CODEWITHSUNDEEP
haskell
mgt
mgt

Reputation: 87

counting digits in a list of characters

I need to count how many digits appear in a list of characters using haskell. eg. ['2','3','d','r','t','5'] would output 3

I have managed to do it using string but cannot get it to work while using characters

this is my code for string

isNumber' :: String -> Bool
isNumber' s = case (reads s) :: [(Int, String)] of
[(_,"")] -> True
_         -> False

counta :: [String] -> Int 
counta = length . filter isNumber'

e.g. ['2','3','d','r','t','5'] would output 3

Upvotes: 1

Views: 310

Answers (3)

Ravi Chandak
Ravi Chandak

Reputation: 113

You can use Char -> Bool instead of String -> Bool and it should work without the read

Upvotes: 0

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 477607

Since you say it is a list of characters, the type of counta should be:

counta :: [Char] -> Int 
counta = length . filter isNumber'

So that means that isNumber' has as signature Char -> Bool, and thus checks if a character is a digit. We however do not need to implement such function: in the Data.Char module, there is an isDigit :: Char -> Bool function, so we can implement the counta with:

import Data.Char(isDigit)

counta :: [Char] -> Int
counta = length . filter isDigit

Upvotes: 2

Paul Johnson
Paul Johnson

Reputation: 17806

Not far off, but your isNumber' needs to take a character, not a string. So it would be of type

isNumber' :: Char -> Bool

Using read is not necessary.

Upvotes: 1

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