删除字符串中的每个第n个元素

Question

如何删除字符串的每个第n个元素？

我猜你会以某种方式使用drop函数。

像这样滴第一个n，你怎么能改变这个，所以只有第n个，然后是第n个，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

这里的功能是什么：

zip [1..]用于索引的所有项目在列表中，因此如zip [1..] "foo"变成[(1,'f'), (2,'o'), (3,'o')]。

然后使用right fold处理索引列表，该索引列表累积索引不能被n整除的每个元素。

这是一个稍微长一点的版本，它基本上做了同样的事情，但是避免了来自zip [1..]的额外内存分配，并且不需要计算模量。

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

尝试结合take和drop来实现此目的。

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

简单。取（n-1）个元素，然后跳过1，冲洗并重复。

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

或者在节目样式效率的缘故

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

Answer 5

我喜欢以下解决方案：

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

你只需要定义一个函数group哪些群体在长的部分名单ñ。但是，这是很容易的：

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