How to convert list of numbers to list of strings(one string = one number from list) in Haskell.
[Int] -> [String]
Examples: [1,2,3,4] -> ["1","2","3","4"]
3 Answers
If you have a function f :: a -> b, then map f :: [a] -> [b] applies f on all the list elements.
The function show can convert "printable" types in their string representation. In particular, one of the possible types for show is Int -> String.
Use both tools.
Comments
If you need to write a function to print an element from 0, this could be another solution
cp_log :: (Show a) => [a] -> String
cp_log [] = ""
cp_log [x] = (show x)
cp_log (x:xs) = (show x) ++ ", " ++ cp_log xs
A complete example can be the following one
cp_log :: (Show a) => [a] -> String
cp_log [] = ""
cp_log [x] = (show x)
cp_log (x:xs) = (show x) ++ ", " ++ cp_log xs
quick_sort :: (Ord a) => [a] -> [a]
quick_sort [] = []
quick_sort (x:xs) =
let smaller = quick_sort [a | a <- xs, a <= x]
bigger = quick_sort [a | a <- xs, a > x]
in smaller ++ [x] ++ bigger
main =
let sorted = (quick_sort [4, 5, 3, 2, 4, 3, 2])
in putStrLn (cp_log sorted)
6 Comments
madbird
All the TS was needed is simple
map show. What you came up with is a rough equivalent of intercalate "," . map show. So your answer doesn't answers the question, and is excessively complicated at the same time. Please pay attention to what was the question about next time :)
macros
ops, I'm sorry :) I try only to post a craft approach, but you have right
madbird
There's nothing wrong with that, please accept my apologies. Your haskell is impressive, I think you could help somebody with their pending questions, let's forget about this ancient-one :)
macros
don't worry @madbird, I will happy if you remove the downvote :)
madbird
Unfortunately, my downvote is locked until the answer is edited. Add a couple of lines, or, maybe, replace
let with where, so I'll be able to remove my vote. |
Using the list monad:
f :: [Int] -> String
f xs = do
x <- xs
return $ show x
or equivalently:
f' :: [Int] -> [String]
f' = (>>= return.show)
Int -> Stringyou could make a function[Int] -> [String]using map` map :: (a -> b) -> [a] -> [b]`