Iterate and count [String] Haskell

Question

I am trying to learn Haskell. Currently I am making a function that should take a [String] and 'char' and return in how many strings this char presents.

count [] _ = 0
count (x:xs) v
 | elem v x = 1 + count xs
 | otherwise = 0 + count xs

How is it done correctly?

!EDIT

I get this error ::

Occurs check: cannot construct the infinite type: a ~ a1 -> a
  Relevant bindings include
  v :: a1 (bound at prog.hs:7:14)
  xs :: [t a1] (bound at prog.hs:7:10)
  x :: t a1 (bound at prog.hs:7:8)
  count :: [t a1] -> a1 -> a (bound at prog.hs:6:1)
In the second argument of `(+)', namely `count xs'
In the expression: 1 + count xs

So you're trying the total number of occurrences of the character in all strings? Or the amount of strings in which the character occurs at least once? — Glubus
– Glubus, Commented Dec 23, 2015 at 12:09
Amount of strings in which the character occurs at least once — Fiedberg Nichenko
– Fiedberg Nichenko, Commented Dec 23, 2015 at 12:10

Kapol · Accepted Answer · 2015-12-23 12:22:01Z

1

count takes two arguments (a list and a character), but you are invoking it with only one (xs) in each guard. You should change your function to:

count [] _ = 0
count (x:xs) v
 | elem v x  = 1 + count xs v
 | otherwise = count xs v

Notice that I removed adding zero, because it's redundant and doesn't really make the code more readable.

edited Dec 23, 2015 at 12:22

answered Dec 23, 2015 at 12:15

Kapol

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Comments

Xophmeister · Accepted Answer · 2015-12-23 12:22:44Z

1

You've got the right idea. elem is a predicate that answers whether a list contains an element. Given that a String is a list of char, it would probably be better to just match on equality. Something like this:

count :: String -> Char -> Int
count "" _ = 0
count (x:xs) v
  | x == v    = 1 + count xs v
  | otherwise = count xs v

Your count function takes two arguments. In your example code, you're missing the second (the character you're searching for); hence the error.

EDIT The signature is count :: [String] -> Char -> Int; my mistake. In which case, you should use elem and it can be simplified to something like:

count :: [String] -> Char -> Int
count [] _ = 0
count (x:xs) v = c + count xs v
                 where c = if elem v x then 1 else 0

...you always do the count xs v recursive call, so you don't really need the guard.

edited Dec 23, 2015 at 12:22

answered Dec 23, 2015 at 12:15

Xophmeister

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4 Comments

Kapol Over a year ago

The function takes a list of Strings, not a single String.

Xophmeister Over a year ago

@Kapol I didn't spot that! In which case, ignore what I said about elem: that's what should be used!

Fiedberg Nichenko Over a year ago

Yep .. just noticed it myself :D

Xophmeister Over a year ago

Fixed and added a simplification

Netwave · Accepted Answer · 2015-12-23 12:26:37Z

1

Just for show a folding solution:

count :: String -> Char -> Int
count x c = foldr (\x y -> y+1 if x == c else y) s 0

foldr pick each element and perform the desired operation taking the result as the input for the next item, so in this case, we are initializing it to 0, and adding 1 for each element that is equal to the char passed to te function.

answered Dec 23, 2015 at 12:26

Netwave

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Comments

Glubus · Accepted Answer · 2015-12-23 12:21:59Z

0

I would recommend defining the desired signature for count, so you'll get better errors, and a good feeling for the type system (since you're learning haskell)

count :: [String] -> Char -> Int

To make your code work these are in my mind the changes you'd need to make (untested)

count [] _ = 0
count (x:xs) v
 | elem v x = 1 + count xs v
 | otherwise = count xs v

You could also use built in functions like haskell to make your code a little more readable (less guards and lines), though I think it would make it a little worse performance wise.

answered Dec 23, 2015 at 12:21

Glubus

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