (L.tail bs) 0
-- Replaces the first 'X' in a string with the show value of the given value
replaceX :: (Show a) => String -> a -> String
replaceX str x | str == "" = ""
| head str == 'X' = show x ++ tail str
| otherwise = head str : replaceX (tail str) x
This is a lie:
getResultCounter :: String -> Integer
The type signature above is promising that the resulting integer only depends on the input string, when this is not the case: Google can add/remove results from one call to the other, affecting the output.
Making the type more honest, we get
getResultCounter :: String -> IO Integer
This honestly admits it's going to interact with the external world. The code then is easily adapted to:
getResultCounter search = do
let url = "http://www.google.de/search?q=" ++ search
download <- simpleHttp url -- perform IO here
let parse = takeByteStringUntil "<"
$ dropByteStringUntil "id=\"resultStats\">" download
return (read $ filter isNum $ L.unpack parse :: Integer)
Above, I tried to preserve the original structure of the code.
Now, in main we can no longer do
let res = map getResultCounter searches
but we can do
res <- mapM getResultCounter searches
after importing Control.Monad.
I hope this works by just pasting and running it with "runghc euler4.hs 1000". Since I am having a hard time learning Haskell, can someone perhaps tell me how I could improve here? Especially all those "fromIntegral" are a mess.
module Main where
import System.Environment
main :: IO ()
main = do
args <- getArgs
let
hBound = read (args !! 0)::Int
squarePal = pal hBound
lBound = floor $ fromIntegral squarePal /
(fromIntegral hBound / fromIntegral squarePal)
euler = maximum $ takeWhile (>squarePal) [ x | y <- [lBound..hBound],
z <- [y..hBound],
let x = y * z,
let s = show x,
s == reverse s ]
putStrLn $ show euler
pal :: Int -> Int
pal n
| show pow == reverse (show pow) = n
| otherwise = pal (n-1)
where
pow = n^2
If what you want is integer division, you should use div instead of converting back and forth to Integral in order to use ordinary /.
module Main where
import System.Environment
main :: IO ()
main = do
(arg:_) <- getArgs
let
hBound = read arg :: Int
squarePal = pal hBound
lBound = squarePal * squarePal `div` hBound
euler = maximum $ takeWhile (>squarePal) [ x | y <- [lBound..hBound],
z <- [y..hBound],
let x = y * z,
let s = show x,
s == reverse s ]
print euler
pal :: Int -> Int
pal n
| show pow == reverse (show pow) = n
| otherwise = pal (n - 1)
where
pow = n * n
(I've re-written the lbound expression, that used two /, and fixed some styling issues highlighted by hlint.)
Okay, couple of things:
First, it might be better to pass in a lower bound and an upper bound for this question, it makes it a little bit more expandable.
If you're only going to use the first two (one in your previous case) arguments from the CL, we can handle this with pattern matching easily and avoid yucky statements like (args !! 0):
(arg0:arg1:_) <- getArgs
Let's convert these to Ints:
let [a, b] = map (\x -> read x :: Int) [arg0,arg1]
Now we can reference a and b, our upper and lower bounds.
Next, let's make a function that runs through all of the numbers between an upper and lower bound and gets a list of their products:
products a b = [x*y | x <- [a..b], y <- [x..b]]
We do not have to run over each number twice, so we start x at our current y to get all of the different products.
from here, we'll want to make a method that filters out non-palindromes in some data set:
palindromes xs = filter palindrome xs
where palindrome x = show x == reverse $ show x
finally, in our main function:
print . maximum . palindromes $ products a b
Here's the full code if you would like to review it:
import System.Environment
main = do
(arg0:arg1:_) <- getArgs
let [a, b] = map (\x -> read x :: Int) [arg0,arg1]
print . maximum . palindromes $ products a b
products a b = [x*y | x <- [a..b], y <- [x..b]]
palindromes = filter palindrome
where palindrome x = (show x) == (reverse $ show x)
I'm trying to complete the last part of my Haskell homework and I'm stuck, my code so far:
data Entry = Entry (String, String)
class Lexico a where
(<!), (=!), (>!) :: a -> a -> Bool
instance Lexico Entry where
Entry (a,_) <! Entry (b,_) = a < b
Entry (a,_) =! Entry (b,_) = a == b
Entry (a,_) >! Entry (b,_) = a > b
entries :: [(String, String)]
entries = [("saves", "en vaut"), ("time", "temps"), ("in", "<`a>"),
("{", "{"), ("A", "Un"), ("}", "}"), ("stitch", "point"),
("nine.", "cent."), ("Zazie", "Zazie")]
build :: (String, String) -> Entry
build (a, b) = Entry (a, b)
diction :: [Entry]
diction = quiksrt (map build entries)
size :: [a] -> Integer
size [] = 0
size (x:xs) = 1+ size xs
quiksrt :: Lexico a => [a] -> [a]
quiksrt [] = []
quiksrt (x:xs)
|(size [y|y <- xs, y =! x]) > 0 = error "Duplicates not allowed."
|otherwise = quiksrt [y|y <- xs, y <! x]++ [x] ++ quiksrt [y|y <- xs, y >! x]
english :: String
english = "A stitch in time save nine."
show :: Entry -> String
show (Entry (a, b)) = "(" ++ Prelude.show a ++ ", " ++ Prelude.show b ++ ")"
showAll :: [Entry] -> String
showAll [] = []
showAll (x:xs) = Main.show x ++ "\n" ++ showAll xs
main :: IO ()
main = do putStr (showAll ( diction ))
The question asks:
Write a Haskell programs that takes
the English sentence 'english', looks
up each word in the English-French
dictionary using binary search,
performs word-for-word substitution,
assembles the French translation, and
prints it out.
The function 'quicksort' rejects
duplicate entries (with 'error'/abort)
so that there is precisely one French
definition for any English word. Test
'quicksort' with both the original
'raw_data' and after having added
'("saves", "sauve")' to 'raw_data'.
Here is a von Neumann late-stopping
version of binary search. Make a
literal transliteration into Haskell.
Immediately upon entry, the Haskell
version must verify the recursive
"loop invariant", terminating with
'error'/abort if it fails to hold. It
also terminates in the same fashion if
the English word is not found.
function binsearch (x : integer) : integer
local j, k, h : integer
j,k := 1,n
do j+1 <> k --->
h := (j+k) div 2
{a[j] <= x < a[k]} // loop invariant
if x < a[h] ---> k := h
| x >= a[h] ---> j := h
fi
od
{a[j] <= x < a[j+1]} // termination assertion
found := x = a[j]
if found ---> return j
| not found ---> return 0
fi
In the Haskell version
binsearch :: String -> Integer -> Integer -> Entry
as the constant dictionary 'a' of type
'[Entry]' is globally visible. Hint:
Make your string (English word) into
an 'Entry' immediately upon entering
'binsearch'.
The programming value of the
high-level data type 'Entry' is that,
if you can design these two functions
over the integers, it is trivial to
lift them to to operate over Entry's.
Anybody know how I'm supposed to go about my binarysearch function?
The instructor asks for a "literal transliteration", so use the same variable names, in the same order. But note some differences:
the given version takes only 1
parameter, the signature he gives
requires 3. Hmmm,
the given version is not recursive, but he asks for a
recursive version.
Another answer says to convert to an Array, but for such a small exercise (this is homework after all), I felt we could pretend that lists are direct access. I just took your diction::[Entry] and indexed into that. I did have to convert between Int and Integer in a few places.
Minor nit: You've got a typo in your english value (bs is a shortcut to binSearch I made):
*Main> map bs (words english)
[Entry ("A","Un"),Entry ("stitch","point"),Entry ("in","<`a>"),Entry ("time","te
mps"),*** Exception: Not found
*Main> map bs (words englishFixed)
[Entry ("A","Un"),Entry ("stitch","point"),Entry ("in","<`a>"),Entry ("time","te
mps"),Entry ("saves","en vaut"),Entry ("nine.","cent.")]
*Main>
A binary search needs random access, which is not possible on a list. So, the first thing to do would probably be to convert the list to an Array (with listArray), and do the search on it.
here's my code for just the English part of the question (I tested it and it works perfectly) :
module Main where
class Lex a where
(<!), (=!), (>!) :: a -> a -> Bool
data Entry = Entry String String
instance Lex Entry where
(Entry a _) <! (Entry b _) = a < b
(Entry a _) =! (Entry b _) = a == b
(Entry a _) >! (Entry b _) = a > b
-- at this point, three binary (infix) operators on values of type 'Entry'
-- have been defined
type Raw = (String, String)
raw_data :: [Raw]
raw_data = [("than a", "qu'un"), ("saves", "en vaut"), ("time", "temps"),
("in", "<`a>"), ("worse", "pire"), ("{", "{"), ("A", "Un"),
("}", "}"), ("stitch", "point"), ("crime;", "crime,"),
("a", "une"), ("nine.", "cent."), ("It's", "C'est"),
("Zazie", "Zazie"), ("cat", "chat"), ("it's", "c'est"),
("raisin", "raisin sec"), ("mistake.", "faute."),
("blueberry", "myrtille"), ("luck", "chance"),
("bad", "mauvais")]
cook :: Raw -> Entry
cook (x, y) = Entry x y
a :: [Entry]
a = map cook raw_data
quicksort :: Lex a => [a] -> [a]
quicksort [] = []
quicksort (x:xs) = quicksort (filter (<! x) xs) ++ [x] ++ quicksort (filter (=! x) xs) ++ quicksort (filter (>! x) xs)
getfirst :: Entry -> String
getfirst (Entry x y) = x
getsecond :: Entry -> String
getsecond (Entry x y) = y
binarysearch :: String -> [Entry] -> Int -> Int -> String
binarysearch s e low high
| low > high = " NOT fOUND "
| getfirst ((e)!!(mid)) > s = binarysearch s (e) low (mid-1)
| getfirst ((e)!!(mid)) < s = binarysearch s (e) (mid+1) high
| otherwise = getsecond ((e)!!(mid))
where mid = (div (low+high) 2)
translator :: [String] -> [Entry] -> [String]
translator [] y = []
translator (x:xs) y = (binarysearch x y 0 ((length y)-1):translator xs y)
english :: String
english = "A stitch in time saves nine."
compute :: String -> [Entry] -> String
compute x y = unwords(translator (words (x)) y)
main = do
putStr (compute english (quicksort a))
An important Prelude operator is:
(!!) :: [a] -> Integer -> a
-- xs!!n returns the nth element of xs, starting at the left and
-- counting from 0.
Thus, [14,7,3]!!1 ~~> 7.