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

From HaskellWiki

< 99 questions | Solutions

(Added another solution) |
|||

(6 intermediate revisions by 6 users not shown) | |||

Line 6: | Line 6: | ||

[] -> error "removeAt: index too large" |
[] -> error "removeAt: index too large" |
||

x:rest -> (x, front ++ rest) |
x:rest -> (x, front ++ rest) |
||

− | where (front, back) = splitAt k xs |
+ | where (front, back) = splitAt (k - 1) xs |

</haskell> |
</haskell> |
||

− | Simply use the <hask>splitAt</hask> to split after k elements. |
+ | Simply use the <hask>splitAt</hask> to split after k - 1 elements. |

− | If the original list has fewer than k |
+ | If the original list has fewer than k elements, the second list will be empty, and there will be no element to extract. |

− | Note that |
+ | Note that we treat 1 as the first element in the list. |

or |
or |
||

<haskell> |
<haskell> |
||

− | removeAt n xs = (xs!!n,take n xs ++ drop |
+ | removeAt n xs = (xs !! (n - 1), take (n - 1) xs ++ drop n xs) |

+ | </haskell> |
||

+ | |||

+ | or, a safe version of that |
||

+ | |||

+ | <haskell> |
||

+ | removeAt n xs | n > 0 && n <= length xs = (Just (xs !! index), take index xs ++ drop n xs) |
||

+ | | otherwise = (Nothing, xs) |
||

+ | where index = n - 1 |
||

+ | </haskell> |
||

+ | |||

+ | or, a list comprehension based one: |
||

+ | |||

+ | <haskell> |
||

+ | removeAt :: (Enum a) => Int -> [a] -> (a, [a]) |
||

+ | removeAt n xs = ((xs !! (n-1)), [ x | (i, x) <- zip [1..] xs, i /= n ]) |
||

</haskell> |
</haskell> |
||

Line 45: | Line 45: | ||

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) |
||

</haskell> |
</haskell> |
||

+ | |||

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

+ | |||

+ | <haskell> |
||

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

+ | </haskell> |
||

+ | |||

+ | A simple recursive solution: |
||

+ | |||

+ | <haskell> |
||

+ | removeAt 1 (x:xs) = (x, xs) |
||

+ | removeAt n (x:xs) = (l, x:r) |
||

+ | where (l, r) = removeAt (n - 1) xs |
||

+ | </haskell> |
||

+ | |||

+ | |||

+ | [[Category:Programming exercise spoilers]] |

## Latest revision as of 11:56, 26 April 2016

(*) 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 - 1) xs
```

Simply use the `splitAt`

to split after k - 1 elements.
If the original list has fewer than k elements, the second list will be empty, and there will be no element to extract.
Note that we treat 1 as the first element in the list.

or

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

or, a safe version of that

```
removeAt n xs | n > 0 && n <= length xs = (Just (xs !! index), take index xs ++ drop n xs)
| otherwise = (Nothing, xs)
where index = n - 1
```

or, a list comprehension based one:

```
removeAt :: (Enum a) => Int -> [a] -> (a, [a])
removeAt n xs = ((xs !! (n-1)), [ x | (i, x) <- zip [1..] xs, i /= n ])
```

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
```