CODEWITHSUNDEEP
haskellrecursion
James
James

Reputation: 65

Haskell how to count occurance of an element in a list recursively?

I'm trying to recursively count how many times a given element (number) appears in a list.

For example, equals 2 [1,2,3,2,1,2,3] should return 3 for the number of times 2 appears in the list.

What i've done so far:

equals :: Integer -> [Integer] -> Int

equals n [] = 0

equals n (x:xs) =

    if equals n == x
        then 1 + equals n xs
        else equals xs

Upvotes: 1

Views: 774

Answers (2)

Semmel
Semmel

Reputation: 575

You are nearly there. Just keep in mind when you want to call your function and remember to give it all required arguments.

equals :: Integer -> [Integer] -> Int

equals n [] = 0

equals n (x:xs)
    | n == x = 1 + equals n xs
    | otherwise = equals n xs

Upvotes: 3

Silvio Mayolo
Silvio Mayolo

Reputation: 70377

You basically have the solution. There's just a few minor typos to squash.

equals :: Integer -> [Integer] -> Int
equals n [] = 0
equals n (x:xs) =
    if n == x then
        1 + equals n xs
    else
        equals n xs

In particular, whenever you call equals, you need to pass it two parameters. So your recursive call should generally look like equals n xs, not equals xs. Similarly, your comparison between n and x is just that: a comparison. It needn't itself make a recursive call, so it only needs to look like n == x.

Upvotes: 3

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