刪除字符串中的每個第n個元素

Question

如何刪除字符串的每個第n個元素？

我猜你會以某種方式使用drop函數。

像這樣滴第一個n，你怎麼能改變這個，所以只有第n個，然後是第n個，等等，而不是所有的？

dropthem n xs = drop n xs 

Answer 1

remove_every_nth :: Int -> [a] -> [a] 
remove_every_nth n = foldr step [] . zip [1..] 
    where step (i,x) acc = if (i `mod` n) == 0 then acc else x:acc

這裏的功能是什麼：

zip [1..]用於索引的所有項目在列表中，因此如zip [1..] "foo"變成[(1,'f'), (2,'o'), (3,'o')]。

然後使用right fold處理索引列表，該索引列表累積索引不能被n整除的每個元素。

這是一個稍微長一點的版本，它基本上做了同樣的事情，但是避免了來自zip [1..]的額外內存分配，並且不需要計算模量。

remove_every_nth :: Int -> [a] -> [a] 
remove_every_nth = recur 1 
    where recur _ _ []  = [] 
      recur i n (x:xs) = if i == n 
      then recur 1 n xs 
      else x:recur (i+1) n xs

Answer 2

嘗試結合take和drop來實現此目的。

take 3 "hello world" = "hel" 
drop 4 "hello world" = "o world" 

Answer 3

-- groups is a pretty useful function on its own! 
groups :: Int -> [a] -> [[a]] 
groups n = map (take n) . takeWhile (not . null) . iterate (drop n) 

removeEveryNth :: Int -> [a] -> [a] 
removeEveryNth n = concatMap (take (n-1)) . groups n 

Answer 4

簡單。取（n-1）個元素，然後跳過1，沖洗並重復。

dropEvery _ [] = [] 
dropEvery n xs = take (n-1) xs ++ dropEvery n (drop n xs) 

或者在節目樣式效率的緣故

dropEvery n xs = dropEvery' n xs $ [] 
    where dropEvery' n [] = id 
      dropEvery' n xs = (take (n-1) xs ++) . dropEvery n (drop n xs) 

Answer 5

我喜歡以下解決方案：

del_every_nth :: Int -> [a] -> [a]  
del_every_nth n = concat . map init . group n 

你只需要定義一個函數group哪些羣體在長的部分名單ñ。但是，這是很容易的：

group :: Int -> [a] -> [[a]] 
group n [] = [] 
group n xs = take n xs : group n (drop n xs)