如何在 Haskell 中获取无限列表的每个第 N 个元素？

Question

More specifically, how do I generate a new list of every Nth element from an existing infinite list?

Eg if the list is [5, 3, 0, 1, 8, 0, 3, 4, 0, 93, 211, 0 ...] then getting every 3rd element would result in this list [0,0,0,0,0 ...]

Answer 1

My version using drop:

every n xs = case drop (n-1) xs of
              y : ys -> y : every n ys
              [] -> []

Edit: this also works for finite lists. For infinite lists only, Charles Stewart's solution is slightly shorter.

Answer 2

All the solutions using zip and so on do tons of unnecessary allocations. As a functional programmer, you want to get into the habit of not allocating unless you really have to. Allocation is expensive, and compared to allocation, everything else is free. And allocation doesn't just show up in the "hot spots" you would find with a profiler; if you don't pay attention to allocation, it kills you everywhere .

Now I agree with the commentators that readability is the most important thing . Nobody wants to get wrong answers fast. But it happens very often in functional programming that there are multiple solutions, all about equally readable, some of which allocate and some of which do not. It's really important to build a habit of looking for those readable, non-allocating solutions.

You might think that the optimizer would get rid of allocations, but you'd only be half right. GHC, the world's best optimizing compiler for Haskell, does manage to avoid allocating a pair per element; it fuses the filter - zip composition into a foldr2 . The allocation of the list [1..] remains. Now you might not think this is so bad, but stream fusion, which is GHC's current technology, is a somewhat fragile optimization. It's hard even for experts to predict exactly when it's going to work, just by looking at the code. The more general point is that when it comes to a critical property like allocation, you don't want to rely on a fancy optimizer whose results you can't predict . As long as you can write your code in an equally readable way, you're much better off never introducing those allocations.

For these reason I find Nefrubyr's solution using drop to be by far the most compelling. The only values that are allocated are exactly those cons cells (with : ) that must be part of the final answer. Added to that, the use of drop makes the solution more than just easy to read: it is crystal clear and obviously correct.

Answer 3

I don't have anything to test this with at work, but something like:

extractEvery m = map snd . filter (\(x,y) -> (mod x m) == 0) . zip [1..]

should work even on infinite lists.

(Edit: tested and corrected.)

Answer 4

Starting at the first element:

everyf n [] = []
everyf n as  = head as : everyf n (drop n as)

Starting at the nth element:

every n = everyf n . drop (n-1)

Answer 5

The compiler or interpreter will compute the step size (subtract 1 since it's zero based):

f l n = [l !! i | i <- [n-1,n-1+n..]]

The Haskell 98 Report: Arithmetic Sequences

Answer 6

MHarris's answer is great. But I like to avoid using \\ , so here's my golf:

extractEvery n
  = map snd . filter fst
  . zip (cycle (replicate (n-1) False ++ [True]))

Answer 7

摆脱mod替代解决方案：

extractEvery n = map snd . filter ((== n) . fst) . zip (cycle [1..n])

Answer 8

I nuked my version of Haskell the other day, so untested, but the following seems a little simpler than the others (leverages pattern matching and drop to avoid zip and mod):

everynth :: Int -> [a] -> [a]
everynth n xs = y : everynth n ys
         where y : ys = drop (n-1) xs

Answer 9

Another way of doing it:

takeEveryM m lst = [n | (i,n) <- zip [1..] lst, i `mod` m == 0]

Yet Another:

import Data.List

takeEveryMth m = 
  (unfoldr g)  . dr
     where 
       dr = drop (m-1)
       g (x:xs) = Just (x, dr xs)
       g []     = Nothing

Answer 10

Use a view pattern!

{-# LANGUAGE ViewPatterns #-}

everynth n (drop (n-1) -> l)
  | null l = []
  | otherwise = head l : everynth n (tail l)

---

Ugly version of Nefrubyr's answer preserved so comments make sense.

everynth :: Int -> [a] -> [a]
everynth n l = case splitAt (n-1) l of
                 (_, (x:xs)) -> x : everynth n xs
                 _           -> []

Answer 11

Explicit recursion is evil! Use a library construct like map, filter, fold, scan, reproduce, iterate etc instead. Unless it makes the code drastically easier to read even to those au fait with the libraries, then it's just making your code less modular, more verbose and harder to reason about.

Seriously, the explicitly recursive versions need to be voted down to negative for a task as simple as this. Now I guess I should say something constructive to balance my carping.

I prefer to use a map:

every n xs = map (xs!!) [n-1,n-1+n..]

rather than the ja's list comprehension so the reader doesn't have to worry about what i is. But either way it's O(n^2) when it could be O(n), so maybe this is better:

every n = map (!!(n-1)) . iterate (drop n)

Answer 12

extractEvery nl = map head (iterate (drop n) (drop (n-1) l))

我会为自己感到骄傲，直到我看到 Greg 得到了大致相同的答案并且在我之前

Answer 13

A lot of answers here already use Data.List.unfoldr, but I wanted to bring up an interesting way of writing the somewhat annoying unfold that may be helpful in other contexts:

{-# LANGUAGE TupleSections #-}
import Data.List (unfoldr)
import Data.Maybe (listToMaybe)

every n = unfoldr f . drop (n - 1)
    where f xs = (,drop n xs) <$> listToMaybe xs

When the list is null, listToMaybe returns a Nothing , and fmap similarly will then be Nothing . However, if a Just is produced, then the proper tuple is constructed by turning the head of the list into a tuple of the head and the rest of the values to use in the next iteration.

Answer 14

A slightly silly version with foldr :

everyNth n l = foldr cons nil l (n-1) where
  nil _ = []
  cons x rest 0 = x : rest n
  cons x rest n = rest (n-1)

Answer 15

(This was in response to a comment asking for a solution without drop)

I couldn't see this solution, so:

every _ []     = []
every n (x:xs) = every' n (n-1) (x:xs)
                 where every' n c []     = []
                       every' n 0 (x:xs) = x : every' n (n-1) xs
                       every' n c (x:xs) = every' n (c-1) xs

works for finite and infinite list:

take 15 (every 3 [1..])
-- [3,6,9,12,15,18,21,24,27,30,33,36,39,42,45]

Answer 16

This seems a slightly better use of unfold:

everyNth :: Int -> [a] -> [a]
everyNth n = unfoldr g
  where
    g [] = Nothing
    g xs = let (ys, zs) = splitAt n xs in Just (head ys, zs)

Or even better, use Data.List.Spit:

everyNth n = (map head) . chunksOf n

Answer 17

Old question but I'll post what I came up with:

everyNth :: [a] -> Int -> [a]
everyNth xs n = [snd x | x <- (zip [1..] xs), fst x `mod` n == 0]

I put the list argument before the Int so I can curry the function and a list then map that over a list of Int s if ever I need to.

Answer 18

It's more elegant to solve a related problem first: Keep every element whose index is divisible by n.

everyNth n [] = []
everyNth n (x:xs) = x : (everyNth n . drop (n-1)) xs

And then to solve the example, use

everyNthFirst n = everyNth n . drop (n-1)

everyNthFirst 3 [5, 3, 0, 1, 8, 0, 3, 4, 0, 93, 211, 0 ...] gives [0, 0, 0, ...]

Answer 19

Data.List.HT from utility-ht has sieve :: Int -> [a] -> [a] .

See documentation and source :

{-| keep every k-th value from the list -}
sieve, sieve', sieve'', sieve''' :: Int -> [a] -> [a]
sieve k =
   unfoldr (\xs -> toMaybe (not (null xs)) (head xs, drop k xs))

sieve' k = map head . sliceVertical k

sieve'' k x = map (x!!) [0,k..(length x-1)]

sieve''' k = map head . takeWhile (not . null) . iterate (drop k)

propSieve :: Eq a => Int -> [a] -> Bool
propSieve n x =
   sieve n x == sieve'  n x   &&
   sieve n x == sieve'' n x

Answer 20

And a very functional one without apparent conditionals:

everyNth :: Int -> [a] -> [a]
everyNth 0 = take 1
everyNth n = foldr ($) [] . zipWith ($) operations where
  operations = cycle $ (:) : replicate (n-1) (const id)

(Note, this one takes every element whose index is a multiple of n. Which is slightly different from the original question, but easy to convert.)

Answer 21

We can use a list comprehension:

takeEvery :: Int -> [a] -> [a]
takeEvery n xs = [x | (x, i) <- zip xs [1..], i `mod` n == 0]

From all value-index pairs (x, i), take x if i is divisible by n. Note that the index starts from one here.

Answer 22

My solution is:

every :: Int -> [a] -> [[a]]
every _ [] = []
every n list = take n list : (every n $ drop n list)

It do not use zip but I can't tell about it's performances and memory profile.

Answer 23

An uglier, and more limited version of the accepted answer

every :: Eq a => Int -> [a] -> [a]
every n xs = if rest == [] 
                then [] 
                else head rest : every n (tail rest)
    where rest = drop (n-1) xs

For "line golfing" it can be written like this:

every n xs = if rest == [] then [] else head rest : every n (tail rest) 
    where rest = drop (n-1) xs

(It's more limited because it has an unnecessary Eq a constraint.)