# Difference between revisions of "99 questions/Solutions/16"

Citizen428 (talk | contribs) m |
SunjayVarma (talk | contribs) (Added a new approach using mod and composing list functions instead of using foldr) |
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Line 21: | Line 21: | ||

helper (x:xs) count 1 = helper xs count count | helper (x:xs) count 1 = helper xs count count | ||

helper (x:xs) count n = x : (helper xs count (n - 1)) | helper (x:xs) count n = x : (helper xs count (n - 1)) | ||

+ | </haskell> | ||

+ | |||

+ | A similar iterative solution but using a closure: | ||

+ | |||

+ | <haskell> | ||

+ | dropEvery :: [a] -> Int -> [a] | ||

+ | dropEvery xs n = helper xs n | ||

+ | where helper [] _ = [] | ||

+ | helper (x:xs) 1 = helper xs n | ||

+ | helper (x:xs) k = x : helper xs (k-1) | ||

+ | </haskell> | ||

+ | |||

+ | Or, counting up (and using guards instead of pattern matching): | ||

+ | |||

+ | <haskell> | ||

+ | dropEvery :: [a] -> Int -> [a] | ||

+ | dropEvery xs n = helper xs 1 | ||

+ | where helper :: [a] -> Int -> [a] | ||

+ | helper [] _ = [] | ||

+ | helper (x:xs) i | ||

+ | | i == n = helper xs 1 | ||

+ | | i /= n = x:helper xs (i + 1) | ||

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

Line 57: | Line 79: | ||

dropEvery [] _ = [] | dropEvery [] _ = [] | ||

dropEvery xs n = concat (split n xs) | dropEvery xs n = concat (split n xs) | ||

− | where | + | where |

split _ [] = [] | split _ [] = [] | ||

split n xs = fst splitted : split n ((safetail . snd) splitted) | split n xs = fst splitted : split n ((safetail . snd) splitted) | ||

− | where | + | where |

splitted = splitAt (n-1) xs | splitted = splitAt (n-1) xs | ||

safetail xs | null xs = [] | safetail xs | null xs = [] | ||

Line 72: | Line 94: | ||

The filter function can be simplified as seen above: | The filter function can be simplified as seen above: | ||

<haskell> | <haskell> | ||

− | dropEvery xs n = map fst $ filter ((n/=) . snd) $ zip xs [1..] | + | dropEvery xs n = map fst $ filter ((n/=) . snd) $ zip xs (cycle [1..n]) |

+ | </haskell> | ||

+ | |||

+ | And yet another approach using folds: | ||

+ | <haskell> | ||

+ | dropEvery :: Int -> [a] -> [a] | ||

+ | dropEvery n xs = snd $ foldl (\acc e -> if fst acc > 1 then (fst acc - 1, snd acc ++ [e]) else (n, snd acc)) (n, []) xs | ||

+ | </haskell> | ||

+ | |||

+ | Another very similar approach to the previous: | ||

+ | <haskell> | ||

+ | dropEvery :: [a] -> Int -> [a] | ||

+ | dropEvery xs n = fst $ foldr (\x (xs, i) -> (if mod i n == 0 then xs else x:xs, i - 1)) ([], length xs) xs | ||

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

+ | |||

+ | Using mod like the previous solution, but composing list functions instead of using fold: | ||

+ | <haskell> | ||

+ | -- Type signature is slightly different from the problem, but equivalent | ||

+ | dropEvery :: Int -> [a] -> [a] | ||

+ | dropEvery k = snd . unzip . filter (\(i, _) -> i `mod` k /= 0) . zip [1..] | ||

+ | </haskell> | ||

+ | |||

+ | Another foldl solution: | ||

+ | <haskell> | ||

+ | dropEvery :: [a] -> Int -> [a] | ||

+ | dropEvery lst n = snd $ foldl helper (1, []) lst | ||

+ | where helper (i,acc) x = if n == i | ||

+ | then (1,acc) | ||

+ | else (i+1,acc++[x]) | ||

+ | </haskell> | ||

+ | |||

+ | |||

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

## Latest revision as of 19:18, 31 August 2016

(**) Drop every N'th element from a list.

```
dropEvery :: [a] -> Int -> [a]
dropEvery [] _ = []
dropEvery (x:xs) n = dropEvery' (x:xs) n 1 where
dropEvery' (x:xs) n i = (if (n `divides` i) then
[] else
[x])
++ (dropEvery' xs n (i+1))
dropEvery' [] _ _ = []
divides x y = y `mod` x == 0
```

An alternative iterative solution:

```
dropEvery :: [a] -> Int -> [a]
dropEvery list count = helper list count count
where helper [] _ _ = []
helper (x:xs) count 1 = helper xs count count
helper (x:xs) count n = x : (helper xs count (n - 1))
```

A similar iterative solution but using a closure:

```
dropEvery :: [a] -> Int -> [a]
dropEvery xs n = helper xs n
where helper [] _ = []
helper (x:xs) 1 = helper xs n
helper (x:xs) k = x : helper xs (k-1)
```

Or, counting up (and using guards instead of pattern matching):

```
dropEvery :: [a] -> Int -> [a]
dropEvery xs n = helper xs 1
where helper :: [a] -> Int -> [a]
helper [] _ = []
helper (x:xs) i
| i == n = helper xs 1
| i /= n = x:helper xs (i + 1)
```

Yet another iterative solution which divides lists using Prelude:

```
dropEvery :: [a] -> Int -> [a]
dropEvery [] _ = []
dropEvery list count = (take (count-1) list) ++ dropEvery (drop count list) count
```

A similar approach using guards:

```
dropEvery :: [a] -> Int -> [a]
dropEvery xs n
| length xs < n = xs
| otherwise = take (n-1) xs ++ dropEvery (drop n xs) n
```

Using zip:

```
dropEvery = flip $ \n -> map snd . filter ((n/=) . fst) . zip (cycle [1..n])
```

Using zip and list comprehensions

```
dropEvery :: [a] -> Int -> [a]
dropEvery xs n = [ i | (i,c) <- ( zip xs [1,2..]), (mod c n) /= 0]
```

A more complicated approach which first divides the input list into sublists that do not contain the nth element, and then concatenates the sublists to a result list (if not apparent: the author's a novice):

```
dropEvery :: [a] -> Int -> [a]
dropEvery [] _ = []
dropEvery xs n = concat (split n xs)
where
split _ [] = []
split n xs = fst splitted : split n ((safetail . snd) splitted)
where
splitted = splitAt (n-1) xs
safetail xs | null xs = []
| otherwise = tail xs
```

First thing that came to mind:

```
dropEvery xs n = map fst $ filter (\(x,i) -> i `mod` n /= 0) $ zip xs [1..]
```

The filter function can be simplified as seen above:

```
dropEvery xs n = map fst $ filter ((n/=) . snd) $ zip xs (cycle [1..n])
```

And yet another approach using folds:

```
dropEvery :: Int -> [a] -> [a]
dropEvery n xs = snd $ foldl (\acc e -> if fst acc > 1 then (fst acc - 1, snd acc ++ [e]) else (n, snd acc)) (n, []) xs
```

Another very similar approach to the previous:

```
dropEvery :: [a] -> Int -> [a]
dropEvery xs n = fst $ foldr (\x (xs, i) -> (if mod i n == 0 then xs else x:xs, i - 1)) ([], length xs) xs
```

Using mod like the previous solution, but composing list functions instead of using fold:

```
-- Type signature is slightly different from the problem, but equivalent
dropEvery :: Int -> [a] -> [a]
dropEvery k = snd . unzip . filter (\(i, _) -> i `mod` k /= 0) . zip [1..]
```

Another foldl solution:

```
dropEvery :: [a] -> Int -> [a]
dropEvery lst n = snd $ foldl helper (1, []) lst
where helper (i,acc) x = if n == i
then (1,acc)
else (i+1,acc++[x])
```