implementation of unzip function in haskell

Question

I am trying to implement the unzip function, I did the following code but I get error. ``` myUnzip [] =() myUnzip ((a,b):xs) = a:fst (myUnzip xs) b:snd (myUnzip xs) ``` I know that problem is in the right side of the second line but I do know how to improve it . any hint please . the error that I am getting is ``` ex1.hs:190:22: Couldn't match expected type `()' with actual type `[a0]' In the expression: a : fst (myUnzip xs) b : snd (myUnzip xs) In an equation for `myUnzip': myUnzip ((a, b) : xs) = a : fst (myUnzip xs) b : snd (myUnzip xs) ex1.hs:190:29: Couldn't match expected type `(t0 -> a0, b0)' with actual type `()' In the return type of a call of `myUnzip' In the first argument of `fst', namely `(myUnzip xs)' In the first argument of `(:)', namely `fst (myUnzip xs) b' ex1.hs:190:49: Couldn't match expected type `(a1, [a0])' with actual type `()' In the return type of a call of `myUnzip' In the first argument of `snd', namely `(myUnzip xs)' In the second argument of `(:)', namely…

Accepted Answer

You could do it inefficiently by traversing the list twice

```
myUnzip [] = ([], []) -- Defaults to a pair of empty lists, not null
myUnzip xs = (map fst xs, map snd xs)
```

But this isn't very ideal, since it's bound to be quite slow compared to only looping once. To get around this, we have to do it recursively

```
myUnzip [] = ([], [])
myUnzip ((a, b):xs) = (a : ???, b : ???)
    where ??? = myUnzip xs
```

I'll let you fill in the blanks, but it should be straightforward from here, just look at the type signature of `myUnzip` and figure out what you can possible put in place of the question marks at `where ??? = myUnzip xs`

implementation of unzip function in haskell

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