CODEWITHSUNDEEP
haskell
Diana
Diana

Reputation: 1437

Haskell- find element in a list

I want to write a function that takes a number i and a list of numbers xs and returns the position of i in the list xs, counting the first position as 1. If i does not occur in xs, then position returns 0.

So far I have this:

import Data.List
position :: Int -> [Int] -> Int
position i xs 
   | i `elem` xs = i `elemIndex` xs
   | otherwise = 0

But when I compile, it gives the following error:

Couldn't match expected type ‘Int’ with actual type ‘Maybe Int’

I know that elemIndex returns a Maybe Int type and I defined my function to return Int but I don't know how to change it. Any ideas?

Upvotes: 4

Views: 23488

Answers (4)

Eilya Amin
Eilya Amin

Reputation: 1

This works properly: 

=>> findK k l = l !! k

Upvotes: 0

Jaskirat Singh
Jaskirat Singh

Reputation: 82

You could even use the zip function in the standard prelude of haskell :) Just zip it with [1..] and then use the list comprehension provided by it. Check out the code here:

positions n xs = [y | (y,z) <- zip [1..] xs, z==n]

To find the number of occurences, just use length from the standard prelude:

positions n xs = length [y | (y,z) <- zip [1..] xs, z==n]

Upvotes: 1

bmk
bmk

Reputation: 1568

I think, the solution above could be one-liner using the maybe function:

import Data.List

position :: Eq a => a -> [a] -> Int
position i xs = maybe 0 (+1) $ i `elemIndex` xs

Upvotes: 6

Zeta
Zeta

Reputation: 105985

First of all, lists are indexed with 0…. So elemIndex will return Just 0 if i happens to be the first element of your list.

Since elemIndex returns Maybe Int, you could pattern match on its result instead:

import Data.List (elemIndex)

position :: Eq a => a -> [a] -> Int
position i xs =
    case i `elemIndex` xs of
       Just n  -> n + 1
       Nothing -> 0

Upvotes: 6

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