# Difference between revisions of "99 questions/Solutions/10"

m (Added one more simple way with list comprehension) |
m (fix omission of function "group" in offered alternate list comprehension) |
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− | [(length (x:xs), x) | (x:xs) <- xs] |
+ | [(length (x:xs), x) | (x:xs) <- group xs] |

</haskell> |
</haskell> |
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## Latest revision as of 03:45, 19 May 2021

(*) Run-length encoding of a list.

Use the result of problem P09 to implement the so-called run-length encoding data compression method. Consecutive duplicates of elements are encoded as lists (N E) where N is the number of duplicates of the element E.

```
encode xs = map (\x -> (length x,head x)) (group xs)
```

which can also be expressed as a list comprehension:

```
[(length x, head x) | x <- group xs]
```

or

```
[(length (x:xs), x) | (x:xs) <- group xs]
```

Or writing it Pointfree (Note that the type signature is essential here to avoid hitting the Monomorphism Restriction):

```
encode :: Eq a => [a] -> [(Int, a)]
encode = map (\x -> (length x, head x)) . group
```

Or (ab)using the "&&&" arrow operator for tuples:

```
encode :: Eq a => [a] -> [(Int, a)]
encode xs = map (length &&& head) $ group xs
```

Or using the slightly more verbose (w.r.t. `(&&&)`

) Applicative combinators:

```
encode :: Eq a => [a] -> [(Int, a)]
encode = map ((,) <$> length <*> head) . pack
```

Or with the help of foldr (*pack* is the resulting function from P09):

```
encode xs = (enc . pack) xs
where enc = foldr (\x acc -> (length x, head x) : acc) []
```

Or using takeWhile and dropWhile:

```
encode [] = []
encode (x:xs) = (length $ x : takeWhile (==x) xs, x)
: encode (dropWhile (==x) xs)
```

Or without higher order functions:

```
encode [] = []
encode (x:xs) = encode' 1 x xs where
encode' n x [] = [(n, x)]
encode' n x (y:ys)
| x == y = encode' (n + 1) x ys
| otherwise = (n, x) : encode' 1 y ys
```

Or we can make use of zip and group:

```
import List
encode :: Eq a => [a] -> [(Int, a)]
encode xs=zip (map length l) h where
l = (group xs)
h = map head l
```

Or if we ignore the rule that we should use the result of P09,

```
encode :: Eq a => [a] -> [(Int,a)]
encode xs = foldr f final xs Nothing
where
f x r (Just a@(i,q)) | x == q = r (Just (i+1,q))
| otherwise = a : r (Just (1, x))
f x r Nothing = r (Just (1, x))
final (Just a@(i,q)) = [a]
final Nothing = []
```

which can become a good transformer for list fusion like so:

```
{-# INLINE encode #-}
encode :: Eq a => [a] -> [(Int,a)]
encode xs = build (\c n ->
let
f x r (Just a@(i,q)) | x == q = r (Just (i+1,q))
| otherwise = a `c` r (Just (1, x))
f x r Nothing = r (Just (1, x))
final (Just a@(i,q)) = a `c` n
final Nothing = n
in
foldr f final xs Nothing)
```

Just one more way with recursion:

```
encode :: [[t]] -> [(Int, t)]
encode = let f acc [] = acc
f acc (x:xs) = f ((length x, head x): acc) xs
in reverse . f []
```