# Difference between revisions of "99 questions/Solutions/20"

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< 99 questions | Solutions

(point-free style) |
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removeAt :: Int -> [a] -> (a, [a]) |
removeAt :: Int -> [a] -> (a, [a]) |
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removeAt n xs = let (front, back) = splitAt n xs in (last front, init front ++ back) |
removeAt n xs = let (front, back) = splitAt n xs in (last front, init front ++ back) |
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+ | </haskell> |
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+ | |||

+ | Similar, point-free style: |
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+ | |||

+ | <haskell> |
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+ | removeAt n = (\(a, b) -> (head b, a ++ tail b)) . splitAt (n - 1) |
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</haskell> |
</haskell> |
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## Revision as of 14:29, 23 August 2012

(*) Remove the K'th element from a list.

```
removeAt :: Int -> [a] -> (a, [a])
removeAt k xs = case back of
[] -> error "removeAt: index too large"
x:rest -> (x, front ++ rest)
where (front, back) = splitAt k xs
```

Simply use the `splitAt`

to split after k elements.
If the original list has fewer than k+1 elements, the second list will be empty, and there will be no element to extract.
Note that the Prolog and Lisp versions treat 1 as the first element in the list, and the Lisp version appends NIL elements to the end of the list if k is greater than the list length.

or

```
removeAt n xs = (xs!!n,take n xs ++ drop (n+1) xs)
```

Another solution that avoids throwing an error and using ++ operators. Treats 1 as the first element in the list.

```
removeAt :: Int -> [a] -> (Maybe a, [a])
removeAt _ [] = (Nothing, [])
removeAt 1 (x:xs) = (Just x, xs)
removeAt k (x:xs) = let (a, r) = removeAt (k - 1) xs in (a, x:r)
```

Another solution that also uses Maybe to indicate failure:

```
removeAt :: Int -> [a] -> (Maybe a, [a])
removeAt _ [] = (Nothing, [])
removeAt 0 xs = (Nothing, xs)
removeAt nr xs | nr > length xs = (Nothing, xs)
| otherwise = (Just (xs !! nr), fst splitted ++ (tail . snd) splitted)
where splitted = splitAt nr xs
```

And yet another solution (without error checking):

```
removeAt :: Int -> [a] -> (a, [a])
removeAt n xs = let (front, back) = splitAt n xs in (last front, init front ++ back)
```

Similar, point-free style:

```
removeAt n = (\(a, b) -> (head b, a ++ tail b)) . splitAt (n - 1)
```

A simple recursive solution:

```
removeAt 1 (x:xs) = (x, xs)
removeAt n (x:xs) = (l, x:r)
where (l, r) = removeAt (n - 1) xs
```