simple function that shifts specific element of a list

Question

I am new to Haskell and I am trying to figure out how to make a function:

    shift:: Eq a => a -> [a] -> Int -> [a]
    shift x (h:t) z
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Answer 1

For you are new to Haskell, it's better to break the problem into simpler subproblems. One trick to solve your problem is:

thinking of "insert" instead of "shift". Pseudo code:
- remove x from the list
- insert x back to the list at the proper position based on given n

You already implement delete, lets take advantage of it. Enjoy simple code:

-- assume x is in the list, otherwise your problem would not make sense
shift :: Eq a => a->[a]->Int->[a]
shift x lst n = if n == 0 then delete x lst 
                else insert x (n + (position x lst)) (delete x lst)

Sub-problems are:

1. Delete x from the list: you did that
  delete :: Eq a => a->[a]->[a]
  delete x []          = []
  delete x (head:tail) = if x == head then tail else head : delete x tail

2. find the position of x in the list
  -- assume x in the list
  position :: Eq a => a->[a]->Int
  position x (head:tail) = if x == head then 1 else 1 + position x tail

3. -- insert x into the list at nth position
  insert :: a->Int->[a]->[a]
  insert x n (head:tail) = if n <= 1 then x : (head:tail) 
                           else head : (insert x (n-1) tail)
  insert x n lst         = if n >= length lst then lst ++ [x] 
                           else insert x n lst

Your precondition is that x is in the list, otherwise it does not make sense. If you want to include the case x is not in the list, twist the code.

Have fun.