Haskell: Output list of tuples as string output

Question

I'm trying to get this list of tuples:

[(2,"a"), (1,"a"), (1,"b"), (1,"c"), (2,"dd")]

into this string output

a 1,2

b 1

c 1

dd 2

I assume I need to use the unzip and unlines functions. But I also saw some solutions using the show function which makes the integers strings. Any ideas?

Answer 1

If you have this list:

pairs = [ ("a", [1,2]), ("b", [1]), ("c", [1]), ("dd", [2]) ]

then you can get the desired output with:

putStrLn $ unlines [ x ++ " " ++ unwords (map show ys) | (x, ys) <- pairs ]

but you have to figure out how to get the pairs list first.

Answer 2

Break the problem down into steps. What you really want to do first is aggregate all the tuples that have the same string in the second position, so you'll have a function like

aggregate :: [(Int, String)] -> [([Int], String)]

So for your input list you would get the output

[([1, 2], "a"), ([1], "b"), ([1], "c"), ([2], "dd")]

Your hints are

aggregate items = someFunc (map (\(num, str) -> ([num], str)) items)

And take a look at foldr . Before you ask a follow up question about foldr , there are probably hundreds of stackoverflow answers showing how to use it already, take some time to figure it out or it'll get closed immediately as a duplicate.

Then you need a function to convert a single tuple of this form into a single String for outputting:

prettyPrint :: ([Int], String) -> String
prettyPrint (nums, str) = str ++ " " ++ joinWithComma (map show nums)

Where you'll have to implement joinWithComma yourself. Then you need to calculate this and print it for each item in your aggregated list, mapM_ and putStrLn would be preferred, so your main might look like

main :: IO ()
main = do
    let inputList = [(2,"a"), (1,"a"), (1,"b"), (1,"c"), (2,"dd")]
    mapM_ (putStrLn . prettyPrint) (aggregate inputList)

Answer 3

Work step by step. You could start with the groupBy function:

groupBy (\x y-> (snd x)==(snd y)) [(2,"a"), (1,"a"), (1,"b"), (1,"c"), (2,"dd")]

gives you

[[(2,"a"),(1,"a")],[(1,"b")],[(1,"c")],[(2,"dd")]]

The next step would be "totalling" the inner lists, map and foldL (and depending on your requirements maybe sortBy ) should be helpful. If you have this, constructing the output is trivial (using show , as you already mentioned).