# Difference between revisions of "99 questions/Solutions/18"

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Or, a concise solution using list comprehension: |
Or, a concise solution using list comprehension: |
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− | (Incorrect |
+ | (Incorrect - only works on sequential data, needs update.) |

<haskell> |
<haskell> |
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--slice :: (Enum a) => [a] -> Int -> Int -> [a] |
--slice :: (Enum a) => [a] -> Int -> Int -> [a] |

## Latest revision as of 02:14, 4 July 2016

(**) Extract a slice from a list.

Given two indices, i and k, the slice is the list containing the elements between the i'th and k'th element of the original list (both limits included). Start counting the elements with 1.

```
slice xs i k | i>0 = take (k-i+1) $ drop (i-1) xs
```

The same solution as above, but the more paranoid (maybe too paranoid?) version of it (uses guards and Maybe):

```
slice :: [a] -> Int -> Int -> Maybe [a]
slice [] _ _ = Just []
slice xs k n | k == n = Just []
| k > n || k > length xs ||
n > length xs || k < 0 || n < 0 = Nothing
| k == 0 = Just (take n xs)
| otherwise = Just (drop (k-1) $ take n xs)
```

Or, a concise solution using list comprehension:

(Incorrect - only works on sequential data, needs update.)

```
--slice :: (Enum a) => [a] -> Int -> Int -> [a]
--slice [] _ _ = []
--slice xs m n = [(xs !! (m-1)) .. (xs !! (n-1))]
```

Or, an iterative solution:

```
slice :: [a] -> Int -> Int -> [a]
slice lst 1 m = slice' lst m []
where
slice' :: [a]->Int->[a]->[a]
slice' _ 0 acc = reverse acc
slice' (x:xs) n acc = slice' xs (n - 1) (x:acc)
slice' [] _ _ = []
slice (x:xs) n m = slice xs (n - 1) (m - 1)
slice [] _ _ = []
```

Or:

```
slice :: [a] -> Int -> Int -> [a]
slice [] _ _ = []
slice (x:xs) i k
| i > 1 = slice xs (i - 1) (k - 1)
| k < 1 = []
| otherwise = x:slice xs (i - 1) (k - 1)
```

Another way using `splitAt`

, though not nearly as elegant as the `take`

and `drop`

version:

```
slice :: [a] -> Int -> Int -> [a]
slice xs i k = chunk
where chop = snd $ splitAt i' xs -- Get the piece starting at i
chunk = fst $ splitAt (k - i') chop -- Remove the part after k
i' = i - 1
```

A little cleaner, using the previous problem's split (a.k.a. `splitAt`

):

```
slice xs (i+1) k = snd (split (fst (split xs k)) i)
```

A solution using `zip`

, `filter`

then `map`

seems straight-forward to me (*NB: this won't work for infinite lists*):

```
slice xs i j = map snd
$ filter (\(x,_) -> x >= i && x <= j)
$ zip [1..] xs
```

A solution using list comprehension:

```
slice xs i k = [x | (x,j) <- zip xs [1..k], i <= j]
```

Another simple solution using take and drop:

```
slice :: [a] -> Int -> Int -> [a]
slice l i k
| i > k = []
| otherwise = (take (k-i+1) (drop (i-1) l))
```

Zip, filter, unzip:

```
slice :: [a] -> Int -> Int -> [a]
slice xs a b = fst $ unzip $ filter ((>=a) . snd) $ zip xs [1..b]
```

Take and drop can be applied in the opposite order too:

```
slice xs i k = drop (i-1) $ take k xs
```

Using a fold:

```
slice :: [a] -> Int -> Int -> [a]
slice (x:xs) begin end = snd $ foldl helper (1, []) (x:xs)
where helper (i, acc) x = if (i >= begin) && (i <= end) then (i+1, acc ++ [x]) else (i+1, acc)
```