Reputation: 2759
I know you suppose to think differently in Haskell, but can someone give me a quick answer on how to iterate over a list or nested list and print out a character based on the value of the list element.
list1 = [[1 0 0][0 1 0][0 0 1]]
By iterate through this nested list, it should print out x for 0 and y for 1
yxx
xyx
xxy
Thanks
Upvotes: 3
Views: 33964
Reputation: 21
The solutions
cambiar = putStr.unlines.(map (map f)) where f x = if x == 0 then 'x' else 'y'
Upvotes: -1
Reputation: 668
The solutions using map
are the preferred Haskell style. But while you're learning, you may find explicit recursion easier to follow. Like so:
charSub :: Int -> Char
charSub 0 = 'x'
charSub 1 = 'y'
charSub x = error "Non-binary value!"
listSub :: [Int] -> String
listSub [] = []
listSub (x:xs) = (charSub x) : (listSub xs)
nestedSub :: [[Int]] -> String
nestedSub [] = []
nestedSub (y:ys) = (listSub y) ++ "\n" ++ (nestedSub ys)
map
does pretty much the same thing--it applies a function to each element in a list. But it may be easier to see what's going on here.
Upvotes: 2
Reputation: 2327
Try this:
fun :: [[Int]] -> [String]
fun = (map . map) (\x -> if x == 0 then 'x' else 'y')
If you really need printing of result:
printSomeFancyList :: [[Int]] -> IO ()
printSomeFancyList = putStrLn . unlines . fun
Upvotes: 3
Reputation: 11218
iterateList = foldl1 (>>).concat.intersperse [putStrLn ""].(map.map) (\c -> putStr $ if (c==0) then "X" else "Y")
Upvotes: 1
Reputation: 13574
If you are interested in arbitrary nested lists, then you can write something like this (an arbitrary nested list is essentially a tree):
data Nested a = Leaf a | Nest [Nested a] deriving Show
traverse :: Nested Integer -> Nested Char
traverse (Leaf x) = Leaf (valueOf x)
traverse (Nest xs) = Nest (map traverse xs)
valueOf :: Integer -> Char
valueOf 0 = 'x'
valueOf 1 = 'y'
valueOf _ = 'z'
With that you can do:
Main> let nl = Nest [Leaf 1, Leaf 0, Nest [Leaf 0, Leaf 0, Leaf 1, Nest [Leaf 1, Leaf 1, Leaf 0]], Nest [Leaf 1, Leaf 1]]
Main> traverse nl
Nest [Leaf 'y',Leaf 'x',Nest [Leaf 'x',Leaf 'x',Leaf 'y',Nest [Leaf 'y',Leaf 'y',Leaf 'x']],Nest [Leaf 'y',Leaf 'y']]
The function traverse
takes an arbitrary nested list of Integer
s and returns a corresponding nested list of Char
s according to the valueOf
rule
Upvotes: 1
Reputation: 52300
define f by something like
f x = if x == 0 then 'x' else 'y'
then
map (map f) [[1,0,0],[0,1,0],[0,0,1]]
is what you want or if you want it fancier:
map' = map.map
map' f [[1,0,0],[0,1,0],[0,0,1]]
Upvotes: 2
Reputation: 3927
First of all, I think you mean:
list1 :: [[Int]]
list1 = [[1,0,0],[0,1,0],[0,0,1]]
As for what you want:
valueOf :: Int -> Char
valueOf 0 = 'x'
valueOf 1 = 'y'
valueOf _ = 'z'
listValues :: [[Int]] -> [String]
listValues = map (map valueOf)
printValues :: [[Int]] -> IO ()
printValues = putStrLn . unlines . listValues
And then in ghci:
*Main> printValues list1
yxx
xyx
xxy
Upvotes: 17