Foldr issues (Haskell) - haskell

I'm trying to convert from a table which is of type ([Char, Int]) to a string tab2str :: Table -> String (that follows some specific formatting patterns.)
I'm using foldr (as the title implies) but I'm having some issues getting the exact function to work - i.e. It errors. My function looks like this:
tab2str xs = foldr (++) ' ' $ map (\char count -> show char ++ ':' ++ show count ++ '\n') xs
The output should be each letter in the table, a colon, then \n. So a test might be:
tab2str test1 == "F: 1\no: 1\nl: 1\nd: 1\nr: 1\n"
where
test1 == [(F, 1), (o, 1), (l, 1), (d, 1), (r, 1)]
Any help gratefully received.

After minimal corrections this typechecks:
tab2str xs = foldr (++) " " $ map (\(char, count) -> show char ++ ":" ++ show count ++ "\n") xs
– but produces not quite what you want.
You'll probably like this better:
tab2str table = concat $ map formatRow table
where formatRow (char, count) = [char] ++ ": " ++ show count ++ "\n"
Then your test example:
ghci> let test1 = [('F', 1), ('o', 1), ('l', 1), ('d', 1), ('r', 1)]
ghci> tab2str test1
"F: 1\no: 1\nl: 1\nd: 1\nr: 1\n"
ghci> putStr $ tab2str test1
F: 1
o: 1
l: 1
d: 1
r: 1
ghci>

Related

Haskell string/output manipulation

I do have the following code:
suffixes :: [a] -> [[a]]
suffixes [] = [[]]
suffixes l#(_:t) = l : suffixes t
prefixes :: [a] -> [[a]]
prefixes [] = [[]]
prefixes l#x = l : prefixes (init x)
menu :: Char -> [a] -> Either String [[a]]
menu 'p' l = Right (prefixes l)
menu 's' l = Right (suffixes l)
menu x _ = Left ("(" ++ show x ++ ")" ++ "is not supported, use (p)refix or (s)uffix")
I do have the following test function:
testMenuP = "Expected Right [[1,2],[1],[]]; menu 'p' [1,2] returned " ++ show (menu 'p' [1,2] :: Either String [[Int]])
testMenuS = "Expected Right [[1,2],[2],[]]; menu 's' [1,2] returned " ++ show (menu 's' [1,2] :: Either String [[Int]])
testMenuC = "Expected Left \"(d) is not supported, use (p)refix or (s)uffix\"; menu 'd' [1,2] returned " ++ show (menu 'd' [1,2] :: Either String [[Int]])
testMenu = putStr (testMenuP ++ "\n" ++ testMenuS ++ "\n" ++ testMenuC ++ "\n")
My question is now, how do I get rid of the quotes '' in the Char 'd' when I output the string (as shown in the test function testMenuC).
You can replace the part of menu with:
menu x _ = Left ("(" ++ [x] ++ ")" ++ "is not supported, use (p)refix or (s)uffix")
or even
menu x _ = Left . mconcat $ ["(", [x], ")", "is not supported, use (p)refix or (s)uffix"]

Haskell list of int tuples to string

I'm trying to write out my list of tuples from:
[(8,7),(5,6),(3,3),(9,4),(5,4)]
to:
8 7 5 6 3 3 9 4 5 4
this is how far i've come:
(Updated)
showTuples :: Board -> String
showTuples = unwords . fmap show . concatMap (\(x,y) -> [x,y])
I know i'm suppose to map this function to all elements of my list but i can't seem to get it right.
(UPDATE)
It worked, still having problems with the quotation marks though
also type of Board is:
type Board = [(Int,Int)]
Using pattern matching:
type Board = [(Int, Int)]
showTuples :: Board -> String
showTuples [] = ""
showTuples (x:[]) = show(fst(x)) ++ " " ++ show(snd(x))
showTuples (x:xs) = show(fst(x)) ++ " " ++ show(snd(x)) ++ " " ++ showTuples xs
main :: IO ()
main = putStrLn . showTuples $ [(8, 7), (5, 6), (3, 3), (9, 4), (5, 4)] -- 8 7 5 6 3 3 9 4 5 4
This can be done by a foldl too.
Prelude> foldl (\r (x,y) -> r ++ " " ++ show x ++ " " ++ show y) "" [(8,7),(5,6),(3,3),(9,4),(5,4)]
" 8 7 5 6 3 3 9 4 5 4"
If you don't want the preceding whitespace then just do like tail $ foldl (\r (x,y) -> ...

Accessing a list entry and showing its values

I have a function
(.#.) :: [a] -> Integer -> a -- 1-indexing with 'Integer'
xs .#. j = xs !! (fromIntegral $ j-1)
showIntegers :: [Integer] -> String
showIntegers r = let
str = concat $ "List: " : [r (.#.) j | j <- [1..length r]]
How can I show r (.#.) j as a Char/String rather than an integer? I tried using show, but it gave me an error.
Here is an example of how I used show:
str = concat $ "List: " : [show $ r (.#.) j | j <- [1..length r]]
Example input and output:
> showIntegers [1,2,3]
List: 1 2 3
You should just use Data.List.intercalate or even better use unwords.
import Data.List
showIntegers :: [Integer] -> String
showIntegers r = "List: " ++ intercalate " " $ map show r
--showIntegers r = "List: " ++ unwords $ map show r
EDIT: In either case you should avoid using !! especially to enumerate the original list.
First I would get rid of .#. it is just going to confuse you to use a different numbering system, best to rip that bandaid off.
Next realize that [show $ r !! j <- 0 .. length r - 1] is the same as map show r (and the latter is standard).
Now going with that you have: concat $ "List: " : (map show r) which creates List: 123 because we lost the spaces.
We could reproduce the spaces but what is the difference between using intercalate and concat? Honestly the best solution without using intercalate would be to reproduce intercalate (whose source code is available on Hackage).
Just remove the parenthesis around (.#.) and it works.
If you have an infix operator !#$ , with something before and after it, e.g. x !#$ y, you must not use parentheses. In the other cases, add parentheses, like in the type declaration.
(this technically answers the question, but Guvante's advice is better.)

Haskell - how to generate permutations

How can I create a function which lazily makes permutations for the chars '_' and '*' like this:
For example:
Main> function 3
["___","*__","_*_","__*","**_","_**","*_*","***"]
First element is made only from _, the next 3 are permutations that lists: *__, the second 3 are permutations that lists **_, and the last element contains only *.
How can I do that?
Here's another "correct order" version:
function :: Int -> [String]
function c = concatMap helper $ zip (reverse [0..c]) [0..c]
helper :: (Int, Int) -> [String]
helper (c, 0) = [replicate c '_']
helper (0, c) = [replicate c '*']
helper (cUnderscores, cAsterisks) = map ('_' :) (helper (cUnderscores - 1, cAsterisks))
++ map ('*' :) (helper (cUnderscores, cAsterisks - 1))
You might want to look at replicateM.
let k = ["_", "*"]
let p = [ a ++ b ++ c | a <- k, b <- k, c <- k ]
The “correct order” version:
import Data.List
function k = concatMap (nub . permutations . pat) [0..k]
where pat x = replicate x '*' ++ replicate (k-x) '_'
I don’t know how to step from one permutation to another in constant time, though.

Printing the values inside a tuple in Haskell

I have a list of tuples. For example: [("A",100,1),("B",101,2)]. I need to display it in a simple way. For example: "your name is: A", "Your id is: 100".
If anyone can find a solution for this, it would be a great help. Thanks in advance.
The easiest way to do this is to create a function that works for one of the elements in your list. So you'll need something like:
showDetails :: (String, Int, Int) -> String
showDetails (name, uid, _) = "Your name is:" ++ name ++ " Your ID is: " ++ show uid
Then you would apply this function to each element in the list, which means you want to use the mapping function:
map :: (a -> b) -> [a] -> [b]
So, if your list is called xs, you would want something like:
map showDetails xs
This obviously gives you a result of type [String], so you might be interested in the unlines function:
unlines :: [String] -> String
This simply takes a list of strings, and creates a string where each element is separated by a new line.
Putting this all together, then, gives you:
main :: IO ()
main = putStrLn . unlines . map showDetails $ [("A",100,1),("B",101,2)]
For a single tuple, just pattern match all the elements, and do something with them. Having a function that does that, you can use map to transform the entire list.
import Data.List (foldl')
show_tuple :: (Num a, Num b) => (String, a, b) -> String
show_tuple (name, id, something) =
"Your name is: " ++ name ++ "\n" ++
"Your ID is: " ++ (show id) ++ "\n" ++
"Your something: " ++ (show something) ++ "\n\n"
-- transforms the list, and then concatenates it into a single string
show_tuple_list :: (Num a, Num b) => [(String, a, b)] -> String
show_tuple_list = (foldl' (++) "") . (map show_tuple)
The output:
*Main Data.List> putStr $ show_tuple_list [("ab", 2, 3), ("cd", 4, 5)]
Your name is: ab
Your ID is: 2
Your something: 3
Your name is: cd
Your ID is: 4
Your something: 5
Quick and dirty solution
f (x,y,z) = "your id is " ++ (show y) ++ ", your name is " ++ (show x) ++ "\n"
main = putStrLn $ foldr (++) "" (map f [("A",100,1),("B",101,2)])
OR (by #maksenov)
main = putStrLn $ concatMap f [("A",100,1),("B",101,2)]
Please try:
get1st (a,_,_) = a
get2nd (_,a,_) = a
get3rd (_,_,a) = a
showTuples [] = ""
showTuples (x:xs) = "Your name is:" ++ show(get1st(x)) ++ " Your ID is: " ++ show(get2nd(x)) ++ "\n" ++ showTuples xs
main = do
let x = [("A",100,1),("B",101,2)]
putStrLn . showTuples $ x

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