CODEWITHSUNDEEP
haskell
Newbie
Newbie

Reputation: 44

Haskell mapping function to list

I am new to Haskell and I have the following problem. I have to create a list of numbers [f1, f2, f3...] where fi x = x ^ i. Then I have to create a function that applies the fi to a list of numbers. For example if I have a list lis = [4,5,6,7..] the output would be [4^1, 5^2,6^3, 7^4...]. This is what I have written so far :

powers x= [x^y |y<-[1,2,3,4]]

list = [1,2,3,4]

match :: (x -> xs) -> [x] -> [xs]
match f [] = []
match f (x:xs) = (f x) : ( match f xs )

So if I put the list = [1,2,3] the output is [1,1,1,1][2,4,8,16],[3,9,27,81] instead of [1,4,27]

Can you please tell me what is wrong and point me to the right direction?

Upvotes: 2

Views: 2508

Answers (3)

Retief
Retief

Reputation: 3217

The first issue is that powers is of type Int -> [Int]. What you really want, I think, is something of type [Int -> Int] -- a list of Int -> Int functions instead of a function that takes an Int and returns a list of Int. If you define powers like so:

powers = [(^y) | y <- [1..4]]

you can use zipWith to apply each power to its corresponding element in the list, like so:

zipWith ($) powers [1,2,3] -- returns [1,4,27]

The ($) applies its left (first) argument to its right (second) argument.

Note that using powers as defined here will limit the length of the returned list to 4. If you want to be able to use arbitrary length lists, you want to make powers an infinite list, like so:

powers = [(^y) | y <- [1..]]

Of course, as dave4420 points out, a simpler technique is to simply use

zipWith (^) [1,2,3] [1..] -- returns [1,4,27]

Upvotes: 6

3lectrologos
3lectrologos

Reputation: 9662

You are currently creating a list for every input value. What you need to do is recursively compute the appropriate power for each input value, like this:

match f [] = []
match f (x:xs) y = (f x y) : (match f xs y+1)

Then, you can call this as match pow [1, 2, 3] 1. This is equivalent to using zipWith and providing the desired function (pow), your input list ([1, 2, 3]) and the exponent list (a lazy one to infinity list) as arguments.

Upvotes: 1

dave4420
dave4420

Reputation: 47062

Your match is the standard function map by another name. You need to use zipWith instead (which you can think of as mapping over two lists side-by-side).

Is this homework?

Upvotes: 3

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