CODEWITHSUNDEEP
stringhaskellcountintrange
Saulius Kriauza
Saulius Kriauza

Reputation: 15

Haskell, make single string from integer set?

I'd greatly appreciate if you could tell me how to make a single string from a range between two ints. Like [5..10] i would need to get a "5678910". And then I'd have to calculate how many (zeroes, ones ... nines) there are in a string.

For example: if i have a range from [1..10] i'd need to print out

1 2 1 1 1 1 1 1 1 1

For now i only have a function to search for a element in string.

`countOfElem elem list = length $ filter (\x -> x == elem) list`

But the part how to construct such a string is bugging me out, or maybe there is an easier way? Thank you.

I tried something like this, but it wouldn't work. let intList = map (read::Int->String) [15..22]

Upvotes: 1

Views: 371

Answers (1)

leftaroundabout
leftaroundabout

Reputation: 120741

I tried something like this, but it wouldn't work. let intList = map (read::Int->String) [15..22]

Well... the purpose of read is to parse strings to read-able values. Hence it has a type signature String -> a, which obviously doesn't unify with Int -> String. What you want here is the inverse1 of read, it's called show.

Indeed map show [15..22] gives almost the result you asked for – the numbers as decimal-encoded strings – but still each number as a seperate list element, i.e. type [String] while you want only String. Well, how about asking Hoogle? It gives the function you need as the fifth hit: concat.

If you want to get fancy you can then combine the map and concat stages: both the concatMap function and the >>= operator do that. The most compact way to achieve the result: [15..22]>>=show.

1show is only the right inverse of read, to be precise.

Upvotes: 3

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